text:
A Logarithmic Function is an inverse function of an exponential function.

$f(x)=\log&space;_{b}x$       and      $g(x)=\ln&space;x$

Logarithmic functions are widely used to represent quantities that spread out over a wide range (earthquake magnitude, sound level, etc).

Taking the logarithm of a number produces the exponent to which the base of the logarithm was originally raised.

Let’s go over how to graph logarithmic functions and how to solve logarithmic equations.

Graphing Logarithmic Functions

$f(x)=logx$   (base 10, from $g(x)=10^x$)

 $x$ $y$ -1 undefined 0 undefined 1 0 2 0.3 3 0.477

There are no $y$-intercept, as there is a Vertical Asymptote (VA) at $x=0$

The $x$-intercept is at  (1,0)

Domain: { $x&space;\epsilon&space;R&space;|&space;x>0$ }

Range: { $y\epsilon&space;R$ }

$f(x)=lnx$  (base $e$, natural logarithm from $g(x)=e^x$)

 $x$ $y$ -1 undefined 0 undefined 1 0 2 0.693 3 1.0986

There are no $y$-intercept, as there is a Vertical Asymptote (VA) at $x=0$

The $x$-intercept is at  (1,0)

Domain: { x&space;\epsilon&space;R|&space;x>0″ alt=”x \epsilon R| x>0″ align=”absmiddle”> }; Range: { }

Transformations of Log Functions

$f(x)=alog(k(x-c))+d$

$a$ – vertical stretch or compression, reflection in the $x$-axis

$k$ – horizontal stretch or compression, reflection in the $y$-axis

$c$ – horizontal translation left or right

$d$ – vertical translation up or down

 $f(x)=2logx$ $f(x)=\frac{1}{2}logx$
 $f(x)=log2x$ $f(x)=log\frac{1}{2}x$
 $f(x)=-logx$ $f(x)=log(-x)$
 $f(x)=log(x-2)$ $f(x)=log(x+2)$
 $f(x)=logx-2$ $f(x)=logx+2$

$f(x)=-2log(x-1)+3$

• vertical compression by a factor of 2
• reflection in the $x$ – axis

• horizontal translation 1 unit left

• vertical translation 3 units up

Vertical Asymptote (VA): $x=1$

$y$ – intercept: $f(0)=-2log(0-1)+3$      –    NONE

$x$ – intercept(s):

In order to determine the $x$ – intercepts, set the equation to = 0, then solve it for $x$ by performing inverse operations in the reverse BEDMAS order.

At the end, the equation would need to be changed to an exponential form in order to be solved.

$0=-2log(x-1)+3$

$-3=-2log(x-1)$

$\frac{-3}{-2}=log(x-1)$

$\frac{3}{2}=log(x-1)$

at this point, in order to solve this equation, represent it in its exponential form, where base is 10 and the exponent is $\frac{3}{2}$ :

$10^{\frac{3}{2}}=x-1$

$x=10^\frac{3}{2}+1$

$x=32.6$

Therefore, the $x$ – intercept is at (32.6, 0)

Review Exponential Functions

Applications of Logs

Log laws (base 10 logarithm)

$loga&space;+&space;log&space;b&space;=&space;log(ab)$

$loga&space;-&space;logb=log&space;(\frac{a}{b})$

$loga^b=bloga$

$log1=0$

Laws of Logarithms Quiz

Example 1: A new car is purchased for \$20,000. The car’s value after $t$ years is given by  $V=20,000(0.8)^t$. How long will it take for the car to be worth half of its purchase value?

Solution:

$10,000=20,000(0.8)^t$

$\frac{10,000}{20,000}=0.8^t$

$0.5=0.8^t$     convert into logarithmic form

$log0.5=tlog0.8$

$t=\frac{log0.5}{log0.8}$

$t=3.1$

Therefore, it would take just over 3 years for the car do depreciate to half of its original value.

Example 2: On the Richter scale, the magnitude of the earthquake in City A is 2.7 and the magnitude of the earthquake in City is  . How many times stronger is the earthquake in City A?

Formula to compare two earthquakes: $M_{2}-M_{1}=log(\frac{I_{2}}{I_{1}})$ where M is the magnitude and I is the intensity.

Solution:

$2.7-1.3=log(\frac{I_{A}}{I_{B}})$

$1.4=log(\frac{I_{a}}{I_{b}})$    convert into exponential form with base 10

$\frac{I_{A}}{I_{B}}=10^{1.4}$

$\frac{I_{A}}{I_{B}}\approx&space;25$

Therefore, the earthquake in City A is approximately 25 times stronger than the earthquake in City B.

Example 3: Tomato juice has a hydronium ion concentration of approximately 0.0001 mol/L. What is its pH? (acidity level)

Formula to determine the pH: $pH=-log[H^{+}]$

Solution:

$pH=-log(0.0001)$

$pH=4$

Since the pH level of tomato juice is 4, it is acidic.

Solving Logarithmic Equations Quiz


text:
## The Annals of Mathematical Statistics

### Asymptotically Optimum Properties of Certain Sequential Tests

Seok Pin Wong

#### Abstract

Let $X_1, X_2, \cdots$ be independent and identically distributed random variables whose common distribution is of the one-parameter Koopman-Darmois type, i.e., the density function of $X_1$ relative to some $\sigma$-finite nondegenerate measure of $F$ on the real line can be written as $f(x, \theta) = \exp (\theta x - b(\theta))$, where $b(\theta)$ is some real function of the parameter $\theta$. Consider the hypotheses $H_0 = \{\theta \leqq \theta_0\}$ and $H_1 = \{\theta \geqq \theta_1\}$ where $\theta_0 < \theta_1$ and $\theta_0, \theta_1$ are in $\Omega$, the natural parameter space. We want to decide sequentially between the two hypotheses. Suppose $l(\theta)$ is the loss for making a wrong decision when $\theta$ is the true parameter and assume $0 \leqq l(\theta) \leqq 1$ for all $\theta$ and $l(\theta) = 0$ if $\theta$ is in $(\theta_0, \theta_1)$, i.e., $(\theta_0, \theta_1)$ is an indifference zone. Let $c$ be the cost of each observation. It is sufficient to let the decision depend on the sequence $(n, S_n), n \geqq 1$, where $S_n = X_1 + \cdots + X_n$. We shall consider the observed values of $(n, S_n)$ as points in a $(u, v)$ plane. Then, for any test, the region in the $(u, v)$ plane where sampling does not stop is called the continuation region of the test. A test and its continuation region will be denoted by the same symbol. Schwarz [4] introduced an a priori distribution $W$ and studied the asymptotic shape of the Bayes continuation region, say $B_W(c)$, as $c \rightarrow 0$. He showed that $B_W(c)/\ln c^{-1}$ approaches, in a certain sense, a region $B_W$ that depends on $W$ only through its support. Whereas Schwarz's work is concerned with Bayes tests, in this paper the main interest is in characteristics of sequential tests as a function of $\theta$. In particular, it is desired to minimize the expected sample size (uniformly in $\theta$ if possible) subject to certain bounds on the error probabilities. Our approach, like Schwarz's, is asymptotic, as $c \rightarrow 0$. It turns out that an asymptotically optimum test--in the sense indicated above, is $B_W \ln c^{-1}$ if $W$ is a measure that dominates Lebesgue measure. Such a measure will be denoted by $L$ (for Lebesgue dominating) from now on. Thus, Bayes tests, as a tool, will play a significant role in this paper. In order to prove the optimum characteristic of $B_L \ln c^{-1}$, some other results, of interest in their own right, are established. For any $W$ satisfying certain conditions that will be given later, we show that the stopping variable $N(c)$ of $B_W(c)$ approaches $\infty$ a.e. $P_\theta$ for every $\theta$ in $\Omega$. This result together with Schwarz's result that $B_W(c) \ln c^{-1}$ approaches a finite region, leads to the following results: (i) for $B_W(c), E_\theta N(c)/\ln c^{-1}$ tends to a constant for each $\theta$ in $\Omega$ and (ii) the same is true for the stopping variable of $B_W \ln c^{-1}$. Furthermore, it is shown that for $B_L \ln c^{-1}$ the error probabilities tend to zero faster than $c \ln c^{-1}$. Consequently, the contributions of the expected sample sizes of both $B_L \ln c^{-1}$ and $B_L(c)$ to their integrated risks, over any $L$-measure, approach 100%. Moreover $B_L \ln c^{-1}$ is asymptotically Bayes. The last result can be shown without (i) since it is sufficient to show (ii) and to apply the same argument used by Kiefer and Sacks [3] in the proof of their Theorem 1. But we show (i) because of its intrinsic interest and present a different proof using (i). Kiefer and Sacks assumed a more general distribution for $X_1$, constructed a procedure $\delta^{'I}_c$ and showed that it is asymptotically Bayes. Our $B_L \ln c^{-1}$ is somewhat more explicit than their $\delta_c'I$. We would also like to point out that an example of $B_L \ln c^{-1}$, when the distribution is normal, is very briefly discussed in their work. We shall restrict ourselves to a priori distribution $W$ for which $\sup (\mod W)H_0 = \theta_0, \inf (\mod W)H_1 = \theta_1$ and $0 < W(H_0 \cup H_1) < 1$. The phrase "for any $W"$ or "for every $W$" is to be understood in that sense. Any Lebesgue dominating measure satisfies these conditions and also the following type of $W$ that will be used: the support of $W$ consists of $\theta_0, \theta_1$ and a third point $\theta^\ast, \theta_0 < \theta^\ast < \theta_1$. Such a $W$ will be called a $\theta^\ast$-measure, and the corresponding $B_W$ denoted by $B_{\theta^\ast}$. From Schwarz's equations for $B_W$ it follows readily that $B_L \subset B_W$ for every $W$. In particular, $B_L \subset B_{\theta^\ast}$. As a consequence, the statement about the error probabilities as well as others concerning $B_L \ln c^{-1}$ in the last paragraph, remain true when $L$ is replaced by $\theta^\ast$ or any $W$. Those geometric characteristics will be dealt with in Section 2. We shall also show there that $\partial B_{\theta^\ast}$, the boundary of $B_{\theta^\ast}$ (which consists of line segments), is tangent to $\partial B_L$ at some point, and that if $\theta^\ast$ is such that $b'(\theta^\ast) = (b(\theta_1) - b(\theta_0))/(\theta_1 - \theta_0)$ then $\max_{(u, \nu)\text{in}B_L} u = \max_{u,v)\text{in}B_\theta^\ast} u$. Let the ray through the origin and with slope equal to $E_\theta X_1$ intersect $\partial B_L$ at $(m(\theta), m(\theta)E_\theta X_1)$. In Section 3, after proving $\lim_{c\rightarrow 0} N(c) = \infty$ a.e. $P_\theta$, we show $\lim_{c\rightarrow 0} N(c)/\ln c^{-1} = m(\theta)$ a.e. $P_\theta$ and $\lim_{c\rightarrow 0} E_\theta N(c)/\ln c^{-1} = m(\theta)$. It is shown in Section 4 that $\sup_{\theta \text{in} H_0 \mathbf{\cup} H_1} P_\theta$ (error $\mid B_L \ln c^{-1}) = o(c \ln c^{-1})$. The main results are given in Section 5. We first show that after dividing by $c \ln c^{-1}$, the difference of the integrated risks of $B_L \ln c^{-1}$ and $B_W(c)$, for any $W$, tends to zero. It follows from this result that $B_L \ln c^{-1}$ asymptotically minimizes the maximum (over $\theta$ in $\Omega$) expected sample size in $\mathscr{F}(c)$, a family of tests whose error probabilities are bounded by $\max_{i=0,1} P_{\theta_i}$ (error $\mid B_L \ln c^{-1}$). The precise statement is given in Theorem 5.1. A sharper result under a stronger hypothesis is given in Theorem 5.2 which states that $B_L \ln c^{-1}$ asymptotically minimizes the expected sample size $E_\theta N$ for each $\theta, \theta_0 < \theta < \theta_1$, among all procedures of $\mathscr{F}(c)$ for which $E_{\theta_0}N/\ln c^{-1}$ and $E_{\theta_1}N/\ln c^{-1}$ are bounded in $c$.

#### Article information

Source
Ann. Math. Statist., Volume 39, Number 4 (1968), 1244-1263.

Dates
First available in Project Euclid: 27 April 2007

Permanent link to this document
https://projecteuclid.org/euclid.aoms/1177698250

Digital Object Identifier
doi:10.1214/aoms/1177698250

Mathematical Reviews number (MathSciNet)
MR229344

Zentralblatt MATH identifier
0162.49805

JSTOR

format: plain


text:
What does convergence in the $L^2$ sense to a constant mean curvature surface imply? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T09:25:18Z http://mathoverflow.net/feeds/question/89980 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89980/what-does-convergence-in-the-l2-sense-to-a-constant-mean-curvature-surface-imp What does convergence in the $L^2$ sense to a constant mean curvature surface imply? Dorian 2012-03-01T18:09:10Z 2012-04-04T11:33:12Z
I have been thinking about the following question and have been unable to find any literature on the subject.

Question: Assume I have a sequence of smooth, simply connected, compact domains $\Omega_s \subset \mathbb{R}^d$ such that $|\Omega_s|=1$ and

$\int_{\partial \Omega_s} (\kappa - \bar \kappa)^2 dS(y) \to 0$ as $s \to +\infty$,

where here $\kappa$ is the mean curvature of the surface $\partial \Omega_s$ and $\bar \kappa$ denotes the average mean curvature over $\partial \Omega_s$. I can prove that the limit is in fact a ball in the following two cases:

All of the sets $\Omega_s$ are convex. or

I assume the uniform bound $\limsup_{s \to +\infty} |\partial \Omega_s| + \int_{\partial \Omega_s} \kappa^2 dS < +\infty$.

I would however like to remove these restrictions since they seem quite artificial. I have been able to rule out the standard "pinching" counter examples of a long rod with capped ends, but am not sure if there could exist other pathologies. Any direction to results in this direction would be appreciated.


text:
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$t$
$\bf{\text{Solution Outline:}}$ Use the properties of radicals to simplify the given expression, $\sqrt[9]{t^9} .$ $\bf{\text{Solution Details:}}$ Since the index (equal to $9$) is odd, then \begin{array}{l}\require{cancel} \sqrt[9]{t^9} \\\\= t .\end{array}


text:
# Rethinking Depthwise Separable Convolutions in PyTorch

This is a follow-up to my previous post of Depthwise Separable Convolutions in PyTorch. This article is based on the nice CVPR paper titled “Rethinking Depthwise Separable Convolutions: How Intra-Kernel Correlations Lead to Improved MobileNets” by Haase and Amthor. Previously I took a look at depthwise separable convolutions which are a drop-in replacement for standard convolutions, but focused on computational and parameter-based efficiency. Basically, you can gain similar results with a lot less parameters and FLOPs, so they are used in MobileNet style architectures.

# Depthwise Separable Convolutions in PyTorch

In many neural network architectures like MobileNets, depthwise separable convolutions are used instead of regular convolutions. They have been shown to yield similar performance while being much more efficient in terms of using much less parameters and less floating point operations (FLOPs). Today, we will take a look at the difference of depthwise separable convolutions to standard convolutions and will analyze where the efficiency comes from. Short recap: standard convolution In standard convolutions, we are analyzing an input map of height H and width W comprised of C channels.

# Pyramidal Convolution: Rethinking Convolutional Neural Networks for Visual Recognition

Today’s paper: Pyramidal Convolution by Duta et al. This is the third paper of the new series Deep Learning Papers visualized and it’s about using convolutions in a pyramidal style to capture information of different magnifications from an image. The authors show how a pyramidal convolution can be constructed and apply it to several problems in the visual domain. What’s really interesting is that the number of parameters can be kept the same while performance tends to improve.


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# System requirements for the Ruby agent

Before installing the Ruby agent, your system must meet the following requirements:

• There is a deployed application to be analyzed, and the web application technology is supported by Contrast.

• The application can be restarted.

• The web server has network connectivity with Contrast.

• The web server has network connectivity with RubyGems or the agent manually installed.

• The server meets the minimum requirements shown in this table.

Requirement

Versions

Notes

Operating system

• 64-bit Linux (recommended)

• 64-bit OSX

• 64-bit Alpine

Starting with agent version 3.0.0, the gem installation step requires the compilation of C extensions. This process is automatic, but you may need to install the following in the target environment:

• gcc, make, automake and autoconf (Package names may vary. You may install your platform's version of build-essential.)

The agent runs in the Ruby application layer with some C dependencies so it may need to be installed with the --platform ruby flag to allow for compilation in either g libc or musl libc based systems.


text:
# An implicit function theorem question

In my course on multivariate calculus we treat the implicit function theorem and I am stuck on the following question:

Find the values of $a$ and $b$ such that, in a neighbourhood of $(x,y,u,v) = (0,1,1,-1)$, $x$ and $y$ are implicitly defined as $C^1$-functions $f$ and $g$ of $u$ and $v$ by the system of equations : $$\begin{cases} e^x +uy + u^4v-a=0 \\ y\cos(x) +bx +b^2u -2v = 4 \end{cases}$$

NB: check whether all conditions of the implicit function theorem are satisfied and state the conclusions as accurately as possible.

My work so far:

Let $G : \mathbb{R}^4 \to \mathbb{R}^2$ be defined by: $$G(x,y,u,v)= \begin{pmatrix} e^x +uy + u^4v -a\\ y\cos(x) +bx +b^2u -2v \end{pmatrix}$$ Obviously, this function is $C^1$.

After this, I presume I have to form a matrix of partial derivatives and check for what values of $a$ and $b$ that matrix is non-singular.

Could anyone give me some guidance on how to continue?

• Just do what you presume you have to do: this is exactly the way it should be done. – Etienne Feb 26 '14 at 14:06

The Jacobian Matrix of the function $G: \mathbb{R}^4 \to \mathbb{R}^2$ defined by: $$G(x,y,u,v)=\begin{pmatrix} e^x + uy + u^4 \\ y\cos(x) + bx + b^2u-2 \end{pmatrix}$$ is $$DG(x,y,u,v)=\begin{pmatrix} e^x & u & y+4u^3v & u^4 \\ -y\sin(x) + b & \cos(x) &b^2 & -2 \end{pmatrix}.$$ Because all entries are continuous functions, $G$ is a $C^1$ function.
Observe that $$G(0,1,1,-1) = \begin{pmatrix} 1 \\ b^2+3 \end{pmatrix} = \begin{pmatrix} a \\ 4 \end{pmatrix} \iff a = 1 \wedge b^2=1 \Longrightarrow a = 1 \wedge \left( b=-1 \vee b = 1\right).$$ The matrix with partial derivatives of $G$ with respect to $x$ and $y$ at $(0,1,1,-1)$, which is $$\begin{pmatrix} \frac{\partial G}{\partial x}(0,1,1,-1) & \frac{\partial G}{\partial y}(0,1,1,-1) \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ b & 1 \end{pmatrix}$$ is singular if $b=1$ and non-singular if $b=-1$. So for $a=1$ and $b=-1$ all conditions of the implicit function theorem are satisfied and thus there exists an open neighbourhood $B$ of $(1,-1)$, an open neighbourhood $U$ of $(0,1,1,-1)$ and $C^1$ functions $f:B \to \mathbb{R}$ and $g : B \to \mathbb{R}$ such that $$\{ (f(u,v), g(u,v), u, v : (u,v) \in B \} = \{(x,y,u,v) \in U: G(,x,y,u,v) = (1.4) \}$$
That's right. Let $z=[u,v]'$. Then the implicit function theorem implies in your case that
$$\frac{\partial G}{\partial z'} + \begin{bmatrix} \frac{\partial G}{\partial x} \; \frac{\partial G}{\partial y}\end{bmatrix} \begin{bmatrix} \frac{\partial f}{\partial z'} \\ \frac{\partial g}{\partial z'} \end{bmatrix}=0.$$
So if the matrix of partial derivatives (with respect to $x,y$) is invertible then life is wonderful.
• What do you mean with $z=[u,v]'$ and $z'$? – Nigel Overmars Feb 26 '14 at 14:35
• $z$ is a (column) vector with elements $u$ and $v$, $z$ is its transpose. – JPi Feb 26 '14 at 19:40


text:
Czechoslovak Mathematical Journal, Vol. 53, No. 4, pp. 917-924, 2003

# Codimension 1 subvarieties of $\Cal{M}_g$ and real gonality of real curves

## E. Ballico

Dept. of Mathematics, University of Trento, 38050 Povo (TN), Italy, e-mail: ballico@science.unitn.it

Abstract: Let ${\mathcal{M}}_g$ be the moduli space of smooth complex projective curves of genus $g$. Here we prove that the subset of ${\mathcal{M}}_g$ formed by all curves for which some Brill-Noether locus has dimension larger than the expected one has codimension at least two in ${\mathcal{M}}_g$. As an application we show that if $X \in{\mathcal{M}}_g$ is defined over ${\bb{R}}$, then there exists a low degree pencil $u X \to{\bb{P}}^1$ defined over ${\bb{R}}$.

Keywords: moduli space of curves, gonality, real curves, Brill-Noether theory, real algebraic curves, real Riemann surfaces

Classification (MSC 2000): 14H10, 14H51, 14P99

Full text available as PDF (smallest), as compressed PostScript (.ps.gz) or as raw PostScript (.ps).

Access to the full text of journal articles on this site is restricted to the subscribers of Myris Trade. To activate your access, please contact Myris Trade at myris@myris.cz.
Subscribers of Springer (formerly Kluwer) need to access the articles on their site, which is http://www.springeronline.com/10587.


text:
iCub-main
iCub::learningmachine::Prediction Class Reference

A class that represents a prediction result. More...

#include <Prediction.h>

Inheritance diagram for iCub::learningmachine::Prediction:

## Public Member Functions

Prediction ()
Empty constructor. More...

Prediction (const yarp::sig::Vector &prediction)
Constructor for expected value prediction without predictive variance. More...

Prediction (const yarp::sig::Vector &prediction, const yarp::sig::Vector &variance)
Constructor for expected value prediction with predictive variance. More...

unsigned int size ()
Returns the size of the prediction. More...

yarp::sig::Vector getPrediction ()
Accessor for the expected value of the prediction. More...

void setPrediction (const yarp::sig::Vector &prediction)
Mutator for the expected value of the prediction. More...

yarp::sig::Vector getVariance ()
Accessor for the variance of the prediction. More...

void setVariance (const yarp::sig::Vector &variance)
Mutator for the variance of the prediction. More...

bool hasVariance ()
Indicator whether the prediction contains a predicted variance. More...

std::string toString ()
Returns a string represenation of the prediction. More...

bool write (yarp::os::ConnectionWriter &connection) const

virtual void onCompletion ()

## Protected Attributes

yarp::sig::Vector prediction
Expected value of the prediction. More...

yarp::sig::Vector variance
Optional variance of the prediction, measured as a unit standard deviation. More...

bool varianceSet
Indicator whether prediction variance is available. More...

## Detailed Description

A class that represents a prediction result.

The serialization of this class is compatible with PortablePair<Vector,Vector>, such that standard Yarp classes can be used to read a Prediction from the network.

Definition at line 44 of file Prediction.h.

## ◆ Prediction() [1/3]

 iCub::learningmachine::Prediction::Prediction ( )
inline

Empty constructor.

Definition at line 65 of file Prediction.h.

## ◆ Prediction() [2/3]

 iCub::learningmachine::Prediction::Prediction ( const yarp::sig::Vector & prediction )
inline

Constructor for expected value prediction without predictive variance.

Parameters
 prediction the predicted expected value variance the predicted expected variance in terms of a unit standard deviation

Definition at line 77 of file Prediction.h.

## ◆ Prediction() [3/3]

 iCub::learningmachine::Prediction::Prediction ( const yarp::sig::Vector & prediction, const yarp::sig::Vector & variance )
inline

Constructor for expected value prediction with predictive variance.

Parameters
 prediction the predicted expected value

Definition at line 87 of file Prediction.h.

## ◆ getPrediction()

 yarp::sig::Vector iCub::learningmachine::Prediction::getPrediction ( )
inline

Accessor for the expected value of the prediction.

Returns
the predicted expected value

Definition at line 106 of file Prediction.h.

## ◆ getVariance()

 yarp::sig::Vector iCub::learningmachine::Prediction::getVariance ( )
inline

Accessor for the variance of the prediction.

More precisely, this returns a predicted unit standard deviation.

Returns
the predicted variance

Definition at line 125 of file Prediction.h.

## ◆ hasVariance()

 bool iCub::learningmachine::Prediction::hasVariance ( )
inline

Indicator whether the prediction contains a predicted variance.

Returns
true if variance is available

Definition at line 146 of file Prediction.h.

## ◆ onCompletion()

 virtual void iCub::learningmachine::Prediction::onCompletion ( )
inlinevirtual

Definition at line 211 of file Prediction.h.

inline

Definition at line 186 of file Prediction.h.

## ◆ setPrediction()

 void iCub::learningmachine::Prediction::setPrediction ( const yarp::sig::Vector & prediction )
inline

Mutator for the expected value of the prediction.

Parameters
 prediction the predicted expected value

Definition at line 115 of file Prediction.h.

## ◆ setVariance()

 void iCub::learningmachine::Prediction::setVariance ( const yarp::sig::Vector & variance )
inline

Mutator for the variance of the prediction.

Parameters
 variance the predicted expected variance in terms of a unit standard deviation

Definition at line 135 of file Prediction.h.

## ◆ size()

 unsigned int iCub::learningmachine::Prediction::size ( )
inline

Returns the size of the prediction.

Returns
the size

Definition at line 97 of file Prediction.h.

## ◆ toString()

 std::string iCub::learningmachine::Prediction::toString ( )
inline

Returns a string represenation of the prediction.

Returns
a string represenation of the prediction

Definition at line 155 of file Prediction.h.

## ◆ write()

 bool iCub::learningmachine::Prediction::write ( yarp::os::ConnectionWriter & connection ) const
inline

Definition at line 166 of file Prediction.h.

## ◆ prediction

 yarp::sig::Vector iCub::learningmachine::Prediction::prediction
protected

Expected value of the prediction.

Definition at line 49 of file Prediction.h.

## ◆ variance

 yarp::sig::Vector iCub::learningmachine::Prediction::variance
protected

Optional variance of the prediction, measured as a unit standard deviation.

Definition at line 54 of file Prediction.h.

## ◆ varianceSet

 bool iCub::learningmachine::Prediction::varianceSet
protected

Indicator whether prediction variance is available.

Definition at line 59 of file Prediction.h.

The documentation for this class was generated from the following file:
• icub-main/src/libraries/learningMachine/include/iCub/learningMachine/Prediction.h


text:
#### Matrices and Determinants Book Back Questions

11th Standard

Reg.No. :
•
•
•
•
•
•

Maths

Time : 00:45:00 Hrs
Total Marks : 30
5 x 1 = 5
1. If aij =${1\over2}(3i-2j)$ and A=[aij]2x2 is

(a)

$\begin{bmatrix} {1\over 2}& 2 \\ -{1\over2} & 1 \end{bmatrix}$

(b)

$\begin{bmatrix} {1\over 2}& -{1\over2} \\ 2& 1 \end{bmatrix}$

(c)

$\begin{bmatrix} 2& 2\\ {1\over 2}& -{1\over2} \end{bmatrix}$

(d)

$\begin{bmatrix} -{1\over 2}& {1\over2} \\ 1& 2 \end{bmatrix}$

2. What must be the matrix X, if 2x+$\begin{bmatrix} 1& 2 \\ 3 & 4 \end{bmatrix}=\begin{bmatrix} 3 & 8 \\ 7 & 2 \end{bmatrix}?$

(a)

$\begin{bmatrix} 1& 3 \\ 2 &-1 \end{bmatrix}$

(b)

$\begin{bmatrix} 1& -3 \\ 2 &-1 \end{bmatrix}$

(c)

$\begin{bmatrix} 2& 6 \\ 4 &-2 \end{bmatrix}$

(d)

$\begin{bmatrix} 2& -6 \\ 4 &-2 \end{bmatrix}$

3. If A is a square matrix, then which of the following is not symmetric?

(a)

A+ AT

(b)

AAT

(c)

AT A

(d)

A− AT

4. If A=$\begin{bmatrix}a & x \\ y& a \end{bmatrix}$ and if xy =1, then det(A AT ) is equal to

(a)

(a −1)2

(b)

(a2 +1)2

(c)

a2 −1

(d)

(a2 −1)2

5. If the points (x,−2), (5, 2), (8,8) are collinear, then x is equal to

(a)

-3

(b)

${1\over 3}$

(c)

1

(d)

3

6. 3 x 2 = 6
7. Determine 3B + 4C - D if B, C, and D are given by
B= $\begin{bmatrix} 2 & 3 & 0\\1 & -1 & 5\end{bmatrix}$ ,C=$\begin{bmatrix} -1 & -2 & 3\\-1 & 0 & 2\end{bmatrix}$,D= $\begin{bmatrix} 0 & 4 & -1\\5 & 6 & -5\end{bmatrix}$

8. Determine the matrices A and B if they satisfy
$2A-B+\begin{bmatrix} 6 & -6 & 0\\ -4 & 2 & 1\end{bmatrix}=0 \ and \ A-2B=\begin{bmatrix} 3 & 2&8 \\ -2 & 1&-7 \end{bmatrix}$

9. Find |A| if A=$\begin{bmatrix} 0& sin \alpha &cos \alpha \\ sin \alpha & 0 & sin \beta \\ cos \alpha & -sin\beta & 0 \end{bmatrix}$ .

10. 3 x 3 = 9
11. If A =$\begin{bmatrix} 4 & 6 & 2 \\ 0 & 1 & 5 \\ 0 & 3 & 2 \end{bmatrix}$ and B= $\begin{bmatrix} 0 & 1 & -1 \\ 3 & -1 & 4 \\ -1 & 2 & 1 \end{bmatrix}$
verify(AB)T=BTAT

12. If =$\begin{bmatrix} 1 &0 &0 \\0 & 1 & 0 \\a &b &-1 \end{bmatrix}$ , show that A2 is a unit matrix.

13. If AT=$\begin{bmatrix} 4 & 5 \\ -1 & 0 \\ 2 & 3 \end{bmatrix}$ and B= $\begin{bmatrix} 2 & -1&1 \\7 & 5&-2 \end{bmatrix}$,verify (A+B)T=AT+BT=BT+AT

14. 2 x 5 = 10
15. For what value of x, the matrix A=$\begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & x^3 \\ 2 & -3 & 0 \end{bmatrix}$ is skew-symmetric

16. A shopkeeper in a Nuts and Spices shop makes gift packs of cashew nuts, raisins, and almonds.
Pack I contains 100 gm of cashew nuts, 100 gm of raisins and 50 gm of almonds.
Pack-II contains 200 gm of cashew nuts, 100 gm of raisins and 100 gm of almonds.
Pack-III contains 250 gm of cashew nuts, 250 gm of raisins and 150 gm of almonds.
The cost of 50 gm of cashew nuts is Rs.50, 50 gm of raisins is Rs.10, and 50 gm of almonds is Rs.60. What is the cost of each gift pack?


text:
# Is the ideal gas constant not a constant if Celsius is used instead of Kelvin? [closed]

For example, Charles' law does not work with Celsius. Why is this? I can only think that the gas constant is not a constant when Celsius is used rather than Kelvin.

• I'd say, when Celsius is used instead of Kelvin, there is no such law and no such constant. Physical laws are awfully strict in that they apply to certain things and not to others. You just can't expect them to work after you've changed some physical quantity to another at random, even if the names of the said quantities happen to be kinda similar ("temperature"). – Ivan Neretin Nov 17 '16 at 18:32
• Charles law works perfectly well with celsius: the algebra is slightly more complicated but the constant is exactly the same. – matt_black Nov 17 '16 at 19:04
• I'm voting to close this question as off-topic because this is a physics question with hardly any relevance to chemistry. – Jan Nov 17 '16 at 20:10
• I disagree that it is a physics question per se- it is confusion about temperature scales and relative vs absolute. – Jon Custer Nov 17 '16 at 23:25

Charles' Law certainly DOES work with temperature in degrees Centigrade!

If we express it (relative to $T=0$ absolute) as $V/T = k$ and if we convert $T$ (K) to $t$ (°C) as $t = T-273$ then obviously Charles' Law would have the form $V/(t+273) = k$.

More generally if, instead of $T$ we use some linear function of $T$, say $Y= mT + b$, then $$V/T = Vm/(Y-b) = k$$ or $$V/(Y-b) = k/m,$$ where $k$, $b$, and $m$ are constants. We can substitute $k'$ for $k/m$ and get $V/(Y-b) = k'$, but we can't get rid of the $-b$ that way.

The claim that a law (even if it is an approximate one like Charles') "doesn't work" with a different unit of measure confuses a formula with a law. Formulas are great for solving high school level problems, but a deeper understanding is expected in college and beyond. Any strictly increasing function can be used as a temperature scale, for instance $X = π^T$ could be a temperature scale.

The formula we'd use to express Charles' Law would look very different, but the relationship between volume and $\log(X)/\log(π)$ would be just as constant (and since $\log(π)$ is a constant, we'd have $V/\log(X) = k''$.)

If you consider temperature to be the measure of mean kinetic energy, then to get the correct linear scale, you'd need an absolute temperature. The Kelvin scale (or the Rankine scale) provides that absolute temperature where in the limit of zero temperature, the system is in its lowest energy state.

• Why do you need absolute temperature? – JobHunter69 Nov 18 '16 at 0:30
• Because the third law of thermodynamics should work. You want properties to scale directly with temperature in a linear relationship (which they do with an absolute scale) instead of an affine relationship (linear if you shift the scale). You might as well just shift the scale and make the system linear. Then you just can multiply without that pesky addition. – Zhe Nov 18 '16 at 3:01

Charles’ law, for those in need of a refreshment (like me) states that an increase in temperature results in an increase in volume and vice-versa; thus, that temperature and volume are directly proportional.

Typically, this would be an experiment in physics class, measuring the temperature of a gas cylinder and the volume it requires under constant pressure. A number of different temperatures and volumes would be recorded and then filled into a graph. It turns out that that graph gives a linear relationship between the two. Assuming temperature to be on the $x$ axis and volume on the $y$ axis, the equation of the linear relationship would be:

$$V = k\vartheta + V_0\tag{1}$$

Wherein $k$ is a constant (which can be calculated depending on $p, n$ and the gas constant $R$) and $V_0$ is the $y$-axis-bisection of the graph. Since $V_0$ is also a constant, you still observe a linear relationship between $V$ and $\vartheta$. The difficulty is that you need to determine two constants and that you need both for any algebra that you want to perform using this law.

If you extrapolate, you realise that the $x$-axis will be bisected at a temperature of $\vartheta \approx -273~\mathrm{^\circ C}$. You could go ahead and decide to modify your equation so that $V_0 = 0$; that requires modifying what you insert as $\vartheta$ in such a way that you effectively move the graph sideways in the coordinate system. There is a number of general rules for doing so; what you want is to move the graph to the right by $273~\mathrm{^\circ C}$ so the resulting equation becomes:

$$V = k \cdot (\vartheta + 273~\mathrm{^\circ C})\tag{2}$$

And then all some clever people did was to replace $\vartheta + 273~\mathrm{^\circ C}$ with the ‘new’ temperature $T$ using the new unit $\mathrm{K}$; the relationship is such that $0~\mathrm{K}\ \hat{\approx}\ {-273}~\mathrm{^\circ C}$. This leads us to our final equation:

$$V = k \cdot T\tag{3}$$

Equations $(1), (2)$ and $(3)$ are all mathematically equivalent to each other. Thus, they describe the same physical consequence. Thus, unless you learnt equation $(3)$ and equation $(3)$ only as Charles’ law, said law is valid no matter which choice of temperature scale you make. Even Farenheit temperatures would obey the law.

The other answers sort of address your confusion, but not very directly. So let me give it a shot.

For gas laws, the gas constant is always a constant, regardless of which temperature scale you need. It's also a constant regardless of whether you are using it in an ideal gas law or some other law. It's even a constant when it shows up in places that aren't gas laws at all! That's because it's a fundamental constant which relates the statistical properties of molecules to macroscopic phenomena like pressure and temperature.

The catch is, in order to use it, the variables have to be on an absolute scale. The reason for this is a little hard to explain, but I guess an easy way to think about it is that it defines how things change relative to each other. For example, if you double the temperature while holding pressure constant, the volume will also double (Charles's Law). Volume is already on an absolute scale - zero volume means it takes up no space. Pressure is the same way - zero pressure means no force exerted on the walls of the container (on average). Temperature is really the only one that has an arbitrary zero point on a lot of scales, where "zero" doesn't necessarily mean "no kinetic energy."

If you double the temperature as measured in an absolute scale, you are doubling the average kinetic energy of the molecules. If you double the temperature as measured on the (for example) Celsius scale, the kinetic energy doesn't necessarily increase by the same amount. Take a negative temperature. What happens when you double -100? Or look at room temperature. If you go from 20 to 40 degrees Celsius, the absolute temperature only increased by about 7% (20 / 293).

Each combination of pressure, volume and temperature units has its own version of the gas constant, but these are all really just unit conversions of the "fundamental" constant, which is the Boltzmann constant multiplied by Avogadro's number. The key thing to remember is that besides making sure the units match the measurements you have, the temperature scale must be an absolute scale.


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# Isoperimetric type inequality in $\mathbb{R}^2$

Fix $$L \in (0,\infty)$$ and consider $$\mathcal{C}_L$$ defined as follows: \begin{align*} \mathcal{C}_L := \{ \gamma:[0,1] \rightarrow \mathbb{R}^2 |~ \gamma \text{ is smooth and length(\gamma)=L }\}. \end{align*}

I am interested in the "blow-up" of $$\gamma$$, denoted $$\gamma_{+r}$$, defined as follows: For any set $$S \subseteq \mathbb{R}^2$$ and $$r>0$$ \begin{align*} S_{+r} := \cup_{z\in S}(z+r\mathbb{D}), \end{align*} where $$\mathbb{D}$$ is the unit disc in $$\mathbb{R}^2$$ which is centred at the origin. So $$\gamma_{+r}$$ is a bounded open set in $$\mathbb{R}^2$$. My question is for which $$\gamma \in \mathcal{C}_L$$ is $$m(\gamma_{+r})$$ maximised? Here $$m(\cdot)$$ is the Lebesgue measure in $$\mathbb{R}^2$$. I feel that it should be maximised by the line segment with length $$L$$.

If this is a version of some well known result, please do indicate it.

The reason for this title is that sometimes the isoperimetric inequality in $$\mathbb{R}^2$$ is stated as follows: For any Borel subset $$A \subseteq \mathbb{R}^2$$ with $$m(A) < \infty$$ and for every $$\epsilon >0$$, we have $$m(A_{+\epsilon}) \geq m(B_{+\epsilon})$$. Here $$B$$ is a Euclidean ball with $$m(A) = m(B)$$.

Thanks!

• As long as the curvature of $\gamma$ is less than something of the order $1/r$, and there are no "overlaps", the area of the "sausage" $\gamma_{+r}$ will be equal to $2 r L + \pi r^2$, and it is intuitively clear this is the maximal value. I guess this is standard, I vaguely remember having read that somewhere, but unfortunately I do not have a reference. Apr 13, 2021 at 21:45


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# A Functional Continued Fraction ( FCF ) is exp_(cf)(a;a;a)=a^(a+a/a^(a+a/a^(a+...))), a > 1. Choosing a=pi, how do you prove that the 17-sd value of the FCF is 39.90130307286401?

Sep 8, 2016

See details in explanation for the derivation. Some Socratic graphs are now included for graphical verification.

#### Explanation:

Let y = exp_(cf)(pi;pi;pi)=pi^(pi+pi/(pi^(pi+pi/(pi^(pi+...))). Then, The

implicit form for this FCF value y is

$y = {\pi}^{\pi + \frac{\pi}{y}} = {\pi}^{\pi \left(1 + \frac{1}{y}\right)}$.

A discrete anolog for approximating y is the nonlinear difference

equation

y_n = pi^( pi ( 1 + 1/y_(n-1))

Adopting this for iteration, with starter value ${y}_{0} = {\pi}^{\pi}$, and

making 15 iterations in long precision arithmetic,.

$y = {y}_{15} = 39.901130307206401$, nearly,

with the forward difference

$\Delta {y}_{14} = {y}_{15} - {y}_{14} = 0$, for this 17-sd precision.

Here, 0 means smallness of order ${10}^{- 18}$.

Scaled local graphs, for cross check:

Use y = exp_(cf)(x;x;x)=x^(x(1+1/y)) for the graph

x-range encloses $\pi$ for each of the two graphs.

y-ranges are appropriate, for precision levels.

The first is for higher precision.

Read y against x = $\pi$
graph{y-x^(x(1+1/y))=0 [3.141592 3.141593 39.9011 39.90115]}

graph{y-x^(x(1+1/y))=0 [1.6 4 0 60]}


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## 05 July 2014

### Minimal Sets over Monotone Predicates

This post describes a very nice use of binary search. I learned about it from a paper by Joao Marques-Silva, Mikolas Janota, and Anton Belov.

Let's fix the universe to be $U=\{0,1,\ldots,m-1\}$. Whenever I say ‘set’ I mean ‘subset of the universe’.

Problem. Many problems reduce to the following one: Given a non-empty upward-closed family of sets, find one of its minimal sets. As an example, take the family of sets $\{\{0\},\{1\},\{0,1\},\{0,1,2\}\}$. It's upward closed: If $X$ is in the family then so are all $Y\supset X$. Its minimal sets, one of which we want to find, are $\{0\}$ and $\{1\}$. We can ask only questions of the form ‘is set $X$ in the given family?’

A simple solution. I bumped into this problem while working on refinement for parametric analyses. The solution I found was the following. We are given ${\cal A}\subseteq \wp(U)$. We maintain a current candidate $X$. The invariant is that $X\in{\cal A}$; the variant is that $|X|$ decreases. We establish the invariant by setting $X=U$. We decrease the size as follows: for each element, we throw it away if possible. More precisely, we iterate the elements $u$ of $U$. If $X-\{u\}$ is still in $\cal A$, then we remove $u$ from $X$.

Clearly, this algorithm asks exactly $m$ questions. Also, this solution finds the last set from $\cal A$, lexicographically. (The last set from $\cal A$ is also minimal.)

Lower bound. Is $m$ questions the best we can do? In the worst case, yes. Here is why. Consider the families of sets with exactly one minimal set: $${\cal A}_X = \{\,Y\mid X \subseteq Y \subseteq U\,\}$$ There are $2^m$ such families. To identify one of them, we need $m$ bits of information.

At this point I stopped looking for a better solution. I did so despite being told by Mikoláš Janota that better is possible. My attitude was: Improvements to the average case are tricks, not algorithms. I now think that that is a very bad attitude.

A better solution. The previous solution goes through $0,1,\ldots,m-1$ and says

 drop $0$, drop $1$, $\ldots$ drop $u_0-1$, keep $u_0$, drop $u_0+1$, drop $u_0+2$, $\ldots$ drop $u_1-1$, keep $u_1$, $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ drop $u_{k-2}+1$, drop $u_{k-2}+2$, $\ldots$ drop $u_{k-1}-1$, keep $u_{k-1}$ drop $u_{k-1}+1$, drop $u_{k-1}+2$, $\ldots$ drop $m-1$

and then returns the solution $\{u_0,u_1,\ldots,u_{k-1}\}$.

The long runs of drops are wasteful, if the output is small. The trick is to use binary search to find the next element to keep.

We start by finding $u_0$. To do so, we find the biggest $v$ such that $\{v,v+1,\ldots,m-1\}\in{\cal A}$. Why? First: If $\{u_0,u_1,\ldots,u_{k-1}\}\in{\cal A}$, then $\{u_0,u_0+1,\ldots,m-1\}\in{\cal A}$. Second: If $\{v,v+1,\ldots,m-1\}\in{\cal A}$, then $v\le u_0$; otherwise, $\{u_0,u_1,\ldots,u_{k-1}\}$ wouldn't be the last element of $\cal A$.

In general, we found some prefix $u_0,u_1,\ldots,u_{i-1}$ and we search for $u_i$. Now we look at sets of the form $$\{u_0,u_1,\ldots,u_{i-1}\}\cup\{v,v+1,\ldots,m-1\}$$ and identify the biggest $v$ for which the set is in $\cal A$.

The second trick is that we use open-ended binary search. If you are familiar with it, then it's safe to skip this paragraph. The simplest example of open-ended binary search is the following: Consider a function $f:\mathbb{N}\to\{0,1\}$ such that $f(i)\le f(i+1)$ for all $i$. We want to find the smallest $i$ such that $f(i)=1$. In the first phase we find some $i$ such that $f(i)=1$. We do so by evaluating $f(1)$, $f(2)$, $f(4)$, $f(8)$, $f(16)$, and so on. After this first phase, we can use normal binary search. We will need the following observation: If the solution is $i$, then the runtime is $\sim \lg i$.

The time needed to find $u_i$ is roughly $1+\lg(u_i-u_{i-1})$. The total runtime is $1+k+\lg u_0 + \lg(u_1-u_0) + \cdots + \lg(u_{k-1}-u_{k-2})+\lg(m-u_{k-1})$. Let us denote the arguments of the logarithms by $x_0,\ldots,x_k$. With this notation, the worst runtime is \begin{align} &\text{maximum of}&&1+k+\lg(x_0x_1\cdots x_k) \\ &\text{given that}&&x_0+x_1+\cdots+x_k=m \end{align} The maximum is achieved when $x_0=\cdots=x_k=m/(k+1)$. (This can be shown by using Lagrange multipliers.) Thus, the maximum runtime is $\sim k+k\lg(m/k)$. This runtime depends on two parameters: $m$ describes the input, and $k$ describes the output.

I think it is clear now that this is not ‘just a trick’, but a cute algorithm. I learned at least the following:

1. If you can't improve the worst-case, try to improve something else.
2. Average case analyses are sometimes difficult because you lack a distribution that describes the inputs realistically. In that case, try to improve the runtime when seen as a function of both input size and output size.
3. Listen more carefully to what Mikoláš says. :)

References. If you want to find out more, look at the following articles:

1. [Marques-Silva et al, Minimal Sets over Monotone Predicates in Boolean Formulae, 2013]: This is the paper from which I learned the algorithm presented above. In it, you will also find another (older) algorithm with the same asymptotic performance.
2. [Marques-Silva, Janota, Computing Minimal Sets on Propositional Formulae I: Problems & Reductions, 2014]: Read this if you didn't believe me that many other problems reduce to this one.


text:
3 Tutor System
Starting just at 265/hour

# Q.4 Express the number appearing in the following statements in standard form: (a) The distance between Earth and Moon is 384.0. 000 m. (b) Speed of light in vacuum is 300,000,000 m/s. (c) Diameter of the Earth is 1,27,56,000 m. (d) Diameter of the Sun is 1,400,000,000 m. (e) In a galaxy there are an average 100,000,000,000 stars. (f) The universe is estimated to be about 12,000,000,000 years old. (g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300,000,000,000,000,000,000 m. (h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm. (i) The Earth has 1,353,000,000 cubic km of sea water. (j) The population of India was about 1,027,000,000 in March 2001.

(a)$$384,000,000 m = 3.84 × 10^8 m$$
(b) $$300,000,000 m/s = 3 × 10^8 m/s$$
(c)$$1,27,56,000 m = 1.2756 × 10^7 m$$
(d) $$1,400,000,000 m = 1.4 × 10^9 m$$
(e) $$100,000,000,000 stars = 1 × 10^{11} stars$$
(f) $$12,000,000,000 years old = 1.2 × 10^{10} years old$$
(g) $$300,000,000,000,000,000,000 m = 3 × 10^{20} m$$
(h)$$60, 230, 000, 000, 000, 000, 000, 000 molecules = 6.023 × 10^{22}molecules$$
(i) $$1,353,000,000 cubic km = 1.353 × 10^9 cubic km$$
(j)$$1,0,27,000,000 = 1.027 × 10^9$$


text:
# Games and Ramsey's theorem

Here, we identify subsets of $$\mathbb{N}$$ with sequences obtained by listing the members of the set in strictly increasing order. Suppose that we have some set $$\mathcal{F}$$ of sets (sequences) of the form $$(m_n, l_n)_{n=1}^t$$. We let $$\mathcal{U}$$ denote the set of all $$(m_n, l_n)_{n=1}^\infty$$ such that $$(m_n, l_n)_{n=1}^t\in \mathcal{F}$$ for all $$t\in\mathbb{N}$$. We let $$\mathcal{V}$$ denote the set of all $$(m_n)_{n=1}^\infty$$ such that there exists $$(l_n)_{n=1}^\infty$$ such that $$(m_n, l_n)_{n=1}^\infty \in \mathcal{U}$$.

It is easy to see that $$\mathcal{U}$$ is closed and $$\mathcal{V}$$ is analytic (in either the Cantor or Ellentuck topology). Suppose also that for every infinite subset $$K$$ of $$\mathbb{N}$$, there exists a further, infinite subset $$M$$ of $$L$$ such that every infinite subset of $$M$$ lies in $$\mathcal{V}$$.

Is it known that the latter property is equivalent to a winning strategy for a certain player in some two player game?

• I don't get it. In the first sentence you talk about subsets of $\mathbb N$. So how is $(m_n,l_n))_{n=1}^t$ a subset of $\mathbb N$? Isn't it a subset of $\mathbb N\times\mathbb N$? – bof Jun 24 '19 at 4:24
• The object of interest is $\mathcal{V}$, which consists of subsets of $\mathbb{N}$. – user78375 Jun 24 '19 at 11:08
• Voted to close as unclear what you're asking because I had the same question as @bof and the OP refused to answer it. – Steven Landsburg Jun 24 '19 at 17:21

If you mean completely Ramsey sets there is the following characterization using the Kastanas game. Let's start with some definitions :

1. Here the letters $$a, b, c, ...$$ vary over finite subsets of $$\mathbb{N}$$ and $$A, B, C, ...$$ over infinite subsets of $$\mathbb{N}$$.

• Let $$a$$ and $$A$$, we write $$a if $$a=\emptyset$$ or $$max(a).

• For $$a, let $$[a, A]=\{S\in [\mathbb{N}]^{\aleph_{0}} : a\subseteq S\subseteq a\cup S \}= \{a\cup C\in [\mathbb{N}]^{\aleph_{0}} : a Note that $$[\emptyset, A]=[A]^{\aleph_{0}}$$

• A set $$X\subseteq [\mathbb{N}]^{\aleph_{0}}$$ is called Ramsey if there is $$A$$ with $$[\emptyset, A]\subseteq X$$ or $$[\emptyset, A]\subseteq\hspace{0.1cm}\sim X$$, here $$\sim X$$ denotes the complement of $$X$$.

• A set $$X\subseteq [\mathbb{N}]^{\aleph_{0}}$$ is called completely Ramsey if for every $$a there is $$B\subseteq A$$ with $$[a, B]\subseteq X$$ or $$[a, B]\subseteq\hspace{0.1cm} \sim X$$.

2. Let $$A \subseteq [\omega]^{\omega}$$ = the set of infinite sets of integers. Then $$A$$ has a homogeneous set $$H$$ if, by definition, $$H \in [\omega]^{\omega}$$ and either every infinite subset of $$H$$ belongs to $$A$$ or every infinite subset of H belongs to the complement of $$A$$. $$A$$ has the Ramsey property iff it has a homogeneous set.

3. Kastanas's game. For $$\varphi\subseteq [\omega]^{\omega}$$ we define the game $$G_{\varphi}$$ as follows :

Player I wins if $$\{n_{0}, n_{1}, \cdots \}\in\varphi$$

Theorem (Kastanas)

a.) Player I has a winning strategy in $$G_{\varphi}$$, iff there is a homogeneous set in $$\varphi$$ (i.e. an infinite $$H$$ such that every infinite subset of it belongs to $$\varphi$$)

b.) Player II has a winning strategy in $$G_{\varphi}$$ iff for every $$A$$ there is a subset of it homogeneous in $$[\omega]^{\omega}\setminus \varphi$$.

For the information of the bibliography we have the following article "On the Ramsey Property for Sets of Reals" Author: Ilias G. Kastanas.

• This is about as far as I had gotten. The hypothesis that for every infinite subset $K$ of $\mathbb{N}$, there exists a further subset $M$ which is homogeneous in $\mathcal{V}$. If we write $M=(m_n)_{n=1}^\infty$, this means there exists $(l_n)_{n=1}^\infty$ such that $(m_n, l_n)_{n=1}^t\in \mathcal{F}$ for all $t$. However, this is only good enough to let me choose $(l_n)_{n=1}^\infty$ after all $(m_n)_{n=1}^\infty$ have been chosen. – user78375 Jun 24 '19 at 0:47
• Since whether or not a sequence $(m_n, l_n)_{n=1}^\infty$ lies in $\mathcal{V}$ depends only on its initial segments, I would like to know if it is possible to choose the $l_n$ sequence in a game without future information. That is, if every $A$ has a subset homogeneous in $\mathcal{V}$, is it possible to write down game where Players take turns choosing infinite subsets and $m_n, l_n$ values, and one of the players wins if $(m_n, l_n)_{n=1}^\infty\in \mathcal{U}$ (equivalently, $(m_n, l_n)_{n=1}^t\in \mathcal{F}$ for all $t$? Basically, can we choose the $l_n$ concurrently with the $m_n$? – user78375 Jun 24 '19 at 0:51


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# Percentile Formula

What is Percentile?

What is percentile?  Mostly we can define percentile as a number where a certain percentage of scores fall below that given number. Percentile and percentage are often confused but both are different concepts. The percentage used to express fractions of a whole, while percentiles are the values below which a certain percentage of the data in a data set is found. If you want to know where you stand compared to the rest of the crowd, you need a statistic that reports relative standing, and that statistic is called a percentile.

For example, you are the fourth tallest person in a group of 20.80% of people are shorter than you. That means you are at the 80th percentile.

If your height is 5.4inch then "5.4 inch" is the 80th percentile height in that group.

Percentile Formula is given as -

 Percentile = $\frac{n}{N}$ x 100

Where, n =ordinal rank of a given value

N = number of values in the data set,

P = Percentile

Formula for Percentile

In mathematics, we use this term percentile informally to indicate that a certain percentage falls below that percentile value. For example, if we score in the 25th percentile, then 25% of test-takers are below this score. Here 25 is called the percentile rank. Percentile divides a data set into the 100 equal parts. A percentile is a measurement that tells us what percent of the total frequency of a data set was at or below that measure. As an example, let us consider a student’s percentile in some exams.

If on this test, a given student scored in the 60th percentile on the quantitative section, she scored at or better than 60% of the other students. Further, if a total of 500 students took the test, then this student scored at or better than

500×0.60=300 students out of 500 students. In other words, 200 students scored better than that particular student.

Therefore, percentiles are used to understand and interpret the given data. They also indicate the values below which a certain percentage of the data in a data set is found. Percentiles are frequently useful to understand the test scores and biometric measurements in our day-to-day routine.

Percentile is calculated by the ratio of the number of values below ‘x’ to the total number of values.

Formula for Percentile

The Percentile Formula is given as,

 Percentile (P) = (Number of Values Below “x” / Total Number of Values) × 100

Procedure to Calculate Kth Percentile

The kth percentile is a value in a data set that divides the data into two parts. The lower part contains k percent of the data, and the upper part contains the rest of the data.

The following are the steps to calculate the kth percentile (where k is any number between zero and one hundred).

Step 1: Arrange all data values in the data set in ascending order.

Step 2: Count the number of values in the data set where it is represented as 'n'.

Step 3: calculate the value of k /100, where k = any number between zero and one hundred.

Step 4: Multiply 'k' percent by 'n'.The resultant number is called an index.

Step 5: If the resultant index is not a whole number then round to the nearest whole number, then go to Step 7. If the index is a whole number, then go to Step 6.

Step 6: Count the values in your data set from left to right until you reach the number. Then find the mean for that corresponding number and the next number. The resultant value is the kth percentile of your data set.

Step 7: Count the values in your data set from left to right until you reach the number. The obtained value will be the kth percentile of your data set.

Solved Examples

Percentile formula example.

Example 1:

Let us consider the percentile example problem: In a college, a list of grades of 15 students has been declared. Their grades are given as: 85, 34, 42, 51, 84, 86, 78, 85, 87, 69, 74, 65. Find the 80th percentile?

Solution:

Step 1:

Arrange the data in ascending order.

Let us arrange in Ascending Order = 34, 42, 51, 65, 69, 74, 78, 84, 85, 85, 86, 87.

Step 2:

Find Rank,

Rank = Percentile / 100

= 80 / 100

k = 0.80

Step 3:

Find 80th percentile,

80th percentile = 0.80 x 12

= 9.6

Step 4:

Since it is not a whole number, round to the nearest whole number.

Therefore, 9.6 is rounded to 10.

Now, Count the values from left to right in the given data set until you reach the number 10.

From the given data set, 10th number is 85.

Therefore, 80th percentile of given data set is 85

Example 2: The scores for students are 40, 45, 49, 53, 61, 65, 71, 79, 85, 91. What is the percentile for score 71?

Solution: We have,

No. of. scores below 71 = 6 Total no. of. scores = 10

The formula for percentile is given as,

Percentile = (Number of Values Below “x” / Total Number of Values) × 100 = $\frac{n}{N}$x 100

Percentile of 71= 6/10 × 100

= 0.6 × 100 = 60

1. What is the Percentile Rank Formula?

Answer: Percentile Rank is a usual term mainly used in statistics which is arrived from Percentile. Percentile (also referred to as Centile) is the percentage of scores that range between 0 and 100 which is less than or equal to the given set of distribution. Percentiles divide any distribution into 100 equal parts. This is predominantly used for the interpretation of scores with different ranges across various tests. Percentile Rank formula (PR) is arrived based on the total number of ranks and the number of ranks below and above percentile.

A Percentile Rank formula is given by:

Percentile Rank (PR) = [(M + (0.5 * R)) / Y] x 100

Where,

• M = Number of Ranks below x

• R = Number of Ranks equals x

• Y = Total Number of Ranks

2. What is the Percentile Range?

Answer: A percentile range is a difference between two specified percentiles. These could theoretically be any two percentiles, but the 10-90 percentile range is the most common. To find the 10-90 percentile range:

1. Calculate the 10th percentile using the percentile formula.

2. Calculate the 90th percentile using the percentile formula.

3. Subtract the 10th percentile from the 90th percentile.


text:
# Degree of field extension $\mathbb{Q}\left(\sqrt{1 + \sqrt{3}}\right):\mathbb{Q}$

I think the degree

$$\left[\mathbb{Q}\left(\sqrt{1 + \sqrt{3}}\right):\mathbb{Q}\right]$$

is equal to four, but how do I find the minimum polynomial of such an extension? If I square the term I take care of one square root, but squaring again doesn't help me get rid of the $\sqrt{3}$ term. Any advice is very much appreciated!

• Before squaring again, arrange all non square-root terms on the other side!! Oct 18 '16 at 23:37

## 1 Answer

$x = \sqrt{1 + \sqrt 3} \implies x^2-1 =\sqrt{3} \implies x^4-2x^2-2 = 0$.

This polynomial is irreducible by Eisenstein's criterion, since $2$ divides all the coefficients except the first, and $2^2=4$ does not divide the constant. Hence, the degree of the extension, is the degree of the minimal polynomial, which is $4$.

• Thank you @Bernard I have made the correction. Oct 18 '16 at 23:34


text:
Move2EarnAPY
Search…
⌃K

# How The APY Is Calculated

### Compound Interest Equation (Principal + Interest)

$A = P (1+i)^n$
• A = Total Accrued Amount (principal + interest)
• P = Principal Amount
• i = Interest Rates per Epoch
• n = The number of Epochs in the period to be calculated.
Note: MOVEAPY's epoch is 3 Seconds/epoch, so it is necessary to convert the period to calculate the yield to the number of epochs.


text:
How do we observe galaxies further than 13.7 billion light years away?

z8_GND_5296. Credit: V. Tilvi (Texas A&M), S. Finkelstein (UT Austin), the CANDELS team, and HST/NASA

Sorry for such a lengthy title, but the subject came up because of a widely circulated announcement of the discovery of “The Most Distant Galaxy Yet Seen,” a title that’s being constantly revived (Matt Francis over at Universe Today has a nice discussion of why this galaxy really is a big deal).

The galaxy, which for now goes by z8_GND_5296 was discovered by the CANDELS collaboration at a distance of about 30 billion light-years.

“But wait!” my friends said on facebook and twitter, “how can we even see a galaxy 30 billion light years away when the universe is only 13.7 billion years old? Isn’t light the ultimate speed limit in the universe?”

Yes. It is. But the universe was smaller in the past.

What does “size of the universe” mean?

To begin with, the CANDELS team didn’t actually find that the galaxy was 30 billion light-years away directly. Rather, they found that it had a “redshift” of about 7.5. Or, to put it another way, the universe was about
$a_{emitted}=frac{1}{1+z}=frac{1}{1+7.5}=frac{1}{8.5}=11.7%$
the size it is now, and we measure that fact by noting that light that left this distant galaxy has grown by a factor of 8.5 in the time it takes to reach us. That’s what redshift is all about. Long wavelengths of light are “redder” than blue ones.

Now for the misconception. Naively you might suppose that if the universe were 11.7% the size it is now at some point in the past, that must mean that the bounds of the universe were 11.7% smaller than they are now. But the universe has no bounds!

Instead, the standard picture is that the universe is much like a balloon, and as it inflates, the distances between galaxies becomes larger by a fixed rate. “Doubling in size” is shorthand for “galaxies getting twice as far apart from one another,” as well as “gas and dark matter becoming eight times as diffuse.” (8 times, because the universe increases in each of 3 dimensions). Here’s a particularly crude version of the whole shebang that Jeff drew for the User’s Guide:

Light and Space

Imagine that, rather than beaming light to us directly, our pal z8_GND_5296 sent us a signal by means of post-stations or whisper down the lane. It sent a signal to a galaxy a few million light years away from it. Galaxy 1 sent a signal to galaxy 2, and so on, until we got the message. Further, imagine that each of those galaxies are currently 10 million light-years apart from one another. This, by the way, is known as their “comoving distance” if you want the technical term.

But the first signal took much less than 10 million years to transmit. Because the universe was so much smaller then than now, it took about 1.1 million years. The next 10 million light-years of comoving distance might be traversed by a light beam in 1.3 million years, and so on, so that after 13 billion years of whispering down the lane, the total distance now between us and z8_GND_5296 is about 30 billion light years.

I warn you, though, even this 30 billion light year number isn’t terribly useful. After all, if I say that it’s 2 miles to the drug store, the implication is that you’ll have to traverse 2 miles to get there. On the other hand, a galaxy 30 billion light years away will be significantly farther by the time you try to reach it. Indeed, there are galaxies out there that, even traveling at the speed of light, we couldn’t ever reach. That’s just a consequence of living in an expanding universe.

What is the maximum distance?

Though we can see to a distance of more than 13.8 Billion light-years, we can’t see infinitely far. The ultimate limit is what’s known as the particle horizon, which for us is a (comoving) distance of about 45 billion light-years. Anything further than that and we have no hope of seeing it.

On the other hand, the universe also has an “event horizon” (yes, just like a black hole), the maximum distance we ever could reach. Because our universe is accelerating, the limit maxes out at a certain point, and there are regions of space that are forever inaccessible to us. Those are systems more than about 60 billion light years from here.

Beyond that is anyone’s guess.

-Dave

ps If there’s interest, I may do a followup on this as to why it is that cosmologists claim there must be cosmic inflation. It’ll even use Zeno’s paradox to talk about how the smallness of the early universe was trumped by its youngness.

Posted in Uncategorized | Tagged , | 10 Comments

Blastoff!

Good news, everyone! I was on the Professor Blastoff Podcast this week talking about “The Universe” (which may win the prize for best episode name) with Tig, David and Kyle. We talk about time travel, symmetry, antimatter (and Uncley matter, a joke that I failed to get for about 10 minutes), and much more.

Check it out, and don’t forget to subscribe via iTunes. Oh, and a warning for the faint of heart. There’s a fair bit of saucy language.

-Dave

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Higgs!

Francois Englert and Peter Higgs

In a move that surprised almost no one in the physics community, the Nobel Prize in Physics was awarded this morning to François Englert and Peter W. Higgs “for the theoretical discovery of a mechanism that contributes to our understanding of the origin of mass of subatomic particles, and which recently was confirmed through the discovery of the predicted fundamental particle, by the ATLAS and CMS experiments at CERN’s Large Hadron Collider”

This is a huge deal, and very well deserved. I won’t go on and on about the Higgs right now, in large part because I’ve already written a ton on the subject, including:

Also, of course, there’s a whole chapter about the Higgs and the Standard Model in my new book, The Universe in the Rearview Mirror.

Congratulations again to Drs. Englert and Higgs!

-Dave

What do you really need to know about the universe?

Just a quicky. I have a new, somewhat navel-gazing post up at io9 on what you really need to know about the universe, and more generally, why people embrace pop-sci at all. I’m curious to hear what you think, and of course, I’m always looking for more questions for the column.

-Dave

Posted in Uncategorized | Tagged , , | 2 Comments

Hi. Would you like to see me give a talk on symmetry? Of course you would. Here’s a professionally produced video of my presentation to the good folks at Google entitled, “Why Symmetry Matters.” Enjoy!

-Dave

ps I also had a really nice visit from a former student who’s now off to grad school. She made me a little Thank You gift that just about made me cry.

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Slate: 4 Reasons you Shouldn't Exist

Greetings, true believers. I have a piece up at Slate today: Four Reasons You Shouldn’t Exist: Physics says you’re an impurity in an otherwise beautiful universe.. The headline might be a bit over the top, but it’s a fun little piece about the importance of symmetry breaking. Take a look, and then when you’re done, be sure to check out a piece I did a few years ago: A Physicist Looks at the Time-Traveler’s Wife.

-Dave

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New York, New York

I’m in the Big Apple! And I’m pleased to say that, having made it here, I’m confident that I can make it anywhere.

Today I was on the Leonard Lopate Show for a pretty long interview. We talked about symmetry, my book, a bit about time travel, and finished it off with a nice chat about Emmy Noether. Do check it out. I think it went pretty well.

I also checked out the Union Square B&N to make sure they prominently placed my book (they did), saw some friends, and had some delicious brooklyn ice cream.

Tomorrow, I’ll be giving a google talk at the NY HQ. The YouTube video will be linked when it’s up.

Thanks for your continued support!

-Dave

Posted in Uncategorized | Tagged , , | 1 Comment

Welcome, Wired visitors!

Hello there.

If you happened to find yourself clicking in from Wired, let me welcome you to my blog. There are some interesting things here (including slightly more technical versions of some pieces from io9, observations about the publishing industry, and lots and lots of updates about my new book. But if you’re looking for something a bit more specific:

1. Go here to read endorsements and reviews of “The Universe in the Rearview Mirror,” including a very nice one from Sean Carroll, with whom I had my Wired dialogue.
2. Go here to check out my io9 “Ask a Physicist” columns. They’re some of my better stuff on the web. Way better than the blog.
3. Go here to follow me on facebook and find out about talks and the like.
4. Go here if you’d like links, comments on physics, and occasional silly observations, limited to 140 characters.

And to my regular readers who’d like to know about what this Wired chat is all about. Sean and I had an exchange on what the public needs to know about science (if anything) and what obligations science popularizers (such as ourselves) have to the public. It was a lot of fun, and you should check it out.

-Dave

Posted in Uncategorized | Tagged , | 1 Comment

Two weeks in

What a couple of weeks! In the last two weeks I’ve:

• Done about 20 radio interviews, with several more on the way (WAMC). I’m doing WHYY’s NewsWorks Tonight in about a week, and WNYC’s Leonard Lopate Show on August 20. I also had a really nice interview on philly.com.
• Had lots and lots of great press. Universe Today said:

At most times in Rearview Mirror, Goldberg’s style feels more like a discussion than a book – it’s as if your delightfully nerdy friend from college (the one with a knack for identifying stars, he’s convinced it’s a total turn-on) came over for dinner one night to talk about his favorite topic – the mysteries of the cosmos.

There’s also a major review coming out in New Scientist next week. It’s generally quite positive with a few criticisms (some valid, a few not, but it’s British, so you have to correct for that), with a very strong finish:

Fortunately, good sense, humour and inventive explanationsmanage, like matter, to triumph over their evil twins. Goldberg has a fine ear for the absurd, and is deft at revealing why things we take for granted, such as the equality of gravitational and inertial mass, are strange and not obvious at all.

In the end, Goldberg has a good stab at fulfilling his ambitions. His enthusiasm may sink into scattiness at times but it carries you on a ride through picturesque bits of science. We zoom by Maxwell’s demon, quantum teleporters and black holes, then plunge down for a glimpse of the secret underground, where physics and reality rest on foundations of pure symmetry.

It pains me to say it this way, but it’s a bit like a kick-ass rollercoaster built through Dwarrowdelf in Middle-earth.

(where the last probably requires some context).

• Sold lots of books! We were the #1 Science book at Barnes and Noble last week, at least according to my editor! Woot! Two weeks in and we’re going to a second printing!

Thanks to everyone for giving this book such an enthusiastic reception. Please continue to tell your friends, review online, and on and on.

-Dave

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Pub Day!

Hurray! The day has finally come. If you haven’t already bought up your copy what are you waiting for?

Seriously, though, thanks to everyone who helped with the book (and to the many, many io9 readers who’ve asked questions over the years). As a gesture, here are my formal acknowledgements from the book:

Writing a book is fun, but you know what? It’s also hard work, and I wouldn’t have been able to get through it without the love, support, and feedback of a lot of people. First, and foremost, my wife, Emily Joy, has read every line and has given me the tough love that I so desperately need. I thank my family, and (though they didn’t give me any comments at all) my daughters, Willa and Lily.

Thanks to all of my friends and colleagues who provided comments and feedback on various parts of the draft and useful conversations: Sean Carroll, Rich Gott, Richard Henretta, Andy Hicks, Lynn Hoffman, Josh Kamensky, Adrienne Leonard, Sean Lynch, Kate Mason, Kevin Owens, John Peacock, Dawn Peterson, Tina Peters, Som Tyagi, Liz Fekete Trubey, and Enrico Vesperini.

I also want to acknowledge Annalee Newitz, Charlie Jane Anders, and everyone at io9.com. Annalee and Charlie Jane were very supportive throughout, and many of these topics were road tested at io9. I would also like to thank all of the readers whose comments helped identify roadblocks to comprehension (theirs and mine).

Not every academic environment smiles on popular science writing as a serious pursuit. I’d like to thank my department head, Michel Vallieres, and my dean, Donna Murasko, for taking the larger view of scholarly work.

I’d like to thank my agent, Andrew Stuart. Andrew did more than simply sell this book; he was an excellent sounding board and advocate and was instrumental in helping me find my voice. My editor, Stephen Morrow, and his outstanding assistant, Stephanie Hitchcock, were tough but fair with their virtual red pen and in stemming my natural verbosity (and trimming some of the more terrible jokes) they improved the book immensely. Thanks also to my amazing illustrator, Herb Thornby, who so deftly and beautifully conveyed ideas that my words may have otherwise obfuscated.

I also gratefully acknowledge Michael Blanton and the Sloan Digital Survey Collaboration for use of their data.

Finally, here are some fun places to go on the web and in real life today:

• At 2pm EDT (though you can start early), I’ll be doing a Reddit “Ask me Anything.” Please do.
• At 7pm EDT, I’ll be doing a reading and signing at the Barnes and Noble in Rittenhouse Square. Please come.
• Some very nice book blogs have had some very nice things to say about the Universe in the Rearview Mirror, including:

Goldberg is a simply wonderful guide to and through all the thickets of complexity and obscurity that one encounters along the road of trying to understand what makes everything tick.


text:
# News Sandra Day O'Connor to Retire From Supreme Court

#### Astronuc

Staff Emeritus
WASHINGTON (AP, July 1) -
Sandra Day O'Connor, the first woman on the Supreme Court and a swing vote on abortion as well as other contentious issues, announced her retirement Friday. A bruising Senate confirmation struggle loomed as President Bush selects a successor. "It has been a great privilege indeed to have served as a member of the court for 24 years," the 75-year-old justice wrote Bush in a one-paragraph resignation letter. "I will leave it with enormous respect for the integrity of the court and its role under our constitutional structure."

Related General Discussion News on Phys.org

#### Ivan Seeking

Staff Emeritus
Gold Member
The possibility that two or even three justices could retire during GWB's term was our greatest fear. This is a disaster for the U.S.

#### faust9

Bush is a Lame Duck right now. He'll have problems pushing an ideologue through the Senate IMO. His appointee will be conservative I'm sure (Warren was 'supposed' to be conservative) but anything too extreme will be tough. It's sad that the main swing voter in court is the one retiring though.

#### McGyver

Juustice O'Connor should be commended for her willingness to assess each case before her in terms of "present times and standards," including a occassion or two where she re-assessed and opposed her own opinions written years earlier. In this way, her opinions were typically not pre-disposed to any bias before she was pesented with applicable facts, points, and arguments. I would hope that her replacement is of a similar philosophy of practice. Appointing an agenda prejudiced justice for this type of decision-making post, would do poorly in interpreting the complex needs of the American public and U.S. Constitution today, tomorrow, and in the years to come.

The public must understand that the Supreme Court exists to interpret the U.S. Constitution and best extrapololate what they believe our founding fathers would have invoked upon present day matters if they were alive today. Such ultimate interpretations must take into account the present knowledge of our times.

#### Ivan Seeking

Staff Emeritus
Gold Member
faust9 said:
Bush is a Lame Duck right now. He'll have problems pushing an ideologue through the Senate IMO.
You know, I tend to forget that his presidency is pretty much over. You make a good point. Whewwwww. :yuck:

Thanks for reminding me of that.

#### Astronuc

Staff Emeritus
Ivan Seeking said:
You know, I tend to forget that his presidency is pretty much over. You make a good point. Whewwwww. :yuck:

Thanks for reminding me of that.
Bush will not be out of office until January 20, 2009 - inauguration day of the next president. That's 3 years and 7 months - plenty of time to wreak havoc. Can Rehnquist or Stevens survive that time?

What if he were to put Judge Pryor on the Court, the one who placed a religions symbol (the 10 commandments) in his court?

#### pattylou

Harry Reid has suggested conservative judges to Bush, that will meet with Democratic support.

A good thing! I doubt Bush will give the suggestions much consideration, but pretty much a win:win for Reid.

(I'll dig up the names.)

#### pattylou

"Reid later offered four names of people he said would be good for the court: GOP Sens. Mel Martinez of Florida, Mike DeWine of Ohio, Mike Crapo of Idaho and Lindsey Graham of South Carolina. They are people who serve in the Senate now who are Republicans who I think would be outstanding Supreme Court members,'' Reid said. "

#### Astronuc

Staff Emeritus
I could go with Lindsey Graham of South Carolina.

I have been impressed by former Senator John Danforth, except for the fact that he said that GW Bush was a good president.

Another person I admire is former Senator Warren Rudman.

I just want to see a fair and impartial jurist, who does not have a religious agenda.

#### wasteofo2

pattylou said:
"Reid later offered four names of people he said would be good for the court: GOP Sens. Mel Martinez of Florida, Mike DeWine of Ohio, Mike Crapo of Idaho and Lindsey Graham of South Carolina. They are people who serve in the Senate now who are Republicans who I think would be outstanding Supreme Court members,'' Reid said. "
How often have senators been appointed to the Supreme Court? It would seem that Reid should suggest some Judges that he thinks would make good Supreme Court judges, not Senators...

#### pattylou

wasteofo2 said:
How often have senators been appointed to the Supreme Court? It would seem that Reid should suggest some Judges that he thinks would make good Supreme Court judges, not Senators...
The article went on to discuss that. You could google in quotes, some part of the quote above - if you want to find the exact number. Otherwise, I think it was about 14 (justices that had been senators).

Many senators have experience with the law, (law degrees, a practice, etc) although I don't know how extensively those 14 (?) were involved with sitting as justices.

#### Ivan Seeking

Staff Emeritus
Gold Member
Astronuc said:
Bush will not be out of office until January 20, 2009 - inauguration day of the next president. That's 3 years and 7 months - plenty of time to wreak havoc. Can Rehnquist or Stevens survive that time?

What if he were to put Judge Pryor on the Court, the one who placed a religions symbol (the 10 commandments) in his court?
Okay, lets say that I am extremely hopefully that GWBs re-election was his last hurrah. His popularity, and the popularity of his programs is plummeting. The Sunday political shows paint a picture of RW moderates putting distance between themselves and GWBs policies. And he has already lost several key issue; like personal accounts for SS. That is a dead issue, as is..uh hum...Mars.

#### Astronuc

Staff Emeritus
Ivan Seeking said:
Okay, lets say that I am extremely hopefully that GWBs re-election was his last hurrah. His popularity, and the popularity of his programs is plummeting. The Sunday political shows paint a picture of RW moderates putting distance between themselves and GWBs policies. And he has already lost several key issue; like personal accounts for SS. That is a dead issue, as is..uh hum...Mars.
Let's hope so.

I bet GW will leave office and become a board member of several companies which are making big bucks in Iraq and the War on Terror. Nice little racket there. Use your office to start a war, for personal reasons, and then make a huge profit from it. And the Vets - well they have the privilege of serving their country.

Cheney will probably go back to Halliburton, which I believe were notified sometime in 2000 to come up with a plan (hence they were the only company with a plan and got an exclusive contract), and he will get a multi-million dollar bonus.

I know - it reeks of unsubstantiated conspiracy - but just watch in 4 years time.

#### Ivan Seeking

Staff Emeritus
Gold Member
Astronuc said:
I know - it reeks of unsubstantiated conspiracy - but just watch in 4 years time.
That's the sad part. You have to justify your statement in spite of the obvious likelihood. Oh, did you hear that they renamed the oil tanker Condoleeza Rice to avoid any appearance of impropriety or misplaced loyalties.

#### pattylou

Alberto Gonzalez???

he got flack for his nomination to AG. ("The Geneva convention is "quaint." "torture may sometimes be reasonable.")

I see he is getting flack from the conservative base - presumably on his stand on abortion.

Opinions here?

#### plover

Homework Helper
As far as I can tell from discussions I've seen, Sen. Lindsay Graham is perhaps the most favored choice by Dems, with second place actually going to Gonzales. There's a perception that Gonzales would be fairly independent once he was no longer directly working for the President (although it's also been suggested that he would become Scalia's patsy). He's also apparently not anti-pro-choice (I don't know anything concrete enough to put this less equivocally , however I understand one-issue, anti-abortion types are quite decidedly against him). Lots of Dem discussion here, and in other threads on the same site.

One intriguing suggestion was 4th Circuit judge Allyson Duncan (Info here and here.) She's a Bush appointee who was confirmed without a hitch (93-0).

At a GOP site, one person handicapped some possible choices like this:
17-1 Samuel A. Alito Jr.
29-1 Janice R. Brown
24-1 Edith Brown Clement
06-1 Emilio Garza
06-1 Alberto R. Gonzales
08-1 Edith Jones
07-1 J. Michael Luttig
12-1 Michael McConnell
17-1 Theodore Olson
22-1 William Pryor
28-1 John G. Roberts Jr.
36-1 Larry Thompson
11-1 James Harvie Wilkinson III
Discussion then moved to Sen. John Cornyn, and later returned to Emilio Garza (5th Circ.) and Edith Brown Clement (5th Circ.).

The AP on some of the http://www.sfgate.com/cgi-bin/article.cgi?f=/n/a/2005/06/18/national/w092159D57.DTL [Broken].

The NYT on possible nominees and flavors of conservative jurists.

Last edited by a moderator:

#### Ivan Seeking

Staff Emeritus
Gold Member
pattylou said:
Alberto Gonzalez???

he got flack for his nomination to AG. ("The Geneva convention is "quaint." "torture may sometimes be reasonable.")

I see he is getting flack from the conservative base - presumably on his stand on abortion.

Opinions here?
No way, I hope...

#### Pengwuino

Gold Member
Astronuc said:
I know - it reeks of unsubstantiated conspiracy - but just watch in 4 years time.
No, we can find out right now

*travels into the future*

.
.
.

*arrives back*

Sorry guys, meteor hits earth next week.

#### motai

pattylou said:
"Reid later offered four names of people he said would be good for the court: GOP Sens. Mel Martinez of Florida, Mike DeWine of Ohio, Mike Crapo of Idaho and Lindsey Graham of South Carolina. They are people who serve in the Senate now who are Republicans who I think would be outstanding Supreme Court members,'' Reid said. "
If I remember correctly, states can choose to either have judges in their local courts elected on a non-partisan basis, or selection by merit processes or so forth. I wonder why Supreme Court nominees are suspect to political bickering over the current party in power, especially with such an important position at stake.

But Mel Martinez? He isn't even a judge, and he is deep in the political power play. I remember before the recent elections around 25% (at least) of the commercials on TV were his rather nasty political attack ads against his opponent. He definitely rode Bush's coattails to the top.

Biases that are formed politically should not be introduced into the courtroom... it tends to interfere with the concept of fair justice.

#### Rev Prez

motai said:
I wonder why Supreme Court nominees are suspect to political bickering over the current party in power, especially with such an important position at stake.
Because it is an important political issue.

Rev Prez

#### Informal Logic

This is what gets me...

http://www.msnbc.msn.com/id/8443761/site/newsweek/ [Broken]

The Holy War Begins
Bush must choose between the big tent or the revival tent. Inside his Supreme Machine.

July 11 Newsweek issue -

...To those on the religious right, anyone on the list would be preferable to Gonzales, whom they regard as a chilling reincarnation of David Souter, Bush One's moderate pick in 1990. Choosing the attorney general might well doom GOP Senate incumbents, they say, by infuriating the party's fervent, evangelical grass roots. "If the president is foolish enough to nominate Al Gonzales, what he will find is a divided base that will take it out on candidates in 2006," said Manuel Miranda, who heads a coalition of conservative groups called Third Branch Conference. A former legal counsel to Senate Majority Leader Bill Frist, Miranda went on to threaten retribution against First Brother, should he decide to run for president. "We're not Republican patsies," he said. "Jeb Bush can go sell insurance."
Since when is our entire country run by extortion? The feeding-tube fundamentalist fringe can take their thug threats elsewhere.

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#### McGyver

Could GW Be CEO of Fortune 500 Co.?

Astronuc said:
I bet GW will leave office and become a board member of several companies which are making big bucks in Iraq and the War on Terror. Nice little racket there. Use your office to start a war, for personal reasons, and then make a huge profit from it. And the Vets - well they have the privilege of serving their country.
I don't think GW would do well at a publicly traded company. He doesn't do well when challenged by outsiders, and without Carl Rove at his side. His more suitable position would be with a privately-held company within his inner circle.

#### Astronuc

Staff Emeritus
McGyver said:
I don't think GW would do well at a publicly traded company. He doesn't do well when challenged by outsiders, and without Carl Rove at his side. His more suitable position would be with a privately-held company within his inner circle.
Like the Carlyle group - http://www.carlyle.com/eng/index.html [Broken]

For 14 years now, with almost no publicity, the company has been signing up an impressive list of former politicians - including the first President Bush and his secretary of state, James Baker; John Major; one-time World Bank treasurer Afsaneh Masheyekhi and several south-east Asian powerbrokers - and using their contacts and influence to promote the group. Among the companies Carlyle owns are those which make equipment, vehicles and munitions for the US military, and its celebrity employees have long served an ingenious dual purpose, helping encourage investments from the very wealthy while also smoothing the path for Carlyle's defence firms.
The ex-presidents' club

Last edited by a moderator:

#### Art

Pengwuino said:
Sorry guys, meteor hits earth next week.
Lucky it wasn't a meteorite or an asteroid

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# Introduction to Real Analysis

This is a collection of lecture notes I’ve used several times in the two-semester senior/graduate-level real analysis course at the University of Louisville. They are an ongoing project and are often updated. They are here for the use of anyone interested in such material. In return, I only ask that you tell me of mistakes, make suggestions for improvements and, when sharing with others, please give me credit or blame.

I’m very interested in feedback of any type, so don’t be shy about contacting me!

• All chapters as one file (Updated 17-October-2017)
• This file has the advantage that it is filled with hyperlinks, making it easier to track references.

Following are the chapters broken out into individual files. The advantage of these files is that when small updates are done, they show up here first. The disadvantage is they are not hyperlinked.

• Contents (Updated 23-June-2017)

• Chapter 1: Basic Ideas (Updated 23-June-2017)
• Basic set theory
• notation
• Schröder-Bernstein Theorem
• countability, uncountability
• cardinal numbers
• Chapter 2: The Real Numbers (Updated 23-June-2017)
• axioms of a complete ordered field
• basic properties of $$\mathbb{R}$$
• uncountability of $$\mathbb{R}$$
• Chapter 3: Sequences (Updated 17-October-2017)
• monotone sequences
• Nested Interval Theorem
• Cauchy sequences
• contractions
• Chapter 4: Series (Updated 23-June-2017)
• positive series
• convergence tests
• absolute convergence
• conditional convergence
• alternating series
• Riemann’s rearrangement theorem
• Chapter 5: The Topology of $$\mathbb{R}$$ (Updated 23-June-2017)
• open and closed sets
• limit points
• relative topologies
• compactness
• Baire Category Theorem
• measure zero sets
• Cantor middle-thirds set
• Chapter 6: Limits of Functions (Updated 13-July-2016)
• limits
• unilateral limits
• continuity
• uniform continuity
• Chapter 7: Differentiation (Updated 13-July-2016)
• differentiation of functions
• Darboux property
• Mean Value Theorem
• Taylor’s Theorem
• l’Hôspital’s rules
• Chapter 8: Integration (Updated 13-July-2016)
• Riemann-Darboux integral
• Fundamental Theorem of Calculus.
• change of variables
• Chapter 9: Sequences of Functions (Updated 13-July-2016)
• pointwise convergence
• uniform convergence and its relation to continuity, integration and differentiation
• Weierstrass Approximation Theorem
• power series.
• Chapter 10: Fourier Series (Updated 13-July-2016)
• Dirichlet and Fejér kernels
• Césaro convergence
• pointwise convergence.
• continuous function with divergent Fourier series
• Gibbs phenomenon
• Bibliography (Updated 13-July-2016)

• Index (Updated 13-July-2016)

Contact me at Lee Larson (lee.larson@louisville.edu)


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# Small question about Bloom RT formats

This topic is 1958 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

When doing the whole bloom process do the render targets have to be FP16 like the original light accumulation target ?
Do I perform the brightpass and gaussian blur with FP16 format buffer ?
I realize I need to keep it in HDR until the tonemapping is performed, right ?

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There's no right way to do Bloom because it's basically an hack, so you do it anyway you want to achieve the result you want ;)

In my opinion using FP16 render targets in the bright/blur passes produces better results because you might get some color information during the blur that would be clipped in the bright pass if you used 8-bit render targets. Edited by TiagoCosta

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[quote name='TiagoCosta' timestamp='1346954150' post='4977288']
There's no right way to do Bloom because it's basically an hack
[/quote]
Bloom is no hack. It's actually just a part of the flares lenses produce. Because no glass is perfect a really small amount of light always gets refracted onto nearby pixels. Normal distribution (aka gaussian filter) doesn't produce the most physically correct results though. An exponential distribution is better suited for Bloom, but it's not separable thus being not well suited for a realtime application. You can approximate an exponential distribution by using multiple gaussians though which is what they are doing in CryEngine 3 and Unreal Engine 4.
Also it doesn't really make any sense to implement a bright pass to subtract a certain amount of light so that only the bright light gets blurred. That's often a sign that the engine is not actually High Dynamic Range. Light always scatters onto surrounding pixels, just a tiny amount of it, but it's completely independent of its brightness. So a proper exponential filter and maybe a multiplicative factor should be about it.

BTT: If you want to have High Dynamic Range Rendering, you should keep the Bloom in High Dynamic Range and later perform Tone Mapping to resolve it to Low Dynamic Range. Also there's no need for an actual bright pass. Just do that in the Bloom Pass to save Bandwidth.

Also a cool trick is to use a R10G10B10A2 resource as a backbuffer. Most monitors can actually display this when the application runs in full screen. So you can actually have a higher dynamic range than the usual B8G8R8A8 backbuffer. Edited by CryZe

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A R10G10B10A2 backbuffer won't give you any extra dynamic range, it will just give you more precision for the standard displayable range. Either way there's no guarantee that the display won't just convert it to something else when it outputs. Edited by MJP

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[quote name='CryZe' timestamp='1346956985' post='4977319']
Also a cool trick is to use a R10G10B10A2 resource as a backbuffer. Most monitors can actually display this when the application runs in full screen.[/quote]
Most consumer monitors (cheap TN and IPS panels) have 6-bit colour depth, and achieve 8-bit equivalence via dithering. Even mid-range IPS panels are only 8-bit (plus 2-bit dither).

Basically, if you are spending less than $1,000 per monitor, you aren't getting a truly 10-bit panel. And even in the over$1,000 range, you have to be pretty selective to find the panels specifically aimed at graphics professionals.

And that's ignoring the fact that a DVI cable only carries 8-bit colour, so you are limited to monitors that offer DisplayPort or full HDMI 1.3 support (and a graphics card to match)...


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# Rotational Kinematics Question

#### Extremist223

1. The problem statement, all variables and given/known data
The wheel of fortune wheel, whose radius is 3.00m, has 20 different numbers, at equal intervals, for a contestant to land on. The second contestant can spin the wheel, and wishes to land on the $1.00. The wheel starts off 4 positions away from the$1.00. Assuming a deceleration of 0.400 rad/s2, what must be the minimum tangential velocity for the contestant to reach the \$1.00, if they pull the wheel down?

2. Relevant equations
2pi rads is a full circle
w^2 = wo^2 + 2(alpha)(theta)
alpha= angular acceleration
theta is distance in radians travelled
w^2 is final angular velocity squared
wo^2 is initial angular velocity squared
Tangential Velocity = Radius x w

3. The attempt at a solution
the goal is 4 intervals away from the start therefore 4x0.314= 1.257 rads travelled.
it wants to land 4 intervals away from the start so the final angular velocity = 0 rad/s
therefore w^2 = wo^2 + 2 (alpha)(theta)
so at this point i took vt= rw vt= 3m x 1.00265rad/s vt = 3.008m/s

I'm getting the answer wrong and I don't know why. Can anyone help me please?

#### Mindscrape

It could be that you are really 16 spaces away because of the direction you are required to spin in. Looks okay otherwise.

#### Extremist223

thanks it seems to be the case one of the multiple choice answers is very close to this, but since it doesnt specify where on the wheel it is being pulled down from left side or right side this question is ambiguous

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We Value Civility
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Comparing absolute values

Write an inequality to compare absolute values.

Problem

Fill in the blank.
open vertical bar, 4, close vertical bar __ minus, 1
Please choose from one of the following options.