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$F$ is $m\times m$ diagonal, with real non negative elements
$D$ is $n \times m$ complex
$P$ is $n \times 1$ complex
$A$ is $m \times 1$ complex.
Minimize $\Gamma(A)$, with respect to $A$.
$$\Gamma(A) = \frac{m^2(DA-P)^H (DA-P) + (FA)^H(FA)}{A^HA}$$
It is known that both numerator and denominator of $\Gamma(A)$ are convex and non negative. Also both the terms of the numerator are individually convex and non negative.
Question: An numerical optimization algorithm to find global minimum.
Apart from a reagular solution, I am also interested in a gradient descent based method if possible as the matrices are large. Also $m >> n$.
PS : This question is a specific version of this question.
EDIT : more known information
No constraints on problem but
$\sum P = 0$, I mean sum of elements of matrix $P$ is zero. Diagonal elements of $F$ are not all zeros. Also $P^HP \ne 0$. Rows of $D$ are orthogonal to each other. Also they are linearly independent. |
Without any loss of generality we can draw a tangent to a circle with centre $C\equiv(C_x,C_y)$ and radius $r$ and choose a point$P\equiv(P_x,P_y)$ on the tangent so that its distance from the origin is$\overline{CP}\equiv d$ with $d \ge r$.
From the sketch above we recognize that the tangency points can bedescribed as the sums of vectors, i.e.,
$$ \overrightarrow{OT_1} = \overrightarrow{OC}+ \frac{a}{d}\,\overrightarrow{CP}+ \frac{b}{d}\,\overrightarrow{CP}',\qquad \overrightarrow{OT_2} = \overrightarrow{OC}+ \frac{a}{d}\,\overrightarrow{CP}- \frac{b}{d}\,\overrightarrow{CP}'$$
where the prime in $\overrightarrow{CP}'$ represents a 90⁰anticlockwise rotation of the vector.
Because our task is now to determine $a/d$ and $b/d$, we observe thatthe triangles $CT_1C'$ and $CPT_1$ are similar so that, with the position$\rho=r/d$, we have
$$\frac{a}{r} = \frac{r}{d} \implies \frac{a}{d}=\left(\frac{r}{d}\right)^2 =\rho^2$$
and
$$\frac{b}{r} = \frac{R}{d} \implies \frac{b}{d}=\frac{rR}{d^2}=\rho\frac Rd$$
but $R=\sqrt{d^2-r^2}$ and eventually we can write
$$ \frac{b}{d}= \rho\sqrt{1-\rho^2}.$$
From the last relation we recognize 1) that a real solution requires$\rho\le1$ (i.e., $P$ is not inside the circle) and 2) that for$\rho=1$ the point $P$ is on the circumference and the tangency pointsdegenerate into $P$.
The OP can now compute the tangency points, etc.
E.g., in Python 3 (that is as close to pseudo-code as it could possibly be...)
from math import sqrt
# Data Section, change as you need #
Cx, Cy = -2, -7 #
r = 5 #
Px, Py = 4, -3 #
# ################################ #
dx, dy = Px-Cx, Py-Cy
dxr, dyr = -dy, dx
d = sqrt(dx**2+dy**2)
if d >= r :
rho = r/d
ad = rho**2
bd = rho*sqrt(1-rho**2)
T1x = Cx + ad*dx + bd*dxr
T1y = Cy + ad*dy + bd*dyr
T2x = Cx + ad*dx - bd*dxr
T2y = Cy + ad*dy - bd*dyr
print('The tangent points:')
print('\tT1≡(%g,%g), T2≡(%g,%g).'%(T1x, T1y, T2x, T2y))
if (d/r-1) < 1E-8:
print('P is on the circumference')
else:
print('The equations of the lines P-T1 and P-T2:')
print('\t%+g·y%+g·x%+g = 0'%(T1x-Px, Py-T1y, T1y*Px-T1x*Py))
print('\t%+g·y%+g·x%+g = 0'%(T2x-Px, Py-T2y, T2y*Px-T2x*Py))
else:
print('''\
Point P≡(%g,%g) is inside the circle with centre C≡(%g,%g) and radius r=%g.
No tangent is possible...''' % (Px, Py, Cx, Cy, r))
Executing the program yields
The tangent points:
T1≡(-1.1139,-2.07914), T2≡(2.88314,-8.0747).
The equations of the lines P-T1 and P-T2:
-5.1139·y-0.920857·x-11.6583 = 0
-1.11686·y+5.0747·x-23.6494 = 0 |
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Yeah it does seem unreasonable to expect a finite presentation
Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.
How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...
Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants.Then $A$ makes a move, then $B$, and so on...The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...
Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...
The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms
This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.
Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$
Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$
So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$
Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$
But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$
For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube
Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.
Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$
You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point
Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).
Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...
@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.
This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.
You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.
I'll use the latter notation consistently if that's what you're comfortable with
(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)
@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)
Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms
So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.
Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.
That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection
Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$
Voila, Riemann curvature tensor
Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature
Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?
Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.
Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.
Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?
Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.
You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form
(The cotangent bundle is naturally a symplectic manifold)
Yeah
So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.
But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!
So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up
If someone has the time to quickly check my result, I would appreciate. Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$ Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?
Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty
@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method
I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.
My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?
In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.
Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group
Everything about $S_4$ is encoded in the cube, in a way
The same can be said of $A_5$ and the dodecahedron, say |
I know special relativity says that traveling at high speeds (or really any speed) causes time dilation; and General relativity says that gravity also causes time dilation. I was wondering if relative time dilation (where two observers each measure the other's time to be slow) was caused not by time dilation, but instead because with the relative velocity difference between them, if they became increasingly far from each other, light would take longer and longer to reach them from the other. This would result in them both observing each other to have a slower time, though neither would necessarily experience the time dilation.
It's a sensible thought but no. "A sees B's clock running slow." , which you meet in introductory relativity explanations, is shorthand for "A sees the ticks of B's clock arrive at a certain rate. A knows that with every tick, the clock is getting further away (or, in some cases, nearer) so each light signal has further (or less far) to travel, and A compensates for that in working out the rate at which B's clock had to be ticking in order to arrive at the rate they perceive. This calculated rate is slow."
Time dilation in special relativity does not depend on the distance of the two observers. It depends on their relative velocity $v$ only, according to $\Delta t'=1/\sqrt(1-v^2/c^2)\Delta t$. This shows that $\Delta t'$ is in fact delayed compared to $\Delta t$.
In SR (special relativity) two inertial reference frames in relative motion experience the time dilation each vs. the other. The well known relation states
$\Delta t = \gamma \Delta \tau$ where: $t$ stationary observer time $\tau$ moving observer proper time $\gamma = 1 / \sqrt{1 - v^2/c^2}$ Lorentz factor $v$ relative velocity of the moving frame vs. stationary
The time dilation in SR is symmetrical between the two frames in relative motion and depends on the norm of the relative velocity. It does not depend on the sign of the relative velocity and not on the spatial distance between the observers either.
The time dilation in GR (general relativity) is caused by gravity (curvature of spacetime) and it is not symmetrical between two stationary observers at different radial distances from a massive object. |
We present the first observation of exclusive $e^+e^-$ production in hadron-hadron collisions, using $p\bar{p}$ collision data at \mbox{$\sqrt{s}=1.96$ TeV} taken by the Run II Collider Detector at Fermilab, and corresponding to an integrated luminosity of \mbox{532 pb$^{-1}$}. We require the absence of any particle signatures in the detector except for an electron and a positron candidate, each with transverse energy {$E_T>5$ GeV} and pseudorapidity {$|\eta|<2$}. With these criteria, 16 events are observed compared to a background expectation of {$1.9\pm0.3$} events. These events are consistent in cross section and properties with the QED process \mbox{$p\bar{p} \to p + e^+e^- + \bar{p}$} through two-photon exchange. The measured cross section is \mbox{$1.6^{+0.5}_{-0.3}\mathrm{(stat)}\pm0.3\mathrm{(syst)}$ pb}. This agrees with the theoretical prediction of {$1.71 \pm 0.01$ pb}.
We summarize a search for the top quark with the Collider Detector at Fermilab (CDF) in a sample of $\bar{p}p$ collisions at $\sqrt{s}$= 1.8 TeV with an integrated luminosity of 19.3pb$~{-1}$. We find 12 events consistent with either two $W$ bosons, or a $W$ boson and at least one $b$ jet. The probability that the measured yield is consistent with the background is 0.26\%. Though the statistics are too limited to establish firmly the existence of the top quark, a natural interpretation of the excess is that it is due to $t\bar{t}$ production. Under this assumption, constrained fits to individual events yield a top quark mass of $174 \pm 10~{+13}_{-12}$ GeV/c$~2$. The $t\bar{t}$ production cross section is measured to be $13.9~{+6.1}_{-4.8}$pb. (Submitted to Physical Review Letters on May 16, 1994).
We report the first measurement of the cross section for Z boson pair production at a hadron collider. This result is based on a data sample corresponding to 1.9 fb-1 of integrated luminosity from ppbar collisions at sqrt{s} = 1.96 TeV collected with the CDF II detector at the Fermilab Tevatron. In the llll channel, we observe three ZZ candidates with an expected background of 0.096^{+0.092}_{-0.063} events. In the llnunu channel, we use a leading-order calculation of the relative ZZ and WW event probabilities to discriminate between signal and background. In the combination of llll and llnunu channels, we observe an excess of events with a probability of $5.1\times 10^{-6}$ to be due to the expected background. This corresponds to a significance of 4.4 standard deviations. The measured cross section is sigma(ppbar -> ZZ) = 1.4^{+0.7}_{-0.6} (stat.+syst.) pb, consistent with the standard model expectation.
We have measured the differential cross section for the inclusive production of psi(2S) mesons decaying to mu^{+} mu^{-1} that were produced in prompt or B-decay processes from ppbar collisions at 1.96 TeV. These measurements have been made using a data set from an integrated luminosity of 1.1 fb^{-1} collected by the CDF II detector at Fermilab. For events with transverse momentum p_{T} (psi(2S)) > 2 GeV/c and rapidity |y(psi(2S))| < 0.6 we measure the integrated inclusive cross section sigma(ppbar -> psi(2S)X) Br(psi(2S) -> mu^{+} mu^{-}) to be 3.29 +- 0.04(stat.) +- 0.32(syst.) nb.
Azimuthal decorrelations between the two central jets with the largest transverse momenta are sensitive to the dynamics of events with multiple jets. We present a measurement of the normalized differential cross section based on the full dataset (L=36/pb) acquired by the ATLAS detector during the 2010 sqrt(s)=7 TeV proton-proton run of the LHC. The measured distributions include jets with transverse momenta up to 1.3 TeV, probing perturbative QCD in a high energy regime.
This letter describes the observation of the light-by-light scattering process, $\gamma\gamma\rightarrow\gamma\gamma$, in Pb+Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 5.02 TeV. The analysis is conducted using a data sample corresponding to an integrated luminosity of 1.73 nb$^{-1}$, collected in November 2018 by the ATLAS experiment at the LHC. Light-by-light scattering candidates are selected in events with two photons produced exclusively, each with transverse energy $E_{\textrm{T}}^{\gamma} > 3$ GeV and pseudorapidity $|\eta_{\gamma}| < 2.37$, diphoton invariant mass above 6 GeV, and small diphoton transverse momentum and acoplanarity. After applying all selection criteria, 59 candidate events are observed for a background expectation of 12 $\pm$ 3 events. The observed excess of events over the expected background has a significance of 8.2 standard deviations. The measured fiducial cross section is 78 $\pm$ 13 (stat.) $\pm$ 7 (syst.) $\pm$ 3 (lumi.) nb.
Using 1.8 fb-1 of pp collisions at a center-of-mass energy of 7 TeV recorded by the ATLAS detector at the Large Hadron Collider, we present measurements of the production cross sections of Upsilon(1S,2S,3S) mesons. Upsilon mesons are reconstructed using the di-muon decay mode. Total production cross sections for p_T<70 GeV and in the rapidity interval |Upsilon|<2.25 are measured to be 8.01+-0.02+-0.36+-0.31 nb, 2.05+-0.01+-0.12+-0.08 nb, 0.92+-0.01+-0.07+-0.04 nb respectively, with uncertainties separated into statistical, systematic, and luminosity measurement effects. In addition, differential cross section times di-muon branching fractions for Upsilon(1S), Upsilon(2S), and Upsilon(3S) as a function of Upsilon transverse momentum p_T and rapidity are presented. These cross sections are obtained assuming unpolarized production. If the production polarization is fully transverse or longitudinal with no azimuthal dependence in the helicity frame the cross section may vary by approximately +-20%. If a non-trivial azimuthal dependence is considered, integrated cross sections may be significantly enhanced by a factor of two or more. We compare our results to several theoretical models of Upsilon meson production, finding that none provide an accurate description of our data over the full range of Upsilon transverse momenta accessible with this dataset.
We study charged particle production (pT>0.5 GeV/c, |η|<0.8) in proton-antiproton collisions at total center-of-mass energies s=300 GeV, 900 GeV, and 1.96 TeV. We use the direction of the charged particle with the largest transverse momentum in each event to define three regions of η-ϕ space: “toward”, “away”, and “transverse.” The average number and the average scalar pT sum of charged particles in the transverse region are sensitive to the modeling of the “underlying event.” The transverse region is divided into a MAX and MIN transverse region, which helps separate the “hard component” (initial and final-state radiation) from the “beam-beam remnant” and multiple parton interaction components of the scattering. The center-of-mass energy dependence of the various components of the event is studied in detail. The data presented here can be used to constrain and improve QCD Monte Carlo models, resulting in more precise predictions at the LHC energies of 13 and 14 TeV. |
Defining parameters Dimensions
The following table gives the dimensions of various subspaces of \(M_{1}(\Gamma_1(80))\).
Total New Old Modular forms 57 15 42 Cusp forms 1 1 0 Eisenstein series 56 14 42
The following table gives the dimensions of subspaces with specified projective image type.
Decomposition of \(S_{1}^{\mathrm{new}}(\Gamma_1(80))\)
We only show spaces with odd parity, since no modular forms exist when this condition is not satisfied. Within each space \( S_k^{\mathrm{new}}(N, \chi) \) we list the newforms together with their dimension.
Label \(\chi\) Newforms Dimension \(\chi\) degree 80.1.b \(\chi_{80}(31, \cdot)\) None 0 1 80.1.e \(\chi_{80}(39, \cdot)\) None 0 1 80.1.g \(\chi_{80}(71, \cdot)\) None 0 1 80.1.h \(\chi_{80}(79, \cdot)\) 80.1.h.a 1 1 80.1.i \(\chi_{80}(13, \cdot)\) None 0 2 80.1.k \(\chi_{80}(19, \cdot)\) None 0 2 80.1.m \(\chi_{80}(57, \cdot)\) None 0 2 80.1.p \(\chi_{80}(17, \cdot)\) None 0 2 80.1.r \(\chi_{80}(11, \cdot)\) None 0 2 80.1.t \(\chi_{80}(53, \cdot)\) None 0 2 |
I'm reading "Microelectronics" by Millman, Grabel.
Premise:
The book developed, for a pn step-graded junction, the contact potential as $$ V_0 = V_T \text{ln} \frac{p_{p0}}{p_{n0}} = - V_T \text{ln} \frac{n_{p0}}{n_{n0}} \qquad (1) $$ where $p_{p0}$ is the hole concentration in the p-side, $p_{n0}$ is the hole concentration in the n-side, $n_{p0}$ the electron concentration in the p-side and $n_{n0}$ the electron in the n-side.
To develop the previous equation, it made use of the following $$J_p = q \mu_p p \mathscr{E} - q D_p \frac{\text{d} p}{\text{d} x} \\ D_p = \mu_p V_T \\ \mathscr{E}(x) = - \frac{\text{d}V}{\text{d} x} $$ where $\mu$ is the mobility, $\mathscr{E}$ the electric field, $D$ the diffusion constant and $p$ ($n$) the concentration of holes (electrons). For electrons, the formulas are very similar, substituting $p$ with $n$ and changing the sign of the diffusion term in $J_n$.
I was wondering if one could compute the contact potential of a metal-semiconductor junction (terminals of the diode) with the same formula $(1)$. The derivation of the formula $(1)$ does not make any particular assumption regarding the conduction medium (at least in my opinion), provided the Einstein relation is valid for a metal. Is the Einstein relation $\frac{D_n}{\mu_n} = V_T$ valid for a metal conductor? What is a typical value for mobility of electrons in a metal conductor?
In the Example 1.1 the book worked out the electron concentration for a conducting line of an IC chip, that is $4.38 \times 10^{21} ~ \text{cm}^{-3}$. Let's say that the electron concentration of the p-type silicon is a typical $4.2 \times 10^{5}$. Using the formula $(1)$
$$ V_0 = − 0,0259× \text{ln} \frac{4,2×10^5}{4,38×10^{21}} = 0,955 \text{V} $$
It seems to be an extremely big number considering that the cut-in voltage for a diode is $0.6 \text{V}$ and that the contact potential for a metal-semiconductor junction should be negligible.
In the same example it states that the mobility of electrons in the conducting line is $500 ~ \text{cm}^2 / (V \cdot s)$. How can be that small considering that the electron mobility in silicon is $1500 ~ \text{cm}^2 / (V \cdot s)$?
Thanks, Luca |
I am trying to come to grip with some solid state theory. Bloch waves, energy eigenstates for hamiltonians with lattice periodic potential in $\mathbb R^d$, are frequently written as$$\phi_{n,k}(r)=e^{2\pi ik\cdot r}u_{n,k}(r)$$ with $u_{n,k}$ periodic in lattice vectors $R\in\mathbb R^d$. This is 1 version of Bloch's theorem. It implies that $k$ can be restricted to the 1st Brillouin zone.
Thus multiplying $\phi_{n,k}(r)$ by a plane wave with frequency $K$ in the reciprocal lattice yields another Bloch wave with larger momentum -i.e. $p=\partial_r$ expectation value- corresponding to some other index $n'$, that is$$\phi_{n,k+K}(r)=e^{2\pi iK\cdot r}\phi_{n,k}(r)=\phi_{n',k}(r).$$My question is then why do people (e.g. Ashcroft-Mermin) often set $\phi_{n,k+K}(r)=\phi_{n,k}(r)$ against the notation that seems much more natural to me that I used above $\phi_{n,k+K}(r)=e^{2\pi iK\cdot r}\phi_{n,k}(r)$ which attributes essentially momentum $k+K$ to $\phi_{n,k+K}(r)$?
According to their notation all $\phi_{n,K}(r)$ get equal momenta near 0 (more precisely $\langle u_{n,0},pu_{n,0}\rangle$) but if we turn momentum on from 0 to $K$, and beyond to multiples of $K$, continuously in momentum space our Bloch wave varying adiabatically will get larger and larger momentum and they will correspond to my notation for $\phi_{n,k+K}(r)$, not to nearly 0 momentum states. That implies also that to increase momentum while keeping the wavevector $k$ in the 1st Brillouin zone one must increase the index $n$.
A related question is: It seems that Fermi energy is ground state energy if we assume that arbitrarily high momenta get the same energy. We simply pick $\phi_{0,K}(r)$ as wavefunctions for our electrons with arbitarily high momentum $K$ in the reciprocal lattice with $\phi_{0,0}=u_0$ the ground state. Those all have the same energy, increasingly negative potential energy compensates increasing kinetic energy. This is related to having standing waves as energy eigenstates in perfect crystals -i.e. superconductivity when many electrons in are such states. The way out seems to be to consider that real crystals have boundaries. There is also scattering of electrons on phonons.
In real crystals the increasing kinetic energy with increasing momenta $K$ would not be compensated by an infinite periodic potiential. After the boundary the kinetic energy would not be compensated by further potential barriers, thus large momenta Bloch waves would actually be more energetic and electrons would tend to keep low momenta, essentially in the first Brillouin zone.
I find that many lectures on Bloch waves/band structure do not treat momentum crystal(-)clearly. Thanks for any hint.
Update: I am really curious so please share if anybody has any comment. This is related to umklapp processes and other things which are often explained handwavingly.
Update 2: Actually I have thought about my questions. The problem is that momentum cannot be increased continuously. A way to see this is in a simple 1D Kronig-Penney model. Approaching a reciprocal wavevector $K$ there is no Bloch wave with that wavevector so starting from $k=0$ if 1 passes above $K$, $u_{n',k}$ will change drastically/discontinuously to account for the increased momentum while $k$ is taken in the 1st Brillouin zone but is then farther from momentum. To have an accurate picture would require solving Schrödinger equations dynamically. The adiabatic varying eigenstate method I tried to picture myself fails. I may write more with more time.
Update 3: Actually I think I understand what puzzled me. It is that as wavevector $k$ is increased $u_k$ changes so as to compensate oscillations increase due to the $e^{ikr}$ factor. I was visualizing multiplying an almost fixed $u_k$ by $e^{ikr}$ with large k. But continuing from 0 to K in the same band yields exactly $\phi_{n,0}$ physically because $u_k$ unwinds/compensates the more oscillating factor. |
A lens has two surfaces. One of these surfaces still has refracting power, due to the difference in refractive index and angle of incidence.
Refraction on one surface
Let parallel rays start in glas. Radius sign convention results $R=-200\,mm$ since incident light is on same side as center of curvature.
Radius of curvature and the two media influence the focal point. For your rays propagate from right to left (air) the FFL (front focal length, as depicted) is$$f=R\cdot \frac{n_1}{n_1-n_2}=R\cdot \frac{1}{1-n_2}$$This is the reciprocal element of your transfer matrix. Focal length is proportional to radius of curvature. Increasing refractive index $n_2$ of glas shortens it.$$f=-200\,mm \cdot \frac{1}{-0.5}=400\,mm$$
Remark
The cited above lensmaker equation with $R_1=\infty$ will return the same result. Since on-axis parallel rays are not refracted on plane surface, as can be seen from Snell's law. |
The problem is described as follows:
I need to generate a basis(matrix) in lexicographic order.
For two different basis vector $\{n_1,n_2,\cdots,n_M\}$ and $\{t_1,t_2,\cdots,t_M\}$, there must exist a certain index $1\leq k \leq M-1$ such that $n_i=t_i$ for $1\leq i \leq k-1$, while $n_k \neq t_k$. We say $\{n_1,n_2,\cdots,n_M\}$ is superior to $\{t_1,t_2,\cdots,t_M\}$ if $n_k>t_k$.
The whole algorithm can be described as follows, supposing $n_1+n_2+\cdots+n_M=N$,
Starting from $\{N,0,0,\cdots,0\}$, supposing $n_k \neq 0$ while ni=0 for all $k+1 \leq i \leq M-1$, then the next basis vector is $\{t_1,t_2,\cdots,t_M\}$ with
(1) $t_i=n_i$, for $1\leq i \leq k-1$;
(2) $t_k=n_k -1$;
(3) $t_{(k+1)}$=
N-Sum[ni,{i,1,k}]
(4) $t_i=0$ for $i\geq k+2$
The generating procedure shall stop at the final vector
{0,0,...,0,N}
My code is presented as follows
baseGenerator[M_Integer, N_Integer] := NestList[ Block[{k, base}, k = Part[{M}-FirstPosition[Reverse[#[[1 ;; M - 1]]], x_ /; x > 0], 1]; base = #; base[[k]] = #[[k]] - 1; base[[k + 1]] = N - Total[#[[1 ;; k]]] + 1; base[[k + 2 ;; M]] = 0; base] &, Join[{N}, ConstantArray[0, M - 1]], (N + M - 1)!/(N!*(M - 1)!) - 1]
Each line corresponds to a specfic step described above. I don't think my code is efficient since it cost much more time than MATLAB writing with loop. Is there any method to improve it a lot ?
In fact, most of the time is costed when calculating position $k$ with command
FirstPosisiton. |
As far as I understand, Noether's theorem for fields works, as explained in David Tong's QFT lecture notes (page 14) for example, by saying that a transformation $\phi(x) \mapsto \phi(x) + \delta \phi (x)$ is called a symmetry if it produces a change in the Lagrangian density which can be expressed as a four divergence, $$\delta \mathcal{L} = \partial_{\mu} F^{\mu}\tag{1.35} $$ for some 4-vector field $F^{\mu}$.
We thengo onto show that the change in this Lagrangian density may also be expressed
for an arbitrary transformation as
$$\delta \mathcal{L} = \partial_{\mu}\bigg(\frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}\delta \phi\bigg)\tag{1.37}.$$
Which
is a 4-divergence. So how could we say any transformation is not a symmetry in the sense above? |
I am self learning GR. This is a rather long post but I needed to clarify few things about the effect of general coordinate transformations on the global symmetries of metric. Any comments, insights are much appreciated.
To be concrete let's consider that $g_{\mu\nu}$ represents an axially symmetric spacetime i.e a Kerr black hole. In Boyer-Lindquist coordinates ($t,\, r,\, \theta,\, \phi $) metric has no dependence on $t$ and $\phi$ (they are cyclic coordinates):
$$ g_{\mu\nu} dx^{\mu}dx^{\nu}=-(1-\frac{2Mr}{r^2+a^2\cos^2{\theta}})\,dt^2-\frac{r^2+a^2\cos^2{\theta}}{r^2+2Mr +a^2} dr^2 +(r^2+a^2\cos^2{\theta})\, d\theta^2 \\ \,\,\,+ \left( r^2+a^2+ \frac{2Ma^2 r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}}\right) \sin^2{\theta}\, d\phi^2 - \frac{2Ma r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}} dt\, d\phi $$
Consequently particle's energy $E_0$ and angular momentum $L_0$ are conserved. Suppose I make a coordinate transformation: $d \bar{x}^{\mu} = \Lambda^{\mu}_{\,\,\nu}\,dx^{\nu}$ where, if I am not mistaken, $\Lambda$ is a matrix, which is an element of the general linear group GL$\,(4,\,R)$ with a non-zero determinant. In the new coordinates metric may not have cyclic coordinates, a good example would be representation of the above metric in Kerr-Schild coordinates:
$$\begin{align} g_{\mu\nu} dx^{\mu}dx^{\nu}=&-d\bar{t}^2+dx^2+dy^2+dz^2 \\ &+ \frac{2Mr^3}{r^4+a^2z^2} \left[d\bar{t} + \frac{r(x \,dx+y \,dy)+a(x\,dy-y\,dx)}{r^2+a^2} + \frac{z\,dz}{r} \right]^2\\ &\text{where}\qquad r^4+ (x^2+y^2+z^2)\, r^2 -a^2 z^2=0 \end{align} $$
Now there is only one cyclic coordinate, $\bar{t}$. I presume we may in principle introduce new coordinates where none of the coordinates appear to be cyclic in functional form of $g_{\mu\nu}\, dx^{\mu}dx^{\nu}$. Suppose I make such a coordinate transformation.
My questions are:
Does the new metric still have global symmetries i.e conserved quantities that correspond to $E_0$ and $L_0$?
If the answer to the first question is yes, then suppose I gave this new metric to someone without telling about the coordinate transformation and ask if there are any symmetries. Will she/he able to find symmetries by finding closed orbits in the geodesic flow?
To me the answer to both questions above seemed to be yes. I think that the integrability (in the Liouville sense) of the metric should not depend on one's definition of the coordinates. In other words, because of the global symmetries, we expect bounded geodesics around the black hole. In the old coordinates we may easily calculate such closed trajectories and according to my thinking closed geodesics exist (objects revolve around the black hole) regardless of how we label the coordinates.
But I could not be sure of this. To explain my confusion let me write:
$$ \frac{d x^{\mu}}{ds^2} + \Gamma^{\mu}_{\nu \lambda} \frac{d x^{\nu}}{ds} \frac{d x^{\lambda}}{ds}=0 $$
which gives the geodesic flow. The existence of the bounded trajectories here depends on the number of zeros and poles of the Christoffel symbol, and also on their relative locations. The thing is, one can mess with these parameters by making an arbitrary but an invertible coordinate transformation to the metric and therby change the flow. Also, perhaps somewhat related to this, it was pointed out here that a coordinate transformation, when viewed as a diffeomorphism, does not always map geodesics to geodesics unless it is an isometry.
So what is the correct way to think about this? Do all coordinate transformations/diffeomorphisms conserve global symmetries? or a subgroup of them? |
The maximum height that a man can jump is a bit over 70cm for the very best athletes. This is because the only thing that matters is how high your centre of gravity rises. I will assume 80cm.
Work is force times distance. Weight is the force $W = gm_0 $ and when you jump a height $h$ under Earth gravity of 9.8m/s with a weight $m_0$, $E = hgm_0$ is the work that gravity did to bring you to a stop at your maximum height (all of the kinetic energy being converted to potential energy).
The work you performed to initiate the jump is also force times distance. It equals the gravitational potential energy at the top of the jump exactly. When jumping on an asteroid, the length of your legs and your strength we will assume don't change (a fabric-based space suit has a springiness in the legs, so this is an approximation). You perform an equal amount of work to what you did on Earth, namely $E = 0.8m \times 9.8 m/s^2 \times m_0$.
Escape velocity is given by (see Space Mission Engineering: The New SMAD edited by Wertz, Everett & Puschell, 2011, p. 201) $V_e = \sqrt{2GM/R}$ where $G = 6.674 \times 10^{-11}m^3 kg^{-1}s^{-2}$ is the gravitational constant, M the mass of the central body and R your distance from its centre (I assume for simplicity it is circular).
Density $\rho$ is in the region of $1000kg/m^3$ for icy bodies and $3000kg/m^3$ for rocky ones. In the case of a nickel-iron body we will approximate the overall density as $8000kg/m^3$.
We want to calculate the diameter R of a body where your kinetic energy at escape velocity equals $E$ above.
$$0.8m \times 9.8m/s \times m_0 = E = KE = 0.5m_0 \times V_e^2$$$$0.8m \times 9.8m/s^2 = 0.5 \times V_e^2 $$$$ 15.68 m^2/s^2 = V_e^2 = 2GM/R $$still assuming a sphere$$ M = \rho \times V = \rho \times 4/3 \pi R^3 $$$$15.68 m^2/s^2 = 8/3 G \rho \pi R^2 $$$$ \sqrt{(2.8 \times 10^{10} / \rho) m^{-1} kg } = R$$
Substituting the densities from above, we get roughly $R_{icy} \approx 5300m$, $R_{rocky} \approx 3100m$, $R_{iron} \approx 1900m$.
Now let's see what happens with your numbers. The astronaut can still do an equal amount of work to jump but their mass will be $m_1 = m_0 + m_{suit}$ on the asteroid, $\rho = 2000kg/m^3$. A NASA study from 1996 (see The Origins and Technology of the Advanced Extravehicular Space Suit by G.L. Harris, American Astronautical Society History Series Volume 24, 2001, p. 455) specified a maximum pressure suit assembly mass of 27kg for missions to Mars. This excludes the life support systems, for comparison Apollo had (see p. 440) 63.2kg of life support on a suit of 100kg total. We'll assume the future suit is a form-fitting elastic suit with almost all mass in life support, giving no springiness in the legs and massing just $m_{suit} = 30kg$ total. Note: You gave the astronaut weight as 30kg. This is a small child so I assume 70kg is intended instead.
$$ 0.8m \times 9.8m/s^2 \times m_0 = E = KE = 0.5 m_1 \times V_e^2 $$$$ 0.8m \times 9.8m/s^2 \times 70kg = 0.5 \times 100kg \times V_e^2 $$$$ 10.976 m^2/s^2 = V_e^2 = 2GM/R = 8/3 G \rho \pi R^2 = 8/3 \times 6.674 \times 10^{-11} m^3 kg^{-1} s^{-2} \times 2000 kg/m^3 \times \pi \times R^2$$$$R^2 = 9.815 \times 10^6 m^2 $$$$R \approx 3100 m $$
It only happens to be close to the rocky body approximation above because the increase in mass was made up by the decrease in gravity.
Edit: Finally, since you want mass:$$ M = 4/3 \pi R^3 \times \rho = 4/3 \times 2000kg/m^3 \times \pi \times (3100m)^3 = 2.50 \times 10^{14} kg $$
Really finally this time: Since you updated the mass to 80kg and we may assume a suit mass of 10kg included in that, here's how the answer would change from my 100kg with 30kg suit mass answer:
$$V_e^2 = 10.976m^s/s^2 \times 100kg/80kg = 13.720m^2/s^2$$$$R \approx 3100m \times \sqrt{100kg/80kg} \approx 3500m$$$$M = 2.50 \times 10^{14}kg \times (\sqrt{100/80})^3 = 3.6 \times 10^{14}kg$$ |
Krishnan, R and Seshadri, TP (1990)
Structure of dipotassium galactose 1-phosphate pentahydrate. In: Acta Crystallographica, Section C: Crystal Structure Communications, 46 (12). pp. 2299-2302.
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Abstract
$2K^+.C_6H_{11} O_9 P_{2-}. 5H_2O, M_r = 426.4$, monoclinic, $P2_1$, a=6"228 (1), b = 14.600(2), c = 8.982 (1) $\AA,\beta$ = 102.01 $(1)^0$, V= 798.4 $\AA^3$, Z= 2, $D_m$ = 1.756, $D_x = 1.773 Mg m^{-3}$, $\lambda(Cu K\alpha) = 1.5418 \AA, \mu = 7 mm^{-l}$, F(000) = 444, T= 293 K, R = 0.033, wR=0.035 for 1549 unique reflections with $I> 3\sigma(I)$. The ugar ring adopts a $^4C_1$ conformation. The conformation about the exocyclic C5-C6 bond is gauche-trans. The phosphate group exists as adianion with charges localized on 07 and 08 atoms. One of the metal ions binds to the phosphate group directly. All the hydroxyl O atoms, except O4-H, are involved in hydrogen bonding with the water O atoms
Item Type: Journal Article Additional Information: Copyright of this article belongs to International Union of Crystallography Keywords: Dipotassium Galaetose;1-Phosphate Pentahydrate;enzymatic reactions Department/Centre: Division of Biological Sciences > Molecular Reproduction, Development & Genetics
Division of Physical & Mathematical Sciences > Physics
Depositing User: Srinivasa Naika Date Deposited: 21 Dec 2007 Last Modified: 19 Sep 2010 04:41 URI: http://eprints.iisc.ac.in/id/eprint/12534 Actions (login required)
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LaTeX:Symbols
LaTeX About - Getting Started - Diagrams - Symbols - Downloads - Basics - Math - Examples - Pictures - Layout - Commands - Packages - Help
This article will provide a short list of commonly used LaTeX symbols.
Contents Common Symbols Operators Finding Other Symbols
Here are some external resources for finding less commonly used symbols:
Detexify is an app which allows you to draw the symbol you'd like and shows you the code for it! MathJax (what allows us to use on the web) maintains a list of supported commands. The Comprehensive LaTeX Symbol List. Operators Relations
Symbol Command Symbol Command Symbol Command \le \ge \neq \sim \ll \gg \doteq \simeq \subset \supset \approx \asymp \subseteq \supseteq \cong \smile \sqsubset \sqsupset \equiv \frown \sqsubseteq \sqsupseteq \propto \bowtie \in \ni \prec \succ \vdash \dashv \preceq \succeq \models \perp \parallel \mid \bumpeq
Negations of many of these relations can be formed by just putting \not before the symbol, or by slipping an n between the \ and the word. Here are a few examples, plus a few other negations; it works for many of the others as well.
Symbol Command Symbol Command Symbol Command \nmid \nleq \ngeq \nsim \ncong \nparallel \not< \not> \not= \not\le \not\ge \not\sim \not\approx \not\cong \not\equiv \not\parallel \nless \ngtr \lneq \gneq \lnsim \lneqq \gneqq
To use other relations not listed here, such as =, >, and <, in LaTeX, you may just use the symbols on your keyboard.
Greek Letters
Symbol Command Symbol Command Symbol Command Symbol Command \alpha \beta \gamma \delta \epsilon \varepsilon \zeta \eta \theta \vartheta \iota \kappa \lambda \mu \nu \xi \pi \varpi \rho \varrho \sigma \varsigma \tau \upsilon \phi \varphi \chi \psi \omega
Symbol Command Symbol Command Symbol Command Symbol Command \Gamma \Delta \Theta \Lambda \Xi \Pi \Sigma \Upsilon \Phi \Psi \Omega Arrows
Symbol Command Symbol Command \gets \to \leftarrow \Leftarrow \rightarrow \Rightarrow \leftrightarrow \Leftrightarrow \mapsto \hookleftarrow \leftharpoonup \leftharpoondown \rightleftharpoons \longleftarrow \Longleftarrow \longrightarrow \Longrightarrow \longleftrightarrow \Longleftrightarrow \longmapsto \hookrightarrow \rightharpoonup \rightharpoondown \leadsto \uparrow \Uparrow \downarrow \Downarrow \updownarrow \Updownarrow \nearrow \searrow \swarrow \nwarrow
(For those of you who hate typing long strings of letters, \iff and \implies can be used in place of \Longleftrightarrow and \Longrightarrow respectively.)
Dots
Symbol Command Symbol Command \cdot \vdots \dots \ddots \cdots \iddots Accents
Symbol Command Symbol Command Symbol Command \hat{x} \check{x} \dot{x} \breve{x} \acute{x} \ddot{x} \grave{x} \tilde{x} \mathring{x} \bar{x} \vec{x}
When applying accents to i and j, you can use \imath and \jmath to keep the dots from interfering with the accents:
Symbol Command Symbol Command \vec{\jmath} \tilde{\imath}
\tilde and \hat have wide versions that allow you to accent an expression:
Symbol Command Symbol Command \widehat{7+x} \widetilde{abc} Others Command Symbols
Some symbols are used in commands so they need to be treated in a special way.
Symbol Command Symbol Command Symbol Command Symbol Command \textdollar or \&\%\#\_\{\}\backslash$ \backslash
(Warning: Using $ for will result in . This is a bug as far as we know. Depending on the version of this is not always a problem.)
European Language Symbols
Symbol Command Symbol Command Symbol Command Symbol Command {\oe} {\ae} {\o} {\OE} {\AE} {\AA} {\O} {\l} {\ss} !` {\L} {\SS} Bracketing Symbols
In mathematics, sometimes we need to enclose expressions in brackets or braces or parentheses. Some of these work just as you'd imagine in LaTeX; type ( and ) for parentheses, [ and ] for brackets, and | and | for absolute value. However, other symbols have special commands:
Symbol Command Symbol Command Symbol Command \{ \} \| \backslash \lfloor \rfloor \lceil \rceil \langle \rangle
You might notice that if you use any of these to typeset an expression that is vertically large, like
(\frac{a}{x} )^2
the parentheses don't come out the right size:
If we put \left and \right before the relevant parentheses, we get a prettier expression:
\left(\frac{a}{x} \right)^2
gives
And with system of equations:
\left\{\begin{array}{l}x+y=3\\2x+y=5\end{array}\right.
Gives
See that there's a dot after
\right. You must put that dot or the code won't work.
And, if you type this
\underbrace{a_0+a_1+a_2+\cdots+a_n}_{x}
Gives
Or
\overbrace{a_0+a_1+a_2+\cdots+a_n}^{x}
Gives
\left and \right can also be used to resize the following symbols:
Symbol Command Symbol Command Symbol Command \uparrow \downarrow \updownarrow \Uparrow \Downarrow \Updownarrow Multi-Size Symbols
Some symbols render differently in inline math mode and in display mode. Display mode occurs when you use \[...\] or $$...$$, or environments like \begin{equation}...\end{equation}, \begin{align}...\end{align}. Read more in the commands section of the guide about how symbols which take arguments above and below the symbols, such as a summation symbol, behave in the two modes.
In each of the following, the two images show the symbol in display mode, then in inline mode.
Symbol Command Symbol Command Symbol Command \sum \int \oint \prod \coprod \bigcap \bigcup \bigsqcup \bigvee \bigwedge \bigodot \bigotimes \bigoplus \biguplus |
Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..
There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university
Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.
What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?
Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.
Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P
Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?
Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?
Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"
@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.
Although not the only route, can you tell me something contrary to what I expect?
It's a formula. There's no question of well-definedness.
I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.
It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.
Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.
You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.
@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.
@Eric: If you go eastward, we'll never cook! :(
I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.
@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)
@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.
@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator |
According the work by Wang & Zhu, on toric Fano manifolds there exist Kaehler-Ricci solitons. If Futaki=0, there also exist CSCK metrics. But if the Futaki invariant does not vanish, what about extremal metrics? Is there some counterexample?
Here is an example where the Futaki invariant does not vanish, but with extremal metrics. The Futaki invariant on $Bl_p\mathbb{P}^2$, the blow up of $\mathbb{P}^2$, which is toric, does not vanish (Canonical Metrics in Kaehler geometry by Tian, Example 3,10). On the other hand, $Bl_p\mathbb{P}^2$ admits an extremal metric in every Kaehler class (I believe this is in Calabi's Extremal Kaehler Metrics). Is this what you are looking for?
Conjecturally, the existence of an extremal metric is equivalent to relative K-stability, see the work of Székelyhidi. The usual K-stability involved in the Yau-Tian-Donaldson conjecture is a generalisation of the classical Futaki invariant, and relative K-stability is an adaptation of this to the extremal setting. Székelyhidi (in his thesis) has shown that the existence of an extremal metric implies relative K-semistability, so this forms an obstruction. For toric surfaces, this was studied by Bohui Chen, An-Min Li, Li Sheng in "Extremal metrics on Toric Surfaces". In particular, they show relative K-stability is equivalent to the existence of an extremal metric on toric surfaces. I guess not much is known in higher dimensions.
Some additional comment:
Let $(X,J,\omega)$ be Fano Kahler manifold with $\omega_0\in [\omega_0]=\kappa$, then by $\partial\bar\partial$-lemma any other Kahler 2-form can be written as $\omega_0+\sqrt{-1}\partial\bar\partial \varphi$. Now the set of Kahler metrics in $[\omega_0]$ can be identified with $$\mathcal H=\{\varphi \in C^\infty(X,\mathbb R)|\omega_0+\sqrt{-1}\partial\bar\partial \varphi>0\}/\mathbb R$$. Note that $T_\varphi\mathcal H=C^\infty(X,\mathbb R)/\mathbb R$
Now, consider the Calabi- functional(which is sort of special case of Yang-Mills functional)
If we denote $S(\omega_\varphi)$ denotes to be the scalar curvature of the metric $\omega_\varphi$ and $\underline S$ be the average scalar curvature
$$\mathcal C a:\mathcal H\to \mathbb R$$ $$\varphi\to \mathcal Ca(\varphi)=\int_X (S(\omega_\varphi)-\underline S)^2\frac{\omega_\varphi^n}{n!}$$
Now if you take $\psi\in T_\varphi\mathcal H$ , then we can write the variation of Calabi functional via using Lichnerowicz operator $\mathcal D_\varphi=\overline \partial\nabla$ as follows
$$(d\mathcal Ca(\varphi))\psi=2\int_X (\mathcal D_\varphi^*\mathcal D_\varphi S(\omega_\varphi)).\psi\frac{\omega_\varphi^n}{n!}$$ where $\mathcal D_\varphi^*$ is the $L^2(X,\omega_\varphi)$-adjoint of $\mathcal D_\varphi$. As an exercise integrating by part we have $\ker \mathcal D_\varphi^*\mathcal D_\varphi =\ker \mathcal D_\varphi$
This means that $\omega_\varphi$ is extremal metric (which means the critical points of Calabi functional) if the gradient of scalar curvature $S(\omega_\varphi)$ define a vector field .
About your question this recent paper https://arxiv.org/pdf/1610.06865.pdf |
Interested in the following function:$$ \Psi(s)=\sum_{n=2}^\infty \frac{1}{\pi(n)^s}, $$where $\pi(n)$ is the prime counting function.When $s=2$ the sum becomes the following:$$ \Psi(2)=\sum_{n=2}^\infty \frac{1}{\pi(n)^2}=1+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1...
Consider a random binary string where each bit can be set to 1 with probability $p$.Let $Z[x,y]$ denote the number of arrangements of a binary string of length $x$ and the $x$-th bit is set to 1. Moreover, $y$ bits are set 1 including the $x$-th bit and there are no runs of $k$ consecutive zer...
The field $\overline F$ is called an algebraic closure of $F$ if $\overline F$ is algebraic over $F$ and if every polynomial $f(x)\in F[x]$ splits completely over $\overline F$.
Why in def of algebraic closure, do we need $\overline F$ is algebraic over $F$? That is, if we remove '$\overline F$ is algebraic over $F$' condition from def of algebraic closure, do we get a different result?
Consider an observer located at radius $r_o$ from a Schwarzschild black hole of radius $r_s$. The observer may be inside the event horizon ($r_o < r_s$).Suppose the observer receives a light ray from a direction which is at angle $\alpha$ with respect to the radial direction, which points outwa...
@AlessandroCodenotti That is a poor example, as the algebraic closure of the latter is just $\mathbb{C}$ again (assuming choice). But starting with $\overline{\mathbb{Q}}$ instead and comparing to $\mathbb{C}$ works.
Seems like everyone is posting character formulas for simple modules of algebraic groups in positive characteristic on arXiv these days. At least 3 papers with that theme the past 2 months.
Also, I have a definition that says that a ring is a UFD if every element can be written as a product of irreducibles which is unique up units and reordering. It doesn't say anything about this factorization being finite in length. Is that often part of the definition or attained from the definition (I don't see how it could be the latter).
Well, that then becomes a chicken and the egg question. Did we have the reals first and simplify from them to more abstract concepts or did we have the abstract concepts first and build them up to the idea of the reals.
I've been told that the rational numbers from zero to one form a countable infinity, while the irrational ones form an uncountable infinity, which is in some sense "larger". But how could that be? There is always a rational between two irrationals, and always an irrational between two rationals, ...
I was watching this lecture, and in reference to above screenshot, the professor there says: $\frac1{1+x^2}$ has a singularity at $i$ and at $-i$, and power series expansions are limits of polynomials, and limits of polynomials can never give us a singularity and then keep going on the other side.
On page 149 Hatcher introduces the Mayer-Vietoris sequence, along with two maps $\Phi : H_n(A \cap B) \to H_n(A) \oplus H_n(B)$ and $\Psi : H_n(A) \oplus H_n(B) \to H_n(X)$. I've searched through the book, but I couldn't find the definitions of these two maps. Does anyone know how to define them or where there definition appears in Hatcher's book?
suppose $\sum a_n z_0^n = L$, so $a_n z_0^n \to 0$, so $|a_n z_0^n| < \dfrac12$ for sufficiently large $n$, so $|a_n z^n| = |a_n z_0^n| \left(\left|\dfrac{z_0}{z}\right|\right)^n < \dfrac12 \left(\left|\dfrac{z_0}{z}\right|\right)^n$, so $a_n z^n$ is absolutely summable, so $a_n z^n$ is summable
Let $g : [0,\frac{ 1} {2} ] → \mathbb R$ be a continuous function. Define $g_n : [0,\frac{ 1} {2} ] → \mathbb R$ by $g_1 = g$ and $g_{n+1}(t) = \int_0^t g_n(s) ds,$ for all $n ≥ 1.$ Show that $lim_{n→∞} n!g_n(t) = 0,$ for all $t ∈ [0,\frac{1}{2}]$ .
Can you give some hint?
My attempt:- $t\in [0,1/2]$ Consider the sequence $a_n(t)=n!g_n(t)$
If $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}<1$, then it converges to zero.
I have a bilinear functional that is bounded from below
I try to approximate the minimum by a ansatz-function that is a linear combination
of any independent functions of the proper function space
I now obtain an expression that is bilinear in the coeffcients
using the stationarity condition (all derivaties of the functional w.r.t the coefficients = 0)
I get a set of $n$ equations with the $n$ the number of coefficients
a set of n linear homogeneus equations in the $n$ coefficients
Now instead of "directly attempting to solve" the equations for the coefficients I rather look at the secular determinant that should be zero, otherwise no non trivial solution exists
This "characteristic polynomial" directly yields all permissible approximation values for the functional from my linear ansatz.
Avoiding the neccessity to solve for the coefficients.
I have problems now to formulated the question. But it strikes me that a direct solution of the equation can be circumvented and instead the values of the functional are directly obtained by using the condition that the derminant is zero.
I wonder if there is something deeper in the background, or so to say a more very general principle.
If $x$ is a prime number and a number $y$ exists which is the digit reverse of $x$ and is also a prime number, then there must exist an integer z in the mid way of $x, y$ , which is a palindrome and digitsum(z)=digitsum(x).
> Bekanntlich hat P. du Bois-Reymond zuerst die Existenz einer überall stetigen Funktion erwiesen, deren Fouriersche Reihe an einer Stelle divergiert. Herr H. A. Schwarz gab dann ein einfacheres Beispiel.
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
(Translation: It is well-known that Paul du Bois-Reymond was the first to demonstrate the existence of a continuous function with fourier series being divergent on a point. Afterwards, Hermann Amandus Schwarz gave an easier example.)
It's discussed very carefully (but no formula explicitly given) in my favorite introductory book on Fourier analysis. Körner's Fourier Analysis. See pp. 67-73. Right after that is Kolmogoroff's result that you can have an $L^1$ function whose Fourier series diverges everywhere!! |
I recently started to study maths at university and in the analysis course we started, as usual, by looking at the axioms of $\mathbb R$ as a field. I think, I've understood the underlying intuition of these axioms quite well, but one question remained open for me:
How do we define, what we do when we add two numbers, i.e. how to really calculate the result. Is the intuitive addition the only interesting function from $\mathbb R^2$ to $\mathbb R$ which satisfies these axioms? Is this question even relevant?
I thought, I could give it a try, and made the following rules for addition and multiplication (which are more or less accurate):
(1) If $m,n\in\mathbb N_0$, then there exists to sets $E,F$ with cardinality $m$ and $n$ respectively for which $E\cap F=\{\}$. Then: $$ m+n:=|E\cup F| $$ Now define the inverses.
(2) Now either define $$ mn:=|\{(e,f):e\in E,\space f\in F\}| $$ or $$ mn:=\underbrace{m+m+...+m}_{n} $$ Then define the inverses.
From this we can extend the addition to the rationals by: $$ \frac ab +\frac cd:=\frac{ad+bc}{bd} $$ And even to the reals with cantors limit construction of $\mathbb R$.
The principle that I don't fully understand, is that we never seem to define our proper addition and multiplication. It seems to be always implicit what we mean by $+$ and $*$. How to explain this? |
As every MO user knows, and can easily prove, the inverse of the matrix $\begin{pmatrix} a & b \\\ c & d \end{pmatrix}$ is $\frac{1}{ad - bc} \begin{pmatrix} d & -b \\\ -c & a \end{pmatrix}$. This can be proved, for example, by writing the inverse as $ \begin{pmatrix} r & s \\\ t & u \end{pmatrix}$ and solving the resulting system of four equations in four variables.
As a grad student, when studying the theory of modular forms, I repeatedly forgot this formula (do you switch the $a$ and $d$ and invert the sign of $b$ and $c$... or was it the other way around?) and continually had to rederive it. Much later, it occurred to me that it was better to remember the formula was obvious in a couple of special cases such as $\begin{pmatrix} 1 & b \\\ 0 & 1 \end{pmatrix}$, and diagonal matrices, for which the geometric intuition is simple. One can also remember this as a special case of the adjugate matrix.
Is there some way to just write down $\frac{1}{ad - bc} \begin{pmatrix} d & -b \\\ -c & a \end{pmatrix}$, even in the case where $ad - bc = 1$, by pure thought -- without having to compute? In particular, is there some geometric intuition, in terms of a linear transformation on a two-dimensional vector space, that renders this fact crystal clear?
Or may as well I be asking how to remember why $43 \times 87$ is equal to 3741 and not 3731?
Thank you! --Frank |
I'm currently teaching a couple of courses that have a calculus prerequisite, and within the last week I've had two students make notational mistakes that amount to writing $y\frac{d}{dx}$ rather than $\frac{d}{dx}y$ (although in terms of different variables than $y$ and $x$). E.g., they might write something like
$$ x^2 \frac{d}{dx}=2x.$$
At least one of them actually seemed relatively fluent with calculus, in the sense of knowing facts like $\int x^n dx=x^{n+1}/(n+1)$. I pointed out his mistake and made the analogy with the ungrammatical expression $4\sqrt{}$. I mentioned that I had just had another student make a similar mistake recently, and I asked him if he could explain more about what he was thinking when he wrote it. His response was that that was just the way he had gotten used to seeing it in his calculus book! (I assume this is not actually true.)
Can anyone provide any insight into why this would be a common enough mistake that I would see it this frequently? I wonder if there is some confusion because the $f'$ notation involves a postfix operator, or because in an integral we usually write the $dx$ at the end...?
It would seem obvious to me that their notation wouldn't make sense if you think of the derivative as the ratio of two infinitesimals, but presumably many of them haven't been exposed to that way of thinking. Part of their confusion may also be because they are used to everything being $y(x)$, never $\Psi(x)$ or $v(t)$. They also never seem to have been asked to think about the meaning of notation and why it makes sense, and don't seem to understand attempts to elicit discussion of this kind of thing. E.g., if I ask them whether they've been taught that $d/dx$ is an operator, like $\sqrt{}$ or $\sin{}$, they generally look at me blankly, as though this is not the kind of thing that was ever discussed in their calculus class. |
A new Higgs tadpole cancellation condition reformulating the hierarchy problem Strings 2013 [talks] is underway.The first hep-ph paper today probably got to that exclusive place because the authors were excited and wanted to grab the spot. Andre de Gouvea, Jennifer Kile, and Roberto Vega-Morales of Illinois chose the title \(H\rightarrow \gamma\gamma\) as a Triangle Anomaly: Possible Implications for the Hierarchy ProblemThey point out a curious feature of the diagrams calculating the Higgs boson decay to two photons (yes, it's the process that seemed to have a minor excess at the LHC but this excess went away): while the diagram is finite, one actually gets different results according to the choice of the regularization.
\({\Huge \Rightarrow}\)
In particular, the \(d=4\) direct calculation leads to a finite result but it actually violates the gauge invariance so it can't be right. It should be disturbing for you that wrong results may arise from quantum field theory calculations even if you don't encounter any divergence.
However, the right fix is known: work with a regularization, typically dimensional regularization, that automatically respects the gauge invariance. Using dim reg, in \(d=4-\epsilon\) dimensions, one automatically gets the right result. However, it still disagrees with the wrong result computed directly in \(d=4\).
While this episode doesn't mean that QFT is ill-defined or inconsistent and we actually know how to do things correctly, the finite-yet-wrong result in \(d=4\) surely sounds bizarre. The authors propose a new condition on quantum field theories: this strangeness shouldn't be there. In other words, the wrong, Ward-identity-violating terms in the \(d=4\) calculation should cancel. When they cancel, the \(d=4\) calculation will agree with the correct dim reg \(d=4-\epsilon\) calculation.
The paper suggests that this cancellation is a new general principle of physics that constrains the allowed spectra of particles and fields and that should be added next to the usual triangle diagram gauge anomaly cancellation conditions in the Standard Model and similar gauge theories.
Note that the triangle anomaly diagrams may be blamed on linear divergences in the integrals. Here, the new type of an "anomaly" that should be canceled is also related to the linearly divergent part of certain integrals because they behave differently under the shift of momenta. So even the computational origin of their new "anomaly" resembles the case of the chiral anomaly. In some sense, the "new" anomaly only differs from the well-known triangle anomaly by its replacement of one external gauge boson with the Higgs boson.
These diagrams that have to cancel are close to some Higgs tadpole diagrams – Feynman diagrams you would use to compute the shift of the Higgs vacuum expectation value (vev). The "tadpole cancellation conditions" are well-known to string theorists but they weren't really discussed in the context of ordinary 4D quantum field theories yet. I suppose that there should be a more natural way to phrase and justify the Higgs tadpole cancellation condition. The condition looks like eqn (36)\[
3ge^2M_W + \frac{e^2 g m_H^2}{2M_W} +\sum_{\rm scalars} 2\lambda_S v e_s^2-\!\!\sum_{\rm fermions}\!\! 2\lambda_f^2 v e_f^2 = 0
\] Supersymmetry seems to be the only known natural principle that cancels the new "anomaly". The authors have only checked it by some uninspiring brute force calculation in the MSSM as a function of several parameters. I guess that there's a simple proof that supersymmetry – unbroken or broken at an arbitrary scale – cancels the new "anomaly" condition.
It's probably true and they probably realize that the new condition is mostly equivalent to the usual unbearable lightness and naturalness of the Higgs' being. However, if you might phrase the condition for naturalness as a version of an anomaly cancellation condition, it would probably be (or at least look) much more inevitable than the usual arguments discussing the hierarchy problem. |
Probabilists often work with Polish spaces, though it is not always very clear where this assumption is needed.
Question: What can go wrong when doing probability on non-Polish spaces?
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One simple thing that can go wrong is purely related to the size of the space (Polish spaces are all size $\leq 2^{\aleph_0}$). When spaces are large enough product measures become surprisingly badly behaved. Consider Nedoma's pathology:
Let $X$ be a measurable space with $|X| > 2^{\aleph_0}$. Then the diagonal in $X^2$ is not measurable.
We'll prove this by way of a theorem:
Let $U \subseteq X^2$ be measurable. $U$ can be written as a union of at most $2^{\aleph_0}$ subsets of the form $A \times B$.
Proof: First note that we can find some countable collection $(A_i)_{i\ge 0}$ of subsets of $X$, such that $U \subseteq \sigma(\{A_i \times A_j:i,j\ge 0\})$, where $\sigma(\cdot)$ denotes the $\sigma$-algebra generated by the given subsets (proof: The set of $V$ such that we can find such $A_i$ is a $\sigma$-algebra containing the basis sets).
For $x \in \{0, 1\}^\mathbb{N}$ define $B_x = \bigcap \{ A_i : x_i = 1 \} \cap \bigcap \{ A_i^c : x_i = 0 \}$.
Consider all subsets of $X^2$ which can be written as a (possibly uncountable) union of $B_x \times B_y$ for some $y$. This is a $\sigma$-algebra and obviously contains all the $A_i \times A_j$, so contains $U$.
But now we're done. There are at most $2^{\aleph_0}$ of the $B_x$, and each is certainly measurable in $X$, so $U$ can be written as a union of $2^{\aleph_0}$ subsets of the form $A \times B$.
QED
Corollary: The diagonal is not measurable.
Evidently the diagonal cannot be written as a union of at most $2^{\aleph_0}$ rectangles, as they would all have to be single points, and the diagonal has size $|X| > 2^{\aleph_0}$.
Separability is a key technical property used to avoid measure-theoretic difficulties for processes with uncountable index sets. The general problem is that measures are only countably additive and $\sigma$-algebras are closed under only countably many primitive set operations. In a variety of scenarios, uncountable collections of measure zero events can bite you; separability ensures you can use a countable sequence as a proxy for the entire process without losing probabilistic content. Here are two examples.
Weak convergence: the classical theory of weak convergence utilizes Borel-measurable maps. When dealing with some function-valued random elements, such as cadlag functions endowed with the supremum norm, Borel-measurability fails to hold. See the motivation for
Weak Convergence and Empirical Processes. The $J1$ topology is basically a hack which ensures the function space is separable and thereby avoids measurability issues. The parallel theory of weak convergence described in the book embraces non-measurability.
Existence of stochastic processes with nice properties: a key property of Brownian motion is continuity of the sample paths. Continuity, however, is a property involving uncountably many indices. The existence of a continuous version of a process can be ensured with separable modifications. See this lecture and the one that follows.
Metrizability allows us to introduce concepts such as convergence in probability. Completeness (the Cauchy convergence kind, not the null subsets kind) makes it easier to conduct analysis.
There's already been some good responses, but I think it's worth adding a very simple example showing what can go wrong if you don't use Polish spaces.
Consider $\mathbb{R}$ under its usual topology, and let X be a non-Lebesgue measurable set. e.g., a Vitali set. Using the subspace topology on X, the diagonal $D\subseteq\mathbb{R}\times X$, $D=\{(x,x)\colon x\in X\}$ is Borel Measurable. However, its projection onto $\mathbb{R}$ is X, which is not Lebesgue measurable. Problems like this are avoided by keeping to Polish spaces. A measurable function between Polish spaces always take Borel sets to analytic sets which are, at least, universally measurable.
The space X in this example is a separable metrizable metric space, whereas Polish spaces are separable completely metrizable spaces. So things can go badly wrong if just the completeness requirement is dropped.
Below is a copy of an answer I gave here https://stats.stackexchange.com/questions/2932/metric-spaces-and-the-support-of-a-random-variable/20769#20769
Here are some technical conveniences of separable metric spaces
(a) If $X$ and $X'$ take values in a separable metric space $(E,d)$ then the event $\{X=X'\}$ is measurable, and this allows to define random variables in the elegant way: a random variable is the equivalence class of $X$ for the "almost surely equals" relation (note that the normed vector space $L^p$ is a set of equivalence class)
(b) The distance $d(X,X')$ between the two $E$-valued r.v. $X, X'$ is measurable; in passing this allows to define the space $L^0$ of random variables equipped with the topology of convergence in probability
(c) Simple r.v. (those taking only finitely many values) are dense in $L^0$
And some techical conveniences of complete separable (Polish) metric spaces :
(d) Existence of the conditional law of a Polish-valued r.v.
(e) Given a morphism between probability spaces, a Polish-valued r.v. on the first probability space always has a copy in the second one
(f)
Doob-Dynkin functional representation: if $Y$ is a Polish-valued r.v. measurable w.r.t. the $\sigma$-field $\sigma(X)$ generated by a random element $X$ in any measurable space, then $Y = h(X)$ for some measurable function $h$.
We know by Ulam's theorem that a Borel measure on a Polish space is necessarily tight. If we just assume that the metric space is separable, we have that each Borel probability measure on $X$ is tight if and only if $X$ is universally measurable (that is, given a probability measure $\mu$ on the metric completion $\widehat X$, there are two measurable subsets $S_1$ and $S_2$ of $\widehat X$ such that $S_1\subset X\subset S_2$ and $\mu(S_1)=\mu(S_2)$. So a probability measure is not necessarily tight (take $S\subset [0,1]$ of inner Lebesgue measure $0$ and outer measure $1$), see Dudley's book
Real Analysis and Probability.
An other issue related to tightness. We know by Prokhorov theorem that if $(X,d)$ is Polish and if for all sequence of Borel probability measures $\{\mu_n\}$ we can extract a subsequence which converges in law, then $\{\mu_n\}$ is necessarily uniformly tight. It may be not true if we remove the assumption of "Polishness". And it may be problematic when we want results as "$\mu_n\to \mu$ in law if and only if there is uniform tightness and convergence of finite dimensional laws."
Google "image measure catastrophe" with quotation marks.
It can also be useful to have the set of Borel probability measures on $X$ (with weak* convergence, a.k.a. convergence in law) to be metrizable, for instance to be able to treat the convergences sequentially. For this you need the space $X$ to be separable and metrizable (see Lévy-Prohorov metric).
Fun fact: you can find a non-separable Banach space and a Gaussian probability measure on it which gives measure $0$ to
every ball of radius $1$. (In particular your intuition about notions like the "support" of a measure goes pretty badly wrong.)
Consider i.i.d. $\xi_n$ and take as your norm $|\xi|^2 = \sup_{k\ge 0}2^{-k}\sum_{n=1}^{2^{k}} |\xi_n|^2$. This is almost surely finite by Borel-Cantelli and almost surely at least $1$ by the law of large numbers. The fact that it gives measure $0$ to every ball of radius $1$ is left as an exercise. This norm isn't even very exotic: if you interpret the $\xi_n$'s as Fourier coefficients, then $B$ is really just the Besov space $B^{1/2}_{2,\infty}$.
As far as I remember, the projection of a measurable set may fail to be measurable, so something very natural may become not an event. Besides, constructing conditional probabilities as measures on sections becomes problematic. Perhaps, there are more reasons but these two are already good enough. |
Let $a \in \Bbb{R}$. Let $\Bbb{Z}$ act on $S^1$ via $(n,z) \mapsto ze^{2 \pi i \cdot an}$. Claim: The is action is not free if and only if $a \Bbb{Q}$. Here's an attempt at the forward direction: If the action is not free, there is some nonzero $n$ and $z \in S^1$ such that $ze^{2 \pi i \cdot an} = 1$. Note $z = e^{2 \pi i \theta}$ for some $\theta \in [0,1)$.
Then the equation becomes $e^{2 \pi i(\theta + an)} = 1$, which holds if and only if $2\pi (\theta + an) = 2 \pi k$ for some $k \in \Bbb{Z}$. Solving for $a$ gives $a = \frac{k-\theta}{n}$...
What if $\theta$ is irrational...what did I do wrong?
'cause I understand that second one but I'm having a hard time explaining it in words
(Re: the first one: a matrix transpose "looks" like the equation $Ax\cdot y=x\cdot A^\top y$. Which implies several things, like how $A^\top x$ is perpendicular to $A^{-1}x^\top$ where $x^\top$ is the vector space perpendicular to $x$.)
DogAteMy: I looked at the link. You're writing garbage with regard to the transpose stuff. Why should a linear map from $\Bbb R^n$ to $\Bbb R^m$ have an inverse in the first place? And for goodness sake don't use $x^\top$ to mean the orthogonal complement when it already means something.
he based much of his success on principles like this I cant believe ive forgotten it
it's basically saying that it's a waste of time to throw a parade for a scholar or win he or she over with compliments and awards etc but this is the biggest source of sense of purpose in the non scholar
yeah there is this thing called the internet and well yes there are better books than others you can study from provided they are not stolen from you by drug dealers you should buy a text book that they base university courses on if you can save for one
I was working from "Problems in Analytic Number Theory" Second Edition, by M.Ram Murty prior to the idiots robbing me and taking that with them which was a fantastic book to self learn from one of the best ive had actually
Yeah I wasn't happy about it either it was more than $200 usd actually well look if you want my honest opinion self study doesn't exist, you are still being taught something by Euclid if you read his works despite him having died a few thousand years ago but he is as much a teacher as you'll get, and if you don't plan on reading the works of others, to maintain some sort of purity in the word self study, well, no you have failed in life and should give up entirely. but that is a very good book
regardless of you attending Princeton university or not
yeah me neither you are the only one I remember talking to on it but I have been well and truly banned from this IP address for that forum now, which, which was as you might have guessed for being too polite and sensitive to delicate religious sensibilities
but no it's not my forum I just remembered it was one of the first I started talking math on, and it was a long road for someone like me being receptive to constructive criticism, especially from a kid a third my age which according to your profile at the time you were
i have a chronological disability that prevents me from accurately recalling exactly when this was, don't worry about it
well yeah it said you were 10, so it was a troubling thought to be getting advice from a ten year old at the time i think i was still holding on to some sort of hopes of a career in non stupidity related fields which was at some point abandoned
@TedShifrin thanks for that in bookmarking all of these under 3500, is there a 101 i should start with and find my way into four digits? what level of expertise is required for all of these is a more clear way of asking
Well, there are various math sources all over the web, including Khan Academy, etc. My particular course was intended for people seriously interested in mathematics (i.e., proofs as well as computations and applications). The students in there were about half first-year students who had taken BC AP calculus in high school and gotten the top score, about half second-year students who'd taken various first-year calculus paths in college.
long time ago tho even the credits have expired not the student debt though so i think they are trying to hint i should go back a start from first year and double said debt but im a terrible student it really wasn't worth while the first time round considering my rate of attendance then and how unlikely that would be different going back now
@BalarkaSen yeah from the number theory i got into in my most recent years it's bizarre how i almost became allergic to calculus i loved it back then and for some reason not quite so when i began focusing on prime numbers
What do you all think of this theorem: The number of ways to write $n$ as a sum of four squares is equal to $8$ times the sum of divisors of $n$ if $n$ is odd and $24$ times sum of odd divisors of $n$ if $n$ is even
A proof of this uses (basically) Fourier analysis
Even though it looks rather innocuous albeit surprising result in pure number theory
@BalarkaSen well because it was what Wikipedia deemed my interests to be categorized as i have simply told myself that is what i am studying, it really starting with me horsing around not even knowing what category of math you call it. actually, ill show you the exact subject you and i discussed on mmf that reminds me you were actually right, i don't know if i would have taken it well at the time tho
yeah looks like i deleted the stack exchange question on it anyway i had found a discrete Fourier transform for $\lfloor \frac{n}{m} \rfloor$ and you attempted to explain to me that is what it was that's all i remember lol @BalarkaSen
oh and when it comes to transcripts involving me on the internet, don't worry, the younger version of you most definitely will be seen in a positive light, and just contemplating all the possibilities of things said by someone as insane as me, agree that pulling up said past conversations isn't productive
absolutely me too but would we have it any other way? i mean i know im like a dog chasing a car as far as any real "purpose" in learning is concerned i think id be terrified if something didnt unfold into a myriad of new things I'm clueless about
@Daminark They key thing if I remember correctly was that if you look at the subgroup $\Gamma$ of $\text{PSL}_2(\Bbb Z)$ generated by (1, 2|0, 1) and (0, -1|1, 0), then any holomorphic function $f : \Bbb H^2 \to \Bbb C$ invariant under $\Gamma$ (in the sense that $f(z + 2) = f(z)$ and $f(-1/z) = z^{2k} f(z)$, $2k$ is called the weight) such that the Fourier expansion of $f$ at infinity and $-1$ having no constant coefficients is called a cusp form (on $\Bbb H^2/\Gamma$).
The $r_4(n)$ thing follows as an immediate corollary of the fact that the only weight $2$ cusp form is identically zero.
I can try to recall more if you're interested.
It's insightful to look at the picture of $\Bbb H^2/\Gamma$... it's like, take the line $\Re[z] = 1$, the semicircle $|z| = 1, z > 0$, and the line $\Re[z] = -1$. This gives a certain region in the upper half plane
Paste those two lines, and paste half of the semicircle (from -1 to i, and then from i to 1) to the other half by folding along i
Yup, that $E_4$ and $E_6$ generates the space of modular forms, that type of things
I think in general if you start thinking about modular forms as eigenfunctions of a Laplacian, the space generated by the Eisenstein series are orthogonal to the space of cusp forms - there's a general story I don't quite know
Cusp forms vanish at the cusp (those are the $-1$ and $\infty$ points in the quotient $\Bbb H^2/\Gamma$ picture I described above, where the hyperbolic metric gets coned off), whereas given any values on the cusps you can make a linear combination of Eisenstein series which takes those specific values on the cusps
So it sort of makes sense
Regarding that particular result, saying it's a weight 2 cusp form is like specifying a strong decay rate of the cusp form towards the cusp. Indeed, one basically argues like the maximum value theorem in complex analysis
@BalarkaSen no you didn't come across as pretentious at all, i can only imagine being so young and having the mind you have would have resulted in many accusing you of such, but really, my experience in life is diverse to say the least, and I've met know it all types that are in everyway detestable, you shouldn't be so hard on your character you are very humble considering your calibre
You probably don't realise how low the bar drops when it comes to integrity of character is concerned, trust me, you wouldn't have come as far as you clearly have if you were a know it all
it was actually the best thing for me to have met a 10 year old at the age of 30 that was well beyond what ill ever realistically become as far as math is concerned someone like you is going to be accused of arrogance simply because you intimidate many ignore the good majority of that mate |
The idea of a finite automaton is that each state stands for a combination of following kinds of properties.
properties of the whole string, which basically determine whether the string should be accepted or not. what are the last several letters of a string, which basically determine the transitions between the states when another character is appended to the string.
For the language in the question, the relevant properties are
$\quad$- whether the string contains $ab$ $\quad$- the last character of the string So there are about $2\times 2 = 4$ states. $\quad$ $q_{\epsilon}$: the empty string. $\quad$ $q_{\text{no}, a}$: strings that do not contains $ab$, ending with $a$. $\quad$ $q_{\text{no}, b}$: strings that do not contains $ab$, ending with $b$. $\quad$ $q_{\text{yes}, a}$: strings that contain $ab$, ending with $a$. $\quad$ $q_{\text{yes}, b}$: strings that contain $ab$, ending with $b$. The last two states can be merged as one state $\quad$ $q_{yes}$: strings that contain $ab$.
Note that meaningful subscripts are used to indicate the properties associated to the states. Upon seeing ${q}_{\text{no},a}$, you know immediately the input that reaches this state does not contain $ab$ and it ends with $a$.
Exercise. Define a finite automata that accepts the language $\{w\in\{a,b\}^*\mid w\text{ contains } b\text{ but does not contain } bba\text{ as a subword}\}$. |
So I'm reading Dummit and Foote and they define the discriminant of $x_{1},...,x_{n}$ by $$D=\prod_{i<j}(x_{i}-x_{j})^2$$ and the discriminant of a polynomial to be the discriminant of the roots.
They say that a permutation $\sigma \in S_{n}$ is in the alternating group $A_{n}$ iff $\sigma$ fixes the product $\sqrt{D}$.
It follows by the Fundamental Theorem of Galois Theory that if $F$ has characterstic different from 2 then $\sqrt{D}$ generates the fixed field of $A_{n}$ and generates a quadratic extension of $K$.
I am confused where characteristic $2$ comes into this. |
I have $$F(z) = \phi + i\psi$$
Also, it is given that $$u = \frac{\delta \phi}{\delta x}, \ \ v = \frac{\delta\phi}{\delta y}$$ and $$u = \frac{\delta \psi}{\delta x}, \ \ v = -\frac{\delta\psi}{\delta y}$$
Now I have to prove that $$\frac{dF}{dz} = u - iv$$
But I am getting terms like $\dfrac{dx}{dz}$ and $\dfrac{dy}{dz}$ in the differentiated term. How do I proceed?
EDIT:
As pointed out the function $\psi$ was wrongly defined by me. It's actually
$$u = \frac{\delta \psi}{\delta y}, \ \ v = -\frac{\delta\psi}{\delta x}$$ |
There is in fact a whole bunch of related algorithms. In modern context I believe "time warp" would be called sequence alignment. Depending on whether you want to match complete strings or optimal substrings one gets Needleman-Wunsch and Smith-Waterman.In your latter algorithm the costs seem to vary, that is one can attribute different costs for deletion ...
Sure. There's a straightforward way to convert an unnormalized distance metric into a normalized similarity measure. Basically, use$$S(x,y) = \frac{M - D(x,y)}{M},$$where $D(x,y)$ is the distance between $x$ and $y$, $S$ is the normalized similarity measure between $x$ and $y$, and $M$ is the maximum value that $D(x,y)$ could be.In the case of ...
Yes, you could try applying the LSTM iteratively 20 times. In other words: use the first 200 datapoints to predict the 201th; then use datapoints 2..201 to predict the 202th; and so on, until you predict the 220th. You'll have to evaluate how well this works on a test set; it might work, or it might not.This could still fail badly. It could even be that ...
Yes, your data is "time-series data", since it's a sequence of measurements of the same variable collected over time. Time-series data can be collected continuously or at discrete intervals.Your sample data can be expressed as a function of time - maybe it helps to think of the "function" as the process that produces the measured output, the input to the ...
Consider that:a "sliding window" approach can be used with any standard regression / classification algorithm. E.g. given the following time seriesTime High Low Open Close Volume1 H1 L1 O1 C1 V12 H2 L2 O2 C2 V2...i Hi Li Oi Ci Viyou can feed the ML algorithm with these examples:O0, H1, L1, O1, C1, V1, H2,...
OK. So the problem is as follows (with different notation):Inputs: disjoint intervals $I_1,\dots,I_k$; disjoint intervals $J_1,\dots,J_m$Output: disjoint intervals $K_1,\dots,K_n$ that are a "refinement" of the $I,J$-intervalsThe $I$-intervals are not necessarily disjoint from the $J$-intervals. We want the $K$-intervals to have the following property:...
There is a $O(nmW)$-time algorithm using dynamic programming. Let $A[i,j] = $ the cost of the best matching of $[s_1,\dots,s_i]$ to $[t_1,\dots,t_j]$ such that $s_i$ is matched to $t_j$. Then$$A[i,j] = \min\{c(s_i,t_j) + A[i-1,j-k] : k=0,1,\dots,W\}.$$If you consider $W$ a constant, then you obtain a $O(nm)$-time algorithm.I don't know if the factor ...
I would suggest to make the correlation plot and observe the dependencies. Yes, ARIMA, is so far the best forecasting model for time series and it should work in this data set too.For the implementation, in R, there is auto ARIMA, which automatically takes the best suited valued for the parameters(p: The number of lag observations included in the ...
If the timestamps can take $T$ different values, then it is simple to implement the new, update, and delete operations in time $\Theta(T)$ and the query in time $O(1)$, by maintaining two arrays $val[ ]$ and $q[]$ of size $T$. The array $q[]$ is initially filled with zero values.new(x, v): set $val[x] = v$ and increment each $q[y]$ with $y\ge x$ by $v$;...
You note " but with more than one prototype per class/cluster."We often call this a polymorphic class. Consider the analogue in text..C1 = { dpacekfjklwalkflwalkklpacedalyutekwalksfj}C2 = { jhjhleapashljumpokdjklleaphfleapfjjumpacgd}Here the class 1 prototypes are polymorphic {walk OR pace}, and so is class 2 {leap or jump}.You can learn these ...
You don't need a clever or sophisticated algorithm. It suffices to linearly scan through the time series, in chronological order, and keep track of the length of window of values less than $X$.At time $t$, remember the length of the longest window ending at time $t$ such that all values in the window are less than $X$. At time $t+1$, when you see the ...
You could try symbolic regression (SR).The general idea is finding a function that fits the given data points without making any assumptions about the structure of that function.Genetic programming (GP) is well suited to this sort of task since it makes no such assumptions.SR was one of the earliest applications of GP and continues to be widely studied....
Use nonlinear regression: https://en.wikipedia.org/wiki/Nonlinear_regression. Write down a model, e.g., $f(x) = \sin(\alpha x + \beta)$, and then try to find parameters $\alpha,\beta$ that minimize the regression error (the sum of the squared error at each point; i.e., square of difference between model's prediction and observed value). You can use ...
DTW is designed to handling local changes in timing.Global changes is time are referred to as Uniform Scaling [a].You can create a FOR loop, loop over all possible Uniform Scalings, and record the one that has the lowest distance (under either euclidean distance, or DTW, or both [b])That degree of Uniform Scaling is then your prediction of the speed ...
There is no single "speed difference". The way DTW works is that it checks whether you can find a match by slowing up and/or speeding up one of the signals.For instance, suppose you want to match signal X to signal Y, which last 3 minutes. It's possible that the best match might involve speeding up signal Y for the first minute, slowing it down for the ... |
Any set of positive Lebesgue measure, have a Lebesgue measurable proper subset with the same measure?
For example: if $A$ is a measurable set, with $m(A) > 0$,
I can say that $x \in A$ (since $A$ is not empty because $m(A) > 0$), then can I say that $B = A\setminus \{x\}$ is a measurable subset of $A$ and $m(B) = m(A)$ (since $m({x}) = 0$)?
I want to use this "claim" to solve a part of a problem that I have to show that if $\int_{B} f = 0$ for any measurable subset $B$ of $A$, then $\int_{A}f = 0 = \int_{B}f$ to eventually show that $f = 0$ a.e on $A$.
Is this claim true?
By the way, I also know the Inner approximation: If $E$ is a measurable, then for any $\epsilon > 0$, exits $F \subset E$ such that $m(E\setminus F) < \epsilon$. and $F$ is closed and the similar one for open sets (Outter approximation).
Did the previous theorem says that $m(E\setminus F) = 0$?
Thank you in advance! |
I am working on a problem to prove, but I do not understand it completely. Where should I use inductive method? What is the base case? And so on. Here is my problem:
A truth assignment $M$ is a function that maps propositional variables to $\{0, 1\}$ ($1$ for true and $0$ for false). We write $M\vDash x$ if $x$ is true under $M$. We define a partial order $\leq$ on truth assignments by $M \le M'$ if $M(p) \le M'(p)$ for every propositional variable $p$.
A propositional formula is
positiveif it only contains connectives $\wedge$ and $\vee$ (i.e., no negation $\lnot$ or implication $\Rightarrow$).
Use Proof By Induction to show that for any truth assignments $M$ and $M'$ such that $M\le M'$, and any positive propositional formula $x$, if $M \vDash x$, then $M' \vDash x$.
I am really confused. Any help is welcome. Thank you. |
First note that in the special case $\kappa = \omega_1$, countably closed implies countably directed-closed.
Answer 1: No. In any model of CH, there is a countably closed, $\omega_2$-c.c. forcing that is inequivalent to $\mathbb P = \text{Add}(\omega_1,\omega_2)$.
Proof: Let $\mathbb Q$ be Jensen's partial order for adding a Kurepa tree. Conditions are $(t,b)$, where $t$ is a countable tree of successor height $\alpha$, $b$ is a countable subset of $\mathcal P(\omega_1)$, and for all $x \in b$, $x \cap \alpha \subseteq t$. Ordering is $(t_1,b_1) \leq (t_0,b_0)$ when $t_1$ end-extends $t_0$ and $b_0 \subseteq b_1$. It is not hard to show that $\mathbb Q$ is countably closed and $\omega_2$-c.c. under CH. Any two conditions with the same first coordinate are compatible. The union of the first coordinates in any generic is a Kurepa tree $T$ by a density argument.
Note that every subset of $\omega_1$ added by $\mathbb P$ is added by some $\omega_1$-sized regular suborder. If $\mathbb Q$ were forcing equivalent to $\mathbb P$, then there would be some $\omega_1$-sized regular subalgebra $\mathbb R$ of $ro(\mathbb Q)$ that adds the generic Kurepa tree $T$. $\mathbb Q$ forces an $\omega_2$-size subset of $\mathcal P(\omega_1)^V$ to be contained in the set of branches of $T$. Thus suppose $G * H$ is $\mathbb R * \mathbb Q / \mathbb R$-generic. But in $V[G]$ we can compute whether $(p,q) \in H$, since we just determine whether $p \subseteq T$ and whether all $x \in q$ are branches of $T$. Thus $V[G * H] = V[G]$, and $\mathbb Q$ must have a dense set of size $\omega_1$. But this is false: If $\{ (t_i,b_i) : i < \omega_1 \} \subseteq \mathbb Q$, then pick $x \notin \bigcup b_i$. No $(t_i,b_i)$ can force $x$ to be a branch of $T$ since we can extend $t$ to rule out $x$ as a branch.
Note that $\mathbb Q$ doesn't have infima to countable chains since we have a lot of choices for the "top level" of the countable tree. But...
Answer 2: Not necessarily, even for $\omega_1$-closed with infima. Let $V$ be a model of GCH, and let $G \subseteq \text{Col}(\omega_1,\omega_2)$ be generic. $\omega_3^V = \omega_2^{V[G]}$. I claim that in $V[G]$, $\mathbb P = \text{Col}(\omega_1,<\omega_2)^{V[G]}$ and $\mathbb Q = \text{Col}(\omega_2,<\omega_3)^{V}$ are both countably closed with infima and both $\omega_2$-c.c., but are not forcing equivalent.
The closure follows from the fact that $V$ and $V[G]$ have the same countable sequences. The $\omega_2$-c.c. for $\mathbb Q$ follows from the fact that $\mathbb Q \times \text{Col}(\omega_1,\omega_2)$ is $\omega_3$-c.c. in $V$.
Next, one can prove a similar fact to Mohammad's answer. In $V[G]$,
(1) $\Vdash_{\mathbb P} \exists A \in [\omega_2]^{\omega_2}$ such that $A$ does not contain any $\omega_1$-sized set from $V[G]$.
(2) $\Vdash_{\mathbb Q} \exists A \in [\omega_2]^{\omega_2}$ such that for all $X \in [\omega_2]^{\omega_2} \cap V[G]$, there is $y \in [X]^{\omega_1} \cap V[G]$ such that $y \cap A = \emptyset$.
The justification for
(1) is the same as for Mohammad's answer. To see (2), let $\dot A$ be a name for a code of the generic $H$ for $\mathbb Q$. In $V$, let $\dot B$ be a name for a code for $G \times H$. If $X \in ([\omega_2]^{\omega_2})^{V[G]}$, let $\dot X$ be a name for it in $V$. Let $p \in \text{Col}(\omega_1,\omega_2)$ be arbitrary, and let $\{ p_i : i < \omega_3 \} \subseteq \text{Col}(\omega_1,\omega_2) \restriction p$ be such that $p_i$ decides the $i^{th}$ member of $\dot X$. Let $p'$ be such that $p_i = p'$ for $\omega_3$ many $i$, giving a large set $Y$ such that $p' \Vdash \check Y \subseteq \dot X$. Let $q \in \mathbb Q$ be arbitrary. Interpreting $Y$ as a subset of $\mathbb Q$, find $Y' \subseteq Y$ of size $\omega_3$ such that $q \nleq y$ for all $y \in Y'$. For each such $y$, choose $q_y \leq q$ such that $q_y \perp y$. Now form a $\Delta$-system among the $q_y$, and let $z$ be an infimum of a set $S$ of $\omega_1$ many of them. Then $z$ forces that the generic $H$ misses $y$ for $q_y \in S$. As $p$ and $q$ were arbitrary, we are done.
Likewise, we can show that $\mathbb P$ forces the negation of the the property under the forcing sign in
(2), and vice versa. Some details are in section 2.1.2 of my thesis. I may post more later. |
Take $p$ to be a prime and let $a_1,\dots,a_n\in\mathbb Z$ be some set of integers such that discrepancy of the set of fractional parts $$\{\frac{ma_1}p,\dots,\frac{ma_n}p\}$$ with $m\in\{1,\dots,p-1\}$ being $\mathcal D$.
Then we know there is an integer $t\in\{1,\dots,p\}$ such that the vectors $$\{\frac{ta_1}p,\dots,\frac{ta_n}p\}$$ are in any box of size $c\mathcal D$ for some constant $c>0$.
This means for any box of size $((c\mathcal D)^{1/n}p)^n$ in $[-p/2,p/2]^n$ there is an integer $t$ such that $$t(a_1,\dots,a_n)\bmod p$$ lies in that box. For example there is a $t$ such that $t(a_1,\dots,a_n)\bmod p$ is in $[2(cD)^{1/n}p,3(cD)^{1/n}p]\times [0,(cD)^{1/n}p]^{n-1}$.
Is it possible to reduce the box size from $((c\mathcal D)^{1/n}p)^n$ to something smaller?
Take $p$ to be a prime and let $a_1,\dots,a_n\in\mathbb Z$ be some set of integers such that discrepancy of the set of fractional parts $$\{\frac{ma_1}p,\dots,\frac{ma_n}p\}$$ with $m\in\{1,\dots,p-1\}$ being $\mathcal D_a$ and let $b_1,\dots,b_n\in\mathbb Z$ be some set of integers such that discrepancy of the set of fractional parts $$\{\frac{mb_1}p,\dots,\frac{mb_n}p\}$$ with $m\in\{1,\dots,p-1\}$ being $\mathcal D_b$.
What is the smallest size such that given any such box we have $t_a,t_b\in\{1,\dots,p-1\}$ such that $$(t_a(a_1,\dots,a_n)+t_b(b_1,\dots,b_n))\bmod p$$ lies in that box? |
This post is a sequel to Concatenate and average NetCDFs where I introduced NCO (NetCDF operators), how to install NCO, and provided examples for the essential operations of record and ensemble concatenation and averaging of variables across multiple NetCDF files. If you are new to NCO I suggest familiarizing yourself with that post and experimenting with some NetCDF files as well as reading the NCO user manual. This post explains how to calculate weighted and non-weighted zonal statistics on variables over different dimensions (e.g. temporal and spatial averages) using the ncwa command line tool. Including several examples as well as information regarding open-source tools for visualizing NetCDF variables and NCO commands for accessing NetCDF metadata. These tools and examples will aid you in your work flow as they did for me!
Calculate zonal means using ncwa
In the previous NCO post we used the record averager (ncra) command to average variables over multiple files over their record dimension, typically time. But what if we need to average or use other aggregating statistics on a variable over space, time, or any other combination of dimensions we desire? The NetCDF weighted averager is just the right tool for the task.
Let's say we have 12 NetCDF files that each hold monthly model output for the water year 1992, in this case they contain simulation results from the Community Earth System Model:
Side Note: visualizing data in NetCDFs!
Any NetCDF file can be viewed quickly with the free tool ncview which you can easily get through the aptitude repository:
Using ncview is simple
Alternatively you could use the NASA's free Panoply software to view and create custom high quality figures from your NetCDF files. Panoply has a lot of options and is great for publication plots but I prefer ncview for taking quick views of a NetCDF file on the fly while doing analysis. The figures below were created with Panoply.
For example, here is a plot of simulated ground temperature in the model_out_1992-06.nc file (June ground temperature).
Now, before we do any weighted averaging lets concatenate the monthly output files along the record dimension (time) using ncrcat to make one NetCDF file TG_concated.nc which contains a time series of the monthly data. For simplicity let's only consider the variable ground temperature (TG) shown above. We can specify the variable we want to subset using the
-v option with ncrcat. Alternatively we could have averaged the variables across their record dimension using ncra but then we would no longer have the time dimension in the output file.
The resulting file now only contains the TG variable,
-O just says to overwrite the output file if it already exists. We can get useful information about the ground temperature variable with NCO as well using ncinfo:
We can see TG is a 3D variable in time, latitude, and longitude with a corresponding array shape of (12, 192, 288) (months, latitude nodes, longitude nodes). We can leverage this information if we need to aggregate and reduce the dimensionality for plotting or analysis.
Side Note: viewing metadata of NetCDFs!
You might be wondering how to view meta data of NetCDF files-- there are several different ways: 1)
ncinfo netcdf_file will print most basic and important information of the contents such as all variables, dimensions, file history, etc.; 2)
ncdump -h netcdf_file provides more detailed info for each variable; 3)
ncks -m will produce output pertaining to dimensions, -m is useful for finding length and units of dimensions and you can pass it variables as well with
or -M netcdf_file -v. The -M option will print out information regarding global attributes. Experiment with these as they are all super useful!
As we viewed using ncinfo our variable of interest is 3 dimensional in time, latitude, and longitude so we can average over these dimensions using ncwa (NetCDF weighted averager) thus reducing its dimensionality, actually the default behaviour of ncwa is to average over all dimensions thus reducing the variable to a scalar. For example just calling ncwa on TG_concated.nc:
The resulting scalar value for TG in the output file of 267.074 Kelvin is the global average ground temperature over the 12 months. To avoid averaging over all dimensions you must use the
-a option followed by the dimensions you want to average over. If you used ncwa -a followed by the record dimension of the input file this would be equivalent to the default behaviour of the NCO record averager ncra. By specifying dimensions with -a we can get the latitudinal zonal mean ground temperature as a function of time by only averaging over the longitudinal axis:
The new shape of lat_zonal_mean.nc is 192 which is the size of the latitude array in the original file. What we have created is a file with ground temperature that has been averaged over 12 months and over every longitude cell resulting in a singular value for each latitude node. This output file is useful for making a latitudinal zonal mean plot of mean ground temperature:
It is not an error that the line breaks around -60 degrees latitude, this is because there is no land in this stretch of the world and remember we are viewing
ground temperature.
The NCO weighted average operator ncwa also supports sub-setting the domain of operation over one or more dimensions using the syntax
ncwa -d dim,[min],[max],[stride] -d dim,[min],[max],[stride] -d dim,[min],[max],[stride] in.nc out.nc. The resulting output file will have the weighted average only for the desired dimensional domain. If stride is not given then all data between min and max are included, a stride of 2 will select every other value of the dimension (e.g. every other day if dim is time), 3 every third and so on. You need to include one of the values or both from the min and max parameters, using a comma for an open ended range. There is no limit on how many dimensions and ranges of dimensions you can subset over. For example, say we wanted to calculate the latitudinal zonal mean as before but only for latitudes above the equator and say we wanted to see how the zonal average evolved through time:
The resulting output contains the latitudinal zonal mean over the twelve months of data, the resulting 2D plot shows the mean ground temperature over the latitudinal axis (x axis) versus time (y axis).
The global monthly temperature from the model is out of phase because the order of the concatenated files that we started with was in water year order, that is October through September. I chose not to interpolate the monthly data in the plot, as a result you can see 12 distinct bands for each month starting with October. That is not to say that the concatenation did not keep track of the dates correctly- it did, but this example shows that you have flexibility on which date you would like the time series to start from.
Masking with logical conditionals
In addition to being able to subset our domain of operation by dimensions, ncwa supports using logical conditional statements for selection based on a variable's value and a relational operator ($=, \ne, \gt, \lt, \ge, \le$). There are two methods to do this: the long hard way:)
ncwa -m mask_variable -M mask_value -T Fortran_relational in.nc out.nc; or the short simple way:)
ncwa -B 'mask_variable relational mask_value' in.nc out.nc. Where
-m proceeds the variable used for the masking; -T proceeds a Fortran relational operator ($eq, ne, gt, lt, ge, le$) which correspond to ($=, \ne, \gt, \lt, \ge, \le$) in Fortran; -M proceeds the masking value; and if you use the -B option the valid relational operators are ==, !=, >, <, >=, <=. You should make sure to enclose the conditional expression after -B in quotes.
For example lets say we wanted to look at two variables from the model: soil evaporation rate (QSOIL) and plant transpiration rate (QVEGT). With ncwa we could calculate the evaporation for locations where the transpiration is high or vice versa just to see how this NCO operation works. This will require three steps: first we will concatenate the monthly files this time keeping these two variables, second we will calculate the global mean transpiration, and last we will calculate the mean global mean evaporation for locations where transpiration is above the global mean and plot our results.
The resulting plot of soil evaporation (QSOIL) in the output file looks like:
The logical indexing resulting from
-B 'QVEGT > 9.58266e-06' appears to have masked the worlds largest deserts in the output file. This is an obvious result considering there is a dearth vegetation in these regions; although this example does not clearly illustrate the relationship between evaporation and transpiration (which I expect would be a dynamic relationship seasonally as well), with a well thought-out plan the masking option can be quite useful for scientific data analysis.
Weighted averaging
You might be wondering by now how to calculate
weighted averages with ncwa, after all ncwa stands for NetCDF weighted averager! Well it's simple the only extra switch that is required is -w followed by the variable to use for the weights. Sometimes it is useful to include the -N option which tells ncwa to only use the numerator of the weights, i.e. integrating the variable over the selected weighting variable such as grid cell area. It is often that variables (at least from model output) will be area-weighted to begin with, if so
-w area without
-N will produce the same output as would not weighting by area in the first place. However if you use
-N -w area -v var ncwa will calculate the area$\times$var product in each cell without ever dividing by the sum of the areas thus integrating the variable over area. Here are two examples to clarify.
For the first example we will calculate the annual mean leaf area index (LAI) as simulated by the Community Earth System Model. For comparison we will calculate the same mean weighted by simulated plant transpiration (QVEGT). The result from the latter will cause grid cells that have high LAI and high QVEGT to stand out while lessening the value of LAI where QVEGT is below average.
Here is the corresponding plot for the non-weighted mean LAI:
And here is the corresponding plot for the mean LAI weighted by QVEGT:
There are clear differences between the weighted and non-weighted LAI maps, weighting allows us to visualize the locations where both vegetation is dense and where vegetation is transpiring. One observation I made by viewing both plots is there are large regions in northern North America and Asia that transpire quite a bit yet they do not correspond to the highest LAI, most of this region is home to the Boreal or Taiga coniferous forests. Perhaps the Boreal forests are transpiring significantly but are not as dense as say the tropical rainforests of the world. We could of seen the impact of transpiration on LAI more clearly by simply differencing these two plots however I wanted to keep the interpretation of the weighting operation as simple as possible.
Just to clarify the use of the
-N option this example shows how to integrate a variable over grid cell area and time. This is straightforward.
Now let's take a look at the familiar LAI variable after it has been integrated with respect to grid cell area and time:
By integrating with respect to grid cell area (km$^2$) and time the values of LAI are now very huge and not meaningful, to be precise the values are simply the product of LAI and the grid cell area summed over the time length-12 months: $$\mbox{Values for each grid cell above}~=~ \sum_{n=1}^{12} LAI_{i,j}\times GCA_{i,j}$$ where $GCA_{i,j}$ is the grid cell area (km$^2$) for the cell at the $i^{th}$ latitude index and the $j^{th}$ longitude index on the model grid which is 1$^\circ$ resolution. We see that by doing this the values near the equator increased and as we move towards the poles the LAI values decreased as compared to the non-weighted plot of LAI above. This is simply a result of the spherical grid that the model utilizes which has much higher grid cell areas near the equator.
The
-N option is equivalent to -y ttl which tells ncwa to take the sum as opposed to the mean over the specified dimensions after -a and since area is not a dimension in the files we are working with we needed to list area as the weight variable. Note, integrating some variables may or may not make sense depending on what you are trying to accomplish and remember some variables may have already been area weighted and take special care to understand what units you are dealing with. For example QVEGT above has units of mm/s. If you weighted QVEGT by grid cell area (which happens to be in $km^2$) and summed over time as shown above for LAI you would get units of $12\cdot mm\cdot km^2/s$. You could then calculate the average volumetric flux by dividing by 12 and converting the $km^2$ to $mm^2$. Computing other aggregating statistics besides averaging
The ncwa averages by default but also allows additional aggregating statistic calculations by stating them after the
-y option. These methods also work for ncra, ncea, and nces. The ncra (NetCDF record averager) is a less sophisticated version of ncwa, you can find more about it and ncea (NetCDF ensemble averager) here. As mentioned the -N option is the equivalent to -y ttl and tells ncwa to sum over the specified dimensions listed after the -a option. Here is a list of all the statistics that ncwa, ncra, ncea, and nces support with the operation keyword followed by a colon and then its description:
avg: mean value
ttl: sum of values
sqravg: square of mean
sqrt: square root of the mean
avgsqr: mean of sum of squares
max: maximum value
min: minimum value
mabs: maximum absolute value
mebs: mean of absolute value
mibs: minimum absolute value
rms: root-mean-square (normalized by N)
rmssdn: root-mean-square (normalized by N-1)
For example,
Will calculate the max of the monthly simulated QVEGT for the rectangular region around South America.
Final remarks
As we saw the NCO command line tool for weighted averaging (ncwa) is powerful and flexible, it is not limited to weighted averaging but can do a range of aggregating statistics, with or without weights, including subsetting and logical indexing or masking. This makes ncwa one of my favorite NCO tools because it can do so many jobs. Once you get familiar with it I think you will prefer using it over opening NetCDF files in your language of choice and doing the operations there. This is because the NCO syntax is succinct and the inner workings are optimized for speed and efficient memory use. Once you couple NCO with a scripting language of your choice it becomes even more powerful. If you enjoyed this post or have suggestions please comment like subscribe and keep visiting! Cheers and happy NetCDF-ing :) |
I am aware of the inverse function theorem for Lipschitz maps, which uses the notion of generalised derivative $δ_{x_0} f$ of a Lipschitz map $f$, due to F.H.Clarke:
Teorem. Let $f : U ⊂ \mathbb{R}^n → \mathbb{R}^n$ be Lipschitz, and suppose that every matrix $A ∈ δ_{x_0}f$ is invertible. Then there are neighbourhoods $U$ of $x_0$ and $V$ of $f(x_0)$ in $\mathbb{R}^n$ and a Lipschitz map $g : V → U$ such that $f ◦ g = \mathbb{1}_V$ and $g ◦ f = \mathbb{1}_U$.
I am looking for a reference to an inverse function theorem in the similar setting which provides a bound for the inner radius of the neighbourhoods $U$ and $V$.
The closest I have been able to find is due to Ralph Howard's note. In the case of $B_r$ ball of radius $r$ centered in zero his result reads:
Teorem. Let $f : \overline{B_r} ⊂ \mathbb{R}^n → \mathbb{R}^n$, $f(O)=O$, $L:\mathbb{R}^n → \mathbb{R}^n$ invertible linear map and $\rho<1$ such that$$\forall x_1,x_2\in \overline{B_r},\quad \| L^{-1}(f(x_2)-f(x_1)) - (x_2-x_1) \| \leq \rho \| x_2 - x_1 \|\,.$$ Then $f$ is injective on $\overline{B_r}$ with$$\forall x_1,x_2\in \overline{B_r},\quad \frac{1-\rho}{\|L^{-1}\|} \|x_2-x_1\| \leq \| f(x_2)-f(x_1) \| \leq \|L\| (1+\rho) \| x_2-x_1 \| \,.$$ Furthermore, $f$ restricted to open ball $B_r$ is lipeomorphism onto $V=f[B_r]$, where $V$ contains the open ball $B_{r_1}$ of radius $r_1=\rho\frac{1-\rho}{\|L^{-1}\|}$.
In the case of differentiable $f$, results related to what I am looking for are usually based on Newton's iterates. For example, see Theorem 4.1 of Xinghua. |
Let \(A\) be a \(2\times 2\) matrix. Prove that if \(Av_1=\lambda_1v_1\) and \(Av_2=\lambda_2v_2\) for distinct reals \(\lambda_1\) and \(\lambda_2\) and nonzero vectors \(v_1\) and \(v_2\), then both columns of \(A-\lambda_1 I\) is a multiple of \(v_2\).
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Let \( H \) be an \( N \times N \) real symmetric matrix. Suppose that \( |H_{kk}| < 1 \) for \( 1 \leq k \leq N \). Prove that, if \( |H_{ij}| > 4 \) for some \( i, j \), then the largest eigenvalue of \( H \) is larger than \( 3 \).
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Let A=(a
ij) be an n×n matrix such that a ij=cos(i-j)θ and θ=2π/n. Determine the rank and eigenvalues of A.
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Let A be a 0-1 square matrix. If all eigenvalues of A are real positive, then those eigenvalues are all equal to 1.
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On asymptotically lacunary statistical equivalent functions via ideals Authors
Ekrem Savaş - Department of Mathematics, Usak University, Usak, Turkey. Abstract
The goal of this paper is to introduce \(\mathcal{I}_\theta\)-asymptotically statistical equivalent by taking nonnegative two real-valued Lebesgue measurable functions \( \gamma\left( \nu\right) \) and \( \mu\left( \nu\right)\) in
the interval \(\left( 1,\infty \right)\) instead of sequences and we establish some inclusion relations.
Keywords Asymptotical equivalent functions ideal convergence lacunary sequence \(\mathcal{I}\)-statistical convergence MSC References [1] P. Das, E. Savas, S. K. Ghosal, On generalizations of certain summability methods using ideals, Appl. Math. Lett., 24 (2011), 1509–1514. [2] J. A. Fridy , On Statistical Convergence, Analysis, 5 (1985), 301–313. [3] P. Kostyrko, M. Macaj, T. Salat, M. Sleziak, J–Convergence and Extremal J–Limit Points, Math. Slovaca, 55 (2005), 443–464. [4] P. Kostyrko, W. Wilczyński, T. Salat, J–Convergence, Real Anal. Exchange, 26 (2000), 669–686. [5] M. Marouf , Asymptotic equivalence and summability, I. J. Math. Math. Sci., 16 (1993), 755–762. [6] R. F. Patterson, On Asymptotically Statistically Equivalent Sequences, Demonstr. Math., 36 (2003), 149–153. [7] R. F. Patterson, E. Savas, On Asymptotically Lacunary Statistical Equivalent Sequences, Thai J. Math., 4 (2006), 267–272. [8] E. Savas, On J–Asymptotically Lacunary Statistical Equivalent Sequences, Adv. Difference Equ., 2013 (2013), 7 pages. [9] E. Savas, On Asymptotically J–Lacunary Statistical Equivalent Sequences of order \(\alpha\), The 2014 International Conference on Pure Mathematics–Applied Mathematics Venice, Italy (2014) [10] E. Savas, Generalized summability methods of functions using ideals, AIP Conference Proceedings, V. 1676 (2015) [11] E. Savas, On generalized statistically convergent functions via ideals, Appl. Math., 10 (2016), 943–947. [12] E. Savas, P. Das, A generalized statistical convergence via ideals, Appl. Math. Lett., 24 (2011), 826–830. [13] E. Savas, P. Das, S. Dutta, A note on strong matrix summability via ideals, Appl. Math. Lett., 25 (2012), 733–738. [14] E. Savas, H. Gumus, A generalization on I–asymptotically lacunary statistical equivalent sequences, J. Inequal. Appl., 2013 (2013), 9 pages. |
This is an improved version of the argument in Electromagnetic Unruh effect?
In the quantum vacuum particle pairs, with total energy $E_x$, can come into existence provided they annihilate within a time $t$ according to the uncertainty principle $$E_x\ t \sim \hbar.$$ If we let $t=x/c$ then we have $$E_x \sim \frac{\hbar c}{x}$$ where $x$ is the Compton wavelength of the particle pair.
Let us assume that there is a force field present that immediately gives the particles an acceleration $a$ as soon as they appear out of the vacuum.
Approximately, the extra distance, $\Delta x$, that a particle travels before it is annihilated is $$\Delta x \sim a t^2 \sim \frac{ax^2}{c^2}.$$ Therefore the particle pairs have a new Compton wavelength, $X$, given by $$X \sim x + \Delta x \sim x + \frac{ax^2}{c^2}.$$ Accordingly the energy $E_X$ of the particle pairs, after time $t$, is related to their new Compton wavelength $X$ by \begin{eqnarray} E_X &\sim& \frac{\hbar c}{X}\\ &\sim& \frac{\hbar c}{x(1+ax/c^2)}\\ &\sim& \frac{\hbar c}{x}(1-ax/c^2)\\ &\sim& E_x - \frac{\hbar a}{c}. \end{eqnarray} Thus the particle pair energy $E_X$ needed to satisfy the uncertainty principle after time $t$ is less than the energy $E_x$ that was borrowed from the vacuum in the first place. When the particle pair annihilates the excess energy $\Delta E=\hbar a/c$ produces a photon of electromagnetic radiation with temperature $T$ given by $$T \sim \frac{\hbar a}{c k_B}.$$ Thus we have derived an Unruh radiation-like formula for a vacuum that is being accelerated by a field. If the field is the gravitational field then we have derived the Hawking temperature. By the equivalence principle this is the same as the vacuum temperature observed by an accelerating observer. But this formula should be valid for any force field.
Let us assume that the force field is a static electric field $\vec{E}$ and that the particle pair is an electron-positron pair, each with charge $e$ and mass $m_e$. The classical equation of motion for each particle is then $$e\ \vec{E}=m_e\ \vec{a}.$$ Substituting the magnitudes of the electric field and acceleration into the Unruh formula gives $$T \sim \frac{\hbar}{c k_B}\frac{e|\vec{E}|}{m_e}.$$ If we take the electric field strength $|\vec{E}|=1$ MV/m then the electromagnetic Unruh/Hawking temperature is $$T\approx 10^{-2}\ \hbox{K}.$$ If this temperature could be measured then one could experimentally confirm the general Unruh/Hawking effect.
Is there any merit to this admittedly non-rigorous argument or can the Unruh/Hawking effect only be analyzed using quantum field theory? |
Kitefoil introduction
Foil is a technology that allows a hull (propelled by a motor or in this case a sail) to
emerge totally from the water, thanks to the hydrodynamic action of the submerged surface.
In fact, the pressure of the water under the
wings, combined with the depression that forms above them, generates a force of lift opposed to the weight, and allows a great reduction of resistance to motion and consequently an increase in efficiency.
The curve in figure shows qualitatively the
rapid reduction of the resistance, once, reached a certain speed, the hull comes out of the water.
Kitefoil is composed of the following elements:
Fuselage: It extends in length in the direction of motion and transmits the sustaining force to the hull through the mast, to which it is connected; Mast or Keel: It transmits the sustaining force to the hull, connecting it to the fuselage and to the immersed surfaces that create the lift; Supporting and stabilizing wings:These are the surfaces that create lift. The first is able to give all the lift required to separate the hull from the surface of the water, while the second balances the moment provided by the first, with a consequent stabilization effects. History of the hydrofoil
Hydrofoils have been used in different types of boats for over 100 years.
The first person that designed and built an hydrofoil was an Italian named Enrico Forlanini, in 1906. For his hydrofoil Forlanini used a system of 4 groups of parallel wings (a pair in the bow and a pair in the stert) of decreasing width, unlike the single hydrofoil wings in use today.
Forlanini’s design was resumed and improved by various other inventors over the following decades (in particular Alexander Graham Bell and Casey Baldwin), until around the
50’s the world began to invest massively in boats using hydrofoil fins, for both military and commercial use. The boom was reached in the 60 ‘s- 70’s but since then their use in motor boats has gradually decreased, due to various problems; not only due to construction and maintenance costs, but also safety and environmental issues. Materials for hydrofoils were in fact metallic, the same used for the structure of the boat.
The same problems occurred for hydrofoils used in the sailing or hobby disciplines, that began in the 60s, but was soon abandoned.
Since the turn of the century
investments in this technology have resumed. Mainly because new composite materials made it possible to produce extremely light and resistant appendages, different hydrofoil researches began again, in order to identify the best shape and structure for every hull and wind. A wide interest in hydrofoil sailing technology spread trough the media thanks to its use in the 2010 America’s Cup. Some sectors in which the foil has developed, however, are only now becoming popular. Unfortunately research has already reached a moment of stationarity, because the significant risks involved in the sector do not attract investors’ interest. Hydrofoil in kitesurfing
The application of hydrofoil to kitesurfing dates back to the 2000s. The design of modern hydrofoil for kitesurfing varies in geometry based on its type of use. The main categories are:
beginner; freestyle; racing boards. Kitefoils for beginners are designed to be stable at low speeds.
Those for
freestyle instead are more suitable for performing acrobatics and jumps and therefore have greater maneuverability, in addition to being structurally more resistant, in order to be able to withstand impacts on landing jumps. Racing kitefoils are designed to reach the highest possible speeds with the greatest stability for all the different wind conditions. To do this, the latter have a minimal design and are made of carbon fiber, to be as light and resistant as possible.
Hydrofoil for kitesurfing (also called kitefoil) is a combination of various components, each with a very precise function . Although it is easy to design a single fin suitable to a certain sea condition, it is far more complicated to create a kitefoil that is best suited to a wide range of wind conditions, and therefore to a larger speed range.
To best explain the operation of the components of a hydrofoil fin, it is important initially to understand the most important
moments to which the board is subjected: roll, pitch and yaw.
Pitch Yaw and Roll in a Hydrofoil
To understand how kitesurf works we can consider just the first 2, roll and pitch. Yaw can be ignored because load conditions are approximately simmetric and the mast twist can therefore be denied.
Kitefoils must produce enough lift to rise out of the water, giving support to the kitesurfer, and at the same time produce a moment of such magnitude as to allow balancing. The lift created must be sufficient in a wide range of speeds from the starting speed (“take off” speed) to the maximum speed (“top” speed).
The
take-off speed is the speed at which lift begins to be such as to allow the kitesurfer and the board to separate from the water. As resistance decreases, due to the fact that the board is now no longer in contact with water, but in the air (which density is about 1000 times lower than that of water), there is an increase in speed; this increase in speed corresponds to an increase in lift for the main foil, and a change in lift capacity of the stabilizer, which may vary depending on the type used, as will be discussed below.
There are two different functioning systems of the
stabilizer, which can have a positive bearing capacity and a negative bearing capacity. In the case of the positive flow stabilizer, in order to balance the moment of force Fp (Force weight of the kiter minus the force exerted by the kite) must have a arm smaller than the second case and therefore the kiter must have a greater ability to stay in equilibrium. The balance of the moment becomes evident in the behavior of each kitesurfer who uses kitefoils, which centers the back foot on the mast and uses the front foot to apply a force that balances the moment. In simplified terms, the board represents a lever on which the rider applies a force, while he balances the strength of the stabilizer with his front foot, counteracting the moment generated by load-bearing and resistant forces.
The stabilizer moment and the rider’s need to counterbalance it, leads to a more stable equilibrium, and the rider’s
ability lies in maintaining the balance in situations of variable winds and during maneuvers such as tacks or jibes.
Contrary to its name
n egative flow stabilizer improves stability because the proportion of the fin / lift ratio is reduced and therefore its efficiency decreases. The task of the kitefoil designer is to create a geometry that allows at the same time both: A sufficient bearing capacity in a wide range of windy conditions; Produce a stabilizing moment sufficient to allow the achievement of equilibrium.
So, ultimately it is required to
maximize Lift / Resistance ratio without unduly compromising stability. The design of a kitefoil is subject to a number of constraints that must be considered in the optimization phase. If one wants to design a kitefoil for racing, he should consider the rules imposed by the IKA (International Kitefoil Association) which specifies that the maximum length of a kitefoil, (measured perpendicularly to the board) cannot exceed 5000 mm (in the current state of the art foils are about 1.2m long, that is far from 5 m). Furthermore, the appendices can be up to one, and their purpose must be mainly to create lift. No limitations are imposed regarding the materials. Other limitations that must be considered in the design of a kitefoil are imposed on the structural design, since the kitefoil must have an optimized geometry that has to be easy to build and at the same time must be able to withstand the stresses to which it is subjected. Theoretical bases
To understand the functioning of the hydrofoil, we have to analyze the physics of a easy wing profile. The wing of the hydrofoil creates a lift force, perpendicular to the flow direction, and a drag force, oriented with flow direction. The angle of attack α is the angle between the flow direction and the contour string.
The lift produced by a profile is directly proportional to the area of the wing surface “A_L” and proportional to the square of the relative velocity of the flow “v”; it also depends on the density of the fluid “ρ” and the on the lift coefficient “C_L”:
F_L=lift=\frac{1}{2}\rho A_mC_Lv^2
Resistence is a function of:
wing surface “A_R”; relative speed of the water “v”; drag Coefficient “C_D”; water density “ρ”:
F_D=drag=\frac{1}{2}\rho A_mC_Dv^2
The lift and drag dimensionless coefficients C_L and C_D respectively, could be analytically, numerically or experimentally calculated, and are function of the profile shape.
C_L=\frac{L}{\frac{1}{2}\rhoA_mv^2}
C_D=\frac{D}{\frac{1}{2}\rhoA_mv^2} \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} |
I don't know much of set theory, but is this mixed set allowed in this context?
There's an axiom of set theory that, for any two sets $A,B$ then $A \cup B$ is also a set. So it's definitely defined.
When does NFA ε-transition happen?
Try not to think of NFAs as actual machines, but instead focus on their mathematical definitions.
For every state $q$ of our NFA, there is some set $\epsilon CLOSE(q)$ of sets reachable from $q$ using 0 or more $\epsilon$-labelled transitions.
In an NFA without $\epsilon$-transitions, from a state $q$ and an input letter $a$, then the transition function $\delta(q,a)$ gives a set of states reachable from $q$ on input $a$, specifically all those states with an $a$-transition from $q$.
In an $\epsilon$-NFA, the states you can reach from $q$ on input $a$ is defined as to be $\{q' \mid (q,q'') \in \delta, q' \in \epsilon CLOSE(q) \}$.That is, it's the set of states you can get by following $a$.
The exact definitions are a bit more complicated, because you have to account for also taking $\epsilon$-transitions from whatever state you're starting at.
A word is accepted by an NFA if there's some sequence of transitions you can take to an accepting state where, when you concatenate the labels of all the transitions you take (treating the character inputs as 1-letter words), you get that word.
The point is, the meaning is just this definition. There's no "when" it happens, because it's sets and functions and relations all the way down.
I would have avoided this confusion, if ε would mean "NO_INPUT" symbol, rather an empty string as per definition.
We usually don't say this, because $\epsilon$ doesn't occur in the input words. (There's an infinite number of $\epsilon$s implicitly between every character of the words, but then our representation isn't unique we don't want that.) What you could do is have some special symbol which, when attached to a transition, denotes that it's a transition that can be taken without consuming input. But that's purely a definition of style: the final result is not changed. |
I'm currently trying to minimize the following function: $$\min_{\mathbf x,\mathbf y,s} \|F\mathbf x-\mathbf a\|_2^2 + \|SF\mathbf y-\mathbf b\|_2^2 + \lambda \|\max(\mathbf x, \mathbf y)\|_1,$$ $$\mathbf x>\mathbf 0, \mathbf y>\mathbf 0.$$ Here $F$ and $S$ are both linear operators. $F$ is a discrete Fourier transform, $S$ is "shift by $s$ in Fourier domain" operator: $(Sf)_k = f_k \cdot e^{-2\pi i u_k s}$, and $\max(\mathbf x, \mathbf y)$ is elementwise.
So the problem is basically regularized least squares, and should hopefully lead to sparse solutions $\mathbf x, \mathbf y$ having nonzero elements in about the same locations (up to explicit shift $s$).
Note that it is almost linear but for the discontinuity in regularizer and nonlinear dependency on $s$. For fixed $s$ I get good results ($\mathbf x,\mathbf y$ close to true values) using LBFGS method. When optimizing separately for each $s$ from a range, I see a clear minimum around the true value. But when I include the $s$ variable into optimization, its value basically doesn't change from the initial one, no matter if it is close or far from the truth.
I guess the problem is that the function (of $s$) has many local minima all over the range. How to approach the problem correctly, or where to read about similar ones? I had no success with other common optimization algorithms from scipy I tried. |
Given the advection equation for an incompressible flow field $$\frac{\partial c}{\partial t} + \mathrm{Pe} \frac{\partial c}{\partial x} = 0$$
what would the best method be for discretizing this without introducing any numerical diffusion or oscillations? Specifically when we have step changes in boundary conditions, and time (and space) dependent velocity $v(x,t)$.
The book
Numerical Methods for Problems with Moving Fronts by Bruce A. Finlayson goes into great detail on this problem when you have step changes as boundary conditions.
He recommends filtered leapfrog as the best
finite difference method:$$c^{n+1}_i = \frac{\alpha}{2}\left(c_i^n + c_i^{n+2}\right) + (1-\alpha)c_i^{n-1} - \frac{\mathrm{Pe}\Delta t}{\Delta x}\left(c_{i+1}^n - c_{i-1}^n\right)$$
(
filtered means $\alpha = 1$) and Taylor-Galerkin as the best finite element method:
$$ \frac{1}{6}\left(c_{i+1}^{n+1} - c_{i+1}^{n}\right) + \frac{2}{3}\left(c_{i}^{n+1} - c_{i}^{n}\right) + \frac{1}{6}\left(c_{i-1}^{n+1} - c_{i-1}^{n}\right) = -\frac{\mathrm{Pe}\Delta t}{\Delta x} \left(c^{n}_{i+1} - c^{n}_{i-1}\right) + \frac{\mathrm{Pe}^2\Delta t^2}{\Delta x^2}\left(c^{n}_{i+1} - 2c_i^n + c^{n}_{i-1}\right)$$
but I was wondering what the consensus is today?
I have tested the mentioned methods, and they both suffer from numerical and oscillation. But perhaps this can be managed by adjusting $\Delta t$ and $\Delta x$ in relation to the velocity?
This answer to a related question regarding Crank-Nicholson and the advection equation states that
Crank-Nicolson is not necessarily the best method for the advection equation. It is second order accurate and unconditionally stable, which is fantastic. However it will generate (as with all centered difference stencils) spurious oscillation if you have very sharp peaked solutions or initial conditions.
but gives no alternative methods. |
How to Model Optical Anisotropic Media with COMSOL Multiphysics®
On a bright evening in 1669, Professor Erasmus Bartholinus looked through a piece of an Icelandic calcite crystal he had placed onto a bench. He observed when he covered text on the bench with the stone, it appeared as a double image. The observed optical phenomenon, called birefringence, involves a beam of light that splits into two parallel beams while emerging out of a crystal. Here, we demonstrate a modeling approach for this effect.
Understanding Anisotropic Materials
The beam of light that Erasmus Bartholinus observed traveling straight through the crystal is called an ordinary ray. The other light beam, which bends while traveling through the crystal, is an extraordinary ray. Anisotropic materials, such as the crystal from the stone and bench experiment described above, are found in applications ranging from detecting harmful gases to beam splitting for photonic integrated circuits.
Ordinary and extraordinary rays traveling through an anisotropic crystal.
In a physical context, when an unpolarized electromagnetic beam of light propagates through an anisotropic dielectric material, it polarizes the dielectric domain, leading to a distribution of charges known as
electric dipoles. This phenomenon leads to induced fields within the anisotropic dielectric material, wherein two kinds of waves experience two different refractive indices (ordinary and extraordinary).
The ordinary wave is polarized perpendicular to the principal plane and the extraordinary wave is polarized parallel to the principal plane, where the principal plane is spanned by the optic axis and the two propagation directions in the crystal. Because of this behavior, the waves propagate with different velocities and trajectories.
Introducing Anisotropy Within Silicon Waveguides
In a previous blog post, we discussed silicon and how its derivative, silicon dioxide, is used extensively in photonic integrated chips due to its compatibility with the CMOS fabrication technique. Bulk silicon, which has an isotropic property, is used to develop prototypes for photonic integrated chips. However, due to unique optical properties such as splitting beams and polarization-based optical effects, anisotropy comes into play at a later stage.
Anisotropy in silicon photonics occurs unintentionally due to the annealing process while fabricating the waveguide. The difference in thermal expansion between the core and cladding causes geometry mismatch due to stress optical effects, which results in effects such as mode splitting and pulse broadening. Anisotropy could also be intentionally introduced by varying the porosity of silicon dioxide. This enables researchers to work with a range of effective refractive indices from silicon dioxide (n ~1.44) to air (n ~1), giving them the edge to perform very sensitive optical sensor applications.
Optical Modes of Propagation
To perform qualitative analyses of anisotropic media, researchers investigate how optical energy propagates within planar waveguides (also known as modes of propagation). In planar waveguides, we define modes using E^{x}_{p,q} and E^{y}_{p,q} terminology (Ref. 2), where
x and y depict the direction of polarization and p and q depict the number of maxima in the x– and y-coordinates.
Picture it this way: You are walking on an E^{x}_{2,1} “landscape” (as shown below). The “winds” (polarization) are along the ±x direction, and you encounter two distinct peaks when traveling from the –
x to + x direction. When you move from the – y to + y direction, you observe both of the peaks simultaneously. Mode analysis of the planar waveguide. Top row, left to right: E^{x}_{1,1} and E^{y}_{1,1}. Middle row, left to right: E^{x}_{1,2} and E^{y}_{1,2}. Bottom row, left to right: E^{x}_{2,1} and E^{y}_{2,1}. The arrow plot represents the electric field; contour and surface plot represent out-of-plane power flow (red is high and blue is low magnitude). Analyzing Anisotropic Structures in the COMSOL Multiphysics® Software
Before launching a beam of light through a waveguide using a laser source, it is important to know which optical modes could persist within a specified core/cladding dimension of the waveguide. Performing a mode analysis using a full vectorial finite element tool, such as the COMSOL Multiphysics® software, could be very helpful to qualitatively and quantitatively analyze the optical modes and dispersion curve respectively.
Introducing Diagonal Anisotropy
Performing a modal analysis on any isotropic material requires the definition of a single complex value, while in the case of an anisotropic material, a full tensor relative permittivity approach is required. The electric permittivity essentially relates the electric field with the material property. Here,
tensor refers to a 3-by-3 matrix that has both diagonal (\epsilon xx, \epsilon yy, \epsilon zz) and off-diagonal (\epsilon xy, \epsilon xz, \epsilon yx, \epsilon yz, \epsilon zx, \epsilon zy) terms as shown below.
\epsilon_{xx}&\epsilon _{xy}&\epsilon _{xz}\\
\epsilon _{yx}&\epsilon _{yy}&\epsilon _{yz}\\
\epsilon _{zx}&\epsilon _{zy}&\epsilon _{zz}\\
\end{bmatrix}
However, for all materials, you can find a coordinate system in which you only have nonzero diagonal elements in the permittivity tensor, whereas the off-diagonal elements are all zero. The three coordinate axes in this rotated coordinate system are the principal axes of the material and, correspondingly, the three values for the diagonal elements in the permittivity tensor are called the principal permittivities of the material.
There are basically two kinds of anisotropic crystal: uniaxial and biaxial crystal. With a suitable choice of coordinate system, where only the diagonal elements of the permittivity tensor are nonzero, in terms of optical properties,
uniaxial crystal considers only the diagonal terms, that is \epsilon = \epsilon xx = ( yy n) o 2, \epsilon = ( zz n) e 2, where nand o nare the ordinary and extraordinary refractive indices. However, when \epsilon_{xx}\neq \epsilon_{yy} \neq \epsilon_{zz}, it is known as a e biaxial crystal.
To put this argument into a modeling perspective, we can extend the buried rib waveguide example from this blog post on silicon photonics design. We can perform a modal analysis on the 2D cross section of the waveguide with the square core and cladding length of 4 um and 20 um, respectively (shown below). The operating wavelength for all the cases is considered as 1.55[um].
Schematic of 3D buried rib optical waveguide where the mode analysis was performed at the inlet 2D cross section. The intensity plot and arrow plot representing the mode and polarization of E-fields respectively. Core of the rib waveguide depicting the optic axis (red) along the x -axis and the principal axis (blue).
In the classic case of a uniaxial material, we assume the optic axis (i.e.,
c-axis) is along the principal x-axis (as shown above) and consider the diagonal relative permittivity \epsilon and \epsilon yy terms (which are orthogonal to the zz c-axis) as the square of ordinary refractive index (~1.5199 2~ 2.31). The \epsiloncomponent element that is along the xx c-axis is considered as the square of extraordinary refractive index (~1.4799 2~ 2.19) (as per Ref. 3). In addition, the off-diagonal terms are considered zero (as shown below) and the cladding has an isotropic relative permittivity (~1.4318 2). The optical modes derived are the 6 modes shown above. Note the difference in the refractive indices: “ n– xx n” is known as birefringence, where yy n= \sqrt{\epsilon_{xx}} and xx n= \sqrt{\epsilon_{yy}}. yy
\begin{bmatrix}
2.19 & 0 & 0 \\
0 & 2.31 & 0\\
0 & 0 & 2.31\\
\end{bmatrix}
Relative permittivity tensor with diagonal elements. Dispersion Curves
By evaluating the optical modes, we can visually comprehend the behavior of the optical waveguide. However, the dispersion curves could also be handy for performing quantitative analyses. A dispersion curve represents the variation of the effective refractive index with respect to the length of the waveguide or the operating frequency.
Diagonal Anisotropy
A modal analysis is performed while parametrically sweeping the length of the waveguide from 0.5 um to 4 um to derive the dispersion curve for the anisotropic core, as shown in the figure below. We assume the earlier case stated, with diagonal anisotropy terms of the core (i.e., \epsilon
= 2.19, \epsilon xx = \epsilon yy = 2.31 and all of the diagonal elements are zero). The results are compared with Koshiba et al. (Ref. 3). zz Dispersion curve with transverse anisotropic core. Off-Diagonal Transverse Anisotropy (XY Plane)
When the optic axis (i.e.,
c-axis) lies in XY plane and makes an angle of \theta with the x-axis, the diagonal components \epsilon , \epsilon xx , \epsilon yy and off-diagonal components \epsilon zz and \epsilon xy are nonzero, while the rest of the components are zero. The full relative permittivity tensor could be evaluated by using the rotation matrix [R] as shown below, where the rotation matrix [R] is specifically for rotating the yz c-axis in the XY plane. \epsilon is the square of the extraordinary refractive index (~2.19), because the xx c-axis lies along the principal x-axis, while \epsilon and \epsilon yy are the square of the ordinary refractive index (~2.31). The off-diagonal elements \epsilon zz and \epsilon xy are derived from the multiplication of the matrices as stated below. yz The c -axis lying in the XY plane and making an angle of \theta with the x -axis.
\begin{bmatrix}
cos(\theta) & -sin(\theta) & 0 \\
sin(\theta) & cos(\theta) & 0\\
0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
\epsilon_{xx} & 0 & 0 \\
0 & \epsilon_{yy} & 0 \\
0 & 0 & \epsilon_{zz} \\
\end{bmatrix}
\begin{bmatrix}
cos(\theta) & sin(\theta) & 0 \\
-sin(\theta) & cos(\theta) & 0\\
0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
(\epsilon_{xx}) cos^2(\theta) + (\epsilon_{yy}) sin^2(\theta) & (\epsilon_{xx}) sin(\theta) cos(\theta)-(\epsilon_{yy}) sin(\theta) cos(\theta) & 0 \\
(\epsilon_{xx}) sin(\theta)cos(\theta)-(\epsilon_{yy})sin(\theta)cos(\theta) & (\epsilon_{yy}) cos^2(\theta) + (\epsilon_{xx}) sin^2(\theta) & 0\\
0 & 0 & \epsilon_{zz}\\
\end{bmatrix}
The relative permittivity tensor ε is treated along with a rotation matrix, rotating the c -axis in the XY plane with angle \theta.
Finally, the modal analysis of the waveguide with off-diagonal anisotropic core and isotropic cladding, where the optic axis makes angles of 0, 15, 30, and 45 degrees with respect to the principal
x-axis, as shown below. Here, it could be observed that the direction of the in-plane magnetic field changes according to the change in the angle of the optic axis. The dispersion curve could also be plotted by parametrically sweeping the length of the core and cladding from 0.5 um to 4 um, while considering the angle \theta as 45°. The dispersion curve tends to be similar to the dispersion curve of the diagonal anisotropy, as discussed above. Mode analysis, including off-diagonal terms, for θ = 0° (top-left), θ = 15° (top-right), θ = 30° (bottom-left), and θ = 45° (bottom-right). The figure represents the magnetic field lines within the core for different rotation angles. Off-Diagonal Longitudinal Anisotropy (YZ Plane)
Finally, when considering the longitudinal anisotropy where the optic axis (i.e.,
c-axis) lies in the YZ plane and makes an angle of \phi with the y-axis, the diagonal components \epsilon , \epsilon xx , \epsilon yy and the off-diagonal components \epsilon zz and \epsilon yz are nonzero, while the rest of the components are zero. The relative permittivity tensor could be evaluated by using the rotation matrix [R] as shown below, where the rotation matrix [R] is specifically for rotating the zy c-axis in the YZ plane. \epsilon is the square of the extraordinary refractive index (~2.19), because the yy c-axis lies along the principal y-axis, while \epsilon , \epsilon xx is the square of the ordinary refractive index (~2.31). The off-diagonal elements \epsilon zz and \epsilon yz are derived from the multiplication of the matrices as stated below. zy The c -axis lying in the YZ plane and making an angle of \phi with the x -axis.
\begin{bmatrix}
1 & 0 & 0 \\
0 & cos(\phi) & -sin(\phi)\\
0 & sin(\phi) & cos(\phi) \\
\end{bmatrix}
\begin{bmatrix}
\epsilon_{xx} & 0 & 0 \\
0 & \epsilon_{yy} & 0 \\
0 & 0 & \epsilon_{zz} \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & cos(\phi) & sin(\phi)\\
0 & -sin(\phi) & cos(\phi) \\
\end{bmatrix}
\begin{bmatrix}
\epsilon_{xx} & 0 & 0 \\
0 & (\epsilon_{yy}) cos^2(\phi) + (\epsilon_{zz}) sin^2(\phi) & (\epsilon_{yy})sin(\phi)cos(\phi)-(\epsilon_{zz}) sin(\phi)cos(\phi)\\
0 & (\epsilon_{yy})sin(\phi)cos(\phi)-(\epsilon_{zz}) sin(\phi)cos(\phi) & (\epsilon_{zz}) cos^2(\phi) + (\epsilon_{yy}) sin^2(\phi)\\
\end{bmatrix}
The relative permittivity tensor ε is treated along with a rotation matrix, rotating in the YZ plane with angle \phi.
A modal analysis is then performed where the length of the waveguide is parametrically swept from 0.5 um to 4 um to derive the dispersion curve for the longitudinal anisotropic core, as shown in the figure below. In this case, \phi = 45° (i.e., the
c-axis lies in the YZ plane and makes 45° with the y-axis) (Ref. 3). Dispersion curve with longitudinal anisotropic core. Final Thoughts on Modeling Anisotropic Materials
In this blog post, we performed qualitative analyses (modes of propagation) and quantitative analyses (dispersion curves) of the anisotropic optical waveguide using modal analysis in COMSOL Multiphysics. Diagonal anisotropy as well as off-diagonal transverse and longitudinal anisotropy were considered to derive their dispersion relationships. These types of analyses give us more flexibility when carrying out optimization of material and geometric parameters to help us gain an in-depth and intuitive understanding of the physics of anisotropic materials.
A simple tutorial model to help you started would be the Step-Index Fiber, which involves mode analysis over a 2D cross section of the 3D optical fiber.
Next Steps
To try these models, click the button below. This will take you to the Application Gallery, where you can download the related MPH-files as long as you have a COMSOL Access account and valid software license.
Updated List of Blog Posts in the Silicon Photonics Series References E. Hecht, Optics, Pearson. E.A.J. Marcatili, “Dielectric rectangular waveguide and directional coupler for integrated optics”, Bell Syst. Tech. J., vol. 48, pp. 2071–2102, 1969. M. Koshiba, K. Hayata, and M. Suzuki, “Finite-element solution of anisotropic waveguides with arbitrary tensor permittivity,” Journal of Lightwave Technology, vol. 4, no. 2, pp. 121–126, 1986. Comments (4) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science |
Consider
$$\frac{a}{a}\pmod a,\ \ \ a\in\mathbb Z\setminus\{-1,0,1\}$$
There are two cases:
$1)$ $\frac{a}{a}$ is the notation for the real number $1$.
Then the expression is equivalent to $1$ modulo $a$.
E.g., see this, where the expression on the LHS would not even exist if what would be meant by $99$ in the denominator is modular inverse instead of regular integer division.
$2)$ $\frac{a}{a}$ is the notation for $ax$, where $x$ is in the class of solutions to $ax\equiv 1\pmod{a}$.
Since $x$ does not exist, $\frac{a}{a}$ does not exist too in this case.
So $\frac{a}{a}\mod a$ can be thought of as either equivalent to $1$ modulo $a$ or not existing at all.
Now, $aa^{-1}$ is possibly more commonly used than $\frac{a}{a}$ in modular arithmetic and $\frac{a}{a}$ is more commonly used than $aa^{-1}$ in division in integers, but surely not always.
Such notation was probably created due to its similarities to division in integers, but I think less ambiguous notation should've been created instead.
Also see this.
There's a comment there that points out my problem. Consider:
$$\frac{ac}{bc}\pmod m,$$
where $$\gcd(b,m)=1,\ \ \ \gcd(c,m)>1,\ \ \ m\in\mathbb Z\setminus\{-1,0,1\},\ \ \ c\in\mathbb Z\setminus\{0\},\ \ \ b\in\mathbb Z\{0\},\ \ \ a\in\mathbb Z$$
can either be equivalent to $\frac{a}{b}\pmod{m}$ if the fractional notation denotes integer division or it could not exist at all if the fractional notation denotes modular inverses. |
For "attenuation coefficient" as it applies to electromagnetic theory and telecommunications see propagation constant. For the "mass attenuation coefficient", see the article mass attenuation coefficient.
The
attenuation coefficient is a quantity that characterizes how easily a material or medium can be penetrated by a beam of light, sound, particles, or other energy or matter. A large attenuation coefficient means that the beam is quickly "attenuated" (weakened) as it passes through the medium, and a small attenuation coefficient means that the medium is relatively transparent to the beam. Attenuation coefficient is measured using units of reciprocal length.
The attenuation coefficient is also called
linear attenuation coefficient, narrow beam attenuation coefficient. Although all four terms are often used interchangeably, they can occasionally have a subtle distinction, as explained below. Overview
The attenuation coefficient describes the extent to which the intensity of an energy beam is reduced as it passes through a specific material. This might be a beam of electromagnetic radiation or sound.
It is used in the context of X-rays or Gamma rays, where it is represented using the symbol \mu, and measured in cm −1. It is used in the context of neutrons and nuclear reactors, where it called macroscopic cross section (although actually it is not a section dimensionally speaking) being represented using the symbol \Sigma, and measured in m −1. It is also used for modeling solar and infrared radiative transfer in the atmosphere, albeit usually denoted with another symbol (given the standard use of \mu = \cos(\theta) for slant paths). In the case of ultrasound attenuation it is usually denoted as \alpha and measured in dB/cm/MHz. [1] [2] The attenuation coefficient is widely used in acoustics for characterizing particle size distribution. [1] [2] A common unit in this contexts is inverse metres, and the most common symbol is the Greek letter \alpha. It is also used in acoustics for quantifying how well a wall in a building absorbs sound. Wallace Sabine was a pioneer of this concept. A unit named in his honor is the sabin: the absorption by a 1-square-metre (11 sq ft) slab of perfectly absorptive material (the same amount of sound loss as if there were a 1-square-metre window). [3] Note that the sabin is not a unit of attenuation coefficient; rather, it is the unit of a related quantity.
A small linear attenuation coefficient indicates that the material in question is relatively transparent, while a larger value indicates greater degrees of opacity. The linear attenuation coefficient is dependent upon the type of material and the energy of the radiation. Generally, for electromagnetic radiation, the higher the energy of the incident photons and the less dense the material in question, the lower the corresponding linear attenuation coefficient will be.
Definitions and formulae
The measured intensity I of transmitted through a layer of material with thickness z is related to the incident intensity I_0 according to the inverse exponential power law that is usually referred to as Beer–Lambert law:
I(z) = I_{0} \, e^{-\int_0^z \alpha(z') \, \operatorname dz'},
where z denotes the path length. The attenuation coefficient is \alpha (z'). If it is uniform, the situation is referred to as linear attenuation and the law simplifies:
I(z) = I_{0} \, e^{-\alpha \, z},
The Half Value Layer (HVL) signifies the thickness of a material required to reduce the intensity of the emergent radiation to half its incident magnitude. It is from these equations that engineers decide how much protection is needed for "safety" from potentially harmful radiation. The attenuation factor of a material is obtained by the ratio of the emergent and incident radiation intensities I/I_0.
The linear attenuation coefficient and mass attenuation coefficient are related such that the mass attenuation coefficient is simply \alpha/\rho, where \rho is the density in g/cm
3. When this coefficient is used in the Beer-Lambert law, then "mass thickness" (defined as the mass per unit area) replaces the product of length times density.
The linear attenuation coefficient is also inversely related to mean free path. Moreover, it is very closely related to the absorption cross section.
Attenuation versus absorption
The terms "attenuation coefficient" and "absorption coefficient" are generally used interchangeably. However, in certain situations they are distinguished, as follows.
[4]
When a narrow (collimated) beam of light passes through a substance, the beam will lose intensity due to two processes: The light can be absorbed by the substance, or the light can be scattered (i.e., the photons can change direction) by the substance. Just looking at the narrow beam itself, the two processes cannot be distinguished. However, if a detector is set up to measure light leaving in different directions, or conversely using a non-narrow beam, one can measure how much of the lost intensity was scattered, and how much was absorbed.
In this context, the "absorption coefficient" measures how quickly the beam would lose intensity due to the absorption
alone, while "attenuation coefficient" measures the total loss of narrow-beam intensity, including scattering as well. "Narrow-beam attenuation coefficient" always unambiguously refers to the latter. The attenuation coefficient is always larger than the absorption coefficient, although they are equal in the idealized case of no scattering. See also References ^ a b ISO 20998-1:2006 "Measurement and characterization of particles by acoustic methods" ^ a b Dukhin, A.S. and Goetz, P.J. "Ultrasound for characterizing colloids", Elsevier, 2002 ^ "Wallace Clement Sabine". Acoustics Engineering. Retrieved 2011-11-08. ^ Bohren,C. F. and Huffman, D.R. "Absorption and Scattering of Light by Small Particles", Wiley, (1983), isbn= 0-471-29340-7 External links
Absorption Coefficients α of Building Materials and Finishes Sound Absorption Coefficients for Some Common Materials Computations for Radio Wave Propagation Tables of X-Ray Mass Attenuation Coefficients and Mass Energy-Absorption Coefficients from 1 keV to 20 MeV for Elements Z = 1 to 92 and 48 Additional Substances of Dosimetric Interest
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Electroproduction of the omega meson was investigated in the p(e,e'p)omega reaction. The measurement was performed at a 4-momentum transfer Q2 ~ 0.5 GeV2. Angular distributions of the virtual photon-proton center-of-momentum cross sections have been extracted over the full angular range. These distributions exhibit a strong enhancement over t-channel parity exchange processes in the backward direction. According to a newly developed electroproduction model, this enhancement provides significant evidence of resonance formation in the gamma* p -> omega p reaction channel.
Differential cross section for an average W of 1.75 GeV.
Differential cross section for an average W of 1.79 GeV.
Momentum spectra of charged pions over nearly full rapidity coverage from target to beam rapidity have been measured in the 0-5% most central Au+Au collisions in the beam energy range from 2 to 8 AGeV by the E895 Experiment. Using a thermal parameterization to fit the transverse mass spectra, rapidity density distributions are extracted. The observed spectra are compared with predictions from the RQMD v2.3 cascade model and also to a thermal model including longitudinal flow. The total 4$\pi$ yields of the charged pions are used to infer an initial state entropy produced in the collisions.
The differential cross section for the gamma +n --> pi- + p and the gamma + p --> pi+ n processes were measured at Jefferson Lab. The photon energies ranged from 1.1 to 5.5 GeV, corresponding to center-of-mass energies from 1.7 to 3.4 GeV. The pion center-of-mass angles varied from 50 degree to 110 degree. The pi- and pi+ photoproduction data both exhibit a global scaling behavior at high energies and high transverse momenta, consistent with the constituent counting rule prediction and the existing pi+ data. The data suggest possible substructure of the scaling behavior, which might be oscillations around the scaling value. The data show an enhancement in the scaled cross section at center-of-mass energy near 2.2 GeV. The differential cross section ratios at high energies and high transverse momenta can be described by calculations based on one-hard-gluon-exchange diagrams.
Differential cross section for the process GAMMA N --> PI- P for an incident electron energy of 5.614 GeV.
Differential cross section for the process GAMMA N --> PI- P for an incident electron energy of 4.236 GeV.
Differential cross section for the process GAMMA N --> PI- P for an incident electron energy of 3.400 GeV.
A precise measurement of the anomalous g value, a_mu=(g-2)/2, for the positive muon has been made at the Brookhaven Alternating Gradient Synchrotron. The result a_mu^+=11 659 202(14)(6) X 10^{-10} (1.3 ppm) is in good agreement with previous measurements and has an error one third that of the combined previous data. The current theoretical value from the standard model is a_mu(SM)=11 659 159.6(6.7) X 10^{-10} (0.57 ppm) and a_mu(exp)-a_mu(SM)=43(16) X 10^{-10} in which a_mu(exp) is the world average experimental value.
The anomalous G value is related to the gyromagnetic ratio by MOM(N=A_MU) =(G-2)/2.
We present a measurement of the pion form factor based on e+e- annihilation data from the CMD-2 detector in the energy range 0.6<sqrt(s)<1.0 GeV with a systematic uncertainty of 0.8%. A data sample is five times larger than that used in our previous measurement.
Measured values of the pion form factor. The errors are statistical only.
Measured value of the bare PI+ PI- cross section including corrections for radiative effects but excluding corrections for vacuum polarization effects. The errors are statistical only.
Real part of amplitude extracted using a more precise UA4 measurement. (1 +RE(AMP)/IM(AMP)**2)SIG(TOT) = 63.5 +- 1.5 MB (Bozzo et al. PL 147B(1984)392).
Result extrapolated to all lambda/c energies has large model dependent uncertainties.
The cross section of the process e+e- -> pi+pi- has been measured using about 114000 events collected by the CMD-2 detector at the VEPP-2M e+e- collider in the center-of-mass energy range from 0.61 to 0.96 GeV. Results of the pion form factor determination with a 0.6% systematic uncertainty are presented. Implications for the hadronic contribution to the muon anomalous magnetic moment are discussed.
Updated measured values of the pion form factor and 'bare' cross section.
Measured cross section.
The cross section of the process e+ e- ---> eta gamma has been measured in the 600-1380 MeV c.m. energy range with the CMD-2 detector. The following branching ratios have been determined: B(rho ---> eta gamma) = (3.28 +- 0.37 +- 0.23) 10^{-4}, B(omega ---> eta gamma) = (5.10 +- 0.72 +- 0.34) 10^{-4}, B(phi --> eta gamma) = (1.287 +- 0.013 +- 0.063) 10^{-2}. Evidence for the rho'(1450) ---> eta gamma decay has been obtained for the first time.
The measured Born cross section for the ETA GAMMA final state.
CORRECTED FOR ACCEPTANCE. AUTHORS NAMED THIS SPECTRUM 'INVARIANT'.
Radiative decays of the $\phi$ meson have been studied using a data sample of about 19 million $\phi$ decays collected by the CMD-2 detector at VEPP-2M collider in Novosibirsk. From selected $e^+e^-\to\pi^{0}\pi^{0}\gamma$ and $e^+e^-\to\eta\pi^{0}\gamma$ events the following model independent results have been obtained: \par $Br(\phi\to\pi^{0}\pi^{0}\gamma) = (0.92\pm 0.08\pm0.06)\times10^{-4}$ for $M_{\pi^{0}\pi^{0}}>700$ MeV, \par $Br(\phi\to\eta\pi^{0}\gamma) = (0.90\pm 0.24\pm 0.10)\times10^{-4}$. It is shown that the intermediate mechanism $f_{0}(980)\gamma$ dominates in the $\phi\to\pi^{0}\pi^{0}\gamma$ decay and the corresponding branching ratio is \par $Br(\phi\to f_{0}(980)\gamma)=(2.90\pm 0.21\pm1.54)\times10^{-4}$. The systematic error is dominated by the possible model uncertainty. \par Using the same data sample the upper limit has been obtained for the P- and CP-violating decay of $\eta$ at 90% CL: \par $Br(\eta\to\pi^{0}\pi^{0}) < 4.3\times10^{-4}$ >.
Non-resonant cross section.
The cross section of the process e+ e- --> pi0 pi0 gamma has been measured in the c.m. energy range 600--970 MeV with the CMD-2 detector. The following branching ratios have been determined: B(rho --> pi0 pi0 gamma) =(5.2^{+1.5}_{-1.3} +- 0.6)x10^{-5} and B(omega --> pi0 pi0 gamma) =(6.4^{+2.4}_{-2.0} +- 0.8)x10^{-5}. Evidence for the rho --> f0(600) gamma decay has been obtained: B(rho --> f0(600) gamma) = (6.0^{+3.3}_{-2.7}\pm 0.9)x10^{-5}. From a search for the process e+ e- --> eta pi0 gamma the following upper limit has been obtained: B(omega --> eta pi0 gamma) < 3.3 10^{-5} at 90% CL.
The Born and 'bare' cross sections.
The e+e- -> pi+pi-pi+pi- cross section has been measured using 5.8/pb of integrated luminosity collected with the CMD-2 detector at the VEPP-2M collider. Analysis of the data with a refined efficiency determination and use of both three- and four-track events allowed doubling of a data sample and reduction of systematic errors to 5-7%.
Cross sections with statistical error only.
The processes e+ e- --> eta gamma, e+ e- --> pi0 gamma --> 3 gamma have been studied in the c.m. energy range 600--1380 MeV with the CMD-2 detector. The following branching ratios have been determined: Br(rho --> eta gamma) = (3.21 +- 1.39 +- 0.20)x 10^{-4}; Br(omega --> eta gamma) = (4.44 + 2.29 -1.83 +- 0.28)x 10^{-4}; Br(phi --> eta gamma) = (1.373 +- 0.014 +- 0.085)x 10^{-2}; Br(rho --> pi0 gamma) = (6.21 +1.28 - 1.18 +- 0.39)x 10^{-4}; Br(omega --> pi0 gamma) = (9.06 +- 0.20 +- 0.57)x 10^{-2}; Br(phi --> pi0 gamma) = (1.258 +- 0.037 +- 0.077)x 10^{-3};
Born cross section for the process E+ E- --> ETA GAMMA.
Born cross section for the process E+ E- --> PI0 GAMMA.
In experiments with the CMD-2 detector at the VEPP-2M electron-positron collider at Novosibirsk about 150000 $e^+e^-\to\pi^+\pi^-$ events were recorded in the center-of-mass energy range from 0.61 up to 0.96 GeV. The result of the pion form factor measurement with a 1.4% systematic error is presented. The following values of the $\rho$-meson and $\rho-\omega$ interference parameters were found: $M_\rho=(775.28\pm 0.61\pm 0.20) MeV, \Gamma_\rho=(147.70\pm 1.29 \pm 0.40) MeV, \Gamma(\rho\to e^+e^-)=(6.93\pm 0.11\pm 0.10) keV, Br(\omega\to\pi^+\pi^-) = (1.32\pm 0.23)%$. |
Your loss function would not work because it incentivizes setting $\theta_1$ to any finite value and $\theta_0$ to $-\infty$.
Let's call $r(x,y)=\frac{1}{m}\sum_{i=1}^m {h_\theta\left(x^{(i)}\right)} -y$ the
residual for $h$.
Your goal is to make $r$
as close to zero as possible, not just minimize it. A high negative value is just as bad as a high positive value.
EDIT: You can counter this by artificially limiting the parameter space $\mathbf{\Theta} $(e.g. you want $|\theta_0| < 10$). In this case, the optimal parameters would lie on certain points on the boundary of the parameter space. See https://math.stackexchange.com/q/896388/12467. This is not what you want. Why do we use the square loss
The squared error forces $h(x)$ and $y$ to match. It's minimized at $u=v$, if possible, and is always $\ge 0$, because it's a square of the real number $u-v$.
$|u-v|$ would also work for the above purpose, as would $(u-v)^{2n}$, with $n$ some positive integer. The first of these is actually used (it's called the $\ell_1$ loss; you might also come across the $\ell_2$ loss, which is another name for squared error).
So, why is the squared loss better than these? This is a
deep question related to the link between Frequentist and Bayesian inference. In short, the squared error relates to Gaussian Noise.
If your data does not fit all points exactly, i.e. $h(x)-y$ is not zero for some point no matter what $\theta$ you choose (as will always happen in practice), that might be because of
noise. In any complex system there will be many small independent causes for the difference between your model $h$ and reality $y$: measurement error, environmental factors etc. By the Central Limit Theorem(CLT), the total noise would be distributed Normally, i.e. according to the Gaussian distribution. We want to pick the best fit $\theta$ taking this noise distribution into account. Assume $R = h(X)-Y$, the part of $\mathbf{y}$ that your model cannot explain, follows the Gaussian distribution $\mathcal{N}(\mu,\sigma)$. We're using capitals because we're talking about random variables now.
The Gaussian distribution has two parameters, mean $\mu = \mathbb{E}[R] = \frac{1}{m} \sum_i h_\theta(X^{(i)})-Y^{(i))}$ and variance $\sigma^2 = E[R^2] = \frac{1}{m} \sum_i \left(h_\theta(X^{(i)})-Y^{(i))}\right)^2$. See here to understand these terms better.
Consider $\mu$, it is the
systematic error of our measurements. Use $h'(x) = h(x) - \mu$ to correct for systematic error, so that $\mu' = \mathbb{E}[R']=0$ (exercise for the reader). Nothing else to do here.
$\sigma$ represents the
random error, also called noise. Once we've taken care of the systematic noise component as in the previous point, the best predictor is obtained when $\sigma^2 = \frac{1}{m} \sum_i \left(h_\theta(X^{(i)})-Y^{(i))}\right)^2$ is minimized. Put another way, the best predictor is the one with the tightest distribution (smallest variance) around the predicted value, i.e. smallest variance. Minimizing the the least squared loss is the same thing as minimizing the variance! That explains why the least squared loss works for a wide range of problems. The underlying noise is very often Gaussian, because of the CLT, and minimizing the squared error turns out to be the right thing to do!
To simultaneously take both the mean and variance into account, we include a
bias term in our classifier (to handle systematic error $\mu$), then minimize the square loss.
Followup questions:
Least squares loss = Gaussian error. Does every other loss function also correspond to some noise distribution? Yes. For example, the $\ell_1$ loss (minimizing absolute value instead of squared error) corresponds to the Laplace distribution (Look at the formula for the PDF in the infobox -- it's just the Gaussian with $|x-\mu|$ instead of $(x-\mu)^2$). A popular loss for probability distributions is the KL-divergence.-The Gaussian distribution is very well motivated because of the Central Limit Theorem, which we discussed earlier. When is the Laplace distribution the right noise model? There are some circumstances where it comes about naturally, but it's more commonly as a regularizer to enforce sparsity: the $\ell_1$ loss is the least convex among all convex losses. As Jan mentions in the comments, the minimizer of squared deviations is the mean and the minimizer of the sum of absolute deviations is the median. Why would we want to find the median of the residuals instead of the mean? Unlike the mean, the median isn't thrown off by one very large outlier. So, the $\ell_1$ loss is used for increased robustness. Sometimes a combination of the two is used. Are there situations where we minimize both the Mean and Variance? Yes. Look up Bias-Variance Trade-off. Here, we are looking at a set of classifiers $h_\theta \in H$ and asking which among them is best. If we ask which set of classifiers is the best for a problem, minimizing both the bias and variance becomes important. It turns out that there is always a trade-off between them and we use regularization to achieve a compromise. Regarding the $\frac{1}{2}$ term
The 1/2 does not matter and actually, neither does the $m$ - they're both constants. The optimal value of $\theta$ would remain the same in both cases.
The expression for the gradient becomes prettier with the $\frac{1}{2}$, because the 2 from the square term cancels out.
When writing code or algorithms, we're usually concerned more with the gradient, so it helps to keep it concise. You can check progress just by checking the norm of the gradient. The loss function itself is sometimes omitted from code because it is used only for validation of the final answer.
The $m$ is useful if you solve this problem with gradient descent. Then your gradient becomes the average of $m$ terms instead of a sum, so its' scale does not change when you add more data points.
I've run into this problem before: I test code with a small number of points and it works fine, but when you test it with the entire dataset there is loss of precision and sometimes over/under-flows, i.e. your gradient becomes
nan or
inf. To avoid that, just normalize w.r.t. number of data points.
These aesthetic decisions are used here to maintain consistency with future equations where you'll add
regularization terms. If you include the $m$, the regularization parameter $\lambda$ will not depend on the dataset size $m$ and it will be more interpretable across problems. |
Using this theorem:
Let $(a_k)$ be a sequence in $\Bbb R$ and let $x_0\in \Bbb R$:
The power series: $$\sum_{k=0}^\infty a_k(x-x_0)^k \qquad \text{and} \qquad \sum_{k=0}^\infty ka_k(x-x_0)^{k-1} $$ converge uniformly in $[x_0-r,x_0+r]$ for all $r\in (0,R)$ where $R$ is the radius of convergence of $\sum_{k=0}^\infty a_k(x-x_0)^k $.
The function $$f(x):= \sum_{k=0}^\infty a_k(x-x_0)^k $$ is continously diferentiable in $(x_0-R,x_0+R)$, and its derivative is given by: $$f'(x)=\sum_{k=0}^\infty ka_k(x-x_0)^{k-1} $$
calculate the following: $$\tag {1} \sum_{k=0}^\infty \frac {k}{2^{k-1}}$$
So, I'm confused on what do I have to do. I'm
gessing that we have to find the $x_0$ of this: $$\sum_{k=0}^\infty \frac {k}{2^{k-1}} (x-x_0)^k$$I know that $\mathbf {(1)}$ converges to $4$, but I didn't used the theorem, and using the definition of radius of convergence, (superficially) I found that $R:=\sup\{r\in(-2,2) \mid \sum_{k=0}^\infty \frac {k}{2^{k-1}}r^k<\infty \text{ (the sum converges in $\Bbb R$)} \}$.I don't see what they're asking me to do, and why/how to use the theorem. |
Suppose after some calculations I arrive at an equation of the form: \begin{equation} [\nabla^2 +V(\mathbf{r})]\psi(\mathbf{r})=k^2\psi(\mathbf{r}) \tag{1} \end{equation} As you can see here the Laplacian has the same sign as the eigenvalue. The eigenvalue $k^2$ will then be used to determine an energy defined as $E=-\dfrac{\hbar^2 k^2}{2m}$, which needs to be negative (in other words $k$ needs to be real).
(For anyone wondering, we're in the context of the Born Oppenheimer approximation, but I didn't want that to divert the attention for the main issue).
I cannot possibly regard equation (1) as a Schrodinger equation, even by changing the sign of the potential $V(r)$, because of the sign in the Laplacian.
My question is if there's a way to map this equation to a Schrodinger equation, in order to use known methods to solve it, and still obtain real values for $k$. (whereas an obvious choice would be to substitute $k$ with $ik$). |
I am studying how to get the normalization factor algebraically in the exited states of the harmonic oscillator using the beautiful ladder operators technique.
I am stuck at the point where is stated that the product of the ladder operators yields the energy levels (in $\widehat{a}^\dagger\widehat{a}$ case) and the energy levels + 1 (in $\widehat{a}\widehat{a}^\dagger$ case).
$$\widehat{a}\widehat{a}^\dagger\psi_n= (n+1)\psi_n$$
$$\widehat{a}^\dagger\widehat{a}\psi_n=n\psi_n$$
The thing is that I know how to get them mathematically using:
$$\widehat{a}^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle$$
$$\widehat{a}|n\rangle=\sqrt{n}|n-1\rangle$$
But I am not satisfied with this due to the fact that my textbook (Griffiths) does not obtain them like that but using both the Schrodinger Equation and the solution to the exited states of the harmonic oscillator:
$$\hbar\omega\left(a\pm a\mp \pm \frac{1}{2}\right)\psi = E \psi$$
$$\psi_n = A_n (\widehat{a}^\dagger )^n \psi_0$$
But I am not getting them using the stated way. May you please prove it?
I am also wondering why the product of the two ladder operators do yield energy levels without a constant of proportionality. In fact; when just one operator is multiplied by the eigenstates you get a constant of proportionality multiplied by the eigenstates (+1 or -1; depending on the operator we used). But this may be another question so please let me know whether I should post it separately: |
Difference between revisions of "Image Dimensions"
(Created oage, added disclaimer)
(Added a bit of description of the properties dialog)
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<div style="background-color:#DDFFDD; border:thin solid green; padding:1em">
<div style="background-color:#DDFFDD; border:thin solid green; padding:1em">
'''Disclaimer:''' This page's content is not official and not guaranteed to be free of mistakes. At the moment, it's even only a sum of personal thoughts to cast a bit of light onto synfig's image dimensions handling.</div>
'''Disclaimer:''' This page's content is not official and not guaranteed to be free of mistakes. At the moment, it's even only a sum of personal thoughts to cast a bit of light onto synfig's image dimensions handling.</div>
+ + + + + + + + + + Revision as of 14:48, 18 September 2008 Describing the fields of the Canvas Properties Dialog
The user access the image dimensions in the Canvas Properties Dialog.
The 'Others' tab
Here some properties can simply be locked (such that they can't be changed) and linked (so that changes in one entry simultaneously change other entries as well).
The 'Image' tab
Obviously here the image dimensions can be set. There seem to be basically three groups of fields to edit:
The on-screen size(?) The fields Widthand Heighttell synfigstudio how many pixels the image shall cover at a zoom level of 100%. The physical size The physical width and height should tell how big the image is on some physical media. That could be when printing out images on paper, or maybe even on transparencies or film. Not all file formats can save this on exporting/rendering images. The mysterious Image Area Given as two points (upper-left and lower-right corner) which also define the image span (Pythagoras: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): {\displaystyle \text{span}=\sqrt{\Delta x^2 + \Delta y^2}}). The unit seems to be not pixels but units, which are at 60 pixels each. |
Why do we choose these numbers and not for e.g $e$ and $\pi$ for success and failure in an experiment?
What is the logic and how badly would it affect my calculations if I choose other values instead of 0 and 1?
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Why do we choose these numbers and not for e.g $e$ and $\pi$ for success and failure in an experiment?
What is the logic and how badly would it affect my calculations if I choose other values instead of 0 and 1?
As already noted by
Mark L. Stone, it is used because of tradition and convenience. It could have been $-1$ and $+1$ as with Rademacher distribution, or any other values.
However there are also other reasons that make $0$ and $1$ a convenient choice. First, expected value of Bernoulli distributed random variable is
$$ E[X] = 0\times(1-p)+ 1\times p = p $$
...so it is instantly obvious from the distributions definition. Sample mean is being maximum likelihood estimator of $p$ since taking the arithmetic mean of zeros and ones leads to proportion of ones in the whole sample.
Moreover, we can easily extend Bernoulli distribution to binomial distribution, i.e. from modelling single success in single draw, change to modelling $k$ successes in $n$ draws -- it is quite helpful that you can just sum $1$'s to get number of successes.
There are also multiple other reasons why such coding is convenient for computations, to learn more see Why is gender typically coded 0/1 rather than 1/2, for example?
Mathematical convenience. Here is an example: if you set the success case to be $1$ and the failure case to be $0$, then you have that the binomial variable $X \sim B(n, p)$, which counts the number of successes in $n$ independent Bernoulli trials $X_i \sim B(1, p)$, with $i = 1, 2, ..., n$, can be written as
$$ X = \sum_i X_i $$
Therefore,
$$ E[X] = \sum_i E[X_i] = np$$
In addition,
$$ V[X] = \sum_i V[X_i] = np(1 - p)$$ |
This is a post regarding the paper Efficient Preconditioning for the $p$-Version Finite Element Method in Two Dimensions. In Lemma 3.3, the basis required is the bubble (or interior) polynomials, the set of edge functions orthogonal to the interiors, and linear (of bilinear) functions.
The edge functions orthogonality is explicitly stated as $\hat a(u, v) = 0$ for all $u \in \Gamma_i$ (edge space) and $v \in \mathcal{I}$ interior space. A very natural question is why the vertex functions does not need to be orthogonal to the interior functions. The fun fact is that it secretly is.
Note that the paper is for the $H^1$ semi-norm, hence $\hat a(u, v) = \int_T \nabla u \nabla v \, dx$. Now let $u$ be a hat (or bilinear) function, and let $v \in \mathcal{I}$. Then we have that
\begin{align*} \int_T \nabla u \nabla v = \int_{\partial T} u \partial_n v – \int_T \Delta u v = 0. \end{align*} The first term is 0 due to the bubble functions vanishing on the boundary, and that $\Delta u = 0$ because it is linear. |
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. |
Hey guys! I built the voltage multiplier with alternating square wave from a 555 timer as a source (which is measured 4.5V by my multimeter) but the voltage multiplier doesn't seem to work. I tried first making a voltage doubler and it showed 9V (which is correct I suppose) but when I try a quadrupler for example and the voltage starts from like 6V and starts to go down around 0.1V per second.
Oh! I found a mistake in my wiring and fixed it. Now it seems to show 12V and instantly starts to go down by 0.1V per sec.
But you really should ask the people in Electrical Engineering. I just had a quick peek, and there was a recent conversation about voltage multipliers. I assume there are people there who've made high voltage stuff, like rail guns, which need a lot of current, so a low current circuit like yours should be simple for them.
So what did the guys in the EE chat say...
The voltage multiplier should be ok on a capacitive load. It will drop the voltage on a resistive load, as mentioned in various Electrical Engineering links on the topic. I assume you have thoroughly explored the links I have been posting for you...
A multimeter is basically an ammeter. To measure voltage, it puts a stable resistor into the circuit and measures the current running through it.
Hi all! There is theorem that links the imaginary and the real part in a time dependent analytic function. I forgot its name. Its named after some dutch(?) scientist and is used in solid state physics, who can help?
The Kramers–Kronig relations are bidirectional mathematical relations, connecting the real and imaginary parts of any complex function that is analytic in the upper half-plane. These relations are often used to calculate the real part from the imaginary part (or vice versa) of response functions in physical systems, because for stable systems, causality implies the analyticity condition, and conversely, analyticity implies causality of the corresponding stable physical system. The relation is named in honor of Ralph Kronig and Hans Kramers. In mathematics these relations are known under the names...
I have a weird question: The output on an astable multivibrator will be shown on a multimeter as half the input voltage (for example we have 9V-0V-9V-0V...and the multimeter averages it out and displays 4.5V). But then if I put that output to a voltage doubler, the voltage should be 18V, not 9V right? Since the voltage doubler will output in DC.
I've tried hooking up a transformer (9V to 230V, 0.5A) to an astable multivibrator (which operates at 671Hz) but something starts to smell burnt and the components of the astable multivibrator get hot. How do I fix this? I check it after that and the astable multivibrator works.
I searched the whole god damn internet, asked every god damn forum and I can't find a single schematic that converts 9V DC to 1500V DC without using giant transformers and power stage devices that weight 1 billion tons....
something so "simple" turns out to be hard as duck
In peskin book of QFT the sum over zero point energy modes is an infinite c-number, fortunately, it's experimental evidence doesn't appear, since experimentalists measure the difference in energy from the ground state. According to my understanding the zro pt energy is the same as the ground state, isn't it?
If so, always it is possible to substract a finite number (higher exited state for e.g.) from this zero point enrgy (which is infinte), it follows that, experimentally we always obtain infinte spectrum.
@AaronStevens Yeah, I had a good laugh to myself when he responded back with "Yeah, maybe they considered it and it was just too complicated". I can't even be mad at people like that. They are clearly fairly new to physics and don't quite grasp yet that most "novel" ideas have been thought of to death by someone; likely 100+ years ago if it's classical physics
I have recently come up with a design of a conceptual electromagntic field propulsion system which should not violate any conservation laws, particularly the Law of Conservation of Momentum and the Law of Conservation of Energy. In fact, this system should work in conjunction with these two laws ...
I rememeber that Gordon Freeman's thesis was "Observation of Einstein-Podolsky-Rosen Entanglement on Supraquantum Structures by Induction Through Nonlinear Transuranic Crystal of Extremely Long Wavelength (ELW) Pulse from Mode-Locked Source Array "
In peskin book of QFT the sum over zero point energy modes is an infinite c-number, fortunately, it's experimental evidence doesn't appear, since experimentalists measure the difference in energy from the ground state. According to my understanding the zro pt energy is the same as the ground state, isn't it? If so, always it is possible to substract a finite number (higher exited state for e.g.) from this zero point enrgy (which is infinte), it follows that, experimentally we always obtain infinte spectrum.
@ACuriousMind What confuses me is the interpretation of Peskin to this infinite c-number and the experimental fact
He said, the second term is the sum over zero point energy modes which is infnite as you mentioned. He added," fortunately, this energy cannot be detected experm., since the experiments measure only the difference between from the ground state of H".
@ACuriousMind Thank you, I understood your explanations clearly. However, regarding what Peskin mentioned in his book, there is a contradiction between what he said about the infinity of the zero point energy/ground state energy, and the fact that this energy is not detectable experimentally because the measurable quantity is the difference in energy between the ground state (which is infinite and this is the confusion) and a higher level.
It's just the first encounter with something that needs to be renormalized. Renormalizable theories are not "incomplete", even though you can take the Wilsonian standpoint that renormalized QFTs are effective theories cut off at a scale.
according to the author, the energy differenc is always infinite according to two fact. the first is, the ground state energy is infnite, secondly, the energy differenc is defined by substituting a higher level energy from the ground state one.
@enumaris That is an unfairly pithy way of putting it. There are finite, rigorous frameworks for renormalized perturbation theories following the work of Epstein and Glaser (buzzword: Causal perturbation theory). Just like in many other areas, the physicist's math sweeps a lot of subtlety under the rug, but that is far from unique to QFT or renormalization
The classical electrostatics formula $H = \int \frac{\mathbf{E}^2}{8 \pi} dV = \frac{1}{2} \sum_a e_a \phi(\mathbf{r}_a)$ with $\phi_a = \sum_b \frac{e_b}{R_{ab}}$ allows for $R_{aa} = 0$ terms i.e. dividing by zero to get infinities also, the problem stems from the fact that $R_{aa}$ can be zero due to using point particles, overall it's an infinite constant added to the particle that we throw away just as in QFT
@bolbteppa I understand the idea that we need to drop such terms to be in consistency with experiments. But i cannot understand why the experiment didn't predict such infinities that arose in the theory?
These $e_a/R_{aa}$ terms in the big sum are called self-energy terms, and are infinite, which means a relativistic electron would also have to have infinite mass if taken seriously, and relativity forbids the notion of a rigid body so we have to model them as point particles and can't avoid these $R_{aa} = 0$ values. |
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Orbital elements are the parameters required to uniquely identify a specific orbit. In celestial mechanics these elements are generally considered in classical two-body systems, where a Kepler orbit is used (derived from Newton's laws of motion and Newton's law of universal gravitation). There are many different ways to mathematically describe the same orbit, but certain schemes, each consisting of a set of six parameters, are commonly used in astronomy and orbital mechanics.
A real orbit (and its elements) changes over time due to gravitational perturbations by other objects and the effects of relativity. A Keplerian orbit is merely an idealized, mathematical approximation at a particular time.
The traditional orbital elements are the six Keplerian elements, after Johannes Kepler and his laws of planetary motion. When viewed from an inertial frame, two orbiting bodies trace out distinct trajectories. Each of these trajectories has its focus at the common center of mass. When viewed from the non-inertial frame of one body only the trajectory of the opposite body is apparent; Keplerian elements describe these non-inertial trajectories. An orbit has two sets of Keplerian elements depending on which body is used as the point of reference. The reference body is called the primary, the other body is called the secondary. The primary does not necessarily possess more mass than the secondary, and even when the bodies are of equal mass, the orbital elements depend on the choice of the primary.
The main two elements that define the shape and size of the ellipse:
Two elements define the orientation of the orbital plane in which the ellipse is embedded:
And finally:
The mean anomaly is a mathematically convenient "angle" which varies linearly with time, but which does not correspond to a real geometric angle. It can be converted into the true anomaly \nu\,\!, which does represent the real geometric angle in the plane of the ellipse, between periapsis (closest approach to the central body) and the position of the orbiting object at any given time. Thus, the true anomaly is shown as the red angle \nu\,\! in the diagram, and the mean anomaly is not shown.
The angles of inclination, longitude of the ascending node, and argument of periapsis can also be described as the Euler angles defining the orientation of the orbit relative to the reference coordinate system.
Note that non-elliptic trajectories also exist, but are not closed, and are thus not orbits. If the eccentricity is greater than one, the trajectory is a hyperbola. If the eccentricity is equal to one and the angular momentum is zero, the trajectory is radial. If the eccentricity is one and there is angular momentum, the trajectory is a parabola.
Given an inertial frame of reference and an arbitrary epoch (a specified point in time), exactly six parameters are necessary to unambiguously define an arbitrary and unperturbed orbit.
This is because the problem contains six degrees of freedom. These correspond to the three spatial dimensions which define position (the x, y, z in a Cartesian coordinate system), plus the velocity in each of these dimensions. These can be described as orbital state vectors, but this is often an inconvenient way to represent an orbit, which is why Keplerian elements are commonly used instead.
Sometimes the epoch is considered a "seventh" orbital parameter, rather than part of the reference frame.
If the epoch is defined to be at the moment when one of the elements is zero, the number of unspecified elements is reduced to five. (The sixth parameter is still necessary to define the orbit; it is merely numerically set to zero by convention or "moved" into the definition of the epoch with respect to real-world clock time.)
Keplerian elements can be obtained from orbital state vectors (x-y-z coordinates for position and velocity) by manual transformations or with computer software.[1]
Other orbital parameters can be computed from the Keplerian elements such as the period, apoapsis, and periapsis. (When orbiting the earth, the last two terms are known as the apogee and perigee.) It is common to specify the period instead of the semi-major axis in Keplerian element sets, as each can be computed from the other provided the standard gravitational parameter, GM, is given for the central body.
Instead of the mean anomaly at epoch, the mean anomaly M\,\!, mean longitude, true anomaly \nu_o\,\!, or (rarely) the eccentric anomaly might be used.
Using, for example, the "mean anomaly" instead of "mean anomaly at epoch" means that time t must be specified as a "seventh" orbital element. Sometimes it is assumed that mean anomaly is zero at the epoch (by choosing the appropriate definition of the epoch), leaving only the five other orbital elements to be specified.
Different sets of elements are used for various astronomical bodies. The eccentricity, e, and either the semi-major axis, a, or the distance of periapsis, q, are used to specify the shape and size of an orbit. The angle of the ascending node, Ω, the inclination, i, and the argument of periapsis, ω, or the longitude of periapsis, ϖ, specify the orientation of the orbit in its plane. Either the longitude at epoch, L0, the mean anomaly at epoch, M0, or the time of perihelion passage, T0, are used to specify a known point in the orbit. The choices made depend whether the vernal equinox or the node are used as the primary reference. The semi-major axis is known if the mean motion and the gravitational mass are known.[2][3]
It is also quite common to see either the Mean Anomaly (M) or the Mean Longitude (L) expressed directly, without either M0 or L0 as intermediary steps, as a polynomial function with respect to time. This method of expression will consolidate the mean motion (n) into the polynomial as one of the coefficients. The appearance will be that L or M are expressed in a more complicated manner, but we will appear to need one fewer orbital element.
Mean motion can also be obscured behind citations of the orbital period P.
The angles \Omega , i, \omega are the Euler angles (\alpha , \beta, \gamma with the notations of that article) characterizing the orientation of the coordinate system
where:
\hat{I},\hat{J} is in the equatorial plane of the central body. \hat{I} is in the direction of the vernal equinox. \hat{J} is perpendicular to \hat{I} and with \hat{I} defines the reference plane. \hat{K} is perpendicular to the reference plane.
\hat{x},\hat{y} are in the orbital plane and with \hat{x} in the direction to the pericenter (periapsis). \hat{z} is perpendicular to the plane of the orbit. \hat{y} is mutually perpendicular to \hat{x} and \hat{z}.
Then, the transformation from the \hat{I},\hat{J},\hat{K} coordinate frame to the \hat{x},\hat{y},\hat{z} frame with the Euler angles \Omega , i, \omega is:
where
The transformation from \hat{x},\hat{y},\hat{z} to Euler angles \Omega , i, \omega is:
Unperturbed, two-body, Newtonian orbits are always conic sections, so the Keplerian elements define an ellipse, parabola, or hyperbola. Real orbits have perturbations, so a given set of Keplerian elements accurately describes an orbit only at the epoch. Evolution of the orbital elements takes place due to the gravitational pull of bodies other than the primary, the nonsphericity of the primary, atmospheric drag, relativistic effects, radiation pressure, electromagnetic forces, and so on.
Keplerian elements can often be used to produce useful predictions at times near the epoch. Alternatively, real trajectories can be modeled as a sequence of Keplerian orbits that Hill.
Keplerian elements parameters can be encoded as text in a number of formats. The most common of them is the NASA/NORAD "two-line elements"(TLE) format [2], originally designed for use with 80-column punched cards, but still in use because it is the most common format, and can be handled easily by all modern data storages as well.
Depending on the application and object orbit, the data derived from TLEs older than 30 days can become unreliable. Orbital positions can be calculated from TLEs through the SGP/SGP4/SDP4/SGP8/SDP8 algorithms.[4]
Example of a two line element:[5]
1 27651U 03004A 07083.49636287 .00000119 00000-0 30706-4 0 2692 2 27651 039.9951 132.2059 0025931 073.4582 286.9047 14.81909376225249
Gnome, Object Request Broker, C , Python (programming language), Perl
Logic, Statistics, Probability distribution, Linguistics, Mathematics
Solar System, Year, Red dwarf, Proxima Centauri, Sirius
Japan, Orbital elements, Asteroid, Minor planet, Semimajor axis
Comet, Asteroid, Orbital elements, Orbital mechanics, Celestial mechanics |
Regression (OLS) - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Regression (OLS)
$z$ test for the difference between two proportions
Paired sample $t$ test
Sign test
Friedman test
Independent variables Independent variable Independent variable Independent variable Independent variable One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variables One categorical with 2 independent groups 2 paired groups 2 paired groups One within subject factor ($\geq 2$ related groups) Dependent variable Dependent variable Dependent variable Dependent variable Dependent variable One quantitative of interval or ratio level One categorical with 2 independent groups One quantitative of interval or ratio level One of ordinal level One of ordinal level Null hypothesis Null hypothesis Null hypothesis Null hypothesis Null hypothesis $F$ test for the complete regression model: $\pi_1 = \pi_2$
$\pi_1$ is the unknown proportion of "successes" in population 1; $\pi_2$ is the unknown proportion of "successes" in population 2
$\mu = \mu_0$
$\mu$ is the unknown population mean of the difference scores; $\mu_0$ is the population mean of the difference scores according to the null hypothesis, which is usually 0
The scores in any of the related groups are not systematically higher or lower than the scores in any of the other related groups
Note: usually, the related groups are the different measurement points
Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
Alternative hypothesis Alternative hypothesis Alternative hypothesis Alternative hypothesis Alternative hypothesis $F$ test for the complete regression model: Two sided: $\pi_1 \neq \pi_2$
Right sided: $\pi_1 > \pi_2$
Left sided: $\pi_1 < \pi_2$
Two sided: $\mu \neq \mu_0$
Right sided: $\mu > \mu_0$
Left sided: $\mu < \mu_0$
The scores in some of the related groups are systematically higher or lower than the scores in other related groups Assumptions Assumptions Assumptions Assumptions Assumptions all individuals in the population. Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another Sample of 'blocks' (usually the subjects) is a simple random sample from the population. That is, blocks are independent of one another Test statistic Test statistic Test statistic Test statistic Test statistic $F$ test for the complete regression model:
Note 2: if only one independent variable ($K = 1$), the $F$ test for the complete regression model is equivalent to the two sided $t$ test for $\beta_1$
$z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
$p_1$ is the sample proportion of successes in group 1: $\dfrac{X_1}{n_1}$, $p_2$ is the sample proportion of successes in group 2: $\dfrac{X_2}{n_2}$, $p$ is the total proportion of successes in the sample: $\dfrac{X_1 + X_2}{n_1 + n_2}$, $n_1$ is the sample size of group 1, $n_2$ is the sample size of group 2
Note: we could just as well compute $p_2 - p_1$ in the numerator, but then the left sided alternative becomes $\pi_2 < \pi_1$, and the right sided alternative becomes $\pi_2 > \pi_1$
$t = \dfrac{\bar{y} - \mu_0}{s / \sqrt{N}}$
$\bar{y}$ is the sample mean of the difference scores, $\mu_0$ is the population mean of the difference scores according to H0, $s$ is the sample standard deviation of the difference scores, $N$ is the sample size (number of difference scores).
The denominator $s / \sqrt{N}$ is the standard error of the sampling distribution of $\bar{y}$. The $t$ value indicates how many standard errors $\bar{y}$ is removed from $\mu_0$
$W = $ number of difference scores that is larger than 0 $Q = \dfrac{12}{N \times k(k + 1)} \sum R^2_i - 3 \times N(k + 1)$
Here $N$ is the number of 'blocks' (usually the subjects - so if you have 4 repeated measurements for 60 subjects, $N$ equals 60), $k$ is the number of related groups (usually the number of repeated measurements), and $R_i$ is the sum of ranks in group $i$.
Remember that multiplication precedes addition, so first compute $\frac{12}{N \times k(k + 1)} \times \sum R^2_i$ and then subtract $3 \times N(k + 1)$.
Note: if ties are present in the data, the formula for $Q$ is more complicated.
Sample standard deviation of the residuals $s$ n.a. n.a. n.a. n.a. $\begin{aligned} s &= \sqrt{\dfrac{\sum (y_j - \hat{y}_j)^2}{N - K - 1}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned} $ - - - - Sampling distribution of $F$ and of $t$ if H0 were true Sampling distribution of $z$ if H0 were true Sampling distribution of $t$ if H0 were true Sampling distribution of $W$ if H0 were true Sampling distribution of $Q$ if H0 were true Sampling distribution of $F$: Approximately standard normal $t$ distribution with $N - 1$ degrees of freedom The exact distribution of $W$ under the null hypothesis is the Binomial($n$, $p$) distribution, with $n =$ number of positive differences $+$ number of negative differences, and $p = 0.5$.
If $n$ is large, $W$ is approximately normally distributed under the null hypothesis, with mean $np = n \times 0.5$ and standard deviation $\sqrt{np(1-p)} = \sqrt{n \times 0.5(1 - 0.5)}$. Hence, if $n$ is large, the standardized test statistic $$z = \frac{W - n \times 0.5}{\sqrt{n \times 0.5(1 - 0.5)}}$$ follows approximately a standard normal distribution if the null hypothesis were true.
If the number of blocks $N$ is large, approximately the chi-squared distribution with $k - 1$ degrees of freedom.
For small samples, the exact distribution of $Q$ should be used.
Significant? Significant? Significant? Significant? Significant? $F$ test: Two sided: Two sided: If $n$ is small, the table for the binomial distribution should be used:
Two sided:
If $n$ is large, the table for standard normal probabilities can be used:
Two sided:
If the number of blocks $N$ is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$: $C\%$ confidence interval for $\beta_k$ and for $\mu_y$; $C\%$ prediction interval for $y_{new}$ Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$ $C\%$ confidence interval for $\mu$ n.a. n.a. Confidence interval for $\beta_k$: Regular (large sample): $\bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}}$
where the critical value $t^*$ is the value under the $t_{N-1}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20)
The confidence interval for $\mu$ can also be used as significance test.
- - Effect size n.a. Effect size n.a. n.a. Complete model: - Cohen's $d$:
Standardized difference between the sample mean of the difference scores and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{s}$$ Indicates how many standard deviations $s$ the sample mean of the difference scores $\bar{y}$ is removed from $\mu_0$
- - n.a. n.a. Visual representation n.a. n.a. - - - - ANOVA table n.a. n.a. n.a. n.a. - - - - n.a. Equivalent to Equivalent to Equivalent to n.a. - When testing two sided: chi-squared test for the relationship between two categorical variables, where both categorical variables have 2 levels One sample $t$ test on the difference scores
Repeated measures ANOVA with one dichotomous within subjects factor
Two sided sign test is equivalent to - Example context Example context Example context Example context Example context Can mental health be predicted from fysical health, economic class, and gender? Is the proportion smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic. Is the average difference between the mental health scores before and after an intervention different from $\mu_0$ = 0? Do people tend to score higher on mental health after a mindfulness course? Is there a difference in depression level between measurement point 1 (pre-intervention), measurement point 2 (1 week post-intervention), and measurement point 3 (6 weeks post-intervention)? SPSS SPSS SPSS SPSS SPSS Analyze > Regression > Linear... SPSS does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Analyze > Descriptive Statistics > Crosstabs...
Analyze > Compare Means > Paired-Samples T Test... Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples... Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples... Jamovi Jamovi Jamovi Jamovi Jamovi Regression > Linear Regression Jamovi does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Frequencies > Independent Samples - $\chi^2$ test of association
T-Tests > Paired Samples T-Test Jamovi does not have a specific option for the sign test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the two sided $p$ value that would have resulted from the sign test. Go to:
ANOVA > Repeated Measures ANOVA - Friedman
ANOVA > Repeated Measures ANOVA - Friedman
Practice questions Practice questions Practice questions Practice questions Practice questions |
Here is a possible suggestion how such an approach might work. The mechanism follows from rather general considerations and does not require quantum mechanics. So I'll use a more general mathematical notation.
Imagine I want to know the time evolution $f(x,t)$ when I only have knowledge of the initial condition $f(x,0)$. The assumption is that there is some dynamics that uniquely fixes $f(x,t)$, given $f(x,0)$. One can express the dynamics by some linear operator (linear equation of motion).
One way, as you pointed out, is to expand the solution in terms of the eigen-functions (let's briefly review it)$$ f(x,t) = \int F(\omega) \phi(x,\omega,t)\ d\omega . $$Here $\phi(x,\omega,t)$ are the eigen-functions and they are indexed by $\omega$ (associated with the energy). These eigen-functions are also orthogonal in the sense that$$ \int \phi(x,\omega,t) \phi^*(x,\omega',t)\ dx = \delta(\omega-\omega'). $$
This now allows us to obtain the spectrum for $f(x,0)$ at $t=0$ using $$ F(\omega) = \int f(x,0) \phi^*(x,\omega,0)\ dx . $$Then we can substitute $F(\omega)$ into the original expansion to get a general expression for $f(x,t)$ that is valid for all time.
Now for the alternative approach. Imagine we do the substitution, just mentioned and then change the order of integration$$ f(x,t) = \int f(x',0) \int \phi^*(x',\omega,0) \phi(x,\omega,t)\ d\omega \ dx'. $$The inner integral now represents a Green function or propagator for the process$$ K(x,x',t) = \int \phi^*(x',\omega,0) \phi(x,\omega,t)\ d\omega. $$If the dynamics is translation invariant in $x$, we'll get$$ K(x-x',t) = \int \phi^*(x',\omega,0) \phi(x,\omega,t)\ d\omega. $$If we substitute this back we obtain a convolution integral$$ f(x,t) = \int f(x',0) K(x-x',t)\ dx'. $$
So this provides an alternative approach. However, one needs to know the Green function, or alternatively derive it from knowledge of the eigen-functions. |
Set Difference is Subset
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Theorem $S \setminus T \subseteq S$
\(\displaystyle x \in S \setminus T\) \(\leadsto\) \(\displaystyle x \in S \land x \notin T\) Definition of Set Difference \(\displaystyle \) \(\leadsto\) \(\displaystyle x \in S\) Rule of Simplification
The result follows from the definition of subset.
$\blacksquare$
\(\displaystyle S \setminus T\) \(=\) \(\displaystyle S \cap \complement_S \left({T}\right)\) Set Difference as Intersection with Relative Complement \(\displaystyle \) \(\subseteq\) \(\displaystyle S\) Intersection is Subset
$\blacksquare$
Sources 1971: Allan Clark: Elements of Abstract Algebra... (previous) ... (next): Chapter $1$: The Notation and Terminology of Set Theory: $\S 8$ 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory(2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets |
Learning Objectives
Draw and interpret scatter diagrams. Use a graphing utility to find the line of best fit. Distinguish between linear and nonlinear relations. Fit a regression line to a set of data and use the linear model to make predictions.
A professor is attempting to identify trends among final exam scores. His class has a mixture of students, so he wonders if there is any relationship between age and final exam scores. One way for him to analyze the scores is by creating a diagram that relates the age of each student to the exam score received. In this section, we will examine one such diagram known as a scatter plot.
Drawing and Interpreting Scatter Plots
A scatter plot is a graph of plotted points that may show a relationship between two sets of data. If the relationship is from a
linear model, or a model that is nearly linear, the professor can draw conclusions using his knowledge of linear functions. Figure \(\PageIndex{1}\) shows a sample scatter plot.
Notice this scatter plot does not indicate a
linear relationship. The points do not appear to follow a trend. In other words, there does not appear to be a relationship between the age of the student and the score on the final exam.
Example \(\PageIndex{1}\): Using a Scatter Plot to Investigate Cricket Chirps
Table shows the number of cricket chirps in 15 seconds, for several different air temperatures, in degrees Fahrenheit[1]. Plot this data, and determine whether the data appears to be linearly related.
Table \(\PageIndex{1}\)
Chirps 44 35 20.4 33 31 35 18.5 37 26 Temperature 80.5 70.5 57 66 68 72 52 73.5 53 Solution
Plotting this data, as depicted in Figure \(\PageIndex{2}\) suggests that there may be a trend. We can see from the trend in the data that the number of chirps increases as the temperature increases. The trend appears to be roughly linear, though certainly not perfectly so.
Finding the Line of Best Fit
Once we recognize a need for a linear function to model that data, the natural follow-up question is “what is that linear function?” One way to approximate our linear function is to sketch the line that seems to best fit the data. Then we can extend the line until we can verify the y-intercept. We can approximate the slope of the line by extending it until we can estimate the \(\frac{\text{rise}}{\text{run}}\).
Example \(\PageIndex{2}\): Finding a Line of Best Fit
Find a linear function that fits the data in Table \(\PageIndex{1}\) by “eyeballing” a line that seems to fit.
Solution
On a graph, we could try sketching a line.
Using the starting and ending points of our hand drawn line, points \((0, 30)\) and \((50, 90)\), this graph has a slope of
\[m=\dfrac{60}{50}=1.2\]
and a y-intercept at 30. This gives an equation of
\[T(c)=1.2c+30\]
where \(c\) is the number of chirps in 15 seconds, and \(T(c)\) is the temperature in degrees Fahrenheit. The resulting equation is represented in Figure \(\PageIndex{3}\).
Analysis
This linear equation can then be used to approximate answers to various questions we might ask about the trend.
Recognizing Interpolation or Extrapolation
While the data for most examples does not fall perfectly on the line, the equation is our best guess as to how the relationship will behave outside of the values for which we have data. We use a process known as
interpolation when we predict a value inside the domain and range of the data. The process of extrapolation is used when we predict a value outside the domain and range of the data.
Figure \(\PageIndex{4}\) compares the two processes for the cricket-chirp data addressed in Example \(\PageIndex{2}\). We can see that interpolation would occur if we used our model to predict temperature when the values for chirps are between 18.5 and 44. Extrapolation would occur if we used our model to predict temperature when the values for chirps are less than 18.5 or greater than 44.
There is a difference between making predictions inside the domain and range of values for which we have data and outside that domain and range. Predicting a value outside of the domain and range has its limitations. When our model no longer applies after a certain point, it is sometimes called
model breakdown. For example, predicting a cost function for a period of two years may involve examining the data where the input is the time in years and the output is the cost. But if we try to extrapolate a cost when \(x=50\), that is in 50 years, the model would not apply because we could not account for factors fifty years in the future.
Note: Interpolation and Extrapolation
Different methods of making predictions are used to analyze data.
The method of interpolationinvolves predicting a value inside the domain and/or range of the data. The method of extrapolationinvolves predicting a value outside the domain and/or range of the data. Model breakdownoccurs at the point when the model no longer applies.
Example \(\PageIndex{3}\): Understanding Interpolation and Extrapolation
Use the cricket data from Table \(\PageIndex{1}\) to answer the following questions:
Would predicting the temperature when crickets are chirping 30 times in 15 seconds be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable. Would predicting the number of chirps crickets will make at 40 degrees be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable. Solution
a. The number of chirps in the data provided varied from 18.5 to 44. A prediction at 30 chirps per 15 seconds is inside the domain of our data, so would be interpolation. Using our model:
\[\begin{align} T(30)&=30+1.2(30) \\ &=66 \text{ degrees} \end{align}\]
Based on the data we have, this value seems reasonable. b. The temperature values varied from 52 to 80.5. Predicting the number of chirps at 40 degrees is extrapolation because 40 is outside the range of our data. Using our model:
\[\begin{align} 40&=30+1.2c \\ 10&=1.2c \\ c&\approx8.33 \end{align}\]
We can compare the regions of interpolation and extrapolation using Figure \(\PageIndex{5}\).
Analysis
Our model predicts the crickets would chirp 8.33 times in 15 seconds. While this might be possible, we have no reason to believe our model is valid outside the domain and range. In fact, generally crickets stop chirping altogether below around 50 degrees.
\(\PageIndex{1}\): According to the data from Table \(\PageIndex{1}\), what temperature can we predict it is if we counted 20 chirps in 15 seconds? Solution
54°F
Finding the Line of Best Fit Using a Graphing Utility
While eyeballing a line works reasonably well, there are statistical techniques for fitting a line to data that minimize the differences between the line and data values[2]. One such technique is called
least squares regression and can be computed by many graphing calculators, spreadsheet software, statistical software, and many web-based calculators[3]. Least squares regression is one means to determine the line that best fits the data, and here we will refer to this method as linear regression. Given data of input and corresponding outputs from a linear function, find the best fit line using linear regression. Enter the input in List 1 (L1). Enter the output in List 2 (L2). On a graphing utility, select Linear Regression (LinReg).
Example \(\PageIndex{4}\): Finding a Least Squares Regression Line
Find the least squares regression line using the cricket-chirp data in Table \(\PageIndex{1}\).
Solution
Enter the input (chirps) in List 1 (L1).
Enter the output (temperature) in List 2 (L2). See Table \(\PageIndex{2}\). Table \(\PageIndex{2}\)
L1 44 35 20.4 33 31 35 18.5 37 26 L2 80.5 70.5 57 66 68 72 52 73.5 53
On a graphing utility, select Linear Regression (LinReg). Using the cricket chirp data from earlier, with technology we obtain the equation:
\[T(c)=30.281+1.143c\]
Analysis
Notice that this line is quite similar to the equation we “eyeballed” but should fit the data better. Notice also that using this equation would change our prediction for the temperature when hearing 30 chirps in 15 seconds from 66 degrees to:
\[\begin{align} T(30)&=30.281+1.143(30) \\ &=64.571 \\ &\approx 64.6 \text{ degrees} \end{align}\]
The graph of the scatter plot with the least squares regression line is shown in Figure \(\PageIndex{6}\).
Will there ever be a case where two different lines will serve as the best fit for the data? No. There is only one best fit line. Distinguishing Between Linear and Non-Linear Models
As we saw above with the cricket-chirp model, some data exhibit strong linear trends, but other data, like the final exam scores plotted by age, are clearly nonlinear. Most calculators and computer software can also provide us with the
correlation coefficient, which is a measure of how closely the line fits the data. Many graphing calculators require the user to turn a ”diagnostic on” selection to find the correlation coefficient, which mathematicians label as \(r\). The correlation coefficient provides an easy way to get an idea of how close to a line the data falls.
We should compute the correlation coefficient only for data that follows a linear pattern or to determine the degree to which a data set is linear. If the data exhibits a nonlinear pattern, the correlation coefficient for a linear regression is meaningless. To get a sense for the relationship between the value of \(r\) and the graph of the data, Figure \(\PageIndex{7}\) shows some large data sets with their correlation coefficients. Remember, for all plots, the horizontal axis shows the input and the vertical axis shows the output.
Note: Correlation Coefficient
The
correlation coefficient is a value, \(r\), between –1 and 1. \(r>0\) suggests a positive (increasing) relationship \(r<0\) suggests a negative (decreasing) relationship The closer the value is to 0, the more scattered the data. The closer the value is to 1 or –1, the less scattered the data is.
Example \(\PageIndex{5}\): Finding a Correlation Coefficient
Calculate the correlation coefficient for cricket-chirp data in Table \(\PageIndex{1}\).
Solution
Because the data appear to follow a linear pattern, we can use technology to calculate \(r\). Enter the inputs and corresponding outputs and select the Linear Regression. The calculator will also provide you with the correlation coefficient, \(r=0.9509\). This value is very close to 1, which suggests a strong increasing linear relationship.
Note: For some calculators, the Diagnostics must be turned "on" in order to get the correlation coefficient when linear regression is performed: [2nd]>[0]>[alpha][x–1], then scroll to DIAGNOSTICSON.
Predicting with a Regression Line
Once we determine that a set of data is linear using the correlation coefficient, we can use the regression line to make predictions. As we learned above, a regression line is a line that is closest to the data in the scatter plot, which means that only one such line is a best fit for the data.
Example \(\PageIndex{6}\): Using a Regression Line to Make Predictions
Gasoline consumption in the United States has been steadily increasing. Consumption data from 1994 to 2004 is shown in Table \(\PageIndex{3}\). Determine whether the trend is linear, and if so, find a model for the data. Use the model to predict the consumption in 2008.
Table \(\PageIndex{3}\)
Year '94 '95 '96 '97 '98 '99 '00 '01 '02 '03 '04 Consumption (billions of gallons) 113 116 118 119 123 125 126 128 131 133 136
The scatter plot of the data, including the least squares regression line, is shown in Figure \(\PageIndex{8}\).
We can introduce new input variable, \(t\),representing years since 1994.
The least squares regression equation is:
\[C(t)=113.318+2.209t\]
Using technology, the correlation coefficient was calculated to be 0.9965, suggesting a very strong increasing linear trend.
Using this to predict consumption in 2008 \((t=14)\),
\[\begin{align} C(14)&=113.318+2.209(14) \\ &=144.244 \end{align}\]
The model predicts 144.244 billion gallons of gasoline consumption in 2008.
\(\PageIndex{2}\): Use the model we created using technology in Example 2.4.6 to predict the gas consumption in 2011. Is this an interpolation or an extrapolation? Solution
150.871 billion gallons; extrapolation
Key Concepts Scatter plots show the relationship between two sets of data. Scatter plots may represent linear or non-linear models. The line of best fit may be estimated or calculated, using a calculator or statistical software. Interpolation can be used to predict values inside the domain and range of the data, whereas extrapolation can be used to predict values outside the domain and range of the data. The correlation coefficient, \(r\), indicates the degree of linear relationship between data. A regression line best fits the data. The least squares regression line is found by minimizing the squares of the distances of points from a line passing through the data and may be used to make predictions regarding either of the variables. |
When does $\int_Af(x,y)dA$ represent a surface area geometrically, and when does it represent a volume? In my lecture notes I'm told it represent the volume underneath the surface $z=f(x,y)$, but I've found examples online computing double integrals to find the surface area of a surface. Thanks a bundle
$\iint_S f \operatorname d S$ is the Surface Integral of scalar field $f$ over surface $S$.
When the curve can be described by a parameterised vector $S: \vec r(s,t)$, we have
$$\iint_S f(\vec r(s,t))\operatorname d S = \iint_S f(\vec r(s,t))\begin{Vmatrix} \frac{\partial \vec r(s,t)}{\partial s}\times \frac{\partial \vec r(s,t)}{\partial t}\end{Vmatrix} \operatorname d s \operatorname d t$$
Note: the scalar field takes a three dimensional vector as its argument, not simply the curvilinear coordinates.
The surface area of the curve $\;\vec r=\begin{bmatrix}x \\ y\\g(x,y)\end{bmatrix}$ projected above some section $T$ of the x,y plane we use:
$$\begin{align} \iint_S \operatorname dS & = \iint_T \begin{Vmatrix}\dfrac{\partial}{\partial x}\begin{bmatrix}x \\ y\\g(x,y)\end{bmatrix}\times\dfrac{\partial\;}{\partial y}\begin{bmatrix}x \\ y\\g(x,y)\end{bmatrix}\end{Vmatrix}\operatorname d x \operatorname d y \\[1ex] & = \iint_T \begin{Vmatrix}\begin{bmatrix}1 \\ 0\\ g_x(x,y)\end{bmatrix}\times\begin{bmatrix}0 \\ 1\\g_y(x,y)\end{bmatrix}\end{Vmatrix}\operatorname d x \operatorname d y \\[1ex] & = \iint_T \sqrt{g_x(x,y)^2 +g_y(x,y)^2 + 1}\operatorname d x \operatorname d y \end{align}$$
Note: here we are using a unit scalar field $f = 1$.
To find the volume between a surface $z=g(x,y)$ and a section $T$ of the x,y plane, we just use:
$$\iint_T g(x,y) \operatorname d x \operatorname d y$$
Which is a totally different beast altogether.
$\iint_A f(x,y)dA$ represents the volume underneath the surface $z=f(x,y)$. When $f(x,y)=1$ you get the area of $A$, which is the volume under the plane $z=1$ intersected with $A$. Note that $\iint_AdA$ represents a volume, and the value of this volume (the scalar part) is the area of $A$. |
Let $(X_1, d_1), (X_2, d_2)$ be metric spaces and define $(X_1 \times X_2, d)$, $(X_1 \times X_2, p)$ where $d[(x_1, x_2),(y_1, y_2)] = d_1(x_1, y_1) + d_2(x_2,y_2)$ and $p[(x_1, x_2),(y_1, y_2)] = \max\{d_1(x_1,y_1), d_2(x_2,y_2)\}$
My attempt:
Notice that $p[(x_1, x_2),(y_1, y_2)] \leq d[(x_1, x_2),(y_1, y_2)]$, that is because the metric $d$ is equal to the metric $p$ + $\min\{d_1(x_1,y_1), d_2(x_2,y_2)\}$.
Let $G$ be an open subset of $X_1 \times X_2$ with the metric $d$, so, for any element $t = (t_1, t_2)$ of $G$ there exists an $r > 0$ such that $B(t; r) \subseteq G$, that is $\{(z_1, z_2) \in X_1 \times X_2: d[(t_1, t_2),(z_1, z_2)] < r\}$, since $p[(t_1, t_2),(z_1, z_2)] \leq d[(t_1, t_2),(z_1, z_2)]$, we have $p[(t_1, t_2),(z_1, z_2)] < r$, therefore for any element in $G$, you can choose the same $r$ that guarantees that the open ball with metric $d$ is contained in $G$. That same $r$ is also going to guarantee that the open ball with metric p is contained in $G$.
Is that correct? I haven't been able to prove it on the other direction, I tried this.
let $d_1(x_1, y_1) > d_2(x_2, y_2)$, and consider an open subset with metric $p$, so $d_1(x_1, y_1) < r$ I tried adding $d_2(x_2, y_2)$ to the inequality and I got $r + d_2(x_2, y_2)$, but this doesn't guarantee that the open ball with metric $d$ is going to be contained in the open subset. That's where I'm stuck. Any ideas? |
Are the Milnor's seven dimensional exotic spheres parallelizable?
A much more general result is true.
Theorem: Let $\Sigma$ be a homotopy sphere and $f: S^n \to \Sigma $ be a homotopy equivalence. Then $f^{\ast} T \Sigma \cong T S^n$.
It says that exotic spheres cannot be distinguished by looking at the tangent bundle. This result is one of the hidden gems of the golden age of topology and the proof invokes the whole plethora of topology of the 1950s.
The argument can be recollected from the old literature, but I do not know a coherent reference.
To start with, there are several invariants of the tangent bundles that do not depend on the smooth structure. Let $\Sigma$ be a homotopy sphere. Then:
the Euler class $\chi(T\Sigma^{n})$ is $2$ if $n$ is even (Gauss-Bonnet, relatively easy). $T \Sigma^n \oplus \mathbb{R}$ is trivial. This is a deep result by Kervaire and Milnor (not in the Annals paper, but a small note published before). $T \Sigma \oplus \mathbb{R}$ is given by an element of $\pi_{n-1} (O)$, which is known, by Bott periodicity, to be either $Z$, $0$ or $Z/2$. The $Z$ groups are detected by the Pontrjagin class, which has to vanish by Hirzebruchs signature formula because the sphere evidently has signature $0$. In the $Z/2$ case, the argument is more delicate. Essentially, the normal spherical fibration of a manifold does not depend on the smooth structure. Since the normal fibration of the standard sphere is trivial, so is that of $T \Sigma$. The process of associating to a vector bundle its spherical fibration is the J-homomorphism which is injective in these dimensions by Adams' $J(X)$ paper.
Now look at the homotopy sequence of the fibration $O(n)\to O(n+1) \to S^n$, i.e. the piece
$$ \mathbb{Z}=\pi_n (S^n) \to \pi_{n-1} (O(n)) \to \pi_{n-1} (O(n+1)) = \pi_{n-1} (O). $$
It is known that $TS^n$ (for the standard smooth structure) is the image of the generator of $\pi_n (S^n)$ (not hard, see Steenrods book). By the above deep result, $T \Sigma$ lies in the kernel of $\pi_{n-1}(O(n)) \to \pi_{n-1} (O(n+1))$, i.e. it also comes from $Z=\pi_n (S^n) $. The image of $Z \to \pi_{n-1} (O(n)$ can be computed. It is $Z$ if $k$ is even (using the Euler class), it is $0$ if $n=1,3,7$ (follows directly from Adams' result on the parallelizability of the standard spheres) and it is $Z/2$ in the remaining cases. You can find the (not so hard, but clever) argument for the last assertion in Levine's lectures on homotopy spheres (which can be viewed as the sequel to the Kervaire-Milnor paper).
How to proceed? If $n=1,3,7$, it follows that $T \Sigma$ is trivial, as $TS^n$. If $n$ is even, then the kernel of $\pi_{n-1}(O(n))\to \pi_{n-1}(S^{n+1})$ is detected by the Euler class, and by Gauss-Bonnet, this characterizes $T \Sigma$.
It remains the case of odd $n$ apart from the "Adams dimensions". One has to argue that in these dimensions, $T \Sigma$ is nontrivial. In the introduction to his Hopf invariant paper, Adams attributes to Dold the result ''$T \Sigma$ parallel implies that $\Sigma$ (and hence $S^n$) is an H-space''. But he (Adams) proved that is not the case $n=1,3,7$. Adams does not give a reference for Dolds result, but in his answer to this question (and the subsequent comments), John Klein sketches a proof that looks like a 1950s argument.
EDIT: I supervised a Master's thesis (written by Julia Heller), who worked out the details of this argument. Here is the argument for the fact that if a homotopy sphere $\Sigma^n$ is parallelizable, then $n=0,1,3,7$.
Consider the diagonal $\Sigma \subset \Sigma \times \Sigma$. Its normal bundle $N$ is isomorphic to $T \Sigma$,
hence trivial. We have the Pontrjagin-Thom collapse $c:\Sigma \times \Sigma \to Th (N) $. Any trivialization $N \cong \Sigma \times \mathbb{R}^n$ induces a map $a: Th (N) \to S^n$. Let $$ \mu: \Sigma \times \Sigma \to Th (N) \to S^n $$ be the composition of the maps just explained. By counting intersections, it can be seen that the restriction of $\mu$ to the submanifolds $\{x\} \times \Sigma$ and $\Sigma \times \{x\}$ has degree $\pm 1$, hence is a homotopy equivalence. By composing with $(h,k):S^n \times S^n \to \Sigma \times \Sigma$ for suitably chosen homotopy equivalences, we get a map $$ \mu' : S^n \times S^n \to S^n $$ which restricts to degree 1 maps $S^n \times \{x\} \to S^n$ and $\{x\} \times S^n \to S^n$. Since maps of degree 1 are homotopic, it follows that $\mu'$ gives $S^n $ the structure of an $H$-space. By Adams' theorem, $n=0,1,3,7$.
Note: it is not important for Adams' theorem that the $H$-space structure is homotopy associative, but homotopy unitality is essential.
The tangent bundle to a smooth structure on $S^7$ is classified by a map $S^7 \to G_7(R^{\infty})$. By the exact sequence for a fibration for the fiber bundle $O(7)\to V_7(R^\infty)\to G_7(R^\infty)$, we see that $\pi_7(G_7(R^\infty)) = \pi_6(O(7))$. But $\pi_6(O(7))=0$ (I found a table A1.1.3.2 of homotopy groups of orthogonal groups here(pdf), since this isn't in the stable range of Bott periodicity), so the tangent bundle is trivial, i.e. parallelizable.
Addendum: From the fibration $O(7)\to O(8)\to S^7$,we have the fibration sequence $$\pi_7(O(7))\to \pi_7(O(8))\to \pi_7(S^7)\to \pi_6 O(7)\to \pi_6 O(8)\to \pi_6 S^7=0.$$
Since $S^7$ is parallelizable (as may be shown via the octonians for example), there is a splitting $\pi_7(S^7)\to \pi_7(O(8))$. Hence we have an isomorphism $\pi_6 O(7)\to \pi_6 O(8) = \pi_6 O(\infty)$, since this is in the stable range. By Bott periodicity, $\pi_6 O(\infty)=0$, so $\pi_6 O(7)=0$.
Here's another way to answer the original question. There is a theorem of Bredon and Kosinski (Annals, 1966) which says that if a manifold $M^n$ is stably parallelizable, then either $M^n$ is parallelizable or the maximum number of linearly independent vector fields on $M^n$ is the same as on $S^n$. Since $S^7$ is parallelizable, this implies that exotic 7-spheres are parallelizable (since they are stably parallelizable).
The following is just an expansion of Johannes' last paragraph.
I went to Adams' paper where he attributes to Dold the statement that $S^n$ parallelizable implies $S^n$ is an $H$-space. No reference is given, so I had to think of why the statement is true (and also its converse).
Here's is one possible argument (that avoids Hopf invariant one considerations).
Consider the diagram (whose horizontal sequences are fibrations) $\require{AMScd}$ $$ \begin{CD} O_n @>>> O_{n+1} @>>> S^n \\ @VVV @VVV @VVV\\ F_n @>>> G_{n+1} @>>> S^n\\ @VVV @VVV @VVV\\ F_n @>>> F_{n+1} @>>> F_{n+1}/F_n \end{CD} $$
where $G_n$ is the unbased self homotopy equivalences of $S^{n-1}$ and $F_n$ is the based homotopy equivalences of $S^n$. The map $O_n \to G_n$ is given by restricting an isometry to its unit sphere and the map $G_n \to F_n$ is given by unreduced suspension.
Then $S^n$ is parallelizable iff the top fibration has a section which implies that the middle fibration has a section $S^n \to G_{n+1}$. We can assume without loss in generality that this map sends the base point of $S^n$ to the identity.
The adjoint of this section is of the form $S^n \times S^n \to S^n$ which is an $H$-space structure.
Conversely, if there's an $H$-space structure, then the middle fibration has a section. The map $S^n \to F_{n+1}/F_n$ is approximately $2n$-connected (this is a consequence of the EHP sequence). Consequently, there's a section of the bottom fibration up to around the $2n$-skeleton of the basespace $F_{n+1}/F_n$.
But, the square
$$ \begin{CD} O_n @>>> O_{n+1} \\ @VVV @VVV \\ F_n @>>> F_{n+1} \end{CD} $$
is about $2n$-cartesian.
This implies that $O_{n+1} \to S^n$ has a section iff and only if the pullback of $F_{n+1} \to F_{n+1}/F_n$ to $S^n$ has one and that's if and only if $G_{n+1} \to S^n$ has a section.
Is this a correct argument? |
"Definition 5 Let X be a nonempty set. By a partician P of X we mean a set of nonempty subsets of X such that:
(a) If A, B$\in$P and A$\neq$B, then A$\bigcap$B=$\emptyset$ (b) $\bigcup \limits_{C \in P}C$ = X"
Source: Set Theory by You-Feng. Lin and Shwu-Yeng T. Lin
"3.7 Definition Let A be a set; by
a partition of A we mean a family {$A_i$}$_{i \in I}$ of nonempty subsets of A with the following properties:
P1. $\forall i,\space j\in I, A_i \bigcap A_j = \emptyset \space or \space A_i=A_j$.
P2. A = $\bigcup \limits_{i \in I}A_i$"
Source: Set Theory Charles C. Pinter.
Why the definition in the first book defines the partition "(a) A, B$\in$P and A$\neq$B then A$\bigcap$B=$\emptyset$", while the second defines the partition as " P1. $\forall i,\space j\in I, A_i \bigcap A_j = \emptyset \space or \space A_i=A_j$" even if they are defining about the same concept, a partition?
If "$A\neq B$ then $A$" is in the former book, shouldn't "$A_i=A_j$" in the second book be "$A_i \neq A_j$" since they are the definins of a partition? When I compare the second book by Charles C. Pinter with the book by Susann Epp, I think "$A_i=A_j$" is not correct, since $A_i$, $A_j$ should be disjoints. |
My question revolves around finding a function based on its derivative of the type below :
Problem :The limit below represents the derivative of some real-valued function $f$ at some real-number $a$. State such an $f$ and $a$ in this case.
$$\lim_{h \ \to \ 0} \frac{\sqrt{9+h}-3}{h}$$
Now this type of problem is slightly unique. In general we can always find a function based on it's derivative by taking the indefinite integral of the derivative, however in this case we don't have the derivative in a general form, we only have the value for the derivative function at some point $a$, and there are a large number of $f$'s which can produce the value for the derivative at that point.
Am I correct in saying this?
This problem above is easily solvable, anyone can see already that the function is obviously going to be $f(x) = \sqrt{x}$, but the way I seem to solve it is a very heuristic method, which bothers me greatly (i.e. similar to guessing a function and working from there). I'm trying to find a methodical way of solving problems of this type, as the way I solve it (shown below) will definitely break down for harder examples.
This is my solution :
By the definition of a derivative :
$$f'(x) = \lim_{h \to\ 0} \frac{f(x+h)-f(x)}{h}$$
In the above case we can see that $\ f(a+h) = \sqrt{9+h}\ $ and $\ f(a)=3 = \sqrt{9}$
This implies that $a = 9$ and $f(x) = \sqrt{x}$ or written more formally ($f : x \to \sqrt{x}, \ \forall x\in\mathbb{R^{+}}$)
As you can see that is a very hap-hazard solution, and I would like to find a better way to solve problems of these types, but it eludes me.
Is there a more methodical approach (or formal approach) to solving problems of this type, where we are given a limit representing the derivative of some real-valued function $f$ at some real-number $a$, and asked to find $f$ and $a$?
If you have any other suggestions for solving these types of problems, please comment below. |
Transfinite Induction/Principle 1 Theorem
Let $A$ denote a class.
Suppose that: For all elements $x$ of $\operatorname{On}$, if $x$ is a subset of $A$, then $x$ is an element of $A$. Then $\operatorname{On} \subseteq A$. Proof This page is beyond the scope of ZFC, and should not be used in anything other than the theory in which it resides. If you see any proofs that link to this page, please insert this template at the top. If you believe that the contents of this page can be reworked to allow ZFC, then you can discuss it at the talk page. Aiming for a contradiction, suppose that $\neg \operatorname{On} \subseteq A$.
Then:
$\left({\operatorname{On} \setminus A}\right) \ne \varnothing$
By Epsilon Relation is Strongly Well-Founded on Ordinal Class, $\operatorname{On} \setminus A$ must have a $\in$-minimal element $y$.
However, from the fact that $y$ is a $\in$-minimal element of $\operatorname{On} \setminus A$:
$\left({\operatorname{On} \setminus A}\right) \cap y = \varnothing$.
So by its subsethood of $\operatorname{On}$:
$\left({\operatorname{On} \cap y}\right) \setminus A = \left({y \setminus A}\right) = \varnothing$
Therefore $y \subseteq A$.
However, by the hypothesis, $y$ must also be an element of $A$.
This contradicts the fact that $y$ is an element of $\operatorname{On} \setminus A$.
Therefore $\operatorname{On} \subseteq A$.
$\blacksquare$
Law of the Excluded Middle Sources 1960: Paul R. Halmos: Naive Set Theory... (previous) ... (next): $\S 17$: Well Ordering 1964: Steven A. Gaal: Point Set Topology... (previous) ... (next): Introduction to Set Theory: $3$. The Axiom of Choice and Its Equivalents 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $7.17$ |
Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..
There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university
Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.
What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?
Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.
Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P
Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?
Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?
Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"
@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.
Although not the only route, can you tell me something contrary to what I expect?
It's a formula. There's no question of well-definedness.
I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.
It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.
Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.
You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.
@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.
@Eric: If you go eastward, we'll never cook! :(
I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.
@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)
@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.
@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator |
I have the following velocity vector (in spherical polars): \begin{equation} \textbf{v} = u \hat{\textbf{r}} + v_{\phi}\hat{\boldsymbol\phi} \end{equation} Where $u(r) = u$ and $v_{\phi} (r) = v_{\phi}$. $\theta =0$ in this system.
I am now required to prove the following: \begin{eqnarray} \rho(\textbf{v}\cdot\nabla)\textbf{v}=\rho \frac{u}{r} \frac{d}{dr}(r v_{\phi}). \end{eqnarray} The RHS is just part of the material derivative in the macroscopic equation of motion - and that is what I am ultimately calculating (the equation of motion). I am, though, only required to find the $\phi$-component of this (the RHS term) although I don't quite understand the mathematical implication of this.
Anyway, this is how I've been solving it (wrongly).
$\nabla$ in spherical coordinates is: \begin{equation} \nabla = \frac{\partial}{\partial r} \hat{\boldsymbol r}+ \frac{1}{r} \frac{\partial}{\partial \theta}\hat{\boldsymbol\theta} +\frac{1}{r \sin \theta}\frac{\partial}{\partial \phi}\hat{\boldsymbol\phi} \end{equation}.
That is: \begin{eqnarray} (\textbf{v} \cdot \nabla) &=& (u \hat{\textbf{r}} + v_{\phi}\hat{\boldsymbol\phi})\cdot(\frac{\partial}{\partial r} \hat{\boldsymbol r}+ \frac{1}{r} \frac{\partial}{\partial \theta}\hat{\boldsymbol\theta} +\frac{1}{r \sin \theta}\frac{\partial}{\partial \phi}\hat{\boldsymbol\phi}) \\ &=& u \frac{\partial}{\partial r} + \underbrace{\frac{v_{\phi}}{r \sin \theta}\frac{\partial}{\partial \theta}}_{A}. \end{eqnarray} Note that term $A$ is equal to zero because there is no system dependence on $\phi$ and $\theta = 0$.
That is: \begin{equation} (\textbf{v} \cdot \nabla) = u \frac{\partial}{\partial r}. \end{equation} Therefore: \begin{eqnarray} \rho(\textbf{v} \cdot \nabla)\textbf{v} &=& \rho u \frac{\partial}{\partial r}(u \hat{\textbf{r}} + v_{\phi}\hat{\boldsymbol\phi}) \\ &=& \underbrace{\rho u \frac{\partial u}{\partial r}\hat{\textbf{r}}}_{B} + \rho u\frac{\partial v_{\phi}}{\partial r}\hat{\boldsymbol\phi} \end{eqnarray} I'm assuming now, because of the need for only the $\phi$-component, term $B$ is neglected, i.e.: \begin{equation} \rho(\textbf{v}\cdot\nabla)\textbf{v} = \rho u\frac{\partial v_{\phi}}{\partial r}\hat{\boldsymbol\phi} \end{equation} Which is not right.
This is relatively simple and it's irritated me hugely, and any help would be much appreciated.
Thanks. |
The XTT^2 ALSV(FD) Specification
Author: Grzegorz J. Nalepa, based on the work with Antoni Ligęza
Version: Draft 2008Q3 Introduction to Attributive Logics
Attributive logics constitute a simple yet widely-used tool for knowledge specification and development of rule-based systems.
In fact in a large variety of applications in various areas of Artificial Intelligence (AI) and Knowledge Engineering (KE) attributive languages constitute the core knowledge representation formalism.
The most typical areas of applications include rule-based systems, expert systems (ones based on rule formalism) and advanced database and data warehouse systems with knowledge discovery applications and contemporary business rules and business intelligence components.
Introduction to Attributive Logics
The description of AL presented here is based on several papers, including
ALSV(FD)
After (ali2005thebook)
it is assumed that an
attribute
A_i is a function (or partial function) of the form
Here
O
is a set of objects and
D_i
is the domain of attribute
A_i
.
A
generalized attribute
A_i is a function (or partial function) of the form
where 2^D_i is the family of all the subsets of D_i.
ALSV(FD)
The basic element of the language of
Attribute Logic with Set Values over Finite Domains (ALSV(FD) for short) are attribute names and attribute values.
For simplicity of presentation no objects are considered here; in practice, the same attribute applied to two (or more) different objects can be considered as two (or more) new, different, object-labelled attributes.
Unless two (or more) different objects are considered at the same time, no explicite reference to an object is necessary.
ALSV(FD)
Let us consider:
Let A = A_1, A_2, … ,A_n be all the attributes such that their values define the state of the system under consideration. It is assumed that the overall domain
D is divided into n sets (disjoint or not), D = D_1 u D_2 u … u D_n, where D_i is the domain related to attribute A_i, i=1,2, … ,n.
Any domain D_i is assumed to be a finite (discrete) set. The set can be ordered, partially ordered, or unordered; in case of ordered (partially ordered) sets some modifications of the notation are allowed.
ALSV(FD)
As we consider dynamic systems, the values of attributes can change over time (or state of the system).
We consider both
simple attributes of the form A_i : T → D_i (i.e. taking a single value at any instant of time) and generalized ones of the form A_i: T → 2^D_i (i.e. taking a set of values at a time); here T denotes the time domain of discourse. Syntax Syntax Syntax Syntax
The proposed set of relations is selected for convenience and as such is not completely independent. For example, A_i = V_i can perhaps be defined as A_i \subset V_{i} ^ A_i \supset V_i; but it is much more concise and convenient to use
= directly.
Various notational conventions extending the basic notation can be used. For example, in case of domains being ordered sets, relational symbols such as >, >=, <, =< can be used with the straightforward meaning.
Semantics
In SAL the semantics of A_i=d is straightforward – the attribute takes a single value.
The semantics of A_i=t is that the attribute takes
all the values of t (the so-called internal conjunction) while the semantics of A_i \in t is that it takes one (in case of simple attributes) or some (in case of generalized attributes) of the values of t (the so-called internal disjunction).
As an example for the necessity of SAL one can consider the specification of working days (denoted with
WDay) given as WDay = D, where D is the set of working days, D = { Monday,Tuesday,Wednesday,Thursday,Friday }. Now one can construct an atomic formula like CurrentDay \in D, or a rule of the form: DayOfInterest \in D → Status(OfficeOfInterest) = open. Semantics Inference Rules Inference Rules
The table is to be read as follows: if an atomic formula in the leftmost column holds, and a condition stated in the same row is true, the to appropriate atomic formula in the topmost row is a logical consequence of the one from the leftmost column.
Inference Rules Inference Rules
In Tables 3 and 4 the conditions are
satisfactory ones.
It is important to note that in case of the first rows of the tables (the cases of A=d_i and A=V, respectively) all the conditions are also
necessary ones.
The interpretation of the tables is straightforward: if an atomic formula in the leftmost column in some row
i is true, then the atomic formula in the topmost row in some column j is also true, provided that the relation indicated on intersection of row i and column j is true.
The rules of Table 3 and Table 4 can be used for checking if preconditions of a formula hold or verifying subsumption among rules.
Inference Rules
The interpretation of the Table 5 is straightforward: if the condition specified at the intersection of some row and column holds, then the atomic formulae labelling this row and column cannot simultaneously hold. Note however, that this is a satisfactory condition only.
The Table can be used for analysis of determinism of the system, i.e. whether satisfaction of precondition of a rule implies that the other rules in the same table cannot be fired.
State, State Representation and Inference
When processing information, the current values of attributes form the state of the inference process.
The values of attributes can, in general, be modified in the following three ways:
by an independent, external system,
by the inference process, and
as some clock-dependent functions.
State, State Representation and Inference
The first case concerns attributes which represent some process variables, which are to be incorporated in the inference process, but depend only of the environment and external systems.
As such, the variables cannot be directly influenced by the XTT system.
Examples: external temperature, the age of a client or the set of foreign languages known by a candidate.
Values of such variables are obtained as a result of some measurement or observation process.
They are assumed to be put into the inference system via a
blackboard communication method; in fact they are written directly into the internal memory whenever their values are obtained or changed. State, State Representation and Inference State, State Representation and Inference
The first solution is offered mostly for permanent changes; before asserting new values typically and appropriate retract operation is to be performed so as to keep a consistent state. In this way also the history (trajectory) of the system can be stored, provided that each value of an attribute is stored with a temporal index.
The second solution is offered for value passing and calculations which do not require permanent storage. For example, if a calculated value is to be passed to some next XTT component and it is no longer used after, it is not necessary to store it in the global memory.
The State of the System
The current state of the system is considered as a complete set of values of all the attributes in use at a certain instant of time.
The concept of the state is similar to the one in dynamic systems and state-machines.
The representation of the state should satisfy the following requirements:
the specification is
internally consistent,
the specification is
externally consistent,
the specification is
complete,
the specification is
deterministic,
the specification is
concise. The State of the System
The first postulate says that the specification itself cannot be inconsistent at the syntactic level. For example, a simple attribute (one taking a single value) cannot take two different values at the same time. In general, assuming independence of the attributes and no use of explicit negation, each value of an attribute should be specified once.
The second postulate says, that only
true knowledge (with respect to the external system) can be specified in state. In other words, facts that are syntactically correct but false cannot occur in the state formula. The State of the System
The third postulate says, that
all the knowledge true at a certain instant of time should be represented within the state.
The four postulate says that there can be no disjunctive knowledge specification within the state.
Finally, the fifth postulate says that no unnecessary, dependent knowledge should be kept in the state. In databases and most of the knowledge bases this has a practical dimension: only true facts are represented explicitly.
The State of the System
The current values of all the attributes are specified with the contents of the knowledge-base (including current sensor readings, measurements, inputs examination, etc.).
From logical point of view it is a formula of the form:
where
(
) for simple attributes and
, (
) for complex.
The State of the System
In order to cover realistic cases some explicit notation for covering unspecified, unknown values is proposed; this is so to deal with the data containing the NULL values imported from a database.
The first case refers to unspecified value of an attribute as a consequence of inappropriateness. A formula of the form
means that the attribute
takes an empty set of values (no value at all) at the current instant of time (or forever) for the object under consideration.
For example, the attribute
Maiden_Name or
The_Year_of_Last_Pregnancy for a man is not applicable and hence it takes no value for all men.
The State of the System State and rule firing
In order to fire a rule all the precondition facts defining its LHS must be true within the current state.
The verification procedure consists in matching these fact against the state specification.
A separate procedure concerns simple (single-valued) attributes, and a separate one is applied in case of complex attributes.
State and rule firing State and rule firing ALSV Rules
ALSV(FD) has been introduced with practical applications for rule languages in mind.
In fact, the primary aim of the presented language is to extend the notational possibilities and expressive power of the XTT-based tabular rule-based systems.
An important extension consist in allowing for explicit specification of one of the symbols eq, neq, in, notin, subset, supset, sim, notsim, with an argument in the table.
Rule Format
Consider a set of
n attributes A = A_1,A_2, …, A_nAny rule is assumed to be of the form:
where
alpha_i is one of the admissible relational symbols in ALSV(FD), and RHS is the right-hand side of the rule covering conclusion and perhaps the retract and assert definitions if necessary. Rule Firing
The current values of all the attributes are specified with the contents of the knowledge-base (including current sensor readings, measurements, inputs examination, etc.).
From logical point of view it is a formula of the form:
Eq: state-formulawhere for simple attributes and for complex. ANY and NULL
In case the value of A_i is unspecified we shall write A_i = NULL (a database convention).
Following a Prolog convention and logic, a
ANY attribute value is possible in comparison (see
_ in Prolog).
The semantics can be: „any value”, „not important”, etc.
The solution:
in preconditions, we can only use
ANY, i.e. an atom such as
A=_ can be specified, meaning „any value”, „all possible values of the attribute”, „we don't care”
on the other hand, attribute A unspecified, in the state formula means
A=NULL, so we store NULL in state
here we come to an inference rule:
A=NULL =⇒
A=_. Seems to be valid… This rules should be optionally disabled/enabled in the inference engine. |
The boiling points of group 15 elements increase on going down the group (or, as size increases) but the same is not true for the melting points. The melting points increase from $\ce{N}$ to $\ce{As}$ and then decrease from $\ce{As}$ to $\ce{Bi}$. Why is this the case?
A lower melting point is an indication of a lesser organised solid structure. So the question is actually:
What in the structures of arsenic and phosphorus make them less prone to melting?
Let’s start off by separating nitrogen from the entire group. It is a prime example of a diatomic gas such as hydrogen with a very inert $\ce{N#N}$ triple bond. As throughout the periodic table, multiple bonds with s and p orbitals between atoms of the same element are magnitudes more stable in the second period than anywhere else due to much better p-orbital overlap (smaller size, smaller interatomic distance). Therefore, we can consider phosphorus, arsenic, antimony and bismuth more or less as similar while nitrogen is an odd one out.
Phosphorus has four stable allotropes with vastly different structures. The white phosphorus allotrope, $\ce{P4}$ molecules, basically falls into the same class as dinitrogen and melts at around 40 degrees. Red, purple and black phosphorus build up network-like structures in two dimensions, either in tube-like chains (red) or as ‘wavy graphite’ (black). Only the black modification is clearly defined. Red phosphorus has melting points of around $600~\mathrm{^\circ C}$ while black phosphorus sublimes in vacuum at $400~\mathrm{^\circ C}$ — probably meaning that its melting point at standard pressure is much higher, maybe even higher than arsenic’s sublimation point. I found no references to liquid phosphorus so I have no idea whether the bonds of the network structure need be broken to liquidify or not.
Figure 1: crystal structure of black phosphorus, taken from Wikipedia.
Henceforth, we have established a structure element which we will be sticking to with only moderate modifications. The transition phosphorus — arsenic — antimony — bismuth is more or less the non-metal—metal transition. What explicitly changes is the distance between the sheets as shown in table 1.
$$\textbf{Table 1: }\text{Comparison of bond lengths and inter-sheet contact distances for pnictogens}$$ $$\begin{array}{lrrrr} \hline \text{ } & \ce{P_{black}} & \ce{As} & \ce{Sb} & \ce{Bi} \\ \hline \ce{E-E}\text{-bond/pm} & 223.0 & 251.7 & 290.8 & 307.1 \\ \ce{E\bond{...}E}\text{-contact/pm} & - & 311.9 & 335.4 & 352.8 \\ \text{d}_\text{contact}/\text{d}_\text{bond} & \approx 1.5 & 1.239 & 1.153 & 1.149 \\ \hline \end{array}$$
So while the phosphorus structure can be described well as consisting of semi-isolated sheets stacked above each other; when we have reached bismuth, these sheets are much closer to an overall three-dimensional network. Therefore, bismuth atoms are in a distorted octahedric environment as other metals would be. Arsenic, where the sheets are still well separated, exhibits almost metallic condictivity perpendicular to the sheets.
Putting this together and taking the extreme points, we can say:
The phosphorus structure is highly organised in that two directions are strictly discernable. Melting along one of these is no problem (perpendicular to the sheets) but melting
withina sheet is a huge problem as $\ce{P-P}$ bonds would need to be broken.
The bismuth structure no longer has a true distinction of directions; all three axes are more or less equal. Therefore, it takes similar energies to break bonds in one direction as it does to break them in another direction meaning the overall process is easiert.
Furthermore, liquid phases are generally those that display a
high localordering but a low long-rangeordering. The high local ordering can be preserved in bismuth even if the long-range ordering present in the solid state is lost. Conversely, the local ordering is phosphorus is a lot stricter and cannot easily be transferred into a state where long-range ordering is low.
Finally, note that non-metals whose solid phase does not consist of small, well defined molecules generally have higher melting points than metals. (Compare carbon’s sublimation point of c. $3600~\mathrm{^\circ C}$.)
As we move down the group no.of shells increases so the energy to release the electrons also increase (size is directly proportional to the Vander walls forces ) as we move down the group size increases so Vander walls force also increases therefore we have to boil it for a long time at a high temperature to break the forces and release electrons |
I can turn-the-crank and show that $\frac{1}{2}\otimes \frac{1}{2} = 1\oplus 0$ etc, but what would be a strategy to proving the general statement for spin representations that $j\otimes s =\bigoplus_{l=|s-j|}^{|s+j|} l$.
It's easy to prove the formula if you just look at the individual basis vectors of the tensor product. Let's use $(2j+1)$ and $(2s+1)$ eigenvectors of $j^2, j_z$ and $s^2, s_z$ called $|j,j_z\rangle$ and so on.
Now let's ask about the multiplicity of basis vectors of the tensor product with a given eigenvalue of $J_z = j_z+s_z$. The maximum eigenvalue of $J_z$ in the tensor product is $j+s$: it can be obtained if we choose $$ |J,J_z=j+s\rangle = |j,j\rangle \otimes |s,s\rangle $$ There are no higher eigenvalues of $J_z$; this proves that no representation with $J>j+s$ is included in the tensor product. However, the $J=j+s$ representation must be included exactly once to obtain one basis vector with $J_z=j+s$; representations with lower values of $J<j+s$ wouldn't contribute any vectors with $J_z=j+s$.
So the tensor product $$ Rep(j) \otimes Rep(s) = Rep(j+s)\oplus Rep(rest) $$ I have used the fact that reducible representations of simple compact groups may be written as direct sums. Now, what about the remaining representation(s) $Rep(rest)$? It is the linear envelope of a set of basis vectors in which we have already removed all the $(2J+1)$ basis vectors with $J=j+s$.
Well, in the rest, the maximum allowed $J$ is $j+s-1$. From the original bases, we see that the original space was 2-dimensional: the old basis included $$|j,j-1\rangle \otimes |s,s\rangle, \qquad |j,j\rangle \otimes |s,s-1\rangle $$ But we have already included one combination to $Rep(j+s)$; so the $Rep(rest)$ representation only contains the other one. By the same argument as above, we may see that the multiplicity of $Rep(j+s-1)$ in the tensor product is also one.
By induction, this algorithm may continue: at the beginning, the number of eigenvectors with a given $J_z$ is increasing by one every time we decrease $J_z$ by one. However, this behavior stops once we get to too low values of $J_z$ that would require too negative values of $j_z$, either $j_z<-j$ or $s_z<-s$. When that happens, the number of basis vectors no longer jumps by one; it stays constant. It happens when $$J_z^{max} \equiv J = |j-s|$$ so $J=|j-s$ is the lowest-$J$ representation included in the decomposition of the tensor product. Another way to see that at this moment, we have already written down all components, is either to notice that $J$ can't be smaller than $|j-s|$ because the minimum is obtained by adding "oppositely directed vectors" and can't be further shortened; or, alternatively, we may check that the dimensions of your formula work: $$ (2j+1)(2s+1) = \sum_{J=|j-s|}^{(j+s)} (2J+1) $$
I think the best way is not the textbook way, but the one described in the warm-up problem in this answer: Mathematically, what is color charge?
I will repeat the main point: the irreducible represenations of SU(2) are given by all complex completely symmetric tensors with all indices down, where each index takes two values 0,1. This is because you have invariant $\epsilon_{ij}$ tensor, which you can use as a metric to raise and lower indices, and you can remove the antisymmetric parts using the $\epsilon$ tensor. The fully symmetric k-index tensor is the spin k/2 representation.
When you tensor two of these together of size k and m, you just put the two tensors end to end, which gives a reducible k+m tensor. You need to remove the antisymmetric parts using the $\epsilon$ tensor, and this steps down by 2 each time, producing exactly one representation of every size between k+m (where you start) and k-m (when you run out of indices to contract).
You can solve this by examining characters. Any finite dimensional representation of $SU(2)$ breaks up into a sum of 1-dimensional representations of $U(1)$, which are classified by an integer called weight (you may divide by 2 to get a half-integer if you need to conform to physics conventions). You can then write the decomposition of a representation as a generating function, by adding the monomial $q^n$ for each representation of $U(1)$ of weight $n$ that you see. For example, the representation $\frac12$ corresponds to $q + q^{-1}$, and the representation $1$ corresponds to $q^2 + 1 + q^{-2}$.
The two facts you need are then:
For any non-negative integer $k$, the irreducible representation $k/2$ has character of the form $$\frac{q^{k+1} - q^{-k-1}}{q - q^{-1}} = q^{k} + q^{k-2} + \cdots + q^{-k}$$
The character of a tensor product is the product of characters.
In other words, the tensor product decomposition can be reconstructed by forgetting the $SU(2)$ action, taking the tensor product of the underlying $U(1)$ representations, then remembering that the characters of irreducible $SU(2)$ representations have a special form.
In your example, squaring $q + q^{-1}$ yields $q^2 + 2 + q^{-2}$. You then subtract $q^2 + 1 + q^{-2}$ to get $1$. For your more general question, you want to show that: $$\frac{q^{j+1} - q^{-j-1}}{q - q^{-1}}\frac{q^{s+1} - q^{-s-1}}{q - q^{-1}} = \sum_{\ell = |s-j|}^{s+j} \frac{q^{l+1} - q^{-l-1}}{q - q^{-1}}$$ You can prove this by induction on $s$: Your base cases are $s=0$ and $s=1$, which are relatively easy to check. For larger $s$, you can do a reduction by splitting the sum $\frac{q^{s+1} - q^{-s-1}}{q - q^{-1}}$ as $(q^s + q^{-s}) + \frac{q^{s-1} - q^{-s+1}}{q - q^{-1}}$. Note that $(q^s + q^{-s})(q^j + q^{j-2} + \cdots + q^{-j})$ is expanded as $$(q^{s+j} + q^{s+j-2} + \cdots + q^{-s-j}) + (q^{|s-j|} + q^{|s-j|-2} + \cdots + q^{-|s-j|}).$$ These are the extreme summands in $\sum_{\ell = |s-j|}^{s+j} \frac{q^{l+1} - q^{-l-1}}{q - q^{-1}}$. The remaining summands are what you get by replacing $s$ by $s-2$.
There is also a combinatorial method (which would be easier to communicate if I could draw): view the Laurent polynomial $q^j + \cdots + q^{-j}$ as a set of evenly spaced dots on the number line, and view each of the monomials in $q^{s} + \cdots + q^{-s}$ as a shifting operator. You get a bunch of shifted sets of dots, and you can reorganize them into a sort of symmetric pile. Each layer of the pile is one of the new irreducible representations. |
The book EXACT SPACE-TIMESIN EINSTEIN’SGENERAL RELATIVITY by Podolsky and Griffiths has a section on Taub-Nut space-time metrics and there is defines the singularity made in the Taub metric as quasi-regular singularity $$ \mathrm{d} s^{2}=-f(r)\left(\mathrm{d} t+4 l \sin ^{2} \frac{1}{2} \theta \mathrm{d} \phi\right)^{2}+\frac{\mathrm{d} r^{2}}{f(r)}+\left(r^{2}+l^{2}\right)\left(\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{d} \phi^{2}\right) $$s.
When we set $$ \theta=0 \text { and } \theta=\pi $$
We get a singularity on one of the axises.
What is a quasi-regular singularity as opposed to a curvature singularity? |
The production of J/ψ mesons in proton-proton collisions at a centre-of-mass energy of $ \sqrt{s}=13 $ TeV is studied with the LHCb detector. Cross-section measurements are performed as a function of the transverse momentum p$_{T}$ and the rapidity y of the J/ψ meson in the region p$_{T}$ < 14 GeV/c and 2.0 < y < 4.5, for both prompt J/ψ mesons and J/ψ mesons from b-hadron decays. The production cross-sections integrated over the kinematic coverage are 15.30 ± 0.03 ± 0.86 μb for prompt J/ψ and 2.34 ± 0.01 ± 0.13 μb for J/ψ from b-hadron decays, assuming zero polarization of the J/ψ meson. The first uncertainties are statistical and the second systematic. The cross-section reported for J/ψ mesons from b-hadron decays is used to extrapolate to a total $ b\overline{b} $ cross-section. The ratios of the cross-sections with respect to $ \sqrt{s}=8 $ TeV are also determined.
A search for neutral Higgs bosons in the minimal supersymmetric extension of the standard model (MSSM) decaying to tau-lepton pairs in pp collisions is performed, using events recorded by the CMS experiment at the LHC. The dataset corresponds to an integrated luminosity of 24.6 fb$^{−1}$, with 4.9 fb$^{−1}$ at 7 TeV and 19.7 fb$^{−1}$ at 8 TeV. To enhance the sensitivity to neutral MSSM Higgs bosons, the search includes the case where the Higgs boson is produced in association with a b-quark jet. No excess is observed in the tau-lepton-pair invariant mass spectrum. Exclusion limits are presented in the MSSM parameter space for different benchmark scenarios, m$_{h}^{max}$ , m$_{h}^{mod +}$ , m$_{h}^{mod −}$ , light-stop, light-stau, τ-phobic, and low-m$_{H}$. Upper limits on the cross section times branching fraction for gluon fusion and b-quark associated Higgs boson production are also given.
The production of J/ψ and mesons in pp collisions at TeV is studied with the LHCb detector. The J/ψ and mesons are reconstructed in the μ (+) μ (−) decay mode and the signal yields are determined with a fit to the μ (+) μ (−) invariant mass distributions. The analysis is performed in the rapidity range 2. 0 < y < 4. 5 and transverse momentum range 0 < p ( )T( ) < 14 (15) GeV/c of the J/ψ mesons. The J/ψ and production cross-sections and the fraction of J/ψ mesons from b-hadron decays are measured as a function of the meson p (T) and y.
Measurements are presented of electroweak boson production using data from pp collisions at a centre-of-mass energy of $ \sqrt{s}=8 $ TeV. The analysis is based on an integrated luminosity of 2.0 fb$^{−1}$ recorded with the LHCb detector. The bosons are identified in the W → μν and Z → μ$^{+}$ μ$^{−}$ decay channels. The cross-sections are measured for muons in the pseudorapidity range 2.0 < η < 4.5, with transverse momenta p$_{T}$ > 20 GeV/c and, in the case of the Z boson, a dimuon mass within $ 60 < {M}_{\mu }{{{}_{{}^{+}}}_{\mu}}_{{}^{-}}<120 $ GeV/c$^{2}$. The results are $ {\sigma}_W{{}_{{}^{+}}}_{\to \mu }{{}_{{}^{+}}}_{\nu }=1093.6\pm 2.1\pm 7.2\pm 10.9\pm 12.7\ \mathrm{p}\mathrm{b}, $ $ {\sigma}_W{{}_{{}^{-}}}_{\to \mu }{{}_{{}^{-}}}_{\overline{\nu}}=818.4\pm 1.9\pm 5.0\pm 7.0\pm 9.5\ \mathrm{p}\mathrm{b}, $ $ {\sigma}_{\mathrm{Z}\to \mu }{{{}_{{}^{+}}}_{\mu}}_{{}^{-}}=95.0\pm 0.3\pm 0.7\pm 1.1\pm 1.1\ \mathrm{p}\mathrm{b}, $ $ {\sigma}_{Z\to \mu }{{{}_{{}^{+}}}_{\mu}}_{{}^{-}}=95.0\pm 0.3\pm 0.7\pm 1.1\pm 1.1\ \mathrm{p}\mathrm{b}, $ where the first uncertainties are statistical, the second are systematic, the third are due to the knowledge of the LHC beam energy and the fourth are due to the luminosity determination. The evolution of the W and Z boson cross-sections with centre-of-mass energy is studied using previously reported measurements with 1.0 fb$^{−1}$ of data at 7 TeV. Differential distributions are also presented. Results are in good agreement with theoretical predictions at next-to-next-to-leading order in perturbative quantum chromodynamics.
Production of Bc+ mesons in proton-proton collisions at a center-of-mass energy of 8 TeV is studied with data corresponding to an integrated luminosity of 2.0 fb-1 recorded by the LHCb experiment. The ratio of production cross sections times branching fractions between the Bc+→J/ψπ+ and B+→J/ψK+ decays is measured as a function of transverse momentum and rapidity in the regions 0<pT<20 GeV/c and 2.0<y<4.5. The ratio in this kinematic range is measured to be (0.683±0.018±0.009)%, where the first uncertainty is statistical and the second systematic.
We report on the inclusive production cross sections of ${\mathrm{J}/\psi }$ , ${\psi (\mathrm{2S})}$ , $\mathrm{\Upsilon }$ (1S), $\mathrm{\Upsilon }$ (2S) and $\mathrm{\Upsilon }$ (3S), measured at forward rapidity with the ALICE detector in $\mathrm{pp}$ collisions at a center-of-mass energy $\sqrt{s}=8$ TeV. The analysis is based on data collected at the LHC and corresponds to an integrated luminosity of 1.23 pb$^{-1}$ . Quarkonia are reconstructed in the dimuon-decay channel. The differential production cross sections are measured as a function of the transverse momentum ${p_\mathrm{T}}$ and rapidity y, over the ${p_\mathrm{T}}$ ranges $0<{p_\mathrm{T}}<20$ GeV/c for ${\mathrm{J}/\psi }$ , $0<{p_\mathrm{T}}<12$ GeV/c for all other resonances, and for $2.5<y<4$ . The cross sections, integrated over ${p_\mathrm{T}}$ and y, and assuming unpolarized quarkonia, are $\sigma _{{\mathrm{J}/\psi }} = 8.98\pm 0.04\pm 0.82$ $\upmu $ b, $\sigma _{{\psi (\mathrm{2S})}} = 1.23\pm 0.08\pm 0.22$ $\upmu $ b, $\sigma _{\mathrm{\Upsilon }\mathrm{(1S)}} = 71\pm 6\pm 7$ nb, $\sigma _{\mathrm{\Upsilon }\mathrm{(2S)}} = 26\pm 5\pm 4$ nb and $\sigma _{\mathrm{\Upsilon }\mathrm{(3S)}} = 9\pm 4\pm 1$ nb, where the first uncertainty is statistical and the second one is systematic. These values agree, within at most $1.4\sigma $ , with measurements performed by the LHCb collaboration in the same rapidity range.
A detailed study of pseudorapidity densities and multiplicity distributions of primary charged particles produced in proton-proton collisions, at $\sqrt{s} =$ 0.9, 2.36, 2.76, 7 and 8 TeV, in the pseudorapidity range $|\eta|<2$, was carried out using the ALICE detector. Measurements were obtained for three event classes: inelastic, non-single diffractive and events with at least one charged particle in the pseudorapidity interval $|\eta|<1$. The use of an improved track-counting algorithm combined with ALICE's measurements of diffractive processes allows a higher precision compared to our previous publications. A KNO scaling study was performed in the pseudorapidity intervals $|\eta|<$ 0.5, 1.0 and 1.5. The data are compared to other experimental results and to models as implemented in Monte Carlo event generators PHOJET and recent tunes of PYTHIA6, PYTHIA8 and EPOS.
The inclusive D$_{s}^{±}$ production asymmetry is measured in pp collisions collected by the LHCb experiment at centre-of-mass energies of $ \sqrt{s}=7 $ and 8 TeV. Promptly produced D$_{s}^{±}$ mesons are used, which decay as D$_{s}^{±}$ → ϕπ$^{±}$, with ϕ → K$^{+}$K$^{−}$. The measurement is performed in bins of transverse momentum, p$_{T}$, and rapidity, y, covering the range 2.5 < p$_{T}$ < 25.0 GeV/c and 2.0 < y < 4.5. No kinematic dependence is observed. Evidence of nonzero D$_{s}^{±}$ production asymmetry is found with a significance of 3.3 standard deviations.
The results of a dedicated search for pair production of scalar partners of charm quarks are reported. The search is based on an integrated luminosity of 20.3 fb-1 of pp collisions at s=8 TeV recorded with the ATLAS detector at the LHC. The search is performed using events with large missing transverse momentum and at least two jets, where the two leading jets are each tagged as originating from c quarks. Events containing isolated electrons or muons are vetoed. In an R-parity-conserving minimal supersymmetric scenario in which a single scalar-charm state is kinematically accessible, and where it decays exclusively into a charm quark and a neutralino, 95% confidence-level upper limits are obtained in the scalar-charm–neutralino mass plane such that, for neutralino masses below 200 GeV, scalar-charm masses up to 490 GeV are excluded.
Measurements of the differential and double-differential Drell–Yan cross sections in the dielectron and dimuon channels are presented. They are based on proton–proton collision data at $\sqrt{s} = 8\,\text {TeV} $ recorded with the CMS detector at the LHC and corresponding to an integrated luminosity of 19.7 $\,\text {fb}^{-1}$ . The measured inclusive cross section in the $\mathrm{Z}$ peak region (60–120 $\,\text {GeV}$ ), obtained from the combination of the dielectron and dimuon channels, is $1138 \pm 8\,\text {(exp)} \pm 25\,\text {(theo)} \pm 30\,\text {(lumi)} \text {\,pb} $ , where the statistical uncertainty is negligible. The differential cross section $\mathrm{d}\sigma /\mathrm{d}{}m$ in the dilepton mass range 15–2000 $\,\text {GeV}$ is measured and corrected to the full phase space. The double-differential cross section $\mathrm{d}^2\sigma /\mathrm{d}{}m\,\mathrm{d}|y |$ is also measured over the mass range 20 to 1500 $\,\text {GeV}$ and absolute dilepton rapidity from 0 to 2.4. In addition, the ratios of the normalized differential cross sections measured at $\sqrt{s} = 7$ and 8 $\,\text {TeV}$ are presented. These measurements are compared to the predictions of perturbative QCD at next-to-leading and next-to-next-to-leading (NNLO) orders using various sets of parton distribution functions (PDFs). The results agree with the NNLO theoretical predictions computed with fewz 3.1 using the CT10 NNLO and NNPDF2.1 NNLO PDFs. The measured double-differential cross section and ratio of normalized differential cross sections are sufficiently precise to constrain the proton PDFs.
The W^+W^- and ZZ production cross sections are measured in proton-proton collisions at s=8 TeV with the CMS experiment at the LHC in data samples corresponding to an integrated luminosity of up to 5.3 fb^-^1. The measurements are performed in the leptonic decay modes W^+W^-->@?^'@n@?^''@n and ZZ->2@?2@?^', where @?=e,@m and @?^'(@?^'')=e,@m,@t. The measured cross sections @s(pp->W^+W^-)=69.9+/-2.8(stat.)+/-5.6(syst.)+/-3.1(lum.) pb and @s(pp->ZZ)=8.4+/-1.0(stat.)+/-0.7(syst.)+/-0.4(lum.) pb, for both Z bosons produced in the mass region 60<m_Z<120 GeV, are consistent with standard model predictions. These are the first measurements of the diboson production cross sections at s=8 TeV.
Measurements of inclusive and direct photon production at mid-rapidity in pp collisions at $\sqrt{s}=2.76$ and 8 TeV are presented by the ALICE experiment at the LHC. The results are reported in transverse momentum ranges of $0.4<p_{T}<10$ GeV/$c$ and $0.3<p_{T}<16$ GeV/$c$, respectively. Photons are detected with the electromagnetic calorimeter~(EMCal) and via reconstruction of e$^+$e$^-$ pairs from conversions in the ALICE detector material using the central tracking system. For the final measurement of the inclusive photon spectra the results are combined in the overlapping $p_{T}$ interval of both methods. Direct photon spectra, or their upper limits at 90% C.L. are extracted using the direct photon excess ratio $R_{\gamma}$, which quantifies the ratio of inclusive photons over decay photons generated with a decay-photon simulation. An additional hybrid method, combining photons reconstructed from conversions with those identified in the EMCal, is used for the combination of the direct photon excess ratio $R_{\gamma}$, as well as the extraction of direct photon spectra or their upper limits. While no significant signal of direct photons is seen over the full $p_{T}$ range, $R_{\gamma}$ for $p_{T}>7$ GeV/$c$ is at least one $\sigma$ above unity and consistent with expectations from next-to-leading order pQCD calculations.
The angular distribution of the flavor-changing neutral current decay B$^+$$\to$ K$^+\mu^+\mu^-$ is studied in proton-proton collisions at a center-of-mass energy of 8 TeV. The analysis is based on data collected with the CMS detector at the LHC, corresponding to an integrated luminosity of 20.5 fb$^{-1}$. The forward-backward asymmetry $A_{\mathrm{FB}}$ of the dimuon system and the contribution $F_{\mathrm{H}}$ from the pseudoscalar, scalar, and tensor amplitudes to the decay width are measured as a function of the dimuon mass squared. The measurements are consistent with the standard model expectations.
The inclusive production rates of isolated, prompt photons in p+Pb collisions at sNN=8.16 TeV are studied with the ATLAS detector at the Large Hadron Collider using a dataset with an integrated luminosity of 165 nb −1 recorded in 2016. The cross-section and nuclear modification factor RpPb are measured as a function of photon transverse energy from 20 GeV to 550 GeV and in three nucleon–nucleon centre-of-mass pseudorapidity regions, (−2.83,−2.02) , (−1.84,0.91) , and (1.09,1.90) . The cross-section and RpPb values are compared with the results of a next-to-leading-order perturbative QCD calculation, with and without nuclear parton distribution function modifications, and with expectations based on a model of the energy loss of partons prior to the hard scattering. The data disfavour a large amount of energy loss and provide new constraints on the parton densities in nuclei. |
Consider the flow $\vec u=(uy,0,0)$ between two plates $y=0$ and $y=1$ (chosen out of simplicity). I want to find the stress tensor of such a flow given by: $$\sigma_{ij}=-\delta_{ij}p+\eta \left( \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_k}\right)$$ Although this seems quite easy (e.g. use Navier-Stokes to find $p$ and then simply sub $\vec u$ into the above) I can find no examples (at all) of a calculation of the stress tensor for any flow $\vec u$. Is there a reason behind this? Am I missing something? If not please can you provide such an example.
There are many exact solutions of the Navier-Stokes equation, which you can find in standard text books. In practice we don't usually specify the velocity field. Instead, we have an equation of state $P(\rho,T)$, and a set of boundary and/or initial conditions.
The specific example you provide is very simple (this is known as Poiseulle flow). In order to satisfy no-slip boundary conditions, the top surface has to move at speed $(u,0,0)$. Then your velocity field is a solution to the static continuity and Navier-Stokes equation for a constant pressure. This is the case because the fluid velicity is linear, so the stress tensor, which is build from first derivatives of $\vec{u}$, is constant. $$ \sigma_{xx}=\sigma_{yy}=\sigma_{zz}=-P, \; \; \; \; \sigma_{xy}=\sigma_{yx}=\eta u. $$ Then the derivative of the stress tensor (the RHS of the Navier-Stokes equation) is zero. The LHS is zero because the flow is static, $\partial_t \vec{u}=0$, and the comoving derivative $(\vec{u}\cdot\vec\nabla)\vec{u}$ also vanishes. As a result the Navier-Stokes equation is indeed satisfied.
I think nobody discuss the stress tensor because it is almost always divided into a pressure field and a diffusion term. Also apart from some simple analytical flow as you expressed or in exercises, I cannot see any practical interest in the determination of the stress tensor. Looking at the pressure field gives all the information, and the stress tensor can be reconstructing later if really needed.
I just looked in the Fluid Mechanics book of Landau-Lifshitz and found an equation using directly the stress tensor for a flow between rotating cylinders (§18 in the 2nd English Edition). |
Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..
There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university
Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.
What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?
Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.
Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P
Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?
Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?
Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"
@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.
Although not the only route, can you tell me something contrary to what I expect?
It's a formula. There's no question of well-definedness.
I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.
It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.
Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.
You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.
@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.
@Eric: If you go eastward, we'll never cook! :(
I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.
@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)
@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.
@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator |
This question appears somewhat similar to other questions asking about why wavelength affects diffraction (a concept which I'm still not 100% sure on...) however my query is different and not answered that I can find. (The focus of my question is to what degree slit size affects diffraction in terms of the wavelength, not how or why) I was wondering, to what degree do the wavelength and the size of the slit have to be similar for diffraction to be reasonably observable (for example in the classic wave tank example) Does diffraction become negligible at 100x the wavelength? 1000x it? And is this different for longitudinal and transverse waves?
closed as off-topic by Carl Witthoft, user36790, John Rennie, CuriousOne, honeste_vivere May 4 '16 at 16:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
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Whether the amount of diffraction is 'negligible' depends on how you define this criterion.
The first order minimum in the diffraction pattern from a single slit occurs where $\sin\theta = \lambda/d$ where $d$ is slit width, $\theta$ is diffraction angle and $\lambda$ is wavelength. If $d = \lambda$ the central lobe of the diffraction pattern will spread out $90$ degrees above and below the axis, filling the whole screen. If $d = 2\lambda$ the central lobe will spread to $30$ degrees above and below the axis. To achieve $\theta = 1\ \mathrm{degree}$ ($\sin\theta = 0.01745$) we need $d = 60 \lambda$ approx.
It makes no difference if the wave is longitudinal or transverse. The same formulas apply to both, unless polarisation is involved, because longitudinal waves cannot be polarised. |
Critical superlinear Ambrosetti-Prodi problems
DOI: http://dx.doi.org/10.12775/TMNA.1999.022
Abstract
We consider the existence of multiple solutions for problem (1.1) below
with either $\lambda\neq \lambda$ or $\lambda=\lambda_1$, where $\lambda_k$, $k=1,2,\ldots$ are eigenvalues of $(-\Delta, H^1_0(\Omega))$. The local bifurcation from $\lambda=\lambda_k$ is also investigated.
with either $\lambda\neq \lambda$ or $\lambda=\lambda_1$, where
$\lambda_k$, $k=1,2,\ldots$ are eigenvalues of $(-\Delta, H^1_0(\Omega))$.
The local bifurcation from $\lambda=\lambda_k$ is also investigated.
Keywords
Elliptic differential equations; variational methods; multiplicity of solutions
Full Text:FULL TEXT Refbacks There are currently no refbacks. |
The approach is wrong because you don't say which random variable satisfies which binomial distribution. The answer is that the number $X$ of bits that differ between the two strings satisfies $B(n, \frac{1}{2})$. Hence the probability is$$P(X=d)=\binom{n}{d}(1/2)^d (1/2)^{n-d} = \frac{1}{2^n}\binom{n}{d} $$
Edit As your comment implies that you want a more complete model, here it is: let $K = \{0, 1\}^n$ be the set of n-bit strings. Let $d\in [0, n]$ and $K_d$ the substet of $K$ consisting of the strings with exactly $d$ bits set. Note that $\text{Card}(K_d) = \binom{n}{d}$.
We choose randomly two strings $s_1, s_2\in K$. They can be the same string, so this amounts to choosing an ordered random pair $(s_1, s_2)\in K^2$. There are $2^{2 n}$ possible choices, each of them has the probability $2^{-2n}$. The number $\delta(s_1, s_2)$ of bits that differ between $s_1$ and $s_2$ is the number of bits set in their bitwise
xor sum $s_1 \oplus s_2$, or bitwise sum mod 2. This operation makes $K$ a group where each element is its own opposite. We need to count the number of pairs $(s_1, s_2)$ such that $s_1\oplus s_2 \in K_d$. It is very easy, because this is equivalent to $s_2 = s \oplus s_1$ with $s \in K_d$. It follows that for each $s_1\in K$ and each $s \in K_d$, there is exactly one $s_2$. Hence the number of ordered pairs is $\text{Card}(K\times K_d) = 2^n \binom{n}{d}$. Finally, the probability is$$2^{-2n} 2^n \binom{n}{d} = \frac{1}{2^n}\binom{n}{d}$$ |
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Yeah it does seem unreasonable to expect a finite presentation
Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.
How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...
Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants.Then $A$ makes a move, then $B$, and so on...The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...
Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...
The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms
This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.
Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$
Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$
So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$
Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$
But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$
For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube
Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.
Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$
You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point
Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).
Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...
@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.
This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.
You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.
I'll use the latter notation consistently if that's what you're comfortable with
(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)
@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)
Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms
So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.
Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.
That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection
Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$
Voila, Riemann curvature tensor
Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature
Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?
Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.
Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.
Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?
Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.
You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form
(The cotangent bundle is naturally a symplectic manifold)
Yeah
So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.
But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!
So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up
If someone has the time to quickly check my result, I would appreciate. Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$ Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?
Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty
@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method
I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.
My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?
In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.
Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group
Everything about $S_4$ is encoded in the cube, in a way
The same can be said of $A_5$ and the dodecahedron, say |
I'm a newbie reading quantum mechanics from "Inroduction to Quantum Meachanics" by Griffiths and in the early pages of the book the author defines:
$$\langle x\rangle =\int_{-\infty}^{\infty} x|\Psi(x,t)|\,dx = \int_{-\infty}^{\infty} \Psi^* (x)\Psi \,dx,$$
$$\langle v\rangle = \frac{d}{dt}\left(\langle x\rangle\right)= -\frac{i\hbar}{m}\int_{-\infty}^{\infty} \Psi^*\frac{\partial\Psi}{\partial x} \,dx,$$
$$\langle p\rangle = m\langle v\rangle= -i\hbar\int_{-\infty}^{\infty} \Psi^*\frac{\partial\Psi}{\partial x} \,dx,$$
so to me the author seems to be working out with expectations, which made perfect sense to me. I then googled the expression for kinetic energy and I was expecting to find out that:
$$\langle T\rangle=\frac{\langle p \rangle^2}{2m},$$
but instead, it seems that
$$\langle T\rangle=\frac{\langle p^2 \rangle}{2m}.$$
Why is this? I don't understand what happened in the case of kinetic energy. Why isn't the author now working with expected momentum in the case of expected kinetic energy? Can you perhaps show me a derivation of $\langle T\rangle $ and more importantly, explanation on why it is done like that? In the book, the author says that generally:
$$\langle Q(x, p)\rangle = \int \Psi^*Q(x, \frac{\hbar}{i}\frac{\partial}{\partial x})\Psi\,dx,$$
with advising that every $p$ should be replaced with $\frac{\hbar}{i}\frac{\partial}{\partial x}$ when calculating the expectation of interest. The why-part for this was however a bit non-existing. |
I'm taking an undergrad GR course, and our text (Lambourne) mentions covariant and contravariant vectors and tensors ad-nauseum, but never really gives a formal definition for what they are, and how they are unique from each other in any physical sense (other than their difference in transformations). Is there any physical intuition behind these two labels? There should be, right? If they differ in how they transform with transformation of coordinates, doesnt that indicate that there has to be some way of visualizing their difference, since coordinate transformations are easily visualized?
This whole business of covariant vs contravariant is very old school. Some very old texts go into ways of visualizing this. I would suggest instead learning about tangent vectors (contravariant) and 1-forms (covariant) and the equivalence between tangent vectors and directional derivatives.
Associate the vector $\vec{v}$ with the derivative operator $\vec{\frac{d}{d\lambda}}$ by saying that there is a curve parameterized by $\lambda$ that has $\vec{v}$ as it's tangent vector.
Similarly, associate to the function $f$ the 1-form $df$. A 1-form is a linear map from tangent vectors onto real numbers. A 1-form $df$ maps a tangent vector $\vec{\frac{d}{d\lambda}}$ to the real number $df \left( \vec{\frac{d}{d\lambda}} \right) \equiv \frac{df}{d\lambda}$.
Once you are comfortable with this idea, you will notice that we can introduce a coordinate system $x^i$ and tangent vectors $\frac{\partial}{\partial x^i}$ and one-forms $dx^i$. Note that from our rule, $dx^i \left( \vec{\frac{\partial}{\partial x^j} } \right) = \delta^i_j$.
You can then parameterize your curve with the functions $x^i(\lambda)$. Note that from the chain rule
$\vec{ \frac{d}{d\lambda} } = \frac{\partial x^i}{\partial \lambda} \vec{\frac{\partial}{\partial x^i}}$
and you can use what we've produced so far to show that
$df = \frac{\partial f}{\partial x^i} dx^i$.
When all is said and done, you can prove that
$df \left( \vec{\frac{d}{d\lambda}} \right) = \frac{\partial x^i}{\partial \lambda} \frac{\partial f}{\partial x^j} \delta_i^j = \frac{df}{d\lambda}$
is coordinate independent, as it should be.
From there on, you can define arbitrary tensors as multilinear maps taking $n$ 1-forms and $m$ vectors onto real numbers. The utility of this construction is that it is very geometrical and at the same time not tied to coordinates (abstract). You also never have to wonder which way a thing transforms, because it's always the natural way.
I recommend you pick up a good book on differential geometry for physicists. Geometrical Methods of Mathematical Physics by Schutz is OK, his GR book is probably more useful. The bible by Misner, Thorne and Wheeler goes into great depth into this business and has handy visualizations of n-forms if you are so inclined.
Here is a visualization from
Geometrical Methods of Mathematical Physics by Schutz. The co-vector is here called a "one form". His notation $\langle \tilde{\omega} , \bar{V} \rangle$ is equivalent to $\omega_\alpha V^\alpha$, which you might be used to seeing.
Note that when the magnitude of $\bar{V}$ increases, the arrow gets longer. When the magnitude of $\tilde{\omega}$ increases, the parallel surfaces get closer together.
Physically, vectors and covectors are not meaningfully differet from each other. Mathematically, either can be defined as the dual space of the original, so they have all the same properties. You could switch which one is the arrow and which one the set of parallel lines in that picture. If you choose displacements to be prototypical vectors, then other quantities will be either vectors or covectors depending on how they're related to displacements. For example, velocities are just derivatives of displacements, so they are also vectors. Gradients act on displacements to produce scalars, so they are co-vectors.
Although this is saying the same thing as Lionel's answer and Mark's answer from a different standpoint, another idea that I like in describing the tangent space is to think of the one dimensional $C^1$ space curve (or spacetime curve) within the manifold $M$ as a grounding concept. So our fundamental idea is some function ("A Path" or "A Trail") through the manifold $M$ and centred on some point $p\in M$ which is constant for the present:
$$\sigma:(-\epsilon,\epsilon) \subset \mathbb{R}\to M$$
such that ${\rm d}_t \sigma(t)$ also exists in the same interval $(-\epsilon,\epsilon)$ and such that $\sigma(0) = p\in M$ and $\sigma(\epsilon)\neq p$.
After all one dimensional paths, even if very windy, have made sense to us human and related animals ever since we've needed to find water, food and the way back to our cave!
Then, the tangent space $T_p M$ at point $p\in M$ is the set of equivalence classes of such paths, where we define two such paths $\sigma_1: (-\epsilon,\epsilon)\to M$ and $\sigma_2: (-\epsilon,\epsilon)\to M$ as "equivalent" if their "tangents" are the same at $p$, i.e. if : $\left.{\rm d}_t \sigma_1(t)\right|_{t=0} = \left.{\rm d}_t \sigma_2(t)\right|_{t=0}$. We can then readily define scalar multiples of tangents and additions of tangents: here we must be a little careful because what we are doing of course is implicitly labelling $M$ with one of its atlas's charts so that we are implicity thinking about paths as functions $\sigma:(-\epsilon,\epsilon) \to \mathbb{R}^m$ and their "tangents" ${\rm d}_t \sigma:(-\epsilon,\epsilon) \to \mathbb{R}^m$, where $m$ is the manifold's dimension and this is how we compare paths and declare them to be "equivalent" in the above way. Otherwise, in general, there is of course no notion of the linear operations of scaling and addition in the manifold itself $M$.
So now "contravariant" vectors (or simply plain vectors) are objects that live in such tangent spaces.
Okay, all this is long winded, but my point is that I actually thing of wiggly, squirmy "threads" in families (the latter defined by this equivalence) when I think of tangent vectors and not little arrows. This is something I personally find very helpful as one can imagine something "real" within the manifold itself (and, implicitly through a chart, within our homely and wonted friend $\mathbb{R}^m$) and not simply some idea of "arrows" stuck all over the manifold by some graffiti vandal!
So now, with this concept, we take up Mark's Answer to imagine the one form - or what you're calling a covariant vector (or sometimes covector). Actually, I find the idea of a dual vector space pretty neat, so I generally sit with the mathematician's idea of the one-form. In finite dimensional $\mathbb{R}^m$, a dual vector - a linear functional $\mathbb{R}^m\to\mathbb{R}$ is
always an inner product as in Mark's answer (this assertion is the same as saying that $\mathbb{R}^m$ is a complete metric space) and indeed can be represented by its "components" - the values of the functional for the basis vectors of $T_p M$, with all values in $T_p M$ then following from linearity. So this (co)vector (one form) uniquely defines the vector orthogonal to it (modulo a multiplicative constant). The spacing between the level planes of this linear functional defines the "length" of the covector. if you want to get heavy handed, this is where the Riesz Representation Theorem comes on stage - although you don't need anything like the full strength of this theorem to discuss the ideas here.
Now, if your background is optics like me, you've got a very strong and concrete example of the one form. Namely, the
wave vector $\tilde{k}$. This beast forms inner products $\left<\tilde{k},\,\underset{\sim}{r}\right>$ with position vectors $\underset{\sim}{r}$ to give you the local phase of the plane wave component it represents. The maximum rate of change of phase in radians per metre is the length of the covector $\tilde{k}$.
Indeed, in Minkowsky spacetime, the four-wavevector is a one-form - a covector:
$$\tilde{k} = (\omega,-k_x,-k_t,-k_z);\qquad \omega = \sqrt{k_x^2+k_y^2+k_z^2}\,c$$
Now, to go to arbitrary valence tensors, if you haven't got some of the references in Lionel's or Mark's answers, a great introductory discussion is given in the first chapter of Kip Thorne's physics 136 course which is downloadable from here. He talks about all these ideas in terms linear functionals and "slots" for components, rather like you would go about representing and storing these ideas in computer memory (with a countably infinite word size, of course, to represent real numbers exactly!).
An aside about references: I'm not altogether sure that Schutz's GR book is as good a reference for geometry as it used to be (as Lionel makes out). True, it does still include the discussion of the one form as you and Mark's answer describe them, but stuff like the Lie derivative and much of the other geometrical discussion that used to be in his book has had to make way for expanded chapters of GR experimental evidence and issues. I think Schutz even says something about using his geometry book together with a second reading of his "first course on GR" in the latter's preface. So have a browse carefully at the contents of any book you might be thinking of buying - an older copy of Schutz may fit better with you.
Tensors (or rather tensor fields in case of differential geometry) are very generic and not particularly intuitive objects that can fill a lot of roles - volume elements, endomorphisms, Riemannian metrics are just a few things you can describe with tensors.
However, to get an intuition about co- and contravariance, it's enough to look at tangent vectors and covectors, which
can be visualized and are the building blocks of higher-rank tensors.
In differential geometry as traditionally taught in physics courses (mathematicians stopped doing it this way some decades ago), we
always work in charts (ie local coordinates).
A vector would be a column vector $$ \begin{pmatrix} v^1\\\vdots\\v^n \end{pmatrix} = (v^i)_{i=1}^n=v^i $$ and a covector a row vector $$ \begin{pmatrix} w_1&\cdots&w_n \end{pmatrix} = (w_i)_{i=1}^n=w_i $$ with duality pairing $$ \begin{pmatrix} w^1&\cdots&w^n \end{pmatrix} \begin{pmatrix} v^1\\\vdots\\v^n \end{pmatrix} =\sum_{i=1}^n w_iv^i = w_iv^i $$ As there is in general no global chart, we need to specify transformation laws and make our vectors and covectors equivalence classes with respect to these transformations.
The tranformations are given by the Jacobi matrix of the coordinate switchover and its inverse, which is obviously necessary to keep pairings invariant.
Now, the coordinates of vectors transform opposite to basis vectors - they are
contravariant - whereas the components of covectors transform the same way as basis vectors - they are covariant.
You may encounter the opinion that the difference between vectors and covectors doesn't really matter as in general relativity, you can raise and lower indices as you see fit, which culminates in the idea that there's only a single geometric object - the vector - with co- and contravariant components.
That is in my opinion a pretty bad idea: It only works if there is a distinguished non-degenerate bilinear form available and normally, there's a 'natural' placement of indices due to geometry which might even be relevant to model physical concepts (eg velocity vs momentum in context of Lagrangian and Hamiltonian mechanics).
Besides these technical definitions of vectors and covectors (which
do make some sense geometrically once you introduce principal bundles and associated vector bundles, but that's not something typically done in physics courses), there are of course more geometric ones:
We can consider a vector an equivalence class of curves through the same point with first order contact. Similarly, a covector would an equivalence class of real-valued functions at a given point with first order contact.
If we compose representatives of a covector and a vector, we get a function $\mathbb R\to\mathbb R$ and evaluating its derivative gives us the natural pairing.
There are also abstract definitions available: We can identify vectors with their directional derivatives, ie a vector is a linear functional on the space of real-valued functions that respects the Leibniz rule. For covectors, there's an algebraic definition as the ideal of real-valued functions vanishing at a point factored by the ideal generated by product of such functions.
Instead of coming up with definitions for both vectors and covectors, it's enough to define one of those manually (typically vectors) and define the other one by duality, ie as real-valued linear maps.
Now, if you actually want to
visualize these objects (as in draw meaningful pictures), the obvious representation for vectors is as little arrows in coordinate space. Now, if there's a distinguished non-degenerate bilinear form available, you can represent covectors via their associated vectors (raising its index) and pairing will just be the Euclidean scalar product.
A second way to visualize covectors would be as (oriented) hyperplane fields, with the pairing as described in Mark's answer. This also does not work for arbitrary manifolds - you need a volume form to do so (this is basically a variant of the Hodge dual with the last step from the Wikipedia explanation omitted).
(If you do not have a bilinear or volume form available, you could of course chose an arbitray local one). |
I know how to combine two observed values with different standard deviations into one mean value:
$$ \left(\frac{\sigma_b^2}{\sigma_a^2+\sigma_b^2}\right)x_a + \left(\frac{\sigma_a^2}{\sigma_b^2 + \sigma_a^2}\right)x_b = \mu $$
With a new standard deviation of:
$$ \frac{1}{\sigma_{new}} = \frac{1}{\sigma_a} + \frac{1}{\sigma_b} $$
How do I combine three values, $(x_a, x_b, x_c)$ with three different standard deviations, $(\sigma_a, \sigma_b, \sigma_c)$ into one mean, $\mu$? And what is that value's standard deviation?
Edit, here is some context: I am reading the first chapter of Maybeck (1979) Stochastic models, estimation, and control (pdf). The equations above are 1-3 and 1-4 from page 12. My situation is that there is a red ball 5 ft in front of a robot that has three distance sensors: an IR laser, a visible spectrum laser, and a sonar. Each sensor, no matter how well crafted, will deliver a value estimate of the ball's distance from the sensor that will vary around 5 ft, over time, with a Gaussian distribution. So one sensor might read 4.9 ft, and vary with a std of 0.1 foot, over time. Another is poorly calibrated and reads 5.2 feet with a std of .5 feet, and the last reads 5.14 with a standard of 0.3 feet. (Those are just made up numbers given to illustrate the idea.)
More generically, lets assume there are three distances: x1, x2, and x3, with three standard deviations: sigma1, sigma2, and sigma3. Then you should be able to combine the distances x1, x2, and x3, using the SDs, mathematically to calculate a new mean which will essentially be 5.00001 feet, +/- 0.00002 std. I need the formula that lets me combine all three. |
Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..
There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university
Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.
What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?
Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.
Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P
Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?
Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?
Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"
@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.
Although not the only route, can you tell me something contrary to what I expect?
It's a formula. There's no question of well-definedness.
I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.
It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.
Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.
You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.
@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.
@Eric: If you go eastward, we'll never cook! :(
I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.
@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)
@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.
@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator |
Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..
There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university
Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.
What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?
Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.
Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P
Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?
Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?
Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"
@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.
Although not the only route, can you tell me something contrary to what I expect?
It's a formula. There's no question of well-definedness.
I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.
It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.
Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.
You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.
@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.
@Eric: If you go eastward, we'll never cook! :(
I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.
@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)
@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.
@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator |
A cool generalization of Rédei's method furnishes the following lower bound:
Theorem-lower bound.Let $U\subset \mathbb F_p$ have cardinality $2p$. Then $U$ is contained in the union of two lines or it determines at least $\lfloor\frac{p+4}{3}\rfloor$ rich directions.
The key idea is to consider the
y-derivative of the Rédei polynomial$$ H_{U}(x,y):=\prod_{(a,b)\in U} (x-ay+b) = x^{2p}+h_1(y)x^{2p-1}+\dots+h_{2p}(y)$$and the following proposition on lacunary fully reducible polynomials:
PropositionLet $p\neq 2$ be prime and let $f(x)\in\mathbb F_p[x]$ be a (monic) fully reducible polynomial of degree $2p$. Write $f(x)=x^{2p}+g(x) x^p + h(x)$ with $deg(g)=m$, $deg(h)=n$ and $m,n<p$. Then either $$max(n,m+1)+m>p$$ or $f(x)$ is one of the following: $f(x)=(x^p-x)^2$ or $f(x)=(x-t)^p(x-s)^p$ or $f(x)=(x^p-x)(x-t)^p$ for some $s,t\in\mathbb F_p$.
Fully reducible means that it factors completely, as a product linear factors, over $\mathbb F_p$. The Theorem should be compared to the following Example.
Example-upper bound. For every odd prime $p$ there is a set $P\subseteq \mathbb F_p^2$, with cardinality $|P|=2p$ that determines exactly $(p+3)/2$ rich directions, and which is contained in 3 lines but not in 2 lines.
Remarks The theorem is of course optimal up to multiplicative constant, but I do not know if it is actually close to be sharp. To see if it is, I suggest to think first about this MO problem. Following Szonyi, it is not difficult to get a result valid for $|U|<2p$. Essentially, all is needed is to prove the version of the Proposition corresponding to almost-fully-reducible $f(x)$. The theorem seems to generalize to counting $n$-rich directions, while the proposition suggests interesting generalizations to lacunary almost-fully reducible polynomials of degree $np$. It would be cool to apply them to blocking-set problems. Question: can the Proposition be proved more directly from the corresponding classical result for lacunary polynomials of degree $p$? Last thing: Rédei-s theorem has a proof that avoids lacunary polynomials due to Lovász, L., and A. Schrijver, but I have been unable to generalize their method.
Proofs follow
Proof of the Proposition
Let $F(x) = gcd(f(x),x^p-x)$, then $$F(x)\ |\ xg(x)+h(x)+x^2$$because $xg(x)+h(x)+x^2=f(x)-(x^p-x)(x^p+x+g(x))$ and$$\frac{f(x)}{F(x)}\ |\ f'(x) = g'(x)x^p + h'(x).$$Therefore $f(x) | (g'(x)x^p + h'(x))(xg(x)+h(x)+x^2)$.
Case1: $(g'(x)x^p + h'(x))(xg(x)+h(x)+x^2)\neq 0$.Then $2p\leq m-1+p+N$ with $N:=max(n,m+1,2)$.
Case2: $g'(x)x^p + h'(x)+0$.Then $f(x)=(x^2+g(0)x+h(0))^p=(x-t)^p(x-s)^p$.
Case3: $h(x)=-xg(x)-x^2$.Then $f(x)=(x^p+g(x)+x)(x^p-x)$. By Rédei's proposition on lacunary fully reducible polynomials of degree $p$, we have either $x^p+g(x)+x=(x-t)^p$, or $x^p+g(x)+x=x^p-x$, or $2m\geq p+1$. Here $n=m+1$.
Proof of the TheoremLet $D$ be the set of rich directions determined by $U$. Without loss of generality we may assume that $\infty\in D$ (i.e. that there is a vertical line $X=c$ containing $\geq 3$ points of $U$. See for example Seva's addendum, May 4, 2018). Notice that for fixed slope $y\in \mathbb F_p$ we have that $(x-ay+b)=(x-a'y+b')$ if and only if $y=\tfrac{b-b'}{a-a'}$.
Thus $y\not \in D$ if and only if $$H_U(x,y)=\prod_{k\in\mathbb F_p}(x-k)^2=x^{2p}-2x^{p+1}-x^2.$$In particular for $j\neq p-1,2p-2$ the polynomial $h_j(y)$ vanishes at least at $p+1-|D|$ values of $y$ (i.e. $y\not\in D$). Since $deg(h_j)\leq j$, we have that $h_j(y)\equiv 0$ identically for $j=1,\ldots,p-|D|$.
Now the y-derivative trick to gain vanishingConsider the y-derivative$$ \partial_y H_U(x,y) = h'_1(y)x^{2p-1}+\dots+h'_{2p}(y).$$By Leibniz formula we see that for $y\not \in D$ we have that $H_U(x,y)$ is divisible by $(x-k)$ for all $k\in\mathbb F_p$. In other words$$ H_U(x,y) = (x^p-x)S_y(x) = S_y(x)x^p-S_y(x)x$$for some polynomial with $deg(S_y)\leq p-1$. We deduce for $j=2,\ldots,p-1$ and $y\not \in D$ the equality $h'_j(y)=h'_{j+p-1}(y)$.
Since $h_j(y)\equiv 0$ identically for $j=1,\ldots,p-|D|$, we get that $h_{p+j}$ vanishes at $y\not\in D$ with order 2 for $j=1,\ldots,p-|D|-1$. Therefore $h_{p+j}(y)\equiv 0$ identically as soon as $2(p+1-|D|)\geq p+j+1$. As a consequence, for all $y\in \mathbb F_p$ we have$$ H_U(x,y) = x^{2p} + g(x)x^p+h(x)$$with $deg(g)\leq |D|-1$ and $deg(h)\leq 2(|D|-1)$.
Take $y\in\mathbb F_p$.
$H_U(x,y)=(x^p-x)^2$ when $y\not\in D$; $H_U(x,y)=(x-s)^p(x-t)^p$ when $U$ is the support of the two lines $Y=yX+s$, $Y=yX+t$; $H_U(x,y)=(x^p-x)(x-t)^p$ means in particular that $p$ points of $U$ are on the line $Y=yX+t$.
It is not difficult to show that, if $U$ is not supported on two lines only, then here is a line containing at least 3 points of $U$ but less than $p$ points. Such line corresponds to $y=y_0\in D$ for which 1-2-3 above do not hold.
By the Proposition applied to $f(x)=H_U(x,y_0)$ we get $3(|D|-1)\geq p+1$, so the Theorem.
Proof of the Example. Let $P'=\{(k,k^{\frac{p+1}2}): k\in\mathbb F_p\}$, which is a set contained in the two lines $y=\pm x$ with cardinality $|P'|=p$. It is shown in [1] that $P'$ attains the equality in the theorem of Rédei. In other words, $P'$ determines exactly $(|P'|+3)/2=(p+3)/2$ directions.
Now let $P''=\{(k,k-2): k\in\mathbb F_p\}$, i.e the line parallel to $y=x$ passing through $(1,-1)$. We observe that $P'\cap P''=\emptyset$ and that the direction of $P''$ is one of the directions determined by $P'$.
Finally we let $P=P'\cup P''$ and we observe immediately that the rich directions of $P$ are exactly the directions determined by $P'$.
[1] Lovász, L., and A. Schrijver. "Remarks on a theorem of Rédei." Studia Scient. Math. Hungar 16.449-454 (1981): 15-32. |
A good theoretical analysis of with and without replacement schemas in the context of iterative algorithms based on random draws (which are how many discriminative Deep Neural Networks (DNNs) are trained against) can be found here
In short, it turns out that sampling
without replacement, leads to faster convergence than sampling with replacement.
I will give a short analysis here based on the toy example that they provide: Let's say that we want to optimize the following objective function:
$$x_{\text{opt}} = \underset{x}{\arg\min} \frac{1}{2} \sum_{i=1}^{N}(x - y_i)^2$$
where the target $y_i \sim \mathcal{N}(\mu, \sigma^2)$. In this example, we are trying to solve for the optimal $x$, given $N$ labels of $y_i$ obviously.
Ok, so if we were to solve for the optimal $x$ in the above directly, then we would take the derivative of the loss function here, set it to 0, and solve for $x$. So for our example above, the loss is
$$L = \frac{1}{2} \sum_{i=1}^{N}(x - y_i)^2$$
and it's first derivative would be:
$$ \frac{\delta L}{\delta x} = \sum_{i=1}^{N}(x - y_i)$$
Setting $ \frac{\delta L}{\delta x}$ to 0 and solving for $x$, yields:
$$x_{\text{opt}} = \frac{1}{N} \sum_{i=1}^{N} y_i$$
In other words, the optimal solution is nothing but the sample mean of all the $N$ samples of $y$.
Now, if we couldn't perform the above computation all at once, we would have to do it recursively, via the gradient descent update equation below:
$$x_i = x_{i-1} - \lambda_i \nabla(f(x_{i-1}))$$
and simply inserting our terms here yields:
$$x_{i} = x_{i-1} - \lambda_i (x_{i-1} - y_{i}) $$
If we run the above for all $i \in {1, 2, ... N}$, then we are effectively performing this update
without replacement. The question then becomes, can we get also get the optimal value of $x$ in this way? (Remember that the optimal value of $x$ is nothing but the sample mean of $y$). The answer is yes, if you let $\lambda_i = 1/i$. To see, this we expand:
$$x_{i} = x_{i-1} - \lambda_i (x_{i-1} - y_{i}) \\\x_{i} = x_{i-1} - \frac{1}{i} (x_{i-1} - y_{i}) \\\x_{i} = \frac{i x_{i-1} - (x_{i-1} - y_{i})}{i} \\\x_{i} = \frac{(i - 1)x_{i-1} + y_{i}}{i} \\\i x_{i} = (i - 1)x_{i-1} + y_{i} \\\$$
The last equation however is nothing but the formula for the running average! Thus as we loop through the set from $i=1$, $i=2$, etc, all the way to $i=N$, we would have performed our updates
without replacement, and our update formula gives us the optimal solution of $x$, which is the sample mean!
$$N x_{N} = (N - 1)x_{N-1} + y_{N} ==> x_N = \frac{1}{N}\sum_{i=1}^{N} y_i = \mu$$
In contrast however, if we actually drew
with replacement, then while our draws would then be truly independent, the optimized value $x_N$ would be different from the (optimal) mean $\mu$, and the square error would be given by:
$$\mathop{E}\{(x_N - \mu)^2\}$$
which is going to be a positive value, and this simple toy example can be extended to higher dimensions. This has the consequence that we would want to perform sampling without replacement as a more optimal solution.
Hope this clarifies it some more! |
Learning Outcomes
Solve linear equations for the variable.
It is a common task in algebra to solve an equation for a variable. The goal will be to get the variable on one side of the equation all by itself and have the other side of the equation just be a number. The process will involve identifying the operations that are done on the variable and apply the inverse operation to both sides of the equation. This will be managed in the reverse of the order of operations.
Example \(\PageIndex{1}\)
Solve the following equation for \(x\).
\[3x+4=11 \label{EQ1.1}\]
Solution
We begin by looking at the operations that are done to \(x\), keeping track the order. The first operation is "multiply by 3" and the second is "add 4". We now do everything backwards. Since the last operation is "add 4", our first step is to subtract 4 from both sides of Equation \ref{EQ1.1}.
\[3x \cancel{+ 4} \color{Cerulean}{ \cancel{-4}} \color{black} =11 \color{Cerulean}{ -4} \nonumber\]
which simplifies the equation
\[3x = 7 \nonumber\]
Next, the way to undo "multiply by 3" is to divide both sides by 3. We get
\[ \dfrac{\cancel{3}x}{\color{Cerulean}{\cancel{3}}} \color{black}= \dfrac{7}{\color{Cerulean}{3}} \nonumber\]
or
\[x=\dfrac{7}{3} \nonumber\]
Example \(\PageIndex{2}\)
The rectangle above is a diagram for a uniform distribution from 2 to 9 that asks for the first quartile. The area of the smaller red rectangle that has base from 2 to Q1 and height 1/7 is 1/4. Find Q1.
Solution
We start by using the area formula for a rectangle:
\[\text{Area} = \text{Base} \times \text{Height} \label{EQ1}\]
We have:
Area = \(\frac{1}{4}\) Base = \(Q1-2\) Height = \(\frac{1}{7}\)
Plug this into Equation \ref{EQ1} to get:
\[\frac{1}{4}=\left(Q1-2\right)\left(\frac{1}{7}\right) \label{EQ2}\]
We need to solve for \(Q1\). First multiple both sides of Equation \ref{EQ2} by 7 to get:
\[ \begin{align} \color{Cerulean}{7} \color{black} \left(\dfrac{1}{4}\right) &= \color{Cerulean}{\cancel{7}} \color{black} \left(Q1-2\right) \cancel{ \left(\frac{1}{7}\right)} \nonumber \\[5pt] \dfrac{7}{4} &=Q1-2 \label{EQ4} \end{align}\]
Now add 2 to both sides of Equation \ref{EQ4} to get:
\[ \begin{align*} \dfrac{7}{4} \color{Cerulean} +2 \color{black} & =Q1 \cancel{-2} \color{Cerulean}{\cancel{+2}} \\[5pt] \dfrac{7}{4}+2&=Q1 \end{align*}\]
or
\[Q1=\frac{7}{4}+2 \nonumber\]
Putting this into a calculator gives:
\[Q1=3.75 \nonumber\]
Example \(\PageIndex{3}\): z-score
The z-score for a given value \(x\) for a distribution with population mean \(\mu \) and population standard deviation \(\sigma \) is given by:
\[z=\frac{x-\mu}{\sigma} \nonumber\]
An online retailer has found that the population mean sales per day is $2,841 and the population standard deviation is $895. A value of \(x\) is considered an outlier if the z-score is less than -2 or greater than 2. How many sales must be made to have a z-score of 2?
Solution
First we identify each of the given variables. Since the population mean is 2,841, we have:
\[\mu=2841 \nonumber \]
We are told that the population standard deviation is 895 meters, so:
\[\sigma=895 \nonumber \]
We are also given that the z-score is 2, hence:
\[z=2 \nonumber \]
Now we put the numbers into the formula for the z-score to get:
\[2=\frac{x-2841}{895} \nonumber \]
We can next switch the order of the equation so that the \(x\) is on the left hand side of the equation:
\[\frac{x-2841}{895}=2 \nonumber \]
Next, we solve for \(x\). First multiply both sides of the equation by 895 to get
\[x-2841=2\left(895\right)=1790 \nonumber \]
Finally, we can add 2841 to both sides of the equation to get \(x\) by itself:
\[x=1790+2841=4631 \nonumber \]
We can conclude that if the day's sales is at $4631, the z-score is 2.
Exercise
The rectangle below is a diagram for a uniform distribution from 5 to 11 that asks for the 72
nd percentile. The area of the smaller red rectangle that has base from 5 to the 72 nd percentile, \(x\), and height 1/6 is 0.72. Find \(x\). |
A Fourth-Order-Centered Finite-Volume Scheme for Regular Hexagonal Grids
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A Hybrid Method for a Countable Family of Multivalued Maps, Equilibrium Problems, and Variational Inequality Problems. Watcharaporn Cholamjiak; Suthep Suantai // Discrete Dynamics in Nature & Society;2010, Special section p1
We introduce a new monotone hybrid iterative scheme for finding a common element of the set of common fixed points of a countable family of non expansive multi valued maps, the set of solutions of variational inequality problem, and the set of the solutions of the equilibrium problem in a...
Theoretical Convergence Analysis of a General Division–deletion Algorithm for Solving Global Search Problems. X. Yang; M. Sun // Journal of Global Optimization;Jan2007, Vol. 37 Issue 1, p27
Abstract??Presented in this paper is the prototype of a very general algorithm referred to as Division ? Deletion Algorithm (DDA) for solving the most general global search problem. Various necessary conditions, sufficient conditions, and necessary and sufficient conditions for the convergence...
The Truncated q-Bernstein Polynomials in the Case q>1. Turan, Mehmet // Abstract & Applied Analysis;2014, p1
The truncated q-Bernstein polynomials Bn,m,q(f; x) n ∈ ℕ, and m ∈ ℕ0 emerge naturally when the q-Bernstein polynomials of functions vanishing in some neighbourhood of 0 are considered. In this paper, the convergence of the truncated q-polynomials on [0, 1] is studied. To...
monotone convergence theorem (early 20th century) Mathematics. // Dictionary of Theories;2002, p351
Information on the monotone convergence theorem is presented. The theorem was proved by French mathematician Henri-Leon Lebesgue for the case where the limit function is integrable. The theorem is an increasing sequence of non-integrable measurable functiond which converghes almost everywhere to... |
Article
Keywords: ${\omega }$-covering set; ${\mathcal E}$; hereditarily nonparadoxical set
Summary: We prove the following theorems: There exists an ${\omega }$-covering with the property $s_0$. Under $\mathop {\mathrm cov}\nolimits ({\mathcal N}) = $ there exists $X$ such that $ \forall _{B \in {\mathcal B}or} [B\cap X$ is not an ${\omega }$-covering or $X\setminus B$ is not an ${\omega }$-covering]. Also we characterize the property of being an ${\omega }$-covering.
References:
[BJ] T. Bartoszyński, H. Judah:
Borel images of sets of reals
. Real Anal. Exchange 20(2) (1994/5), 536–558. MR 1348078
[E] R. Engelking:
General Topology, Revised and Completed Edition. Sigma Series in Pure Mathematics, vol. 6
. Heldermann Verlag, Berlin, 1989. MR 1039321
[K1] P. Komjáth:
Large small sets
. Colloq. Math. 56 (1988), 231–233. MR 0991209
[K2] P. Komjáth:
Some remarks on second category sets
. Colloq. Math. 66 (1993), 57–62. MR 1242645
[M] K. Muthuvel:
Certain measure zero, first category sets
. Real Anal. Exchange 17 (1991–92), 771–774. MR 1171418 |
Suppose I take the $0$-skeleton $X_0$ to have a single inhabitant $base$.
Let $S_1$ have a single point $base'$, with attaching map $f(base', -)=base$. If I construct the pushout to the 1-skeleton $X_1$, as below, $$ \require{AMScd} \begin{CD} S_1 \times \mathbb{S}^0 @>{f}>> X_0;\\ @VVV @VVV \\ \mathbf{1} @>>> X_1; \end{CD} $$ I end up with $X_1$ defined by
$inl: \mathbf{1} \rightarrow X_1$ $inr: X_0 \rightarrow X_1 $ for each $c:S_1 \times \mathbb{S}^0$, a path $inl(\star) = inr(base)$
This is the interval type though! Where have I gone wrong in my construction? |
Regression (OLS) - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Regression (OLS)
$z$ test for the difference between two proportions
Paired sample $t$ test
Sign test
One sample $z$ test for the mean
Independent variables Independent variable Independent variable Independent variable Independent variable One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variables One categorical with 2 independent groups 2 paired groups 2 paired groups None Dependent variable Dependent variable Dependent variable Dependent variable Dependent variable One quantitative of interval or ratio level One categorical with 2 independent groups One quantitative of interval or ratio level One of ordinal level One quantitative of interval or ratio level Null hypothesis Null hypothesis Null hypothesis Null hypothesis Null hypothesis $F$ test for the complete regression model: $\pi_1 = \pi_2$
$\pi_1$ is the unknown proportion of "successes" in population 1; $\pi_2$ is the unknown proportion of "successes" in population 2
$\mu = \mu_0$
$\mu$ is the unknown population mean of the difference scores; $\mu_0$ is the population mean of the difference scores according to the null hypothesis, which is usually 0
$\mu = \mu_0$
$\mu$ is the unknown population mean; $\mu_0$ is the population mean according to the null hypothesis
Alternative hypothesis Alternative hypothesis Alternative hypothesis Alternative hypothesis Alternative hypothesis $F$ test for the complete regression model: Two sided: $\pi_1 \neq \pi_2$
Right sided: $\pi_1 > \pi_2$
Left sided: $\pi_1 < \pi_2$
Two sided: $\mu \neq \mu_0$
Right sided: $\mu > \mu_0$
Left sided: $\mu < \mu_0$
Two sided: $\mu \neq \mu_0$
Right sided: $\mu > \mu_0$
Left sided: $\mu < \mu_0$
Assumptions Assumptions Assumptions Assumptions Assumptions all individuals in the population. Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another Test statistic Test statistic Test statistic Test statistic Test statistic $F$ test for the complete regression model:
Note 2: if only one independent variable ($K = 1$), the $F$ test for the complete regression model is equivalent to the two sided $t$ test for $\beta_1$
$z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
$p_1$ is the sample proportion of successes in group 1: $\dfrac{X_1}{n_1}$, $p_2$ is the sample proportion of successes in group 2: $\dfrac{X_2}{n_2}$, $p$ is the total proportion of successes in the sample: $\dfrac{X_1 + X_2}{n_1 + n_2}$, $n_1$ is the sample size of group 1, $n_2$ is the sample size of group 2
Note: we could just as well compute $p_2 - p_1$ in the numerator, but then the left sided alternative becomes $\pi_2 < \pi_1$, and the right sided alternative becomes $\pi_2 > \pi_1$
$t = \dfrac{\bar{y} - \mu_0}{s / \sqrt{N}}$
$\bar{y}$ is the sample mean of the difference scores, $\mu_0$ is the population mean of the difference scores according to H0, $s$ is the sample standard deviation of the difference scores, $N$ is the sample size (number of difference scores).
The denominator $s / \sqrt{N}$ is the standard error of the sampling distribution of $\bar{y}$. The $t$ value indicates how many standard errors $\bar{y}$ is removed from $\mu_0$
$W = $ number of difference scores that is larger than 0 $z = \dfrac{\bar{y} - \mu_0}{\sigma / \sqrt{N}}$
$\bar{y}$ is the sample mean, $\mu_0$ is the population mean according to H0, $\sigma$ is the population standard deviation, $N$ is the sample size.
The denominator $\sigma / \sqrt{N}$ is the standard deviation of the sampling distribution of $\bar{y}$. The $z$ value indicates how many of these standard deviations $\bar{y}$ is removed from $\mu_0$
Sample standard deviation of the residuals $s$ n.a. n.a. n.a. n.a. $\begin{aligned} s &= \sqrt{\dfrac{\sum (y_j - \hat{y}_j)^2}{N - K - 1}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned} $ - - - - Sampling distribution of $F$ and of $t$ if H0 were true Sampling distribution of $z$ if H0 were true Sampling distribution of $t$ if H0 were true Sampling distribution of $W$ if H0 were true Sampling distribution of $z$ if H0 were true Sampling distribution of $F$: Approximately standard normal $t$ distribution with $N - 1$ degrees of freedom The exact distribution of $W$ under the null hypothesis is the Binomial($n$, $p$) distribution, with $n =$ number of positive differences $+$ number of negative differences, and $p = 0.5$.
If $n$ is large, $W$ is approximately normally distributed under the null hypothesis, with mean $np = n \times 0.5$ and standard deviation $\sqrt{np(1-p)} = \sqrt{n \times 0.5(1 - 0.5)}$. Hence, if $n$ is large, the standardized test statistic $$z = \frac{W - n \times 0.5}{\sqrt{n \times 0.5(1 - 0.5)}}$$ follows approximately a standard normal distribution if the null hypothesis were true.
Standard normal Significant? Significant? Significant? Significant? Significant? $F$ test: Two sided: Two sided: If $n$ is small, the table for the binomial distribution should be used:
Two sided:
If $n$ is large, the table for standard normal probabilities can be used:
Two sided:
Two sided: $C\%$ confidence interval for $\beta_k$ and for $\mu_y$; $C\%$ prediction interval for $y_{new}$ Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$ $C\%$ confidence interval for $\mu$ n.a. $C\%$ confidence interval for $\mu$ Confidence interval for $\beta_k$: Regular (large sample): $\bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}}$
where the critical value $t^*$ is the value under the $t_{N-1}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20)
The confidence interval for $\mu$ can also be used as significance test.
- $\bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}}$
where $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval)
The confidence interval for $\mu$ can also be used as significance test.
Effect size n.a. Effect size n.a. Effect size Complete model: - Cohen's $d$:
Standardized difference between the sample mean of the difference scores and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{s}$$ Indicates how many standard deviations $s$ the sample mean of the difference scores $\bar{y}$ is removed from $\mu_0$
- Cohen's $d$:
Standardized difference between the sample mean and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{\sigma}$$ Indicates how many standard deviations $\sigma$ the sample mean $\bar{y}$ is removed from $\mu_0$
n.a. n.a. Visual representation n.a. Visual representation - - - ANOVA table n.a. n.a. n.a. n.a. - - - - n.a. Equivalent to Equivalent to Equivalent to n.a. - When testing two sided: chi-squared test for the relationship between two categorical variables, where both categorical variables have 2 levels One sample $t$ test on the difference scores
Repeated measures ANOVA with one dichotomous within subjects factor
Two sided sign test is equivalent to - Example context Example context Example context Example context Example context Can mental health be predicted from fysical health, economic class, and gender? Is the proportion smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic. Is the average difference between the mental health scores before and after an intervention different from $\mu_0$ = 0? Do people tend to score higher on mental health after a mindfulness course? Is the average mental health score of office workers different from $\mu_0$ = 50? Assume that the standard deviation of the mental health scores in the population is $\sigma$ = 3. SPSS SPSS SPSS SPSS n.a. Analyze > Regression > Linear... SPSS does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Analyze > Descriptive Statistics > Crosstabs...
Analyze > Compare Means > Paired-Samples T Test... Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples... - Jamovi Jamovi Jamovi Jamovi n.a. Regression > Linear Regression Jamovi does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Frequencies > Independent Samples - $\chi^2$ test of association
T-Tests > Paired Samples T-Test Jamovi does not have a specific option for the sign test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the two sided $p$ value that would have resulted from the sign test. Go to:
ANOVA > Repeated Measures ANOVA - Friedman
- Practice questions Practice questions Practice questions Practice questions Practice questions |
My best guess is that the series $$ \sum_{i=1}^n i(n-(i-1)) $$ becomes $$ 2 \Bigg[ n + 2(n-1) + ... + \frac{n}{2} \bigg(n-\bigg(\frac{n}{2}-1\bigg)\Bigg)\Bigg] $$ So the highest term is $n^2$ and there are $n$ terms. Does that mean its $O(n^3)$? That seems high. Intuitively, it seems like it should be closer to $O(n^2)$ but I can't find a way to bring it down mathematically.
We can lower bound the sum roughly by $$ \sum_{i=1}^n i(n+1-i) \geq \sum_{i=n/3}^{2n/3} (n/3)^2 \geq (n/3)^3. $$ This shows that it is $\Omega(n^3)$. Since each summand is at most $n^2$, the sum is also $O(n^3)$.
We can also compute the sum explicitly: $$ \sum_{i=1}^n i(n+1-i) = (n+1) \sum_{i=1}^n i - \sum_{i=1}^n i^2 = (n+1) \frac{(n+1)n}{2} - \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(n+2)}{6}. $$ This expression equals the binomial coefficient $\binom{n+2}{3}$, and we can prove this bijectively. The binomial coefficient counts triples $(a,b,c)$ of elements satisfying $0 \leq a < b < c \leq n+1$. Notice that $1 \leq b \leq n$, and this corresponds to summation over $i$. Given $b$, we have $b$ choices for $a$ and $n+1-b$ choices for $c$.
$$\begin{align*} T(n)&=1n + 2(n-1) + 3(n-2) + ... + (n-1)2 + n\\ &=n(1+2+...+n)-(1+2+3+...+n)\\ &=(n-1)(1+2+3+...+n)\\ &=\Theta(n^3) \end{align*}$$
From ryan's comment, the summation, when you distribute $i$ and split it into more summations comes to $$\sum_{i=1}^n i(n-(i-1)) = \sum_{i=1}^n (i*n) + \sum_{i=1}^n (-(i^2)-i)$$ which is roughly $\frac{1}{2}n^3 - \frac{1}{3}n^3$ so the asymptotic bound is, in fact $O(n^3)$ |
Given a real symmetric matrix $M$, ostensibly of "low rank", efficiently find an expression $M = \sum \alpha_i u_i u_i^T$ using the number of terms rank($M$).
A 2011 StackOverflow Question Dense Cholesky Update in Python asked about doing low rank updates to Cholesky decompositions, but was answered largely with ways to do rank 1 updates in Python.
Naturally a rank one symmetric term would have the form $\alpha u u^T$, and if we have a rank two symmetric update of the form $uv^T + vu^T$, then we can rewrite it:
$$ uv^T + vu^T = \frac{1}{2} ((u+v)(u+v)^T - (u-v)(u-v)^T) $$
but this does not seem to lead to an efficient general method for similarly expanding an arbitrary symmetric real matrix into as many such terms as the rank requires.
Existence of such an expresssion is guaranteed by eigenbasis expansion:
$$ M = \sum \alpha_i u_i u_i^T $$
where the $u_i$ are an orthonormal basis of eigenvectors with corresponding eigenvalues $\alpha_i$ (and the limitation to as many terms as the rank requires is met by ignoring zero eigenvalues). But a full eigenbasis expansion would call for a disproportionate amount of work relative to the Cholesky updates.
We could easily expand a rank $k$ matrix as a sum of $k$ not necessarily symmetric rank one terms and use the symmetry of the sum to rewrite it as $2k$ symmetric rank one terms:
$$ \sum u_i v_i^T = \frac{1}{2} \sum (u_i v_i^T + v_i u_i^T) = \frac{1}{4} \sum ((u_i+v_i)(u_i+v_i)^T - (u_i-v_i)(u_i-v_i)^T) $$
But this uses twice the number of rank one updates as ought to be necessary. |
I am trying to show the asymptotic expansion for $$\sum_{n\le x}\frac1{\sqrt n}=2\sqrt x+\zeta(1/2)+O(x^{-1/2}).$$
(The exact identity of the zeta term is not important, it need only be some $c$.) To that end, I am attempting to prove the following slightly stronger theorem, which is supported by numerical evidence:
Let $F(x)=2\sqrt x-\sum_{n\le x}\frac1{\sqrt n}$. If $0<a\le b$, then $|F(a)-F(b)|\le\frac1{\sqrt a}$.
So far, I have only been able to show the weaker statement $|F(a)-F(b)|\le\dfrac2{\sqrt{\lfloor a\rfloor}}$, which is good enough for the asymptotic statement but can clearly be improved to work for all positive reals. My method:
$$\sum_{a<n\le b}\int_{n-1}^{n}(t^{-1/2}-n^{-1/2})\,dt=\sum_{a<n\le b}\left[2\sqrt n-2\sqrt{n-1}-\frac1{\sqrt n}\right]=F(\lfloor b\rfloor)-F(\lfloor a\rfloor)$$
Since the integrand is less than $\frac1{\sqrt{n-1}}-\frac1{\sqrt{n}}$ on its domain, we have:
$$|F(\lfloor b\rfloor)-F(\lfloor a\rfloor)|\le\sum_{a<n\le b}\frac1{\sqrt{n-1}}-\frac1{\sqrt{n}}=\frac1{\sqrt{\lfloor a\rfloor}}-\frac1{\sqrt{\lfloor b\rfloor}}$$
Finally, $F$ is differentiable on $(\lfloor x\rfloor,x)$ with $F'(y)=y^{-1/2}\le\lfloor x\rfloor^{-1/2}$, hence $|F(x)-F(\lfloor x\rfloor)|\le\frac1{\sqrt{\lfloor x\rfloor}}$. Putting it all together, we have
\begin{align} |F(b)-F(a)|&\le|F(a)-F(\lfloor a\rfloor)|+|F(b)-F(\lfloor b\rfloor)|+|F(\lfloor b\rfloor)-F(\lfloor a\rfloor)|\\ &\le\frac1{\sqrt{\lfloor a\rfloor}}+\frac1{\sqrt{\lfloor b\rfloor}}+\frac1{\sqrt{\lfloor a\rfloor}}-\frac1{\sqrt{\lfloor b\rfloor}}=\frac2{\sqrt{\lfloor a\rfloor}}. \end{align}
How can I avoid the use of $\frac1{\sqrt{\lfloor a\rfloor}}$, which is not even defined for $a<1$? |
To solve the Poisson equation for the Newton Potential, say $\phi$, one can use the divergence theorem, such that $$\int_U \nabla^2 \phi \sqrt{g}~ \mathrm dV= \int_{\partial U} \langle \nabla \phi,n\rangle \sqrt{g_\Sigma}~ \mathrm d\Sigma=1,$$ where $\sqrt{g_\Sigma}$ represents the induced metric on the border.
Considering that $\phi$ is classic it will only depend on the relative position between the source and the point where we want to compute the potential, so, in order to use the divergence theorem, we need to find the set of points that are at the same distance of a given point (where we consider the source of the field to be) and that will be the border, $\partial U$, of the considered volume that contains the source.
Well, if space is the three dimensional Euclidean space then the volume that contains the source is straightforwardly the 3-ball whose center is at the source position and the border $\partial U$ the corresponding 2-sphere.
But if space is a different 3 Riemannian Manifold things are a "little" bit harder and finding the set of points that are at the same (geodesic) distance from the source is in general quite difficult or even impossible... But when searching for the geodesic curves on a manifold one can, in some circumstances, instead of searching for the curve that minimizes the distance
$$d~=~\int_a^b \left[g\left(\dot{c}(s),\dot{c}(s)\right)\right]^{1/2}~\mathrm ds,$$
search for the curve that minimizes the energy functional
$$E~=~\int_a^b \left[g\left(\dot{c}(s),\dot{c}(s)\right)\right] ~\mathrm ds,$$
since in those circunstances the curves that minimize the energy are the same as the curves that minimize the distance.
Well, finding the set of points that are at the same energy is much easier...I would then ask: what is the significance of the energy defined above and would it make sense, physically, to say that equipotential surfaces are not the set of points that at the same distance from the source but the set of points that have the same energy.
PS: Note that in the Euclidean Space, the set of points that are at the same distance is the same as the set of points that are at the same energy. |
Regression (OLS) - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Regression (OLS)
$z$ test for the difference between two proportions
Paired sample $t$ test
Sign test
Paired sample $t$ test
Independent variables Independent variable Independent variable Independent variable Independent variable One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variables One categorical with 2 independent groups 2 paired groups 2 paired groups 2 paired groups Dependent variable Dependent variable Dependent variable Dependent variable Dependent variable One quantitative of interval or ratio level One categorical with 2 independent groups One quantitative of interval or ratio level One of ordinal level One quantitative of interval or ratio level Null hypothesis Null hypothesis Null hypothesis Null hypothesis Null hypothesis $F$ test for the complete regression model: $\pi_1 = \pi_2$
$\pi_1$ is the unknown proportion of "successes" in population 1; $\pi_2$ is the unknown proportion of "successes" in population 2
$\mu = \mu_0$
$\mu$ is the unknown population mean of the difference scores; $\mu_0$ is the population mean of the difference scores according to the null hypothesis, which is usually 0
$\mu = \mu_0$
$\mu$ is the unknown population mean of the difference scores; $\mu_0$ is the population mean of the difference scores according to the null hypothesis, which is usually 0
Alternative hypothesis Alternative hypothesis Alternative hypothesis Alternative hypothesis Alternative hypothesis $F$ test for the complete regression model: Two sided: $\pi_1 \neq \pi_2$
Right sided: $\pi_1 > \pi_2$
Left sided: $\pi_1 < \pi_2$
Two sided: $\mu \neq \mu_0$
Right sided: $\mu > \mu_0$
Left sided: $\mu < \mu_0$
Two sided: $\mu \neq \mu_0$
Right sided: $\mu > \mu_0$
Left sided: $\mu < \mu_0$
Assumptions Assumptions Assumptions Assumptions Assumptions all individuals in the population. Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another all individuals in the population. Test statistic Test statistic Test statistic Test statistic Test statistic $F$ test for the complete regression model:
Note 2: if only one independent variable ($K = 1$), the $F$ test for the complete regression model is equivalent to the two sided $t$ test for $\beta_1$
$z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
$p_1$ is the sample proportion of successes in group 1: $\dfrac{X_1}{n_1}$, $p_2$ is the sample proportion of successes in group 2: $\dfrac{X_2}{n_2}$, $p$ is the total proportion of successes in the sample: $\dfrac{X_1 + X_2}{n_1 + n_2}$, $n_1$ is the sample size of group 1, $n_2$ is the sample size of group 2
Note: we could just as well compute $p_2 - p_1$ in the numerator, but then the left sided alternative becomes $\pi_2 < \pi_1$, and the right sided alternative becomes $\pi_2 > \pi_1$
$t = \dfrac{\bar{y} - \mu_0}{s / \sqrt{N}}$
$\bar{y}$ is the sample mean of the difference scores, $\mu_0$ is the population mean of the difference scores according to H0, $s$ is the sample standard deviation of the difference scores, $N$ is the sample size (number of difference scores).
The denominator $s / \sqrt{N}$ is the standard error of the sampling distribution of $\bar{y}$. The $t$ value indicates how many standard errors $\bar{y}$ is removed from $\mu_0$
$W = $ number of difference scores that is larger than 0 $t = \dfrac{\bar{y} - \mu_0}{s / \sqrt{N}}$
$\bar{y}$ is the sample mean of the difference scores, $\mu_0$ is the population mean of the difference scores according to H0, $s$ is the sample standard deviation of the difference scores, $N$ is the sample size (number of difference scores).
The denominator $s / \sqrt{N}$ is the standard error of the sampling distribution of $\bar{y}$. The $t$ value indicates how many standard errors $\bar{y}$ is removed from $\mu_0$
Sample standard deviation of the residuals $s$ n.a. n.a. n.a. n.a. $\begin{aligned} s &= \sqrt{\dfrac{\sum (y_j - \hat{y}_j)^2}{N - K - 1}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned} $ - - - - Sampling distribution of $F$ and of $t$ if H0 were true Sampling distribution of $z$ if H0 were true Sampling distribution of $t$ if H0 were true Sampling distribution of $W$ if H0 were true Sampling distribution of $t$ if H0 were true Sampling distribution of $F$: Approximately standard normal $t$ distribution with $N - 1$ degrees of freedom The exact distribution of $W$ under the null hypothesis is the Binomial($n$, $p$) distribution, with $n =$ number of positive differences $+$ number of negative differences, and $p = 0.5$.
If $n$ is large, $W$ is approximately normally distributed under the null hypothesis, with mean $np = n \times 0.5$ and standard deviation $\sqrt{np(1-p)} = \sqrt{n \times 0.5(1 - 0.5)}$. Hence, if $n$ is large, the standardized test statistic $$z = \frac{W - n \times 0.5}{\sqrt{n \times 0.5(1 - 0.5)}}$$ follows approximately a standard normal distribution if the null hypothesis were true.
$t$ distribution with $N - 1$ degrees of freedom Significant? Significant? Significant? Significant? Significant? $F$ test: Two sided: Two sided: If $n$ is small, the table for the binomial distribution should be used:
Two sided:
If $n$ is large, the table for standard normal probabilities can be used:
Two sided:
Two sided: $C\%$ confidence interval for $\beta_k$ and for $\mu_y$; $C\%$ prediction interval for $y_{new}$ Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$ $C\%$ confidence interval for $\mu$ n.a. $C\%$ confidence interval for $\mu$ Confidence interval for $\beta_k$: Regular (large sample): $\bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}}$
where the critical value $t^*$ is the value under the $t_{N-1}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20)
The confidence interval for $\mu$ can also be used as significance test.
- $\bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}}$
where the critical value $t^*$ is the value under the $t_{N-1}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20)
The confidence interval for $\mu$ can also be used as significance test.
Effect size n.a. Effect size n.a. Effect size Complete model: - Cohen's $d$:
Standardized difference between the sample mean of the difference scores and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{s}$$ Indicates how many standard deviations $s$ the sample mean of the difference scores $\bar{y}$ is removed from $\mu_0$
- Cohen's $d$:
Standardized difference between the sample mean of the difference scores and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{s}$$ Indicates how many standard deviations $s$ the sample mean of the difference scores $\bar{y}$ is removed from $\mu_0$
n.a. n.a. Visual representation n.a. Visual representation - - - ANOVA table n.a. n.a. n.a. n.a. - - - - n.a. Equivalent to Equivalent to Equivalent to Equivalent to - When testing two sided: chi-squared test for the relationship between two categorical variables, where both categorical variables have 2 levels One sample $t$ test on the difference scores
Repeated measures ANOVA with one dichotomous within subjects factor
Two sided sign test is equivalent to One sample $t$ test on the difference scores
Repeated measures ANOVA with one dichotomous within subjects factor
Example context Example context Example context Example context Example context Can mental health be predicted from fysical health, economic class, and gender? Is the proportion smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic. Is the average difference between the mental health scores before and after an intervention different from $\mu_0$ = 0? Do people tend to score higher on mental health after a mindfulness course? Is the average difference between the mental health scores before and after an intervention different from $\mu_0$ = 0? SPSS SPSS SPSS SPSS SPSS Analyze > Regression > Linear... SPSS does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Analyze > Descriptive Statistics > Crosstabs...
Analyze > Compare Means > Paired-Samples T Test... Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples... Analyze > Compare Means > Paired-Samples T Test... Jamovi Jamovi Jamovi Jamovi Jamovi Regression > Linear Regression Jamovi does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Frequencies > Independent Samples - $\chi^2$ test of association
T-Tests > Paired Samples T-Test Jamovi does not have a specific option for the sign test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the two sided $p$ value that would have resulted from the sign test. Go to:
ANOVA > Repeated Measures ANOVA - Friedman
T-Tests > Paired Samples T-Test Practice questions Practice questions Practice questions Practice questions Practice questions |
In econometrics, fixed effects binary choice models are important tools for panel data analysis. Our package provides an approach suggested by Stammann, Heiss, and McFadden (2016) to estimate logit and probit panel data models of the following form:
\[ y_{it} = \mathbf{1}\left[\mathbf{x}_{it}\boldsymbol{\beta} + \alpha_{i} > \epsilon_{it}\right] \;, \]
where \(i = 1, \dots, N\) and \(t = 1, \dots, T_i\) denote different panel indices. In many applications, \(i\) represents individuals, firms or other cross-sectional units and \(t\) represents time in a longitudinal data set. But the setup is also useful for instance if \(i\) represents ZIP code areas and \(t\) is an index of individuals.
We are primarily interested in estimating the parameters \(\boldsymbol{\beta}\), but the model also includes individual fixed effects \(\alpha_{i}\). We assume \(E(\epsilon_{it} | \mathbf{X}_{i}, \alpha_{i}) = 0\) but do not make any assumptions about the marginal distribution of \(\alpha_{i}\) or its correlation with the regressors \(\mathbf{x}_{i1},\dots, \mathbf{x}_{iT_i}\).
The estimator implemented in this package is based on maximum likelihood estimation (ML) of both \(\boldsymbol{\beta}\) and \(\alpha_{1}, \dots, \alpha_{N}\). It actually is equivalent to a generalized linear model (
glm()) for binomial data where the set of regressors is extended by a dummy variable for each individual. The main difference is that
bife() applies a pseudo-demeaning algorithm proposed by Stammann, Heiss, and McFadden (2016) to concentrate out the fixed effects from the optimization problem.
1 Its computational costs are lower by orders of magnitude if \(N\) is reasonably large.
It is well known that as \(N \rightarrow \infty\), the ML estimator is not consistent. This “incidental parameters problem” can be severe if \(T\) is small. To tackle this problem, we provide an analytical bias correction for the structural parameters \(\boldsymbol{\beta}\) and the average partial effects derived by Fernández-Val (2009).
2 Thus this package is well suited to analyse big micro-data where \(N\) and/or \(T\) are large.
In the following we utilize an example from labor economics to demonstrate the capabilities of
bife(). More precisely, we use a balanced micro panel data set from the
Panel Study of Income Dynamics to analyze the intertemporal labor force participation of 1,461 married women observed for nine years. A similar empirical illustration is used in Fernández-Val (2009) and is an adoption from Hyslop (1999).
Before we start, we briefly inspect the data set to get an idea about its structure and potential covariates.
ID and
TIME are individual and time-specific identifiers,
LFP is an indicator equal to one if a woman is in labor force,
KID1 -
KID3 are the number of children in a certain age group,
INCH is the annual income of the husband, and
AGE is the age of the woman.
First, we use a specification similar to Fernández-Val (2009) and estimate a static model of women’s labor supply where we control for unobserved individual heterogeneity (so called individual fixed effects).
## binomial - probit link## ## LFP ~ KID1 + KID2 + KID3 + log(INCH) + AGE + I(AGE^2) | ID## ## Estimates:## Estimate Std. error z value Pr(> |z|) ## KID1 -0.7144874 0.0562414 -12.704 < 2e-16 ***## KID2 -0.4114824 0.0515524 -7.982 1.44e-15 ***## KID3 -0.1298719 0.0415477 -3.126 0.00177 ** ## log(INCH) -0.2417756 0.0541720 -4.463 8.08e-06 ***## AGE 0.2319779 0.0375351 6.180 6.40e-10 ***## I(AGE^2) -0.0028846 0.0004989 -5.781 7.41e-09 ***## ---## Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1## ## residual deviance= 6058.88,## null deviance= 8152.05,## nT= 5976, N= 664## ## ( 7173 observation(s) deleted due to perfect classification )## ## Number of Fisher Scoring Iterations: 6 ## ## Average individual fixed effect= -1.121
As
glm(), the summary statistic of the model provides detailed information about the coefficients and some information about the model fit (
residual deviance and
null deviance). Furthermore, we report statistics that are specific to fixed effects models. More precisely, we learn that only 5,976 observations out of 13,149 contribute to the idenfication of the structural parameters. This is indicated by the message that 7,173 observations are deleted due to perfect classification. With respect to binary choice models those are observations that are related to women who never change their labor force participation status during the nine years observed. Thus those women were either always employed or unemployed. Overall the estimation results are based on 664 women observed for nine years.
Because coefficients itself are not very meaningful, econometricians are usually interested in so called partial effects (also known as marginal or ceteris paribus effects). A commonly used statistic is the average partial effect.
bife offers a post-estimation routine to estimate average partial effects and their corresponding standard errors.
## Estimates:## Estimate Std. error z value Pr(> |z|) ## KID1 -0.0927845 0.0161225 -5.755 8.67e-09 ***## KID2 -0.0534358 0.0096773 -5.522 3.36e-08 ***## KID3 -0.0168654 0.0052808 -3.194 0.00140 ** ## log(INCH) -0.0313974 0.0109470 -2.868 0.00413 ** ## AGE 0.0301250 0.0212329 1.419 0.15596 ## I(AGE^2) -0.0003746 0.0099167 -0.038 0.96987 ## ---## Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
A widespread reason that prevents the use of non-linear fixed effects models in practice is the so-called incidental parameter bias problem (
IPP) first mentioned by Neyman and Scott (1948). Fortunately, for classical panel data sets, like in this example, there already exist several asymptotic bias-corrections tackling the IPP (see Fernández-Val and Weidner (2018) for an overview). Our package provides a post-estimation routine that applies the analytical bias-correction derived by Fernández-Val (2009).
## binomial - probit link## ## LFP ~ KID1 + KID2 + KID3 + log(INCH) + AGE + I(AGE^2) | ID## ## Estimates:## Estimate Std. error z value Pr(> |z|) ## KID1 -0.6309003 0.0555075 -11.366 < 2e-16 ***## KID2 -0.3635502 0.0511328 -7.110 1.16e-12 ***## KID3 -0.1149806 0.0413488 -2.781 0.00542 ** ## log(INCH) -0.2139642 0.0536616 -3.987 6.68e-05 ***## AGE 0.2052755 0.0373054 5.503 3.74e-08 ***## I(AGE^2) -0.0025520 0.0004962 -5.144 2.70e-07 ***## ---## Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1## ## residual deviance= 6058.88,## null deviance= 8152.05,## nT= 5976, N= 664## ## ( 7173 observation(s) deleted due to perfect classification )## ## Number of Fisher Scoring Iterations: 6 ## ## Average individual fixed effect= -0.969
## Estimates:## Estimate Std. error z value Pr(> |z|) ## KID1 -0.1015555 0.0146046 -6.954 3.56e-12 ***## KID2 -0.0585204 0.0090041 -6.499 8.07e-11 ***## KID3 -0.0185083 0.0052532 -3.523 0.000426 ***## log(INCH) -0.0344416 0.0102247 -3.368 0.000756 ***## AGE 0.0330430 0.0186548 1.771 0.076513 . ## I(AGE^2) -0.0004108 0.0086844 -0.047 0.962272 ## ---## Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Whereas analytical bias-corrections for static models get more and more attention in applied work, it is not well known that they can also be used for dynamic models with fixed effects.
Before we can adjust our static to a dynamic specification, we first have to generate a lagged dependent variable.
Contrary to the bias-correction for the static models, we need to additionally provide a bandwidth parameter (
L) that is required for the estimation of spectral densities (see Hahn and Kuersteiner (2011)). Fernández-Val and Weidner (2018) suggest to do a sensitivity analysis and try different values for
L but not larger than four.
## binomial - probit link## ## LFP ~ LLFP + KID1 + KID2 + KID3 + log(INCH) + AGE + I(AGE^2) | ## ID## ## Estimates:## Estimate Std. error z value Pr(> |z|) ## LLFP 1.0025730 0.0473072 21.193 < 2e-16 ***## KID1 -0.4741452 0.0679084 -6.982 2.91e-12 ***## KID2 -0.1958494 0.0625931 -3.129 0.001754 ** ## KID3 -0.0754092 0.0505115 -1.493 0.135461 ## log(INCH) -0.1947087 0.0621154 -3.135 0.001721 ** ## AGE 0.2009782 0.0477735 4.207 2.59e-05 ***## I(AGE^2) -0.0024145 0.0006293 -3.837 0.000125 ***## ---## Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1## ## residual deviance= 4774.57,## null deviance= 6549.14,## nT= 4792, N= 599## ## ( 1461 observation(s) deleted due to missingness )## ( 6896 observation(s) deleted due to perfect classification )## ## Number of Fisher Scoring Iterations: 6 ## ## Average individual fixed effect= -1.939
## Estimates:## Estimate Std. error z value Pr(> |z|) ## LLFP 0.1826348 0.0209169 8.731 < 2e-16 ***## KID1 -0.0752484 0.0114039 -6.598 4.15e-11 ***## KID2 -0.0310820 0.0074600 -4.166 3.09e-05 ***## KID3 -0.0119677 0.0053992 -2.217 0.026653 * ## log(INCH) -0.0309009 0.0091310 -3.384 0.000714 ***## AGE 0.0318959 0.0177688 1.795 0.072646 . ## I(AGE^2) -0.0003832 0.0079898 -0.048 0.961749 ## ---## Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Chamberlain, Gary. 1980. “Analysis of Covariance with Qualitative Data.”
The Review of Economic Studies 47 (1): 225–38.
Fernández-Val, Iván. 2009. “Fixed Effects Estimation of Structural Parameters and Marginal Effects in Panel Probit Models.”
Journal of Econometrics 150 (1): 71–85.
Fernández-Val, Iván, and Martin Weidner. 2018. “Fixed Effects Estimation of Large-T Panel Data Models.”
Annual Review of Economics 10 (1): 109–38.
Greene, William. 2004. “The Behaviour of the Maximum Likelihood Estimator of Limited Dependent Variable Models in the Presence of Fixed Effects.”
Econometrics Journal 7 (1): 98–119.
Hahn, Jinyong, and Guido Kuersteiner. 2011. “Bias Reduction for Dynamic Nonlinear Panel Models with Fixed Effects.”
Econometric Theory 27 (6): 1152–91.
Hahn, Jinyong, and Whitney Newey. 2004. “Jackknife and Analytical Bias Reduction for Nonlinear Panel Models.”
Econometrica 72 (4): 1295–1319.
Hyslop, Dean R. 1999. “State Dependence, Serial Correlation and Heterogeneity in Intertemporal Labor Force Participation of Married Women.”
Econometrica 67 (6): 1255–94.
Neyman, Jerzy, and Elizabeth L. Scott. 1948. “Consistent Estimates Based on Partially Consistent Observations.”
Econometrica 16 (1): 1–32.
Stammann, Amrei, Florian Heiss, and Daniel McFadden. 2016. “Estimating Fixed Effects Logit Models with Large Panel Data.” |
The Numerical Aperture (NA) (for fiber optics) is usually used to denote the acceptance cone for a multi-mode fiber.
Does NA also describe the expansion of light emitted from the end of a fiber?
I have a 1 mm core 0.22 NA PMMA fiber, I would like to collimate the light once it's reached a 1.5" diameter.
$$ n\sin(\theta)=NA \\ \theta = \arcsin(\frac{.22}{1.4914}) \\ \frac{.5 \cdot 38 [mm]}{ \tan(\theta)}=\left \{ \text{focal length} \right \} $$
So, I should need a ~130 mm focusing lens?
edited in response to answer below
Is
n the index of refraction of the fiber core (1.4914), or free space (1)? |
Regression (OLS) - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Regression (OLS)
$z$ test for the difference between two proportions
Paired sample $t$ test
Sign test
Two sample $t$ test - equal variances not assumed
Independent variables Independent variable Independent variable Independent variable Independent variable One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variables One categorical with 2 independent groups 2 paired groups 2 paired groups One categorical with 2 independent groups Dependent variable Dependent variable Dependent variable Dependent variable Dependent variable One quantitative of interval or ratio level One categorical with 2 independent groups One quantitative of interval or ratio level One of ordinal level One quantitative of interval or ratio level Null hypothesis Null hypothesis Null hypothesis Null hypothesis Null hypothesis $F$ test for the complete regression model: $\pi_1 = \pi_2$
$\pi_1$ is the unknown proportion of "successes" in population 1; $\pi_2$ is the unknown proportion of "successes" in population 2
$\mu = \mu_0$
$\mu$ is the unknown population mean of the difference scores; $\mu_0$ is the population mean of the difference scores according to the null hypothesis, which is usually 0
$\mu_1 = \mu_2$
$\mu_1$ is the unknown mean in population 1, $\mu_2$ is the unknown mean in population 2
Alternative hypothesis Alternative hypothesis Alternative hypothesis Alternative hypothesis Alternative hypothesis $F$ test for the complete regression model: Two sided: $\pi_1 \neq \pi_2$
Right sided: $\pi_1 > \pi_2$
Left sided: $\pi_1 < \pi_2$
Two sided: $\mu \neq \mu_0$
Right sided: $\mu > \mu_0$
Left sided: $\mu < \mu_0$
Two sided: $\mu_1 \neq \mu_2$
Right sided: $\mu_1 > \mu_2$
Left sided: $\mu_1 < \mu_2$
Assumptions Assumptions Assumptions Assumptions Assumptions all individuals in the population. Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another Test statistic Test statistic Test statistic Test statistic Test statistic $F$ test for the complete regression model:
Note 2: if only one independent variable ($K = 1$), the $F$ test for the complete regression model is equivalent to the two sided $t$ test for $\beta_1$
$z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
$p_1$ is the sample proportion of successes in group 1: $\dfrac{X_1}{n_1}$, $p_2$ is the sample proportion of successes in group 2: $\dfrac{X_2}{n_2}$, $p$ is the total proportion of successes in the sample: $\dfrac{X_1 + X_2}{n_1 + n_2}$, $n_1$ is the sample size of group 1, $n_2$ is the sample size of group 2
Note: we could just as well compute $p_2 - p_1$ in the numerator, but then the left sided alternative becomes $\pi_2 < \pi_1$, and the right sided alternative becomes $\pi_2 > \pi_1$
$t = \dfrac{\bar{y} - \mu_0}{s / \sqrt{N}}$
$\bar{y}$ is the sample mean of the difference scores, $\mu_0$ is the population mean of the difference scores according to H0, $s$ is the sample standard deviation of the difference scores, $N$ is the sample size (number of difference scores).
The denominator $s / \sqrt{N}$ is the standard error of the sampling distribution of $\bar{y}$. The $t$ value indicates how many standard errors $\bar{y}$ is removed from $\mu_0$
$W = $ number of difference scores that is larger than 0 $t = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{\sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}}$
$\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $s^2_1$ is the sample variance in group 1, $s^2_2$ is the sample variance in group 2, $n_1$ is the sample size of group 1, $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to H0.
The denominator $\sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}$ is the standard error of the sampling distribution of $\bar{y}_1 - \bar{y}_2$. The $t$ value indicates how many standard errors $\bar{y}_1 - \bar{y}_2$ is removed from 0.
Note: we could just as well compute $\bar{y}_2 - \bar{y}_1$ in the numerator, but then the left sided alternative becomes $\mu_2 < \mu_1$, and the right sided alternative becomes $\mu_2 > \mu_1$
Sample standard deviation of the residuals $s$ n.a. n.a. n.a. n.a. $\begin{aligned} s &= \sqrt{\dfrac{\sum (y_j - \hat{y}_j)^2}{N - K - 1}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned} $ - - - - Sampling distribution of $F$ and of $t$ if H0 were true Sampling distribution of $z$ if H0 were true Sampling distribution of $t$ if H0 were true Sampling distribution of $W$ if H0 were true Sampling distribution of $t$ if H0 were true Sampling distribution of $F$: Approximately standard normal $t$ distribution with $N - 1$ degrees of freedom The exact distribution of $W$ under the null hypothesis is the Binomial($n$, $p$) distribution, with $n =$ number of positive differences $+$ number of negative differences, and $p = 0.5$.
If $n$ is large, $W$ is approximately normally distributed under the null hypothesis, with mean $np = n \times 0.5$ and standard deviation $\sqrt{np(1-p)} = \sqrt{n \times 0.5(1 - 0.5)}$. Hence, if $n$ is large, the standardized test statistic $$z = \frac{W - n \times 0.5}{\sqrt{n \times 0.5(1 - 0.5)}}$$ follows approximately a standard normal distribution if the null hypothesis were true.
Approximately a $t$ distribution with $k$ degrees of freedom, with $k$ equal to
$k = \dfrac{\Bigg(\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}\Bigg)^2}{\dfrac{1}{n_1 - 1} \Bigg(\dfrac{s^2_1}{n_1}\Bigg)^2 + \dfrac{1}{n_2 - 1} \Bigg(\dfrac{s^2_2}{n_2}\Bigg)^2}$
or
$k$ = the smaller of $n_1$ - 1 and $n_2$ - 1
First definition of $k$ is used by computer programs, second definition is often used for hand calculations
Significant? Significant? Significant? Significant? Significant? $F$ test: Two sided: Two sided: If $n$ is small, the table for the binomial distribution should be used:
Two sided:
If $n$ is large, the table for standard normal probabilities can be used:
Two sided:
Two sided: $C\%$ confidence interval for $\beta_k$ and for $\mu_y$; $C\%$ prediction interval for $y_{new}$ Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$ $C\%$ confidence interval for $\mu$ n.a. Approximate $C\%$ confidence interval for $\mu_1 - \mu_2$ Confidence interval for $\beta_k$: Regular (large sample): $\bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}}$
where the critical value $t^*$ is the value under the $t_{N-1}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20)
The confidence interval for $\mu$ can also be used as significance test.
- $(\bar{y}_1 - \bar{y}_2) \pm t^* \times \sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}$
where the critical value $t^*$ is the value under the $t_{k}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20)
The confidence interval for $\mu_1 - \mu_2$ can also be used as significance test.
Effect size n.a. Effect size n.a. n.a. Complete model: - Cohen's $d$:
Standardized difference between the sample mean of the difference scores and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{s}$$ Indicates how many standard deviations $s$ the sample mean of the difference scores $\bar{y}$ is removed from $\mu_0$
- - n.a. n.a. Visual representation n.a. Visual representation - - - ANOVA table n.a. n.a. n.a. n.a. - - - - n.a. Equivalent to Equivalent to Equivalent to n.a. - When testing two sided: chi-squared test for the relationship between two categorical variables, where both categorical variables have 2 levels One sample $t$ test on the difference scores
Repeated measures ANOVA with one dichotomous within subjects factor
Two sided sign test is equivalent to - Example context Example context Example context Example context Example context Can mental health be predicted from fysical health, economic class, and gender? Is the proportion smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic. Is the average difference between the mental health scores before and after an intervention different from $\mu_0$ = 0? Do people tend to score higher on mental health after a mindfulness course? Is the average mental health score different between men and women? SPSS SPSS SPSS SPSS SPSS Analyze > Regression > Linear... SPSS does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Analyze > Descriptive Statistics > Crosstabs...
Analyze > Compare Means > Paired-Samples T Test... Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples... Analyze > Compare Means > Independent-Samples T Test... Jamovi Jamovi Jamovi Jamovi Jamovi Regression > Linear Regression Jamovi does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Frequencies > Independent Samples - $\chi^2$ test of association
T-Tests > Paired Samples T-Test Jamovi does not have a specific option for the sign test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the two sided $p$ value that would have resulted from the sign test. Go to:
ANOVA > Repeated Measures ANOVA - Friedman
T-Tests > Independent Samples T-Test Practice questions Practice questions Practice questions Practice questions Practice questions |
For certain vector fields, the amount of work required to move a particle from one point to another is dependent only on its initial and final positions, not on the path it takes. Gravitational and electric fields are examples of such vector fields. This section will discuss the properties of these vector fields.
Definition: Path Independent and Conservative
Let \(\mathbf{F}\) be a vector field defined on an open region
D in space, and suppose that for any two points A and B in D the line integral
\[\int_{C}^{ }\mathbf{F}\cdot \mathit{d}\mathbf{r}\]
along a path
C from A to B in D is the same over all paths from A to B. Then the integral \[\int_{C}^{ }\mathbf{F}\cdot \mathit{d}\mathbf{r}\] is path independent in D and the field Fis conservative on D . Potential Function
Definition: If
F is a vector field defined on D and \[\mathbf{F}=\triangledown f\] for some scalar function f on D, then f is called a potential function for F. You can calculate all the line integrals in the domain F over any path between A and B after finding the potential function f
\[\int_{A}^{B}\mathbf{F}\cdot \mathit{d}\mathbf{r}=\int_{A}^{B}\triangledown f\mathit{d}\mathbf{r}=\mathit{f(B)}-\mathit{f(A)}\]
This can be related back to the Fundamental Theorem of Calculus, since the gradient can be thought of as similar to the derivative. Another important property of conservative vector fields is that the integral of
F around any closed path D is always 0. Assumptions on Curves, Vector Fields, and Domains
For computational sake, we have to assume the following properties regarding the curves, surfaces, domains, and vector fields:
The curves we consider are piecewise smooth, meaning they are composed of many infinitesimally small, smooth pieces connected end to end. We assume that the domain Dis a simply connected open region, meaning that any two points in Dcan be joined by a smooth curve within the region and that every loop in Dcan be contracted to a point in Dwithout ever leaving D.
Theorem 1: Fundamental Theorem of Line Integrals
Let
C be a smooth curve joining the point A to point B in the plane ore in space and parametrized by \(\mathbf{r}(t)\). Let f be a differentiable function with a continuous gradient vector \(\mathbf{F}=\bigtriangledown{f}\) on a domain D containing C. Then \(\int_{C}\mathbf{F}\cdot d\mathbf{r}=f(B)-f(A)\).
Proof
Suppose that
A and B are two points in region D and that the curve C is given by \[\mathbf{r}(t)=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\] is a smooth curve in D that joins points A and B. Along C, f is a differentiable function of t and
\[\begin{align*} \dfrac{\partial f }{\partial t}&=\dfrac{\partial f }{\partial x}\dfrac{\partial x }{\partial t}+\dfrac{\partial f }{\partial y}\dfrac{\partial y }{\partial t}+\dfrac{\partial f }{\partial z}\dfrac{\partial z }{\partial t} \\ &=\bigtriangledown f\cdot \left ( \dfrac{\mathrm{d} x}{\mathrm{d} t}\mathbf{i}+\dfrac{\mathrm{d} y}{\mathrm{d} t}\mathbf{j}\dfrac{\mathrm{d} z}{\mathrm{d} t}\mathbf{k} \right ) \\ &=\bigtriangledown f\cdot \dfrac{\mathrm{d} \mathbf{r}}{\mathrm{d} t} \\ &=\mathbf{F}\cdot \dfrac{\mathrm{d} \mathbf{r}}{\mathrm{d} t} \end{align*}\]
Therefore,
\[\int_{C}\mathbf{F}\cdot d\mathbf{r}=\int_{t=a}^{t=b}\mathbf{F}\cdot \dfrac{\mathrm{d} \mathbf{r}}{\mathrm{d} t}dt=\int_a^b\dfrac{\mathrm{d} f}{\mathrm{d} t}dt \nonumber\]
*Note:
\[\mathbf{r}(a)=A, \; \mathbf{r}(b)=B \nonumber\]
Which means:
\[\left. f(g(t),h(t),k(t))\right|_a^b=f(B)-f(A) \nonumber\]
Thus proving Theorem 1. This shows us that the integral of a gradient field is easy to compute, provided we know the function \(f\).
\(\square\)
As mentioned earlier, this is very similar to the Fundamental Theorem of Calculus both in theory and importance. Like the FTC, it provides us with a way to evaluate line integrals without limits of Riemann sums.
Theorem 2: Conservative Fields are Gradient Fields
Let \(\mathbf{F}=M\hat{\mathbf{i}}+N\hat{\mathbf{j}}+P\hat{\mathbf{k}}\) be a vector field whose components are continuous throughout an open connected region
D in space. Then F is conservative if and only it F is a gradient field \(\bigtriangledown f\) for a differentiable function f.
Proof
If
F is a gradient field, then \(\mathbf{F}=\bigtriangledown f\) for a differentiable function f. By Theorem 1, we know that \[\int_C\mathbf{F}\cdot d\mathbf{r}=f(B)-f(A)\] and that the value of the line integral depends only on the two endpoints, not on the path. The line integral is said to be independent and F is a conservative field.
However, suppose
F is a conservative vector field and we want to find some function f on D such that \(\bigtriangledown f=\mathbf{F}\). First, we must pick a point A in the domain D such that \(f(A)=0\). For any other point B, we must define \(f(B)\) as equal to \[\int_C\mathbf{F}\cdot d\mathbf{r},\] where the curve C is any smooth path in D from A to B. Because F is conservative, we know that \(f(B)\) is not dependant on C and vice versa. In order to show that \(\bigtriangledown f=\mathbf{F},\) we need to show that
\[\dfrac{\partial f}{\partial x}=M, \dfrac{\partial f}{\partial y}=N, \dfrac{\partial f}{\partial z}=P.\nonumber\]
Suppose B has coordinates \((x,y,z)\) and a nearby point \(B_0=(x_0,y,z).\) By definition, then, the value of function f at the nearby point is \[\int_{C_0}\mathbf{F}\cdot d\mathbf{r},\] where \(C_0\) is any path from A to \(B_0.\) We can take path C to be the union between path \(C_0\) and line segment L from B to \(B_0\). Therefore,
\[f(x,y,z)=\int_{C_0}\mathbf{F}\cdot d\mathbf{r}+\int_L\mathbf{F}\cdot d\mathbf{r}\nonumber\]
We can differentiate this integral, arriving at:
\[\dfrac{\partial }{\partial x}f(x,y,z)=\dfrac{\partial}{\partial x}\left ( \int_{C_0}\mathbf{F}\cdot d\mathbf{r}+\int_L\mathbf{F}\cdot d\mathbf{r} \right ) \nonumber\]
Only the last term of the above equation is dependent on x, so
\[\dfrac{\partial }{\partial x}f(x,y,z)=\dfrac{\partial }{\partial x}\int_L\mathbf{F}\cdot d\mathbf{r} \nonumber\]
Now, if we parametrize \(L\) such that
\[\mathbf{r}(t)=t\mathbf{i}+y\mathbf{j}+z\mathbf{k} \nonumber\]
where \(x_0\leq t \leq x\) Then,
\[\dfrac{\mathrm{d} r}{\mathrm{d} t}=\mathbf{i}\] \[\mathbf{F}\cdot \dfrac{\mathrm{d} r}{\mathrm{d} t}=M \nonumber\]
and
\[\int_L\mathbf{f}\cdot d\mathbf{r}=\int_{x_0}^xM(t,y,z)dt] \nonumber\]
Substitution gives us
\[\dfrac{\partial }{\partial x}f(x,y,z)=\dfrac{\partial }{\partial x}\int_{x_0}^xM(t,y,z)dt=M(x,y,z) \nonumber\]
by the FTC. The partial derivatives
\[\dfrac{\partial f}{\partial y}=N \nonumber\]
and
\[\dfrac{\partial f}{\partial z}=P \nonumber\]
follow similarly, showing that
\[\mathbf{F}=\bigtriangledown f \nonumber\]
\(\square\)
In other words, \(\mathbf{F}=\bigtriangledown f\) is only true when, for any two point
A and B in the region D, \(\int_C\mathbf{F}\cdot d\mathbf{r}\) is independent of the path C that joins the two points in D.
Theorem 3: Looper Property of Conservative Fields
The following statements are equivalent:
\(\oint_{C}\mathbf{F}\cdot d\mathbf{r}=0\) around every loop (closed curve C) in D. The field Fis conservativeon D.
Proof
Part 1
We want to show that for any two points
A and B in D, the ingtegral of
\[\mathbf{F}\cdot d\mathbf{r}\nonumber\]
has the same value over any two paths \(C_1\) & \(C_2\) from
A to B.
We reverse the direction of \(C_2\) to make the path \(-C_2\) from
B to A.
Together, the two curves \(C_1\) & \(-C_2\) make a closed loop, which we will call
C.
If you recall from earlier in this section, the integral over a closed loop for a conservative field is always 0:
\[\begin{align*} \int_{C_1}\mathbf{F}\cdot d\mathbf{r}-\int_{C_2}\mathbf{F}\cdot d\mathbf{r}&=\int_{C_1}\mathbf{F}\cdot d\mathbf{r}+\int_{-C_2}\mathbf{F}\cdot d\mathbf{r} \\ &=\int_C\mathbf{F}\cdot d\mathbf{r} \\ &=0 \end{align*}\]
Therefore, the integrals over \(C_1\) & \(C_2\) must be equal.
Part 2
We want to show that the integral over \(\mathbf{F}\cdot d\mathbf{r}\) is zero for any closed loop
C. We pick two points A & B on C and use them to break C into 2 pieces: \(C_1\) from A to B and \(C_2\) from B back to A.
Therefore:
\[\begin{align*} \oint _C\mathbf{F}\cdot d\mathbf{r}&=\int_{C_1}\mathbf{F}\cdot d\mathbf{r}+\int_{C_2}\mathbf{F}\cdot d\mathbf{r} \\ &=\int_A^B\mathbf{F}\cdot d\mathbf{r}-\int_A^B\mathbf{F}\cdot d\mathbf{r} \\ &=0 \end{align*}\]
\(\square\)
Finding Potentials for Conservative Fields Component Test for Conservative Fields: Let \(\mathbf{F}=M(x,y,z)\hat{\textbf{i}} + N(x,y,z) \hat{\textbf{j}}+ P(x,y,z) \hat{\textbf{k}} \) be a field on a connected and simply connected domain whose component functions have continuous first partial derivatives. Then, F is conservative if and only if
\[\dfrac{\partial P }{\partial x}=\dfrac{\partial M}{\partial z}\]
\[\dfrac{\partial P }{\partial y}=\dfrac{\partial N}{\partial z}\]
and
\[\dfrac{\partial N }{\partial x}=\dfrac{\partial M}{\partial y}.\]
*Note: See Example 2
Definition: Exact Differential Forms
Any expression
\[M(x,y,z)dx+N(x,y,z)dy+P(x,y,z)dz\]
is a
differential form. A differential form is exact on a domain D in space if
\[M\,dx+N\,dy+P\,dz=\dfrac{\partial f}{\partial x}dx+\dfrac{\partial f}{\partial y}dy+\dfrac{\partial f}{\partial z}dz=df\]
for some scalar function
f throughout D. Component Test for Exactness of \(Mdx+Ndy+Pdz\): The differential form \(Mdx+Ndy+Pdz\) is exact on a connected and simply connected domain if and only if
\[\dfrac{\partial P }{\partial x}=\dfrac{\partial M}{\partial z}\]
\[\dfrac{\partial P }{\partial y}=\dfrac{\partial N}{\partial z}\]
and
\[\dfrac{\partial N }{\partial x}=\dfrac{\partial M}{\partial y}\]
Notice, this is the same as saying the field \(\mathbf{F}=M\hat{\mathbf{i}}+N\hat{\mathbf{j}}+P\hat{\mathbf{k}}\) is conservative.
Example \(\PageIndex{1}\)
Suppose the force field \(\mathbf{F}=\bigtriangledown f\) is the gradient of the function \(f(x,y,z)=-\dfrac{1}{x^2+y^2+z^2}\). Find the work done by
F in moving an object along a smooth curve C joining \((1,0,0)\) to \((0,0,2)\) that does not pass through the origin. Solution
Since we know that this is a conservative field, we can apply Theorem 1, which shows that regardless of the curve
C, the work done by F will be as follows:
\[\begin{align(} \int_{C}\mathbf{F}\cdot d\mathbf{r}&=f(0,0,2)-f(1,0,0) \\ &=-\dfrac{1}{4}-(-1) \\ &=\dfrac{3}{4} \end{align(}\]
Example \(\PageIndex{2}\)
Show that
\[\mathbf{F}=(e^x \cos y+yz)\hat{\mathbf{i}}+(xz-e^x\sin y)\hat{\mathbf{j}}+(xy+z)\hat{\mathbf{k}} \nonumber\]
is conservative over its natural domain and find a potential function for it.
Solution
The natural domain of
F is all of space, which is connected and simply connected. Let's define the following:
\[M=e^x\cos y+yz \nonumber\]
\[N=xz-e^x\sin y \nonumber\]
\[P=xy+z \nonumber\]
and calculate
\[\dfrac{\partial P }{\partial x}=y=\dfrac{\partial M}{\partial z} \nonumber\]
\[\dfrac{\partial P }{\partial y}=x=\dfrac{\partial N}{\partial z} \nonumber\]
\[\dfrac{\partial N }{\partial x}=-e^x\sin y=\dfrac{\partial M}{\partial y}. \nonumber\]
Because the partial derivatives are continuous,
F is conservative. Now that we know there exists a function f where the gradient is equal to F, let's find f.
\[\dfrac{\partial f }{\partial x}=e^x\cos y+yz \nonumber\]
\[\dfrac{\partial f }{\partial y}=xz-e^x\sin y \nonumber\]
\[\dfrac{\partial f }{\partial z}=xy+z\nonumber \]
If we integrate the first of the three equations with respect to x, we find that
\[f(x,y,z)=\int(e^x/cos y+yz)dx=e^x\cos y+xyz+g(y,z)\]
where \(g(y,z)\) is a constant dependent on \(y\) and \(z\) variables. We then calculate the partial derivative with respect to \(y\) from this equation and match it with the equation of above.
\[\dfrac{\partial }{\partial y}(f(x,y,z))=-e^x/sin y+xz+\dfrac{\partial g}{\partial y}=xz-e^x\sin y \nonumber\]
This means that the partial derivative of \(g\) with respect to \(y\) is 0, thus eliminating \(y\) from \(g\) entirely and leaving it as a function of \(z\) alone.
\[f(x,y,z)=e^x\cos y+xyz+h(z) \nonumber\]
We then repeat the process with the partial derivative with respect to \(z\).
\[\dfrac{\partial }{\partial z}(f(x,y,z))=xy+\dfrac{\mathrm{d} h}{\mathrm{d} z}=xy+z \nonumber\]
which means that
\[\dfrac{\mathrm{d} h}{\mathrm{d} z}= z \nonumber\]
so we can find \(h(z)\) by integrating:
\[h(z)=\dfrac{z^2}{2}+C. \nonumber\]
Therefore,
\[f(x,y,z)=e^x\cos y+xyz+\dfrac{z^2}{2}+C. \nonumber\]
We still have infinitely many potential functions for
F-one at each value of C.
Example \(\PageIndex{3}\)
Show that \( ydx+xdy+4dz\) is exact and evaluate the integral
\[\int_{(1,1,1)}^{(2,3,-1)}ydx+xdy+4dz \nonumber\]
over any path from \((1,1,1)\) to \((2,3,-1)\).
Solution
We let \(M=y\), \(N=x\), and \(P=4\). Apply the Test for Exactness:
\[\dfrac{\partial N}{\partial x}=1=\dfrac{\partial M}{\partial y}\nonumber\]
\[\dfrac{\partial N}{\partial z}=0=\dfrac{\partial P}{\partial y}\nonumber\]
and
\[\dfrac{\partial N}{\partial z}=0=\dfrac{\partial P}{\partial y}.\nonumber\]
This proves that \(ydx+xdy+4dz\) is exact, so
\[ydx+xdy+4dz=df \nonumber\]
for some function
f, and the integral's value is \(f(2,3,-1)-f(1,1,1)\).
We find fo up to a constant by integrating the following equations:
\[\dfrac{\partial f}{\partial x}=y, \dfrac{\partial f}{\partial y}=x, \dfrac{\partial f}{\partial z}=4 \nonumber\]
From the first equation, we get that \(f(x,y,z)=xy+g(y,z)\)
The second equation tells us that \(\dfrac{\partial f}{\partial y}=x+\dfrac{\partial g}{\partial y}=x\)
Therefore,
\[\dfrac{\partial g}{\partial y}=0 \nonumber\]
Hence,
\[f(x,y,z)=xy+h(z) \nonumber\]
The third equation tells us that \(\dfrac{\partial f}{\partial z}=0+\dfrac{d h}{d z}=4\) so \(h(z)=4z+C\)
Therefore,
\[f(x,y,z)=xy+4z+C \nonumber\]
By substitution, we find that:
\[f(2,3,-1)-f(1,1,1)=2+C-(5+C)=-3 \nonumber\]
References Weir, Maurice D., Joel Hass, and George B. Thomas. Thomas' Calculus: Early Transcendentals. Boston: Addison-Wesley, 2010. Print. Contributors Alagu Chidambaram (UCD)
Integrated by Justin Marshall. |
just a few remarks:
note: i use the standard abbreviation $(a,b)$ for the GCD of $a$ and $b$
a. it is possible to approach a problem
in the wrong way. occasionally this may lead to startling insight, but usually it leads to a waste of time, and can even be demotivating. here, OP's introduction of:
$$(a,b) = ax+by$$suggests that someone should explain why this property is unlikely to be helpful in the present case, or, alternatively, find a proof which is based on this approach.
b. sometimes an assertion
smells fishy - an intuition which can be wrong, but is often correct. it directs us to look for a counterexample. finding counterexamples is a highly useful ability, which some have more of than others. psychologically, the heuristics of finding counter-examples are different from those of finding a proof of a result, and different again from those of making a discovery.
c. the assertion in this problem smells fishy, and it might be more interesting if, say, the question had been a request to evaluate:$$\frac{(a,c)(b,c)}{(ab,c)}$$d. sometimes the sense of pain or aversion one feels on examining a problem (if you introspect the workings of the mind, which is inherently lazy, at least in my case) can actually serve as a guide to how the problem should be approached. i think the sense of "Oh, no!" i feel when i see the terms GCD or LCM is due to the fact that i like working with fields, rings, ideals and so on, but feel uncomfortable with boolean algebra. questions to do with division arise in an algebraic context of the ideals of $\mathbb{Z}$, but are often really about lattice properties.
e. so, given my mental unease, and OP's evident lack of experience, perhaps the following approach may generate some insight. i will assume the prime decomposition properties of $\mathbb{Z}$. if anyone objects i suggest they prove the necessary assertions as a lemma, and then continue as here.
let $\mathbb{P}$ be the set of prime numbers, and let $\mathbb{N}^+$ denote the set of positive integers. we define the map:$$\psi :\mathbb{P} \rightarrow \mathbb{N}^{\mathbb{N}^+}$$we use the abbreviation $\mathfrak{p}_n$ for the more unwieldy $\psi(\mathfrak{p})(n)$. intuitively $\mathfrak{p}_n$ is the exponent of the highest power of $\mathfrak{p}$ which divides $n$.
LEMMA we use the symbol $m \land n$ for min$(m,n)$
for any $\mathfrak{p} \in \mathbb{P}$$$\mathfrak{p}_{(a,b)} = \mathfrak{p}_a \land \mathfrak{p}_b$$and$$\mathfrak{p}_{ab} = \mathfrak{p}_a + \mathfrak{p}_b$$the claim in the question can now be expressed as a spurious assertion of distributivity, but which can be improved to:$$\\ $$THEOREM
$\forall \mathfrak{p} \in \mathbb{P}$, and $a,b,c \in \mathbb{N}^+$ we have:$$(\mathfrak{p}_a+\mathfrak{p}_b) \land \mathfrak{p}_c \le\mathfrak{p}_a \land \mathfrak{p}_c + \mathfrak{p}_a \land \mathfrak{p}_c $$perhaps someone can produce an aesthetically satisfying demonstration? |
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