text:
# LOG#106. Basic Cosmology (I).

The next thread is devoted to Cosmology. I will intend to be clear and simple about equations and principles of current Cosmology with a General Relativity background.

First of all…I will review the basic concepts of natural units I am going to use here. We will be using the following natural units:

$\hbar=c=k_B=1$

We will take the Planck mass to be given by

$M_P=\sqrt{8\pi G_N}\approx 1\mbox{.}2\cdot 10^{19}GeV$

The solar mass is $M_\odot=2\cdot 10^{30}kg$ and the parsec is given by the value

$1pc=3\mbox{.}26lyr=3\mbox{.}1\cdot 10^{16}m$

Well, current Cosmology is based on General Relativity. Even if I have not reviewed this theory with detail in this blog, the nice thing is that most of Cosmology can be learned with only a very little knowledge of this fenomenal theory. The most important ideas are: metric field, geodesics, Einstein equations and no much more…

In fact, newtonian gravity is a good approximation in some particular cases! And we do know that even in this pre-relativistic theory

$\mbox{Gravitational force}=\mbox{Matter/Mass density}$

via the Poisson’s equation

$\nabla^2\phi =4\pi G_N\rho$

This idea, due to the equivalence principle, is generalized a little bit in the general relativistic framework

$\mbox{Spacetime geometry}=\mbox{Matter content/Energy-momentum}$

The spacetime geometry is determined by the metric tensor $g_{\mu\nu}(x)$. The matter content is given by the stress-energy-momentum tensor $T_{\mu\nu}$. As we know one of these two elements, we can know, via Eisntein’s field equations the another. That is, given a metric tensor, we can tell how energy-momentum “moves” in space-time. Given the energy-momentum tensor, we can know what is the metric tensor in spacetime and we can guess how the spacetime bends… This is the origin of the famous motto: “Spacetime says matter how to move, energy-momentum says spacetime how to curve”! Remember that we have “deduced” the Einstein’s field equations in the previous post. Without a cosmological constant term, we get

$G_{\mu\nu}=\kappa^2T_{\mu\nu}=8\pi G_NT_{\mu\nu}$

Given a spacetime metric $g_{\mu\nu}$, we can calculate the (affine/Levi-Civita) connection

$\Gamma^\sigma_{\;\;\mu\nu}=\dfrac{1}{2}g^{\sigma\rho}\left(\partial_\mu g_{\nu\rho}+\partial_\nu g_{\rho\mu}-\partial_\rho g_{\mu\nu}\right)$

The Riemann tensor that measures the spacetime curvature is provided by the equation

$R^\rho_{\;\; \sigma\mu\nu}=\partial_\mu \Gamma^\rho_{\;\;\mu\sigma}-\partial_\mu \Gamma^\rho_{\;\; \mu \sigma}+\Gamma^\rho_{\;\;\mu\lambda}\Gamma^\lambda_{\;\;\nu\sigma}-\Gamma^\rho_{\;\;\nu\lambda}\Gamma^\lambda_{\;\;\mu\sigma}$

The Ricci tensor is defined to be the following “trace” of the Riemann tensor

$R_{\mu\nu}=R^\lambda_{\;\;\mu\lambda \nu}$

The Einstein tensor is related to the above tensors in the well-known manner

$G_{\mu\nu}=R_{\mu\nu}-\dfrac{1}{2}Rg_{\mu\nu}$

The Einstein’s equations can be derived from the Einstein-Hilbert action we learned in the previous post, using the action principle and the integral

$\boxed{S_{EH}=\int d^4x \sqrt{-g} \left(\kappa^{-2}R+\mathcal{L}_M\right)}$

The geodesic equation is the path of a freely falling particle. It gives a “condensation” of the Einstein’s equivalence principle too and it is also a generalization of Newton’s law of “no force”. That is, the geodesic equation is the feynmanity

$\boxed{\dfrac{d^2 x^\mu}{d\tau^2}+\Gamma^\mu _{\rho\sigma}\dfrac{dx^\rho}{d\tau}\dfrac{dx^\sigma}{d\tau}=0}$

Finally, an important concept in General Relativity is that of isometry. The symmetry of the “spacetime manifold” is provided by a Killing vector that preserves transformations (isometries) of that manifold. Mathematically speaking, the Killing vector fields satisfy certain equation called the Killing equation

$\boxed{\xi_{\mu ; \nu}+\xi_{\nu ; \mu}=0}$

Maximally symmetric spaces have $n(n+1)/2$ Killing vectors in n-dimensional (nD) spacetime. There are 3 main classes or types of 2D maximally symmetric that can be generalized to higher dimensions:

1. The euclidean plane $E^2$.

2. The pseudo-sphere $H^2$. This is a certain “hyperbolic” space.

3. The spehre $S^2$. This is a certain “elliptic” space.

The Friedmann-Robertson-Walker Cosmology

Current cosmological models are based in General Relativity AND  a simplification of the possible metrics due to the so-called Copernican (or cosmological) principle: the Universe is pretty much the same “everywhere” you are in the whole Universe! Remarkbly, the old “perfect” Copernican (cosmological) principle that states that the Universe is the same “everywhere” and “every time” is wrong. Phenomenologically, we have found that the Universe has evolved and it evolves, so the Universe was “different” when it was “young”. Therefore, the perfect cosmological principle is flawed. In fact, this experimental fact allows us to neglect some old theories like the “stationary state” and many other “crazy theories”.

What are the observational facts to keep the Copernican principle? It seems that:

1st. The distribution of matter (mainly galaxies, clusters,…) and radiation (the cosmic microwave background/CMB) in the observable Universe is homogenous and isotropic.

2nd. The Universe is NOT static. From Hubble’s pioneer works/observations, we do know that galaxies are receeding from us!

Therefore, these observations imply that our “local” Hubble volume during the Hubble time is similar to some spacetime with homogenous and isotropic spatial sections, i.e., it is a spacetime manifold $M=\mathbb{R}\times \Sigma$. Here, $\mathbb{R}$ denotes the time “slice” and $\Sigma$ represents a 3D maximally symmetric space.

The geometry of a locally isotropic and homogeneous Universe is represented by the so-called Friedmann-Robertson-Walker metric

$\boxed{ds^2_{FRW}=-dt^2+a^2(t)\left[\dfrac{dr^2}{1-kr^2}+r^2\left(d\theta^2+\sin\theta^2d\phi^2\right)\right]}$

Here, $a(t)$ is the called the scale factor.  The parameter $k$ determines the geometry type (plane, hyperbolic or elliptical/spherical):

1) If $k=0$, then the Universe is “flat”. The manifold is $E^3$.

2) If $k=-1$, then the Universe is “open”/hyperbolic. The manifold would be $H^3$.

3) If $k=+1$, then the Universe is “closed”/spherical or elliptical. The manifold is then $S^3$.

Remark: The ansatz of local homogeneity and istoropy only implies that the spatial metric is locally one of the above three spaces, i.e., $E^3,H^3,S^3$. It could be possible that these 3 spaces had different global (likely topological) properties beyond these two properties.

Kinematical features of a FRW Universe

The first property we are interested in Cosmology/Astrophysics is “distance”. Measuring distance in a expanding Universe like a FRW metric is “tricky”. There are several notions of “useful” distances. They can be measured by different methods/approaches and they provide something called sometimes “the cosmologidal distance ladder”:

1st. Comoving distance. It is a measure in which the distance is “taken” by a fixed coordinate.

2nd. Physical distance. It is essentially the comoving distance times the scale factor.

3rd. Luminosity distance. It uses the light emitted by some object to calculate its distance (provided the speed of light is taken constant, i.e., special relativity holds and we have a constant speed of light)

4th. Angular diameter distance. Another measure of distance using the notion of parallax and the “extension” of the physical object we measure somehow.

There is an important (complementary) idea in FRW Cosmology: the particle horizon. Consider a light-like particle with $ds^2=0$. Then,

$dt=a(t)\dfrac{1}{\sqrt{1-kr^2}}$

or

$\dfrac{dr}{\sqrt{1-kr^2}}=\dfrac{dt}{a(t)}$

The total comoving distance that light have traveled since a time $t=0$ is equal to

$\boxed{\eta=\int_0^{r_H}\dfrac{dr}{\sqrt{1-kr^2}}=\int_0^t\dfrac{dt'}{a(t')}}$

It shows that NO information could have propagated further and thus, there is a “comoving horizon” with every light-like particle! Here, this time is generally used as a “conformal time” as a convenient tiem variable for the particle. The physical distance to the particle horizon can be calculated

$\boxed{d_H(t)=\int_0^{r_H}\sqrt{g_{rr}}dr=a(t)\int_0^t\dfrac{dt'}{a(t')}=a(t)\eta}$

There are some important kinematical equations to be known

A) For the geodesic equation, the free falling particle, we have

$\Gamma^0_{ij}=\dfrac{\dot{a}}{a}\overline{g}_{ij}$

$\Gamma^i_{0j}=\Gamma^i_{j0}=\dfrac{\dot{a}}{a}\delta_{ij}$

$\Gamma^i_{jk}=\overline{\Gamma}^i_{jk}$

for the FRW metric and, moreover, the energy-momentum vector $P^\mu=(E,\mathbf{p})$ is defined by the usual invariant equation

$P^\mu=\dfrac{dx^\mu}{d\lambda}$

This definition defines, in fact, the proper “time” $\lambda$ implicitely, since

$\dfrac{d}{d\lambda}=\dfrac{dx^0}{d\lambda}\dfrac{d}{dx^0}=E\dfrac{d}{dt}$

and the 0th component of the geodesic equation becomes

$E\dfrac{dE}{dt}=-\Gamma^0_{ij}p^ip^j=-\delta_{ij}a\dot{a}p^ip^j$

$g_{\mu\nu}p^\mu p^\nu=-E^2+a^2\delta_{ij}p^ip^j=-m^2$

$EdE=a^2\vert \mathbf{p}\vert d\vert \mathbf{p}\vert$

$a^2 p\dfrac{dp}{dt}=-a\dot{a} p^2$

$\dfrac{1}{\vert \mathbf{p}\vert }\dfrac{d\vert \mathbf{p}\vert}{dt}+\dfrac{\dot{a}}{a}=0$

Therefore we have deduced that $\vert \mathbf{p}\vert \propto a^{-1}$. This is, in fact, the socalled “redshift”.  The cosmological  redshift parameter is more generally defined through the equation

$\boxed{\dfrac{a(t_0)}{a(t)}=1+z=\dfrac{\lambda_0}{\lambda}}$

B) The Hubble’s law.

The luminosity distance measures the flux of light from a distant object of known luminosity (if it is not expanding). The flux and luminosity distance are bound into a single equation

$\boxed{F=\dfrac{L}{4\pi d^2_L}}$

If we use the comoving distance between a distant emitter and us, we get

$\chi (a)=\int_t^{t_0}\dfrac{dt'}{a(t')}=\int_a^1\dfrac{da'}{a'^2 H(a')}$

for a expanding Universe! That is, we have used the fact that luminosity itself goes through a comoving spherical shell of radius $\chi (a)$. Moreover, it shows that

$F=\dfrac{L (\chi)}{4 \pi \chi (a)^2 a_0^2}=\dfrac{L}{4\pi (\chi (a)/a)^2}$

The luminosity distance in the expanding shell is

$d_L=\dfrac{\chi (a)}{a}=\left(\dfrac{L}{4\pi F}\right)^{1/2}$

and this is what we MEASURE in Astrophysics/Cosmology. Knowing $a(t)$, we can express the luminosity distance in terms of the redshift. Taylor expansion provides something like this:

$H_0d_L=z+\dfrac{1}{2}(1-q_0)z^2+\ldots$

where higher order terms are sometimes referred as “statefinder parameters/variables”. In particular, we have

$\boxed{H_0=\dfrac{\dot{a}_0}{a_0}}$

and

$\boxed{q_0=-\dfrac{a_0\ddot{a}_0}{\dot{a}_0^2}}$

C) Angular diameter distance.

If we know that some object has a known length $l$, and it gives some angular “aperture” or separation $\theta$, the angular diameter distance is given by

$\boxed{d_A=\dfrac{l}{\theta}}$

The comoving size is defined as $l/a$, and the coming distance is again $\chi (a)$. For “flat” space, we obtain that

$\theta=\dfrac{l/a}{\chi (a)}$

that is

$d_A=a\chi (a)=\dfrac{\chi}{1+z}$

In the case of “curved” spaces, we get

$d_A=\dfrac{a}{H_0\sqrt{\vert \omega_k\vert}}\cdot\begin{cases}\sinh \left( \sqrt{\Omega_k}H_0\chi\right),\Omega_k>0\\ \sin \left( \sqrt{-\Omega_k}H_0\chi\right),\Omega_k<0\end{cases}$

FRW dynamics

Gravity in General Relativity, a misnomer for the (locally) relativistic theory of gravitation, is described by a metric field, i.e., by a second range tensor (covariant tensor if we are purist with the nature of components). The metric field is related to the matter-energy-momentum content through the Einstein’s equations

$G_{\mu\nu}=-\kappa^2 T_{\mu\ nu}$

The left-handed side can be calculated for a FRW Universe as follows

$R_{00}=-3\dfrac{\ddot{a}}{a}$

$R_{ij}=(a\ddot{a}+2\dot{a}^2+2k)\overline{g}_{ij}$

$R=6\left(\dfrac{\ddot{a}}{a}+\dfrac{\dot{a}^2}{a^2}+\dfrac{k}{a^2}\right)$

The right-handed side is the energy-momentum of the Universe. In order to be fully consistent with the symmetries of the metric, the energy-momentum tensor MUST be diagonal and $T_{11}=T_{22}=T_{33}=T$. In fact, this type of tensor describes a perfect fluid with

$T_{\mu\nu}=(\rho+p)U_\mu U_\nu+pg_{\mu\nu}$

Here, $\rho, p$ are functions of $t$ (cosmological time) only. They are “state variables” somehow. Moreover, we have

$U_\mu =(1,0,0,0)$

for the fluid at rest in the comoving frame. The Friedmann equations are indeed the EFE for a FRW metric Universe

$3\left(\dfrac{\dot{a}^2}{a^2}+\dfrac{k}{a^2}\right)=\kappa^2\rho$ for the 00th compoent as “constraint equation.

$2\dfrac{\ddot{a}}{a}+\dfrac{\dot{a}^2}{a^2}+\dfrac{k}{a^2}=-\kappa^2p$ for the iith components.

Moreover, we also have

$G_{\mu\nu}^{;\nu}=T_{\mu\nu}^{;\nu}=0$

and this conservation law implies that

$\dot{\rho}+3\dfrac{\dot{a}}{a}(\rho+p)=0$

Therefore, we have got two independent equations for three unknowns $(a, \rho, p)$. We need an additional equation. In fact, the equation of state for $p=p(\rho)$ provides such an additional equation. It gives the “dynamics of matter”!

In summary, the basic equations for Cosmology in a FRW metric, via EFE, are the Friedmann’s equations (they are secretly the EFE for the FRW metric) supplemented with the energy-momentum conservations law and the equation of state for the pressure $p=p(\rho)$:

1) $\boxed{\dfrac{\dot{a}^2}{a^2}+\dfrac{k^2}{a^2}=\dfrac{\kappa^2}{3}\rho}$

2) $\boxed{\dot{\rho}+3\dfrac{\dot{a}}{a}(\rho+p)=0}$

3) $\boxed{p=p(\rho)}$

There are many kinds of “matter-energy” content of our interest in Cosmology. Some of them can be described by a simple equation of state:

$\boxed{p=\omega \rho}$

Energy-momentum conservation implies that $\rho\propto a^{-3(\omega +1)}$. 3 special cases are used often:

1st. Radiation (relativistic “matter”). $\omega=1/3$ and thus, $p=1/3\rho$ and $\rho\propto a^{-4}$

2nd. Dust (non-relativistic matter). $\omega=0$. Then, $p=0$ and $\rho\propto a^{-3}$

3rd. Vacuum energy (cosmological constant). $\omega=-1$. Then, $p=-\rho$ and $\rho=\mbox{constant}$

Remark (I): Particle physics enters Cosmology here! Matter dynamics or matter fields ARE the matter content of the Universe.

Remark (II): Existence of a Big Bang (and a spacetime singularity). Using the Friedmann’s equation

$\dfrac{\ddot{a}}{a}=-\dfrac{\kappa^2}{6}(\rho+3p)$

if we have that $(\rho+3p)>0$, the so-called weak energy condition, then $a=0$ should have been reached at some finite time in the past! That is the “Big Bang” and EFE are “singular” there. There is no scape in the framework of GR. Thus, we need a quantum theory of gravity to solve this problem OR give up the FRW metric at the very early Universe by some other type of metric or structure.

Particles and the chemical equilibrium of the early Universe

Today, we have DIRECT evidence for the existence of a “thermal” equilibrium in the early Universe: the cosmic microwave background (CMB). The CMB is an isotropic, accurate and non-homogeneous (over certain scales) blackbody spectrum about $T\approx 3K$!

Then, we know that the early Universe was filled with a hot dieal gas in thermal equilibrium (a temperature $T_e$ can be defined there) such as the energy density and pressure can be written in terms of this temperature. This temperature generates a distribution $f(\mathbf{x},\mathbf{p})$. The number of phase space elements in $d^3xd^3p$ is

$d^3xd^3p=\dfrac{d^3\mathbf{x}d^3\mathbf{p}}{(2\pi\hbar)^3}$

and where the RHS is due to the uncertainty principle. Using homogeneity, we get that, indeed, $f(x,p)=f(p)$, and where we can write the volume $d^3x=dV$. The energy density and the pressure are given by (natural units are used)

$\rho_i=g_i\int \dfrac{d^3p}{(2\pi)^3}f_i(p)E(p)$

$p_i=g_i\int \dfrac{d^3p}{(2\pi)^3}f_i (p)\dfrac{p^2}{3E(p)}$

When we are in the thermal equilibrium at temperature T, we have the Bose-Einstein/Fermi-Dirac distribution

$f(p)=\dfrac{1}{e^{(E-\mu)/T}\pm 1}$

and where the $+$ is for the Fermi-Dirac distribution (particles) and the $-$ is for the Bose-Einstein distribution (particles). The number density, the energy density and the pressure are the following integrals

$\boxed{\mbox{Number density}=n=\dfrac{N}{V}=\dfrac{g}{2\pi^2}\int_m^\infty \dfrac{(E^2-m^2)^{1/2}}{e^{(E-\mu)/T}\pm 1}dE}$

$\boxed{\mbox{Density energy}=\rho=\dfrac{E}{V}=\dfrac{g}{2\pi^2}\int_m^\infty \dfrac{(E^2-m^2)^{1/2}E^2}{e^{(E-\mu)/T}\pm 1}dE}$

$\boxed{\mbox{Pressure}=p=\dfrac{g}{6\pi^2}\int_m^\infty \dfrac{(E^2-m^2)^{3/2}}{e^{(E-\mu)/T}\pm 1}dE}$

And now, we find some special cases of matter-energy for the above variables:

1st. Relativistic, non-degenerate matter (e.g. the known neutrino species). It means that $T>>m$ and $T>>\mu$. Thus,

$n=\left(\dfrac{3}{4}\right)\dfrac{\zeta (3)}{\pi^2}gT^3$

$\rho=\left(\dfrac{7}{8}\right)\dfrac{\pi^2}{30}gT^4$

$p=\dfrac{1}{3}\rho$

2nd. Non-relativistic matter with $m>>T$ only. Then,

$n=g\left(\dfrac{mT}{2\pi}\right)^{3/2}e^{-(m-\mu)/T}$

$\rho= mn+\dfrac{3}{2}p$, and $p=nT<<\rho$

The total energy density is a very important quantity. In the thermal equilibrium, the energy density of non-relativistic species is exponentially smaller (suppressed) than that of the relativistic particles! In fact,

$\rho_R=\dfrac{\pi^2}{30}g_\star T^4$ for radiation with $p_R=\dfrac{1}{3}\rho_R$

and the effective degrees of freedom are

$\displaystyle{\boxed{g_\star=\sum_{bosons}g_b+\dfrac{7}{8}\sum_{fermions}g_f}}$

Remark: The factor $7/8$ in the DOF and the variables above is due to the relation between the Bose-Einstein and the Fermi-Dirac integral in d=3 space dimensions. In general d, the factor would be

$(1-\dfrac{1}{2^d})=\dfrac{2^d-1}{2^d}$

Entropy conservation and the early Universe

The entropy in a comoving volume IS a conserved quantity IN THE THERMAL EQUILIBRIUM. Therefore, we have that

$\dfrac{\partial p_i}{\partial T}=g_i\int \dfrac{d^3p}{(2\pi)^3}\dfrac{df}{dT}\dfrac{p^2}{3E(p)}=g_i\int \dfrac{4\pi pE dE}{(2\pi)^3}\dfrac{df}{dE}\left(-\dfrac{E}{T}\right)\dfrac{p^2}{3E}$

and then

$\dfrac{\partial p_i}{\partial T}=\dfrac{g_i}{2\pi^2}\int \left(-\dfrac{d}{dE}\left(f\dfrac{p^3E}{3T}\right)+f\dfrac{d}{dE}\left(\dfrac{p^3E}{3T}\right)\right)dE$

or

$\dfrac{\partial p_i}{\partial T}=\dfrac{1}{T}(\rho_i+p_i)$

Now, since

$\dfrac{\partial \rho}{\partial t}+3\dfrac{\dot{a}}{a}(\rho+p)=0$

then

$\dfrac{\partial}{\partial t}\left(a^3(\rho+p)\right)-a^3\dfrac{\partial p}{\partial t}=0$

$\dfrac{1}{a^3}\dfrac{\partial (a^3(\rho +p))}{\partial t}-\dfrac{\partial \rho}{\partial t}=0$

if we multiply by $T$ and use the chain rule for $\rho$, we obtain

$\dfrac{1}{a^3}\dfrac{\partial}{\partial t}\left(\dfrac{a^3(\rho+p)}{T}\right)=0$

but it means that $a^3s=\mbox{constant}$, where $s$ is the entropy density defined by

$\boxed{s\equiv \dfrac{\rho+p}{T}}$

Well, the fact is that we know that the entropy or more precisely the entropy density is the early Universe is dominated by relativistic particles ( this is “common knowledge” in the Stantard Cosmological Model, also called $\Lambda CDM$). Thus,

$\boxed{s=\dfrac{2\pi^2}{45}g_\star T^3}$

It implies the evolution of temperature with the redshift in the following way:

$T\propto g_\star^{-1/3}a^{-1}$

Indeed, since we have that $n\propto a^{-3}$, $s\propto a^{-3}$, the yield variable

$Y_i\equiv \dfrac{n_i}{s}$

is a convenient quantity that represents the “abundance” of decoupled particles.

See you in my next cosmological post!

Advertisements


text:
Can't list bilibliography in order cited using ReVTeX 4.1

The journal I am submitting to requires the bibliography be listed in the order cited in the text. How can I do this with ReVTeX?

I have tried different journal styles but can't get it to work. If I try to override the natbib settings there is an error. I have tried the \bibliographystyle settings as well but those seem to be ignored.

-
A simple test using revtex4-1 and \bibliographystyle{unsrtnat} worked as expected. Can you please add to your question a minimal working example showing the settings relevant to this problem? – Gonzalo Medina May 22 '12 at 16:00
Thank you for the response: my settings are as follows: \documentclass[superscriptaddress,onecolumn,reprint]{revtex4-1} \usepackage{array} \usepackage{amsmath} \usepackage{graphicx} \usepackage{verbatim} \usepackage{color} \usepackage{subfigure} \usepackage{epspdfconversion} \usepackage{hyperref} I am not using a bib file as it is not allowed. I have the bibitems taken from the bbl file. I tried \bibliographystyle{unsrtnat} but the articles are still listed in alphabetical order by author's last name. I need them in the order they are cited in the text. – user14860 May 23 '12 at 5:17

The revtex4-1 document class automatically loads the natbib package with the option sort&compress. For numerical citation methods, this ensures that an otherwise wild-looking list such as [4,2,8,3] will be rendered as [2-4,8]; for author-year citation methods, this option is equivalent to (simple) sorting by year (and, if necessary, within year by affixing a, b, etc).

If the journal you're submitting your paper to does not want the bibliography to be sorted alphabetically by authors' last names, it should also specify which bibliography style file should be used. If the journal for some reason has failed to do so, you should select a style file -- unsrtnat.bst would be a good candidate, given that you're likely using natbib's citation commands -- that does not perform any sorting of the bib entries. Incidentally, after selecting a new style file, it may be necessary to wipe out the auxiliary files and perform the latex-bibtex-latex-latex sequence from scratch.

A separate issue: have you been able to determine which basic citation method is required by the journal: numeric or authoryear? If it's the former, you should include the instruction

\setcitestyle{numbers}

in the preamble. If it's the latter, you may want to play it safe by including the instruction

\setcitestyle{authoryear}

in the preamble. This is actually the default when natbib is set up. However, it may be the case that some of the other files you're loading override this default, in which case it may be necessary to restore the default setting explicitly.

-

I was able to resolve the issue because I made what appears to be a basic LaTeX error that I did not accurately describe above.

The journal I am subbmitting to does not allow bib files, therefore the use of \begin{thebibliography} with \bibitem's is required. It seems that the order of bibitem's is taken directly.

When I replaced these items with a bib file, the correct numbering was restored. I then copy/pasted the newly ordered bibitem's from the bbl file and the order was maintained. Thanks to everyone who commented/answered my question.

-

format: plain


text:
## Elementary Algebra

$-21-12i$
We multiply each term of the first complex number with the second complex number and then simplify: $(-2-3i)(6-3i)$ =$-2(6-3i)-3i(6-3i)$ =$-12+6i-18i+9i^{2}$ =$-12-12i+9(-1)$ [We substitute -1 in place of $i^{2}$ as $i^{2}=-1$] =$-12-9-12i$ =$-21-12i$


text:
It’s always fascinating to see what’s going on in textbooks of the olden days, and National Treasure @mathsjem recently found a beauty of its type. Look at those whences! Check out the subjunctives! It thrills the heart, doesn’t it? ((If one were so inclined, one might even shiver in ecstasy.))

What caught my attention, though, was evolution - in this context, taking square or cube (or, presumably, higher) roots of an expression.

I haven’t studied this method in detail - it looks similar in flavour to finding square roots by long division, which International Legend @colinthemathmo has discussed, at least for numbers .

However, it got me thinking: how would I find the cube root of $c(x) = 27 + 108x + 90x^2 + 80x^3 - 60x^4 + 48x^5 - 8x^6$?

### Quick and dirty

If we’re allowed to assume that $c(x)$ is a perfect cube, then it’s not tricky at all: it’s a degree-6 polynomial, so its cube root must be quadratic; the constant term is clearly 3 and the $x^2$ coefficient is just as clearly -2.

That means $c(x) = \br{ 3 + bx - 2x^2}^3$, for some value of $b$.

Assuming $x$ is small enough that we can ignore $x^2$ terms and higher, expanding the bracket with the binomial expansion gives $c(x) \approx 27 + 27bx$, and $b=4$.

Expanding the whole thing properly gives the same answer - which I leave as an exercise.

### Binomial expansion

As an alternative to assuming the form of the solution, it’s also possible to use the general binomial expansion (at least for small $x$) to figure out the cube root of $c(x)$.

Suppose we write $c(x) = 27 + x\br{108 + 90x + 80x^2 + …}$ and work out $\br{c(x)}^{\frac{1}{3}}$?

The general form for $\br{a+bx}^{\frac{1}{3}}$ is $a^{\frac{1}{3}} + \frac{1}{3}a^{-\frac{2}{3}}bx - \frac{1}{9}a^{-\frac{5}{3}}b^2x^2 + …$

That looks a mess; however, $a$ behaves quite nicely here and simplifies the whole thing down to $3 + \frac{1}{27} bx - \frac{1}{3^{7}}b^2 x^2 + …$

There’s a bit of book-keeping to do here, because $b$ is a bit of a monster, but stick with it: $\frac{1}{27} bx = 4x + \frac{10}{3}x^2 + …$, while $\frac{1}{3^7} b^2x^2 = \frac{108^2}{3^7} x^2 + …$.

What’s $\frac{108^2}{3^7}$? I don’t know off the top of my head. I do know that $108 = 2^2 \times 3^3$, so $108^2 = 2^4 \times 3^6$ …

#### A ninja!

Whoosh

“It’s 11,664, because $(100+k)^2 = 10,000 + 200k + k^2$.”

Whoosh

#### As I was saying

… which leaves us with $\frac{16}{3}$.

So, we have an answer – assuming we only need to go up to the $x^2$ term – of $3 + 4x + \br{\frac{10}{3}- \frac{16}{3}}x^2$, which works out to $3 + 4x - 2x^2$ again.

### Checking

The only problem left is checking whether we really have gone far enough. One option is to cube the answer, but you don’t really want to be doing that every time you extract a term.

Presumably, something in the original method given in Jo’s Victorian textbook checks for a remainder, but I don’t really feel like wading through it to figure it out. Perhaps you, dear reader, can enlighten us all?


text:
Topological Insights into Sparse Neural Networks

Shiwei Liu, T.T. van der Lee, Anil Yaman, Zahra Atashgahi, Davide Ferrar, Ghada Sokar, Mykola Pechenizkiy, Decebal C. Mocanu

Research output: Chapter in Book/Report/Conference proceedingConference contributionAcademicpeer-review

Abstract

Sparse neural networks are effective approaches to reduce the resource requirements for the deployment of deep neural networks. Recently, the concept of adaptive sparse connectivity, has emerged to allow training sparse neural networks from scratch by optimizing the sparse structure during training. However, comparing different sparse topologies and determining how sparse topologies evolve during training, especially for the situation in which the sparse structure optimization is involved, remain as challenging open questions. This comparison becomes increasingly complex as the number of possible topological comparisons increases exponentially with the size of networks. In this work, we introduce an approach to understand and compare sparse neural network topologies from the perspective of graph theory. We first propose Neural Network Sparse Topology Distance (NNSTD) to measure the distance between different sparse neural networks. Further, we demonstrate that sparse neural networks can outperform over-parameterized models in terms of performance, even without any further structure optimization. To the end, we also show that adaptive sparse connectivity can always unveil a plenitude of sparse sub-networks with very different topologies which outperform the dense model, by quantifying and comparing their topological evolutionary processes. The latter findings complement the \textit{Lottery Ticket Hypothesis} by showing that there is a much more efficient and robust way to find winning tickets". Altogether, our results start enabling a better theoretical understanding of sparse neural networks, and demonstrate the utility of using graph theory to analyze them.
Original language English Proceedings of the European Conference on Machine Learning and Principles and Practice of Knowledge Discovery in Databases (ECML PKDD) 2020. 17 Published - 4 Jun 2020

Keywords

• sparse neural networks
• neural network sparse topology distance


text:
# Entropy as log of omega (phase space volume)

1. Jan 29, 2015

### diegzumillo

1. The problem statement, all variables and given/known data
I've seen this problem appear in more than one textbook almost without any changes. It goes like this:
Assume the entropy $S$ depends on the volume $\bar{\Omega}$ inside the energy shell: $S(\bar{\Omega})=f(\bar{\Omega})$. Show that from the additivity of $S$ and the multiplicative character of $\bar{\Omega}$, it follows that $S=const \times log \bar{\Omega}$

2. Relevant equations

3. The attempt at a solution
I've found a couple of solutions already (one in Pathria), that consists of considering two subsystems and calculating the derivatives of S in respect to the omegas, plus a bunch of assumptions. But from the problem statement I can't help but think there's got to be a simpler way. I'm trying to expand S as a sum of $f(\Omega_i)$, expand each f as a power series, and then use the fact that omega is the product of the omegas of the subsystems. But that leads nowhere.

Slightly off-topic: I'm taking a grad course on statistical mechanics but my previous knowledge on stat mech is very weak, so I'll probably be on these forums frequently throughout the semester. Is this the appropriate forum for stat mech homework questions?

2. Jan 30, 2015

### Staff: Mentor

I don't see how power series would be easier than the described approach.

Looking at two subsystems together leads to $f(xy)=f(x)+f(y)$ (where x,y are the Ω of two different systems). Calculating the derivative with respect to x gives $yf'(xy)=f'(x)$, the derivative of this with respect to y leads to $f'(xy)+xyf''(xy)=0$ or (using z=xy) $f'(z) = -zf''(z)$. This differential equation is solved by $f'(z)=\frac{c}{z}$ which leads to $f(z)=c \log(z)$.

It is.

3. Jan 31, 2015

### diegzumillo

Thanks :) That makes perfect sense. I solved it in a similar way but made some unnecessary turns here and there and it made things look more complicated.


text:
# Carmichael number

Jump to: navigation, search

A Carmichael number is a composite number named after the mathematician Robert Daniel Carmichael. A Carmichael number $\scriptstyle c\$ divides $\scriptstyle a^c - a\$ for every integer $\scriptstyle a\$. A Carmichael number c also satisfies the congruence $\scriptstyle a^{c-1} \equiv 1 \pmod c$, if $\scriptstyle \operatorname{gcd}(a,c) = 1$. The first few Carmichael numbers are 561, 1105, 1729, 2465, 2821, 6601 and 8911. In 1994 Pomerance, Alford and Granville proved that there exist infinitely many Carmichael numbers.

##  Properties

• Every Carmichael number is square-free and has at least three different prime factors
• For every Carmichael number c it holds that c − 1 is divisible by pn − 1 for every one of its prime factors pn.
• Every absolute Euler pseudoprime is a Carmichael number.

##  Chernick's Carmichael numbers

J. Chernick found in 1939 a way to construct Carmichael numbers[1] [2]. If, for a natural number n, the three numbers $\scriptstyle 6n+1\$, $\scriptstyle 12n+1\$ and $\scriptstyle 18n+1\$ are prime numbers, the product $\scriptstyle M_3(n) = (6n+1)\cdot (12n+1)\cdot (18n+1)$ is a Carmichael number. This condition can only be satisfied if the number $n\$ ends with 0, 1, 5 or 6. An equivalent formulation of Chernick's construction is that if $\scriptstyle m\$, $\scriptstyle 2m-1\$ and $\scriptstyle 3m-2$ are prime numbers, then the product $\scriptstyle m\cdot (2m-1)\cdot (3m-2)$ is a Carmichael number.

This way to construct Carmichael numbers may be extended[3] to

$M_k(n)=(6n+1)(12n+1)\prod_{i=1}^{k-2}(9\cdot 2^i n+1) \,$

with the condition that each of the factors is prime and that $n\$ is divisible by 2k − 4.

## Distribution of Carmichael numbers

Let C(X) denote the number of Carmichael numbers less than or equal to X. Then for all sufficiently large X

$X^{2/7} < C(X) < X \exp(-\log X \log\log\log X / \log\log X) . \,$

The upper bound is due to Erdős(1956)[4] and Pomerance, Selfridge and Wagstaff (1980)[5] and the lower bound is due to Alford, Granville and Pomerance (1994)[6]. The asymptotic rate of growth of C(X) is not known.[7]

## References and notes

1. J. Chernick, "On Fermat's simple theorem", Bull. Amer. Math. Soc. 45 (1939) 269-274
2. (2003-11-22) Generic Carmichael Numbers
3. Paulo Ribenboim, The new book of prime number records, Springer-Verlag (1996) ISBN 0-387-94457-5. P.120
4. Paul Erdős, "On pseudoprimes and Carmichael numbers", Publ. Math. Debrecen 4 (1956) 201-206. MR 18 18
5. C. Pomerance, J.L. Selfridge and S.S. Wagstaff jr, "The pseudoprimes to 25.109", Math. Comp. 35 (1980) 1003-1026. MR 82g:10030
6. W. R. Alford, A. Granville, and C. Pomerance. "There are Infinitely Many Carmichael Numbers", Annals of Mathematics 139 (1994) 703-722. MR 95k:11114
7. Richard Guy, "Unsolved problems in Number Theory" (3rd ed), Springer-Verlag (2004) ISBN 0-387-20860-7. Section A13
 Some content on this page may previously have appeared on Citizendium.


text:
Dear Uncle Colin,

I’ve been challenged to expand $(x-a)(x-b)(x-c)…(x-z)$ so I can experience the ‘joy of mathematics’. I’ve noticed that towards the end, one of the brackets is $(x-x)$ - does that mean the whole thing is zero?

Is that the ‘joy of mathematics they mean’?

- Just Occupied Yesterday Leadenly Expanding Stupid Structures

Dear JOYLESS, thanks for your message!

One person’s ‘joy of mathematics’ is another’s ‘slightly mean trick’. You’re absolutely right that the expected solution to this is 0, and the ‘joy’ is the sense of smugness you get from spotting that rather than expanding the whole thing.

There’s also an argument that it pays to think about a problem before mindlessly bashing away at it, and I’m certainly an advocate of engaging the brain before anything else - but I think this problem is on the dirty side of that.

On the plus side, if you’ve spent the day expanding the stupid thing, you’re not going to make the same mistake again!

-- Uncle Colin


text:
# Average cost Question 2 View Policies Show Attempt History Current Attempt in Progress Sandhill Co. had...

###### Question:

Average cost

Question 2 View Policies Show Attempt History Current Attempt in Progress Sandhill Co. had a beginning inventory of 99 units of Product RST at a cost of $8 per unit. During the year, purchases wer Feb 20 636 units at$9 May 5 520 units at $10 Aug. 12 400 units at$11 Dec. 8 90 units at $12 Sandhill Couses a periodic inventory system. Sales totaled 1.460 units. Your answer is correct. (a) Determine the cost of goods available for sale. 17196 The cost of goods available for sale$ o ella
That Rod Day Tom. Mergieven Chi Pierogies And Saus. Game Over Video Your answer is partially correct. mine the ending inventory and the cost of goods sold under each of the assumed cost flow methods (FIFO, LIFO, re-cost). (Round answers to decimal places, e.. 150.) FIFO LIFO AVERAGE-COST 3 225 $2466 nding inventory S ost of goods sold$ 13971 147305 extbook and Media of Accounts Your answer is correct

#### Similar Solved Questions

##### HO HO O H O H H H H H OH OH OH OH OH H...
HO HO O H O H H H H H OH OH OH OH OH H OH cs a) Circle the glycosidic bond. b) Identify the type of glycosidic bond that is present in this molecule. c) What type of molecule is this? I- H...
##### For the question where should the control limits in an x-bar chart be placed if the...
For the question where should the control limits in an x-bar chart be placed if the design process sets α = 0.01 with the following parameters. average is 100, standard deviation is 20 and n = 25? I need to find zα/2 . the book says α = 0.01 puts the control limits at zα̴...
##### All are multiple choice quesions 1.   Probability is A. Subjective judgement of the observer, how likely...
All are multiple choice quesions 1.   Probability is A. Subjective judgement of the observer, how likely is something.; B. A measure rendered (adjusted to events) satisfying certain rules; C. Empirical, observed result of the (favorable) over (all) cases; D. Any number between 0 and 1; &nb...


text:
Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 12 Feb 2016, 23:58

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Calling All Consortium 2013 Applicants

Author Message
Manager
Joined: 04 Feb 2012
Posts: 84
Followers: 0

Kudos [?]: 13 [0], given: 17

Re: Calling All Consortium 2013 Applicants [#permalink]  07 Jan 2013, 16:29
Fewer consortium peeps. Plus everyone I know, submitted Saturday. Very few of us round 1
Intern
Joined: 02 Dec 2012
Posts: 47
Location: United States
Concentration: General Management, Technology
GPA: 3.6
WE: Law (Law)
Followers: 3

Kudos [?]: 10 [0], given: 5

Re: Calling All Consortium 2013 Applicants [#permalink]  07 Jan 2013, 17:52
I don't know what is going on. I called today and found out my rec met the deadline so I am excited. I did Michigan, UVA, Cornell and UNC. I'm hoping it works out! I wonder if they will maintain the number of fellowships despite a reduced applicant pool.

Its not just the consortium. Here are some interesting articles on the sharp reduction in MBA applications.

http://online.wsj.com/article/SB1000087 ... 32518.html

My philosophy on buying education is the same as anything else. Go when demand is low and pay a reduced price (scholarship $$). Intern Joined: 03 Oct 2012 Posts: 46 Location: United States Concentration: Healthcare, Strategy GMAT 1: 750 Q47 V46 GRE 1: 1540 Q790 V750 GPA: 3.65 WE: Consulting (Consulting) Followers: 1 Kudos [?]: 42 [0], given: 9 Re: Calling All Consortium 2013 Applicants [#permalink] 08 Jan 2013, 06:21 Did anyone else receive an update email from Berkeley recently? We were excited to receive your Consortium application to the Berkeley MBA program. We know how much time and effort you put into the application, and we appreciate your strong interest in Berkeley-Haas. We have started to review your full application file. A Berkeley-Haas application login has been created for you to check your status online and update us on changes, such as if you have a new address or email. To log into the Berkeley-Haas application visit the link below and enter your user login and password information. Intern Joined: 01 May 2012 Posts: 4 Location: United States Concentration: Finance GMAT 1: 710 Q V GPA: 3.2 Followers: 0 Kudos [?]: 5 [0], given: 0 Re: Calling All Consortium 2013 Applicants [#permalink] 08 Jan 2013, 09:16 Has anyone heard from any schools yet?? Manager Joined: 04 Feb 2012 Posts: 84 Followers: 0 Kudos [?]: 13 [0], given: 17 Re: Calling All Consortium 2013 Applicants [#permalink] 08 Jan 2013, 09:18 No. Very anxious Intern Joined: 07 Dec 2012 Posts: 42 Location: United States Followers: 0 Kudos [?]: 19 [0], given: 13 Re: Calling All Consortium 2013 Applicants [#permalink] 08 Jan 2013, 09:42 HinConsulting wrote: Did anyone else receive an update email from Berkeley recently? We were excited to receive your Consortium application to the Berkeley MBA program. We know how much time and effort you put into the application, and we appreciate your strong interest in Berkeley-Haas. We have started to review your full application file. A Berkeley-Haas application login has been created for you to check your status online and update us on changes, such as if you have a new address or email. To log into the Berkeley-Haas application visit the link below and enter your user login and password information. yep. Intern Joined: 10 Oct 2012 Posts: 35 Followers: 0 Kudos [?]: 4 [0], given: 0 Re: Calling All Consortium 2013 Applicants [#permalink] 08 Jan 2013, 12:12 Did anyone else have issues with the Consortium receiving test scores? They only accept paper GRE and have not received mine yet (though it was sent on the 26th and received in my own mailbox). They are holding my application until they get it, and won't verify my scores via phone. I'm super anxious this is going to push my application out too far, even though I ordered new scores as soon as I knew. Any advice? I am really bummed they don't accept online scores, they would be reported immediately! Good luck to all of you! Intern Joined: 02 Dec 2012 Posts: 47 Location: United States Concentration: General Management, Technology GPA: 3.6 WE: Law (Law) Followers: 3 Kudos [?]: 10 [0], given: 5 Re: Calling All Consortium 2013 Applicants [#permalink] 08 Jan 2013, 17:32 Still haven't heard anything... Manager Joined: 03 Oct 2010 Posts: 63 Location: United States (NY) Concentration: Entrepreneurship, Finance GMAT 1: 720 Q48 V40 GPA: 3.23 WE: Asset Management (Other) Followers: 0 Kudos [?]: 10 [0], given: 2 Re: Calling All Consortium 2013 Applicants [#permalink] 08 Jan 2013, 17:36 Nada here either Current Student Joined: 26 Jul 2012 Posts: 63 Followers: 0 Kudos [?]: 8 [0], given: 8 Re: Calling All Consortium 2013 Applicants [#permalink] 08 Jan 2013, 20:03 R2 here. nada. Manager Joined: 19 Aug 2012 Posts: 122 Concentration: Finance, International Business Schools: Booth '15 (M) GMAT 1: 700 Q48 V38 Followers: 2 Kudos [?]: 17 [0], given: 10 Re: Calling All Consortium 2013 Applicants [#permalink] 09 Jan 2013, 04:18 i got the 'thanks for applying we will be sending to yuour schools soon' email thats all. hold on to your horses haha will take a week or more im sure Manager Status: Matriculating at Stern as a Consortium Fellow ! Joined: 28 Jun 2012 Posts: 119 Location: United States (NY) Concentration: Nonprofit, Social Entrepreneurship GMAT 1: 740 Q48 V44 Followers: 3 Kudos [?]: 28 [0], given: 46 Re: Calling All Consortium 2013 Applicants [#permalink] 09 Jan 2013, 08:29 I applied R1. Just TODAY got word that Tuck opened my app... (though I didn't do a self-initiated interview). Also, WL R1 for Darden and confirmed my interest in being on the WL when that happened (in December). last night got another email to REconfirm my interest in the WL.... _________________ The only way to atone for being occasionally a little over-dressed is by being always absolutely over-educated. Oscar Wilde Manager Joined: 04 Feb 2012 Posts: 84 Followers: 0 Kudos [?]: 13 [0], given: 17 Re: Calling All Consortium 2013 Applicants [#permalink] 09 Jan 2013, 09:43 Yea I just same email from Tuck that my app is complete and they will start working on it..... I submitted almost 2 months ago. Intern Joined: 09 Jan 2013 Posts: 2 Followers: 0 Kudos [?]: 0 [0], given: 0 Confirmation on application submission? [#permalink] 09 Jan 2013, 09:46 Hello all - I just submitted my Consortium application on Jan 5th! I still haven't received a confirmation that my application has been processed. Other than the automated email that you get that your application has been submitted, has anyone received any other notifications? This process is nerve racking and exciting at the same time!!! Jessica. Manager Joined: 19 Aug 2012 Posts: 122 Concentration: Finance, International Business Schools: Booth '15 (M) GMAT 1: 700 Q48 V38 Followers: 2 Kudos [?]: 17 [0], given: 10 Re: Calling All Consortium 2013 Applicants [#permalink] 09 Jan 2013, 10:00 i think tuck groups r1 and r2 together, and i only received the automated email -- consortium says theyll notify us when they progress further to the next steps Manager Status: 2013 Joined: 26 Jun 2010 Posts: 62 Followers: 0 Kudos [?]: 17 [0], given: 1 Re: Calling All Consortium 2013 Applicants [#permalink] 09 Jan 2013, 10:09 helpmehelpme wrote: i think tuck groups r1 and r2 together, and i only received the automated email -- consortium says theyll notify us when they progress further to the next steps According to their website http://www.tuck.dartmouth.edu/admission ... ant-dates/ ,R1 Consortium applicants get decision on 2/8 while R2 applicants find out on 3/15. I however believe they wait to see the whole consortium pool applying to Tuck before making decision....just IMO Intern Joined: 07 Dec 2012 Posts: 42 Location: United States Followers: 0 Kudos [?]: 19 [0], given: 13 Re: Calling All Consortium 2013 Applicants [#permalink] 09 Jan 2013, 10:55 have any round 1 applicants received an invitation interview at nyu? Manager Joined: 04 Feb 2012 Posts: 84 Followers: 0 Kudos [?]: 13 [0], given: 17 Re: Calling All Consortium 2013 Applicants [#permalink] 09 Jan 2013, 14:21 No word from NYU yet, but people outside consortium have already been waitlisted/dinged. We've made it this far..... Manager Status: Matriculating at Stern as a Consortium Fellow ! Joined: 28 Jun 2012 Posts: 119 Location: United States (NY) Concentration: Nonprofit, Social Entrepreneurship GMAT 1: 740 Q48 V44 Followers: 3 Kudos [?]: 28 [0], given: 46 Re: Calling All Consortium 2013 Applicants [#permalink] 09 Jan 2013, 18:39 peeweeninja wrote: Still haven't heard anything... Just an FYI for R2 people... I didn't hear from some schools for a few weeks. After you apply, it will probably take 3-5 days for the consortium to "release" your app to the schools. They send you an email at this point. I applied on 11/15. My app went to member schools on 11/20 and it took till 12/15 for Cornell Johnson to acknowledge receipt of it (12/7 for Stern, 12/5 for US, you get the picture). Sit tight and wait it out. On the bright side, some R1 Consortium schools wait for R2 applicants so now we should all hear back soon (ish!). _________________ The only way to atone for being occasionally a little over-dressed is by being always absolutely over-educated. Oscar Wilde Manager Status: Matriculating at Stern as a Consortium Fellow ! Joined: 28 Jun 2012 Posts: 119 Location: United States (NY) Concentration: Nonprofit, Social Entrepreneurship GMAT 1: 740 Q48 V44 Followers: 3 Kudos [?]: 28 [0], given: 46 Re: Calling All Consortium 2013 Applicants [#permalink] 09 Jan 2013, 18:42 peeweeninja wrote: I don't know what is going on. I called today and found out my rec met the deadline so I am excited. I did Michigan, UVA, Cornell and UNC. I'm hoping it works out! I wonder if they will maintain the number of fellowships despite a reduced applicant pool. Its not just the consortium. Here are some interesting articles on the sharp reduction in MBA applications. http://online.wsj.com/article/SB1000087 ... 32518.html http://www.businessweek.com/articles/20 ... on-drought http://www.chicagobusiness.com/article/ ... -and-booth My philosophy on buying education is the same as anything else. Go when demand is low and pay a reduced price (scholarship$$\$).

FYI these stats (at least for WSJ) are for 2011 applicants starting 2012 (second sentence) "Applications for two-year, full-time M.B.A. programs that start this fall "... So these aren't for us. But may still reflect our application climate.
_________________

The only way to atone for being occasionally a little over-dressed is by being always absolutely over-educated.
Oscar Wilde

Re: Calling All Consortium 2013 Applicants   [#permalink] 09 Jan 2013, 18:42

Go to page   Previous    1  ...  5   6   7   8   9   10   11  ...  56    Next  [ 1110 posts ]

Similar topics Replies Last post
Similar
Topics:
66 Calling All 2016 Consortium Applicants!!! 612 14 Aug 2015, 04:05
531 Calling All 2015 Consortium Applicants Thread 1724 08 Aug 2014, 14:23
99 Calling All 2014 Consortium Applicants 559 24 May 2013, 07:15
580 Calling All Fall 2012 Consortium Applicants! 2345 27 Jun 2011, 05:55
62 Calling all Fall 2011 Consortium Applicants! 2638 07 Jun 2010, 23:12
Display posts from previous: Sort by

Calling All Consortium 2013 Applicants

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.


text:
Finding Primes!

$$a$$, $$b$$, $$c$$ and $$d$$ are integers such that $$0<a,b,c,d<999$$ and $$ab=cd$$.

How many ordered sets of integers $$(a,b,c,d)$$ satisfy

$a+b+c+d=p$

where $$p$$ is a prime number?

×


text:
Home > Error Propagation > Propagating Error Logarithm

# Propagating Error Logarithm

## Contents

If you know that there is some specific probability of $x$ being in the interval $[x-\Delta x,x+\Delta x]$, then obviously $y$ will be in $[y_-,y_+]$ with that same probability. We are now in a position to demonstrate under what conditions that is true. JCGM. Chemistry Biology Geology Mathematics Statistics Physics Social Sciences Engineering Medicine Agriculture Photosciences Humanities Periodic Table of the Elements Reference Tables Physical Constants Units and Conversions Organic Chemistry Glossary Search site Search useful reference

Just square each error term; then add them. Plugging this value in for ∆r/r we get: (∆V/V) = 2 (0.05) = 0.1 = 10% The uncertainty of the volume is 10% This method can be used in chemistry as Disadvantages of Propagation of Error Approach Inan ideal case, the propagation of error estimate above will not differ from the estimate made directly from the measurements. Uncertainties can also be defined by the relative error (Δx)/x, which is usually written as a percentage.

## How To Calculate Uncertainty Of Logarithm

more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed At this mathematical level our presentation can be briefer. doi:10.2307/2281592. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Developers Cookie statement Mobile view Appendix A Errors and Uncertainties Course Support Lab Contents Natural Logarithms The absolute uncertainty in a natural log (logarithms

What kind of bugs do "goto" statements lead to? Or in matrix notation, f ≈ f 0 + J x {\displaystyle \mathrm σ 6 \approx \mathrm σ 5 ^ σ 4+\mathrm σ 3 \mathrm σ 2 \,} where J is f = ∑ i n a i x i : f = a x {\displaystyle f=\sum _ σ 4^ σ 3a_ σ 2x_ σ 1:f=\mathrm σ 0 \,} σ f 2 Logarithmic Error Bars Square Terms: $\left(\dfrac{\delta{x}}{\delta{a}}\right)^2(da)^2,\; \left(\dfrac{\delta{x}}{\delta{b}}\right)^2(db)^2, \;\left(\dfrac{\delta{x}}{\delta{c}}\right)^2(dc)^2\tag{4}$ Cross Terms: $\left(\dfrac{\delta{x}}{da}\right)\left(\dfrac{\delta{x}}{db}\right)da\;db,\;\left(\dfrac{\delta{x}}{da}\right)\left(\dfrac{\delta{x}}{dc}\right)da\;dc,\;\left(\dfrac{\delta{x}}{db}\right)\left(\dfrac{\delta{x}}{dc}\right)db\;dc\tag{5}$ Square terms, due to the nature of squaring, are always positive, and therefore never cancel each other out.

Now that we have done this, the next step is to take the derivative of this equation to obtain: (dV/dr) = (∆V/∆r)= 2cr We can now multiply both sides of the Uncertainty Logarithm Base 10 Write an expression for the fractional error in f. The exact covariance of two ratios with a pair of different poles p 1 {\displaystyle p_{1}} and p 2 {\displaystyle p_{2}} is similarly available.[10] The case of the inverse of a These play the very important role of "weighting" factors in the various error terms.

Derivation of Exact Formula Suppose a certain experiment requires multiple instruments to carry out. Absolute Uncertainty Exponents JCGM 102: Evaluation of Measurement Data - Supplement 2 to the "Guide to the Expression of Uncertainty in Measurement" - Extension to Any Number of Output Quantities (PDF) (Technical report). References Skoog, D., Holler, J., Crouch, S. This can aid in experiment design, to help the experimenter choose measuring instruments and values of the measured quantities to minimize the overall error in the result.

## Uncertainty Logarithm Base 10

If you just want a rough-and-ready error bars, though, one fairly trusty method is to draw them in between $y_\pm=\ln(x\pm\Delta x)$. http://chem.libretexts.org/Core/Analytical_Chemistry/Quantifying_Nature/Significant_Digits/Propagation_of_Error John Wiley & Sons. How To Calculate Uncertainty Of Logarithm p.2. Error Propagation Ln External links A detailed discussion of measurements and the propagation of uncertainty explaining the benefits of using error propagation formulas and Monte Carlo simulations instead of simple significance arithmetic Uncertainties and

Uncertainty in logarithms to other bases (such as common logs logarithms to base 10, written as log10 or simply log) is this absolute uncertainty adjusted by a factor (divided by 2.3 see here Also, the reader should understand tha all of these equations are approximate, appropriate only to the case where the relative error sizes are small. [6-4] The error measures, Δx/x, etc. Statistical theory provides ways to account for this tendency of "random" data. The standard form error equations also allow one to perform "after-the-fact" correction for the effect of a consistent measurement error (as might happen with a miscalibrated measuring device). How To Find Log Error In Physics

Sometimes, these terms are omitted from the formula. Example 3: Do the last example using the logarithm method. soerp package, a python program/library for transparently performing *second-order* calculations with uncertainties (and error correlations). http://bsdupdates.com/error-propagation/propagating-error-in-excel.php Contributors http://www.itl.nist.gov/div898/handb...ion5/mpc55.htm Jarred Caldwell (UC Davis), Alex Vahidsafa (UC Davis) Back to top Significant Digits Significant Figures Recommended articles There are no recommended articles.

Generated Mon, 24 Oct 2016 15:40:07 GMT by s_nt6 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.8/ Connection Error Propagation Calculator take upper bound difference directly as the error) since averaging would dis-include the potential of ln (x + delta x) from being a "possible value". log R = log X + log Y Take differentials.

## ISBN0470160551.[pageneeded] ^ Lee, S.

Starting with a simple equation: $x = a \times \dfrac{b}{c} \tag{15}$ where $$x$$ is the desired results with a given standard deviation, and $$a$$, $$b$$, and $$c$$ are experimental variables, each Notice the character of the standard form error equation. In problems, the uncertainty is usually given as a percent. Error Propagation Square Root So long as the errors are of the order of a few percent or less, this will not matter.

Note Addition, subtraction, and logarithmic equations leads to an absolute standard deviation, while multiplication, division, exponential, and anti-logarithmic equations lead to relative standard deviations. One immediately noticeable effect of this is that error bars in a log plot become asymmetric, particularly for data that slope downwards towards zero. These methods build upon the "least squares" principle and are strictly applicable to cases where the errors have a nearly-Gaussian distribution. Get More Info is formed in two steps: i) by squaring Equation 3, and ii) taking the total sum from $$i = 1$$ to $$i = N$$, where $$N$$ is the total number of

Note that even though the errors on x may be uncorrelated, the errors on f are in general correlated; in other words, even if Σ x {\displaystyle \mathrm {\Sigma ^ σ This is desired, because it creates a statistical relationship between the variable $$x$$, and the other variables $$a$$, $$b$$, $$c$$, etc... Correlation can arise from two different sources. Please try the request again.

Retrieved 3 October 2012. ^ Clifford, A. Claudia Neuhauser. Le's say the equation relating radius and volume is: V(r) = c(r^2) Where c is a constant, r is the radius and V(r) is the volume. dR dX dY —— = —— + —— R X Y

This saves a few steps.

Resistance measurement A practical application is an experiment in which one measures current, I, and voltage, V, on a resistor in order to determine the resistance, R, using Ohm's law, R Find an expression for the absolute error in n. (3.9) The focal length, f, of a lens if given by: 1 1 1 — = — + — f p q The term "average deviation" is a number that is the measure of the dispersion of the data set. In a more radical example, if $\Delta x$ is equal to $x$ (and don't even think about it being even bigger), the error bar should go all the way to minus


text:
Proving that $\liminf\limits_{n\to \infty}\int f_n(x)dx\neq \int f(x)dx$

I have a challenge in doing the following proof: if $\{f_n\}$ is a sequence defined by

$f_n(x)=\begin{cases}1, & \text{if}\;n\leq x\leq n+1,\\0 & \text{otherwise},\end{cases}$ then I want to prove that

1. $f_n\to f\equiv 0 \;\;\text{as}\;n\to \infty.$
2. $\liminf\limits_{n\to \infty}\int f_n(x)dx\neq \int f(x)dx.$

The truth is, I don't see the solution coming soon but I believe there are people here who can help me do justice to it. Any help would be highly regarded.

• Do You have problems with 1 or 2 or both? – Logic_Problem_42 May 8 '18 at 5:06
• @ Logic_Problem_42:Thanks for being prompt at these times! Actually, I have a problem with both. I'm new into analysis. – Omojola Micheal May 8 '18 at 5:08

1. Let $x$ be arbitrary but fixed. By definition, we have for all $n>x$: $f_n(x)=0$. This implies $f_n(x)\to 0$. Since $x$ was arbitrary, we have $\lim_n f(x)=0$ for all $x$.
2. $\int_{\infty}^{\infty} f_n(x)dx=\int_n^{n+1}1dx=1$- this holds for all $n$. Therefore the liminf of $\int_{\infty}^{\infty} f_n(x)dx$ is $1$.
For each $x$ eventually (i.e if $n > x$) $f_{n}(x) = 0$. So $f(x) = \lim_{ n\rightarrow \infty} f_{n}(x)= 0$ for all $x$.
$\int f_{n}(x)\,dx = \int_{n}^{n+1}1\,dx = 1$. So $\liminf_{n \rightarrow \infty} \int f_{n}(x)\,dx = \lim_{n \rightarrow \infty} \int f_{n}(x)\,dx = 1$.
For any $x$, there exists $N>x$ so that $f_n(x) = 0$ for all $n>N$. So $f_n\to f \equiv 0$.
Note that $\displaystyle{\int f_n dx = 1}$ for all $n$. So $\displaystyle{\liminf_{n\to\infty} \int f_n dx = 1}$. However, $\displaystyle{\int f dx = 0}$.