text:
# Tag Info

## Hot answers tagged non-perturbative

2

Leaving out numerical factors, we have that $$\mathrm{d}j_A = \mathrm{Tr}(F \wedge F)$$ This already shows that we are dealing with a topological quantity, since the RHS is the second Chern character of the gauge field (or rather, the principal bundle associated to it). Now, there is also the (3D) Chern-Simons form  \omega = \mathrm{Tr}(F \wedge A - ...

2

If you assume separability of the wave function, i.e., $\psi(\mathbf x)=u(x)v(y)w(z)$, you can solve the individual components separately: \begin{align} -\frac{\hbar^2}{2\mu}\frac{d^2u(x)}{dx^2}+V_1(x)u(x)&=E_1u(x)\\ -\frac{\hbar^2}{2\mu}\frac{d^2v(y)}{dy^2}+V_2(y)v(y)&=E_2v(y)\tag{1}\\ ...

2

In the case of QCD there is asymptotic freedom, meaning that though the theory is strongly coupled at low energies (such that we still cannot analytically calculate how the atomic nucleus stays together) the coupling becomes less and less as we go to higher and higher energies. This means that the ultimate picture of the behavior of quarks is a weakly ...

1

Instantons are characterized by the winding number and a set of collective parameters (e.g. location of the centers of the instantons, their sizes and the inequivalent orientations in the global group space / space-time). Quantum fluctuations of a unit winding number instanton can either leave the collective parameters unchanged (non-zero modes), or change ...

1

No. Feynman diagrams are made by summing over the perturbative contributions of quantum amplitudes. They cannot hold non-perturbative information.

1

A vacuum is a field configuration that is pure gauge, i.e. $A = g^{-1}\mathrm{d}g$ for a gauge transformation $g$, and hence $F = \mathrm{d}_A A = 0$ (for $\mathrm{d}_A$ the gauge covariant derivative). An instanton is a local minimum of the action, which is given by an (anti-)self-dual configuration $F = \pm \star F$. It is not a vacuum for non-zero ...

Only top voted, non community-wiki answers of a minimum length are eligible


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# How to approach the inductive step in proving statements like $k! \geq 3^{k-2}$

I am having a really hard time grasping proof by induction and struggling to write consitent thorough proofs which use induction.

For example, proving the following

$k! \geq 3^{k-2}$

Now I understand that we first test the base case $k=1$, and see that it holds and then assume that it is true for any $k$, and if we can show it holds for $k+1$, then we would have proven it for all $k>1$.

However, I am really struggling with the inductive step, someone suggested multiplying both sides by $k+1$ leaving $(k+1)! \geq 3^{k-2}(k+1)$ but I still don't see how it helps, surely we should be putting in $k+1$ instead of $k$?

If anyone has any texts I can read or articles I can read to help improve my skills with proof by induction I would really appreciate it, I'm really struggling with it.

• For some basic information about writing math at this site see e.g. here, here, here and here. – user137731 Nov 16 '14 at 19:36
• rb20, I've edited your text so that the formulas come out properly. Click on "edit" to have a look at the source text I entered and see how it's done. Also, you can correct any mistakes I made in editing. However, it will take some time for the edits I made to be approved, so they won't appear immediately for you. – Mike Nov 16 '14 at 19:40
• HINT#1: Your proof of the inductive step should start with "Assume that $k! \ge 3^{k-2}$ for some $k \in \Bbb N$" and should end with "Therefore $(k+1)!\ge 3^{(k+1)-2}$." HINT#2: For proofs of inequalities, instead of trying to get $P_{n+1} \ge Q_{n+1}$, you often just need to get $P_{n+1} \ge \text{something} \ge Q_{n+1}$. – user137731 Nov 16 '14 at 19:42
• In this case, it doesn't work nicely to take $k = 1$ as your base case. You should check it for $k= 1$ and $k = 2$ separately, and then prove that if it's true for some $k \geq 2$, then it's true for $k + 1$. The point is that in this case $k + 1 \geq 3$, and you can use that in your inequality. – Mike Nov 16 '14 at 19:43
• rb20, here is a suggestion. Write down explicitly what you know at level $k$, and what you're trying to prove at level $k + 1$. You can edit that into the question at the bottom, after your original text. – Mike Nov 16 '14 at 19:48

You can see that that the inequality holds for $k=1$. Now lets assume it holds true for some $k>1$.Thus we have $$k!>3^{k-2}$$ multiply both sides by $(k+1)$, $$(k+1)!>=3^{k-2}.(k+1)$$ we know $k>=2$ thus $k+1>=3$, so $3^{k-2}.(k+1)>=3^{k-2}.3=3^{k-1}$ so we get $(k+1)!>=3^{k-1}$.Thus relation holds true for $k+1$ also. As far as books are concerned you can look up the chapter about induction (chapter 2) in the book A Walk Through Combinatorics by Bona. It has examples plus a lot of solved exercises.

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Original article
2006.03511 Unsupervised Translation of Programming Languages
News
Research
Lample et al.
Researcher
Guillaume Lample
glample@fb.com

## Fully unsupervised neural transcompiler

• Converts languages obtained from GitHub
• BigQuery
• unsupervised
• it looks for previously undetected patterns in data sets without labels and with a minimal amount of human supervision
• outperforms rule-based baselines by a “significant” margin.

### Model of TransCoder

• seq2seq with attention

• composed of
• encoder
• decoder
• transformer architecture
• the same model is used for all programming languages.
• Trained using the three principles of unsupervised machine translation

• initialization,
• LMing, and
• back-translation.

In this section, we summarize these principles and detail how we instantiate them to translate programming languages.

### Cross Programming Language Model pretraining

 1 2 3 4 5 6 7 8  Pretraining A key ingredient of unsupervised machine translation. It ensures that sequences with a similar meaning are mapped to the same latent representation, regardless of their languages.

Originally, pretraining was done by initializing the model with cross-lingual word representations.

In the context of unsupervised English-French translation, the embedding of the word “cat” will be close to the embedding of its French translation “chat”.

Cross-lingual word embeddings can be obtained by training monolingual word embeddings and aligning them in an unsupervised manner.

Subsequent work showed that pretraining the entire model (and not only word representations) in a cross-lingual way could lead to significant improvements in unsupervised machine translation.

### The pretraining strategy of Lample and Conneau

In particular, we follow the pretraining strategy of Lample and Conneau, where a XLM is pretrained with a masked LMing objective on monolingual source code datasets.

First principle: initialization
The first principle initializes the model with

As a result, pieces of code that express the
same instructions are mapped to the same
representation, regardless of the programming
language.

Second principle: denoising
Train the decoder to always generate valid
sequences, even when fed with noisy data, and
increases the encoder robustness to input
noise.
Third (and last) principle: back-translation
Allows the model to generate parallel data
which can be used for training.

Whenever the Python C++ model becomes better,
it generates more accurate data for the C++
Python model, and vice versa.

We obtain cross-lingual embeddings after training.

The cross-lingual nature of the resulting model comes from the significant number of common tokens (anchor points) that exist across languages.

In the context of English-French translation, the anchor points consists essentially of digits and city and people names.

In programming languages, these anchor points come from common keywords (e.g. for, while, if, try), and also digits, mathematical operators, and English strings that appear in the source code.

For the masked LMing (MLM) objective, at each iteration we consider an input stream of source code sequences, randomly mask out some of the tokens, and train TransCoder to predict the tokens that have been masked out based on their contexts.

We alternate between streams of batches of different languages.

This allows the model to create high quality, cross-lingual sequence representations.

Denoising auto-encoding

We initialize the encoder and decoder of the seq2seq model with the XLM model that was pretrained.

The initialization is straightforward for the encoder, as it has the same architecture as the XLM model.

The transformer decoder, however, has extra parameters related to the source attention mechanism .

Following Lample and Conneau, we initialize these parameters randomly.

XLM pretraining
allows the seq2seq model to
generate high quality representations of input
sequences.

However, the decoder lacks the capacity to translate, as it has never been trained to decode a sequence based on a source representation.

To address this issue, we train the model to encode and decode sequences with a Denoising Auto-Encoding (DAE) objective .

The DAE objective operates like a supervised machine translation algorithm, where the model is trained to predict a sequence of tokens given a corrupted version of that sequence.

To corrupt a sequence, we use the same noise model as the one described in Lample et al. .

Namely, we randomly mask, remove and shuffle input tokens.

In practice, the “cross-linguality” of the model highly depends on the amount of anchor points across languages.

As a result, a XLM model trained on English- French will provide better cross-lingual representations than a model trained on English-Chinese, because of the different alphabet which reduces the number of anchor points.

In programming languages, the majority of strings are composed of English words, which results in a fairly high number of anchor points, and the model naturally becomes cross- lingual.

## Arxiv Summary

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54  A transcompiler, also known as source-to- source translator, is a system that converts source code from a high-level programming language (such as C++ or Python) to another. Transcompilers are primarily used for interoperability, and to port codebases written in an obsolete or deprecated language (e.g. COBOL, Python 2) to a modern one. They typically rely on handcrafted rewrite rules, applied to the source code abstract syntax tree. Unfortunately, the resulting translations often lack readability, fail to respect the target language conventions, and require manual modifications in order to work properly. The overall translation process is timeconsuming and requires expertise in both the source and target languages, making code- translation projects expensive. Although neural models significantly outperform their rule-based counterparts in the context of NL translation, their applications to transcompilation have been limited due to the scarcity of parallel data in this domain. In this paper, we propose to leverage recent approaches in unsupervised machine translation to train a fully unsupervised neural transcompiler. We train our model on source code from open source GitHub projects, and show that it can translate functions between C++, Java, and Python with high accuracy. Our method relies exclusively on monolingual source code, requires no expertise in the source or target languages, and can easily be generalized to other programming languages. We also build and release a test set composed of 852 parallel functions, along with unit tests to check the correctness of translations. We show that our model outperforms rule-based commercial baselines by a significant margin.

## GPT-3 summary of Arxiv summary

We train a neural transcompiler using monolingual source code from GitHub, and show that it can translate functions between C++, Java, and Python with high accuracy.


text:
# prevent logo from title page in beamer class

I'm having problems with inserting a logo on my presentation in the beamer class.

\logo{\includegraphics[scale=0.05]{logo.png}}

This is how I inserted the logo, which works perfectly; it is placed in the lower right hand corner as desired.

But I don't want the logo on the title slide. Is there a way to hide the logo on the title slide?

-
You should also upvote Alan Munn's answer (this has to be done separately from accepting). –  lockstep May 22 '12 at 21:49

Just use

\frame[plain]{\titlepage}

instead of \maketitle to make your titlepage slide.

Depending on your needs, the following question might also be of use:

-
Wow, that was fast and very helpful! Thanks also for the link. –  Newby1 May 22 '12 at 21:47

Another option, which only removes the logo but maintains all the other elements, is to locally redefine the logo template for the title page:

\documentclass{beamer}
\usetheme{Berkeley}

\logo{\includegraphics[scale=0.02]{ctanlion}}

\title{The Title}
\author{The Author}
\institute{The Institute}

\begin{document}

{
\setbeamertemplate{logo}{}
\begin{frame}
\maketitle
\end{frame}
}

\begin{frame}
test
\end{frame}

\end{document}

CTAN lion drawing by Duane Bibby.

-
Thanks for this as well. Still amazing how many things are possible with Latex.... –  Newby1 May 22 '12 at 21:59

A simpler solution is to define the logo immediately after the titlepage such that it doesn't appear in that frame.

For example:

\documentclass{beamer}
\usetheme{Berkeley}

\title{The Title}
\author{The Author}
\institute{The Institute}

\begin{document}
\begin{frame}
\titlepage
\end{frame}

% insert the logo here
\logo{\includegraphics[width=1.5cm]{logopolito}} % <= test image
\begin{frame}
test
\end{frame}

\end{document}

will produce the following frames:

Of course, the method holds for all subsequent frames in your presentation.

-


text:
Jamiya F.

# find the value of one side length of an equilateral triangle with the following sides : x+2 , 3x-4 ,13x-34

sides : x+2 , 3x-4 , 13x-34


text:
## Monday, July 7, 2014

### Law of Iterated Expectation

Let $X$ and $Y$ denote continuous, random, real-valued variables with joint probability density function $f(X, Y)$. The marginal density function of $Y$ is $f_Y(y) := \int_{x \in \mathbb R} f(x, y) dx$. The expectation $E(Y)$ of $Y$ can be recovered by integrating against the marginal density function. In particular, $$\label{eq:E(Y)} E(Y)=\int_{y \in \mathbb R} y f_Y(y) dy = \int_{y \in \mathbb R} \int_{x \in \mathbb R} y f(x, y) \ dx \ dy.$$ The conditional probability density function of $Y$ given that $X$ is equal to some value $x$ is defined by $$\label{eq:cdf} f_{Y \mid X} (y \mid X = x):= f_{Y \mid X} (y \mid x)= \frac{f(x, y)}{f_X(x)} = \frac{f(x,y)}{\int_{y \in \mathbb R} f(x, y) \ dy}.$$ The conditional expectation $E(Y \mid X = x)$ of $Y$ given that $X$ has value $x$ is given by $$\label{eq:cond exp} E(Y \mid X = x) = \int_{y \in \mathbb R} y f_{Y \mid X} (y \mid x) \ dy.$$ But $E(Y \mid X = x)$ depends on $X$, so in turn is itself a random variable denoted $E(Y \mid X)$, whence we can compute its expectation. Now \begin{align*} E (E (Y \mid X)) &= \int_{x \in \mathbb R} E(Y \mid x) f_X(x) \ dx \\ & = \int_{x \in \mathbb R} f_X(x) \left( \int_{y \in \mathbb R} y f_{Y \mid X}(x, y) \ dy \right) \ dx \quad \text{(by \ref{eq:cond exp} )} \\ & = \int_{x \in \mathbb R} \int_{y \in \mathbb R} f_X(x) \cdot y \frac{f(x,y)}{f_X(x)} \ dy \ dx \quad \text{(by \ref{eq:cdf})}\\ & = \int_{y \in \mathbb R} \int_{x \in \mathbb R} y f(x, y) \ dx \ dy \\ & = E(Y). \quad \text{(by \ref{eq:E(Y)})} \end{align*} This result is sometimes called the law of the iterated expectation.


text:
1 Notes

• Commands preceded with "$" imply that you should execute the command as a general user - not as root. • Commands preceded with "#" imply that you should be working as root. • Commands with more specific command lines (e.g. "rtrX>" or "mysql>") imply that you are executing commands on remote equipment, or within another program. 2 Exercise The routers are able to send syslog messages to multiple destinations, so that 1 router can send messages to 4 or even 5 destinations. We therefore need to configure the router to send messages to each of the PCs in the group. 2.1 Configure sending of syslog messages from your group's router Configure your virtual router to send syslog messages to every server in your group. Everyone in your group should log into your group's router and do the following (assuming you are already logging in on your virtual machine): $ ssh cisco@rtrX
rtrX> enable
rtrX# config terminal

rtrX(config)# logging 10.10.X.Y

... where X.Y is the IP of your PC (group + number, example pc2 = 10.10.1.2).

rtrX(config)# logging facility local0
rtrX(config)# logging userinfo
rtrX(config)# exit
rtrX# write memory
rtrX# exit

Now run show logging to see the summary of the log configuration.

rtrX# show logging

The other participants in your group will be doing the same thing, so you should not be surprised if you see other destinations as well in the output of "show logging" - Press SPACE to page through the output

Logout from the router (exit):

rtrX# exit

That's it. The router should now be sending UDP SYSLOG packets to your PC on port 514.

$sudo -s # apt-get install tcpdump (if already installed dont worry) # tcpdump -s0 -nv -i eth0 udp port 514 Then have one person in your group log back in on the router and do the following: $ ssh cisco@rtrX
rtrX> enable
rtrX# config terminal
rtrX(config)# exit
rtrX> exit

You should see some output on your PC's screen from tcpdump. It should look something like:

11:20:24.942289 10.10.1.254.63515 > 10.10.1.1.514: SYSLOG local0.notice, length: 110
11:20:24.944376 10.10.1.254.53407 > 10.10.1.1.514: SYSLOG local0.notice, length: 102

When you have seen this, hit Ctrl-C to exit tcpdump.

Now you can configure the logging software on your PC to receive this information and log it to a new set of files.

2.2 Configure rsyslog

Be sure you are logged in to your virtual machine and that you are the root user.

Edit the file /etc/rsyslog.conf:

# editor /etc/rsyslog.conf

...and find and un-comment the following lines (that is, remove the initial '#' only)

#$ModLoad imudp #$UDPServerRun 514

change to:

$ModLoad imudp$UDPServerRun 514

Then change this line:

$PrivDropToGroup syslog change to:$PrivDropToGroup adm

Then save the file and exit.

Now, create a file named "/etc/rsyslog.d/30-routerlogs.conf"

# editor /etc/rsyslog.d/30-routerlogs.conf

... and add the following lines (carefully COPY and PASTE):

$template RouterLogs,"/var/log/network/%$YEAR%/%$MONTH%/%$DAY%/%HOSTNAME%-%$HOUR%.log" local0.* -?RouterLogs & ~ PLEASE double check (verify) that what you have pasted is the SAME as what is above. In particular, make sure that you are using TAB and not SPACE between "template" and "RouterLogs", and also between "local0.*" and "-?RouterLogs". If the above is not pasted correctly, it will NOT work. Save and exit, then do: # mkdir /var/log/network # chown syslog:adm /var/log/network # chmod g+w /var/log/network Restart rsyslog: # service rsyslog restart 2.3 Test syslog To be sure there are some logging messages log back in to the router, and run some "config" commands, then logout. e.g. $ ssh cisco@rtrX
rtrX> enable
rtrX# config terminal
rtrX(config)# exit
rtrX> exit

Be sure you log out of the router when you are finished. If too many people log in without logging out then others cannot gain access to the router.

On your PC, See if messages are starting to appear under /var/log/network/<year>/<month>/<day>/

$cd /var/log/network$ ls
$cd 2013$ ls
... this will show you the directory for the month
... cd into this directory
$ls ... repeat for the next level (the day of the month)$ ls

Then use 'tail' to look at the log file(s) in this directory. The names are dynamic based on the sender and the host, so use the file that you see. It may be something like this:

$ls rtr8-16.log$ tail rtr8-16.log
... logging messages are shown ...

2.4 Troubleshooting rsyslog

If no files are appearing under the /var/log/network directory, then another command to try while logged into the router, in config mode, is to shutdown / no shutdown a Loopback interface, for example:

\$ ssh cisco@rtrX
rtrX> enable
rtrX# conf t
rtrX(config)# interface Loopback 999
rtrX(config-if)# shutdown

wait a few seconds

rtrX(config-if)# no shutdown

Then exit, and save the config ("write mem"):

rtrX(config-if)# exit
rtrX(config)# exit
rtrX# write memory
rtr1# exit

Check the logs under /var/log/network

# cd /var/log/network
# ls
...follow the directory trail

Still no logs?

Try the following command to send a test log message locally:

# logger -p local0.info "Hello World\!"

If a file has not been created yet under /var/log/network, then check your configuration for typos. Don't forget to restart the rsyslog service each time you change the configuration.

What other commands can you think of that you can run on the router (BE CAREFUL!) that will trigger syslog messages? You could try logging in on the router and typing an incorrect password for "enable".

Be sure that you do an "ls" command in your logging directory to see if a new log file has been created at some point.


text:
# How to determine the mole fraction of the liquid phase from a given mole fraction of the vapour phase for a benzene/toluene mixture?

A quantity of benzene, $\ce{C6H6}$, and toluene, $\ce{C6H5CH3}$, is placed in a $1~\mathrm{L}$ evacuated vessel at $25~\mathrm{^\circ C}$. At equilibrium, a small volume of liquid is visible at the bottom of the container. A sample of the vapour phase is analysed and found to contain $53~\mathrm{mol-\%}$ benzene. What is the mole fraction of benzene in the liquid phase?

Vapour pressures at 25 °C: benzene = 0.125 atm, toluene = 0.037 atm

My attempt at the problem:
First of all, I am not even certain if this a Raoult's law problem. I just assumed so, since Henry's constant was not given in the problem.

First I'm getting the total liquid pressure using Raoult's law. \begin{align} P(\text{benzene}) &= X(\text{benzene}) \cdot P_\mathrm{vap}(\text{benzene}) & &= 0.53 \times 0.125 &&= 0.06625~\mathrm{atm}\\ P(\text{toluene}) &= X(\text{toluene}) \cdot P_\mathrm{vap}(\text{toluene}) & &= 0.47 \times 0.037 &&= 0.01739~\mathrm{atm}\\ P_\mathrm{total} &&&= 0.06625 + 0.01739 &&= 0.08364~\mathrm{atm}\\ \end{align}

Then using Dalton's law, I am getting the mole fraction of Benzene in the liquid phase.
\begin{align} P(\text{benzene}) &= Y(\text{benzene}) \cdot P_\mathrm{total}\\ Y(\text{benzene}) &= \frac{P(\text{benzene})}{P_\mathrm{total}}\\ &= \frac{0.06625}{0.08364} = 0.792. \end{align}

However, my answer is incorrect according to the answer key that has $0.25$ as the answer. I have a feeling my whole approach may be wrong since my answer is off by a lot. What am I doing wrong?

The question gives you the mol percentage of benzene ($53 \%$) in the vapour phase, and not in the liquid phase.

It's always good to write down what you have first. (note: I use $y$ to denote mole fractions in the vapour phase and $x$ to denote mole fractions in the liquid state)

First of all, we have $y_\text{benzene} = 0.53$. Additionally, we have been given the vapour pressure of pure benzene and toluene (denoted with a superscript $*$):

$$p^*_\text{benzene} = 0.125\ \mathrm{atm} \qquad p^*_\text{toluene} = 0.037\ \mathrm{atm}$$

and we require $x_\text{benzene}$

We also, know $p_\text{benzene} = x_\text{benzene}\ p^*_\text{benzene}$, and the total vapor pressure is given by $$P = p_\text{benzene} + p_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene} + x_\text{toluene}\ p^*_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}$$

Finally,

$$y_\text{benzene} = \frac{p_\text{benzene}}{P} = \frac{x_\text{benzene}\ p^*_\text{benzene}}{x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}}$$

It is now, trivial to solve this equation. Since I am lazy, and was not in the mood to write down two lines of algebra and reach for a calculator, I chucked it into Mathematica, and the result is:

Thus, $x_\text{benzene} = 0.25$, which is indeed the result given in your answer key.

You solved the problem assuming that 0.53 is the mole fraction of benzene in the liquid phase. However, the problem statement says that "the vapor phase is analyzed and found to contain 53 mole percent benzene." Call P the total pressure. What is the partial pressure of benzene in the vapor? What is the partial pressure of toluene in the vapor? In terms of P, what is the mole fraction of benzene in the liquid? In terms of P, what is the mole fraction toluene in the liquid. In terms of P, what is the sum of the mole fractions of benzene and toluene in the liquid? From this equation, what is the total pressure P?

• Okay, so I retried again and was able to arrive at the correct answer with your steps, but I don't quite understand the chemistry behind it. Partial pressure of benzene in vapour is 0.53P, 0.47P for toluene. Mole fraction of benzene in liquid state is 0.53P/0.125 and for toluene it is 0.47P/0.037. This part, I am confused by why the partial pressure in vapour is used in Raoult's law when I am calculating mole fraction in liquid. – M.M Dec 18 '15 at 5:53
• Raoult's law says that the partial pressure in the vapor is equal to the equilibrium vapor pressure of the pure substance at the given temperature times the mole fraction in the liquid. So, the mole fraction in the liquid is equal to the partial pressure in the vapor divided by the equilibrium vapor pressure of the pure substance. – Chet Miller Dec 18 '15 at 12:18
• Thanks a lot! To clarify on the states, is it always the case that Raoult's law says partial pressure in vapor equals equilibrium vapor pressure times mole fraction of liquid? The partial pressure part is always in vapor right? However the mole fraction part can be liquid/liquid solution or liquid/solid solution? – M.M Dec 18 '15 at 14:36
• Yes, for vapor-liquid equilibrium. For liquid/liquid or liquid/solid solution, a different law applies. – Chet Miller Dec 18 '15 at 22:10

According to Raoults law, it should be 0,25. Lets say, you have the molar fraction of each component in vapor, and lets say you have the saturated pressure at some temperature, than according to Raoults law you can say: 0,53(molar fraction of benzene in vapor)Pi(which is equal to Xbpb(saturated, benzene)+Xtpt(saturated, toluene) = Xbpb than: Xb*0,125=0,53*(Xb*0,125+(1-Xb)*0,037) Xb=0,25


text:
# Rotating Car Platform

1. Jun 22, 2009

### Patta1667

1. The problem statement, all variables and given/known data

I'm not given any signs of a correct answer in the book, so could I get a check here?

A car is driven on a large revolving platform which rotates with constant angular speed w. At t = 0 a driver leaves the origin and follows a line painted radially outward on the platform with constant speed v. The total weight of the car is W, and the coefficient of friction is $$\mu$$ between the car and platform.

a) Find the acceleration of the car as a function of time using polar coordinates.

b) Find the time at which the car starts to slide.

c) Find the direction of the friction force with respect to the instantaneous position vector r just before the car starts to slide.

2. Relevant equations

$$F_{friction} <= \mu W$$
$$a = (r'' - r \theta' ^2) \hat r + (r \theta'' + 2r'\theta') \hat \theta$$

3. The attempt at a solution

a) r' = v, $$\theta'$$ = w, and r'' = $$\theta ''$$ = 0, so $$a = (-v w^2)t \hat r + 2vw \hat \theta$$.

b) In order for the car to follow the correct path (constant velocities in the radial and normal directions), the radial and normal accelerations must be zero. This is due to the frictional force counteracting these accelerations, and when the car starts to slide is the point at which $$m *|a_{net}| = f_{max}$$. This happens when:

$$m|a| = m\sqrt{v^2w^4t^2 + 4v^2w^2} = \mu W = \mu mg \implies t = \sqrt{\frac{u^2g^2 - 4v^2w^2}{v^2w^4}}$$

c) My figuring was that the angle between r (vector) = |r|r (unit vector) and the frictional force is the same as the angle between the r unit vector and the friction. The friction is in the opposite direction as the net acceleration force, and when the car starts to slide, $$|f| = \mu W$$. If I am using the dot product wrong here, please correct me:

$$\hat r * f = |\hat r||f| \cos \theta \implies \cos \theta = \frac{\hat r * f}{|\hat r||f|} = \frac{mvw^2}{\mu W} = \frac{vw^2}{\mu g} \implies \theta = \cos^{-1} \left( \frac{vw^2}{\mu g} \right)$$

This angle, if r is directed down the positive x-axis, should be between r and the down-right frictional force vector, correct? (the net acceleration would be pointing to the upper-left since the radial component is left and the normal component is up, so the frictional force would be pointing opposite that).

I'm not experienced with using accelerations in polar coordinates and finding angles such as in c), so I would greatly appreciate any tips or corrections. Thanks!

PS: In b), when finding |a|, is that way valid in polar vectors also? eg. $$|v| = \sqrt{v_r^2 + v_{\theta}^2}$$


text:
Inhabited Set
Get Inhabited Set essential facts below. View Videos or join the Inhabited Set discussion. Add Inhabited Set to your PopFlock.com topic list for future reference or share this resource on social media.
Inhabited Set

In constructive mathematics, a set A is inhabited if there exists an element ${\displaystyle a\in A}$. In classical mathematics, this is the same as the set being nonempty; however, this equivalence is not valid in intuitionistic logic (or constructive logic).

## Comparison with nonempty sets

In classical mathematics, a set is inhabited if and only if it is not the empty set. These definitions diverge in constructive mathematics, however. A set A is nonempty if it is not empty, that is, if

${\displaystyle \lnot [\forall z(z\not \in A)].}$

It is inhabited if

${\displaystyle \exists z(z\in A).}$

In intuitionistic logic, the negation of a universal quantifier is weaker than an existential quantifier, not equivalent to it as in classical logic.

## Example

Because inhabited sets are the same as nonempty sets in classical logic, it is not possible to produce a model in the classical sense that contains a nonempty set X but does not satisfy "X is inhabited". But it is possible to construct a Kripke model M that satisfies "X is nonempty" without satisfying "X is inhabited". Because an implication is provable in intuitionistic logic if and only if it is true in every Kripke model, this means that one cannot prove in this logic that "X is nonempty" implies "X is inhabited".

The possibility of this construction relies on the intuitionistic interpretation of the existential quantifier. In an intuitionistic setting, in order for ${\displaystyle \exists z\phi (z)}$ to hold, for some formula ${\displaystyle \phi }$, it is necessary for a specific value of z satisfying ${\displaystyle \phi }$ to be known.

For example, consider a subset X of {0,1} specified by the following rule: 0 belongs to X if and only if the Riemann hypothesis is true, and 1 belongs to X if and only if the Riemann hypothesis is false. If we assume that Riemann hypothesis is either true or false, then X is not empty, but any constructive proof that X is inhabited would either prove that 0 is in X or that 1 is in X. Thus a constructive proof that X is inhabited would determine the truth value of the Riemann hypothesis, which is not known, By replacing the Riemann hypothesis in this example by a generic proposition, one can construct a Kripke model with a set that is neither empty nor inhabited (even if the Riemann hypothesis itself is ever proved or refuted).

## References

• D. Bridges and F. Richman. 1987. Varieties of Constructive Mathematics. Oxford University Press. ISBN 978-0-521-31802-0


text:
# Solving Percent Problems

Basically, there are 3 types of percent problems:

• What is ${p}$% of ${m}$?
• ${p}$% of what is ${m}$?
• What percent of ${m}$ is ${n}$?

Now, we will practice in these percent problems.

These types of problems can be easily solved using proportions.

Example 1. What is 20% of 35?

Let ${n}$ represents required number.

Percent can be written as ratio $\frac{{20}}{{100}}$.

From another side the same ratio can be represented as $\frac{{n}}{{35}}$.

We obtained proportion $\frac{{20}}{{100}}=\frac{{n}}{{35}}$.

Solving it, we obtain that ${n}={7}$.

Therefore, 20% of 35 is 7.

We can generalise this result.

${p}$% of ${m}$ is $\frac{{p}}{{100}}\cdot{m}$.

Example 2. 95% of what is 237.5?

Let ${n}$ represents required number.

Percent can be written as ratio $\frac{{95}}{{100}}$.

From another side the same ratio can be represented as $\frac{{237.5}}{{n}}$.

We obtained proportion $\frac{{95}}{{100}}=\frac{{237.5}}{{n}}$.

Solving it, we obtain that ${n}={250}$.

Therefore, 95% of 250 is 237.5.

We can generalise this result.

${p}$% of what is ${m}$? Of $\frac{{m}}{{100}}\cdot{p}$.

Example 3. What percent of 15 is 27?

Let ${p}$ represents required percent.

Percent can be written as ratio $\frac{{p}}{{100}}$.

From another side the same ratio can be represented as $\frac{{27}}{{15}}$.

We obtained proportion $\frac{{p}}{{100}}=\frac{{27}}{{15}}$.

Solving it, we obtain that ${p}={180}$%.

Therefore, 27 is 180% of 15.

We can generalise this result.

${n}$ is $\frac{{n}}{{m}}\cdot{100}$% of ${m}$.

Using above 3 types of percent problems, we can solve some real-world problems.

Example 4. Initially population of some town was 200000 people. Recently it has grown by 15%. What is the current population?

First, we need to find by how many people population has grown?

In other words, what is 15% of 200000? Answer is $\frac{{15}}{{100}}\cdot{200000}={30000}$.

So, the current population is sum of initial population and growth: ${200000}+{30000}={230000}$ people.

Let's see how to solve "backward" problem.

Example 5. Initial price of the dress is $175. Discounted price is$105. What is the discount (in percents)?

First, let's calculate discount in dollars. It is simply $175-$105=$70. Now, we need to find what percent of initial price$175 is $70. Answer is $\frac{{70}}{{175}}\cdot{100}={40}$%. Therefore, discount is 40%. Now, it is time to exercise. Exercise 1. What percent of 150 is 37.5? Answer: 25%. Exercise 2. What is 12% of 57? Answer: 6.84. Exercise 3. 45% of what is 99? Answer: 220. Exercise 4. Discounted price of the toy is$30. It appears, that discount was 4%. What was the initial price?

Answer: \$31.25. Let ${n}$ is initial price. Then discount in dollars is ${n}-{30}$. Percent discount is $\frac{{{n}-{30}}}{{n}}=\frac{{4}}{{100}}$.

Exercise 5. 5 years ago population of small town was 800 people. Current population is 1000 people. By how much percents did the population increase?

Answer: 25%. Increase in people is ${1000}-{800}={200}$. Percent increase is $\frac{{200}}{{800}}\cdot{100}={25}$%.


text:
# Calculating the coordinates of a vertex within a triangle so that the triangle becomes a right-angle

I have three points defining the vertices of a triangle A(1,4) B(6,7) C(5,1)

I have found that vector AC has a slope(m) of 0.6 and vector BC has a slope of 6. From these slopes have the angles of ≈34.39° and ≈73.94° respectively. Based on this, I know that these vectors intersect to form angle B at ≈39.55°. I would like to relocate vertex B so that angle B becomes 90° and therefore, ∠ABC becomes a right-angle.

The solution is constrained by the fact that:

1. I don’t wish to relocate vertex A or C.
2. I would like to adjust the location of B while attempting to keep the degree of adjustment relative to vertex A and C proportionate.

Being a geographer, I attempting to implement this solution using Geographic Information Systems (GIS) so I am not extremely proficient with math. Below is my logic for solving this problem but I haven’t had any success so far, so please post your proposed solution. Thanks!

Essentially, I am attempting to adjust the slopes of vector AC and BC so that angle B is affected and becomes 90°.

-At first I attempted to sum the component angles and find the difference from 90° and identify those points that do not intersect at a right-angle. I then tried to find what the new angles of the vectors should be by dividing each angle by the sum and multiplying by 90. I thought that by doing this I could find a ‘proposed’ new angle of the subject line.

Ex. (angleAC/(angleAC+angleBC))*90 = angleACnew

Using the new ‘proposed’ angle I would convert this back to a slope value and apply this to each of the vector equations and setting the equations equal to each other and algebraically solve for x.

I then planned on subbing back into one of the equations and solving for y. Thereby, giving be the new x,y coordinate for point B where B was a 90° angle. However, I realized that this methodology does not work since the sum of angleAC and angleBC do not accurately reflect the internal angles of the triangle and therefore do not yield a logical “adjustment value”.

Does anyone have some suggestions and/or guidance?

-
what is the degree of adjustment? –  Jorge Fernández Dec 17 '13 at 2:37

take your triangle and look at the side. then the points where the angle at point B' is the circle with side AC as diameter. As to the degree of adjustment I dont know what that is, I suggest you take the point where the angle bisector at B intersects with the circumference, but I will need a better definition of angle of adjustment to give you a better answer.

Hope this helps :)

-
Hi, thanks for replying. I attempted to define what I perceived as the "degree of adjustment". It was simply a term that I dreamt up to define the change in slope that I wanted to apply to the vectors based on my initial logic which proved to be flawed (see above comment). Thanks. –  WD_geo Dec 18 '13 at 1:10

If vertices $A$ and $C$ are fixed, then the possible candidates for the new $B$ lie on the circle with diameter $AC$ (see Thales' Theorem). I'm not sure what the "degree of adjustment" in #2 is, but hopefully this helps.

-
Hi, the "degree of adjustment" was my term for the proposed change I was attempting to calculate for both vector A and vector C. I was attempting to increase or decrease the slopes of the incoming vectors (to B) in order to correct for angle B being either too obtuse or too acute (respectively). In terms of keeping this "adjustment proportional, I simply meant that I did not want to for instance, only adjust vector A to create a right angle at B. –  WD_geo Dec 18 '13 at 1:05
@WD_geo I understand your statement a little better now: you want to adjust both vectors $AB$ and $CB$, rather than just adjust one of them. But this is still rather qualitative. How are you measuring adjustment (angle of rotation, or difference in slope)? How are you quantifying proportionality (minimize the difference in the angles of rotation, etc.)? Regardless, note that your goal is to move your old vertex $B$ to a[ny] point on the boundary of the circle with diameter $AC$. –  angryavian Dec 18 '13 at 1:19
yes I would like to be able to adjust both AB and AC. I realize it's rather qualitative, my best attempt to minimize the changes to the slope was the process I described in the very first italicized paragraph from my initial question. Although, the angles used were not appropriate, my attempt was to allocate the change of slope proportionately to both vectors (I hope this makes sense). Again as you mention, it doesn't really matter since your suggestion of Thales' theorem accomplishes what I need. Incorporating the angle bisector at B (from the other response) would also be useful. –  WD_geo Dec 18 '13 at 2:05


text:
# Documentation

Mathlib.Algebra.Star.Basic

# Star monoids, rings, and modules #

We introduce the basic algebraic notions of star monoids, star rings, and star modules. A star algebra is simply a star ring that is also a star module.

These are implemented as "mixin" typeclasses, so to summon a star ring (for example) one needs to write (R : Type _) [Ring R] [StarRing R]. This avoids difficulties with diamond inheritance.

We also define the class StarOrderedRing R, which says that the order on R respects the star operation, i.e. an element r is nonnegative iff there exists an s such that r = star s * s.

For now we simply do not introduce notations, as different users are expected to feel strongly about the relative merits of r^*, r†, rᘁ, and so on.

Our star rings are actually star semirings, but of course we can prove star_neg : star (-r) = - star r when the underlying semiring is a ring.

## TODO #

• In a Banach star algebra without a well-defined square root, the natural ordering is given by the positive cone which is the closure of the sums of elements star r * r. A weaker version of StarOrderedRing could be defined for this case. Note that the current definition has the advantage of not requiring a topology.
class Star (R : Type u) :
• A star operation (e.g. complex conjugate).

star : RR

Notation typeclass (with no default notation!) for an algebraic structure with a star operation.

Instances
class StarMemClass (S : Type u_1) (R : Type u_2) [inst : Star R] [inst : SetLike S R] :
• Closure under star.

star_mem : ∀ {s : S} {r : R}, r sstar r s

StarMemClass S G states S is a type of subsets s ⊆ G⊆ G closed under star.

Instances
instance StarMemClass.star {R : Type u} {S : Type u} [inst : Star R] [inst : SetLike S R] [hS : ] (s : S) :
Star { x // x s }
Equations
• = { star := fun r => { val := star r, property := (_ : star r s) } }
class InvolutiveStar (R : Type u) extends :
• Involutive condition.

star_involutive :

Typeclass for a star operation with is involutive.

Instances
@[simp]
theorem star_star {R : Type u} [inst : ] (r : R) :
star (star r) = r
theorem star_injective {R : Type u} [inst : ] :
def Equiv.star {R : Type u} [inst : ] :

star as an equivalence when it is involutive.

Equations
theorem eq_star_of_eq_star {R : Type u} [inst : ] {r : R} {s : R} (h : r = star s) :
s = star r
theorem eq_star_iff_eq_star {R : Type u} [inst : ] {r : R} {s : R} :
r = star s s = star r
theorem star_eq_iff_star_eq {R : Type u} [inst : ] {r : R} {s : R} :
star r = s star s = r
class TrivialStar (R : Type u) [inst : Star R] :
• Condition that star is trivial

star_trivial : ∀ (r : R), star r = r

Typeclass for a trivial star operation. This is mostly meant for ℝ.

Instances
class StarSemigroup (R : Type u) [inst : ] extends :
• star skew-distributes over multiplication.

star_mul : ∀ (r s : R), star (r * s) = star s * star r

A *-semigroup is a semigroup R with an involutive operation star such that star (r * s) = star s * star r.

Instances
@[simp]
theorem star_mul' {R : Type u} [inst : ] [inst : ] (x : R) (y : R) :
star (x * y) = star x * star y

In a commutative ring, make simp prefer leaving the order unchanged.

@[simp]
theorem starMulEquiv_apply {R : Type u} [inst : ] [inst : ] (x : R) :
starMulEquiv x = { unop := star x }
def starMulEquiv {R : Type u} [inst : ] [inst : ] :

star as a MulEquiv from R to Rᵐᵒᵖ

Equations
• One or more equations did not get rendered due to their size.
@[simp]
theorem starMulAut_apply {R : Type u} [inst : ] [inst : ] :
∀ (a : R), starMulAut a = star a
def starMulAut {R : Type u} [inst : ] [inst : ] :

star as a MulAut for commutative R.

Equations
• One or more equations did not get rendered due to their size.
@[simp]
theorem star_one (R : Type u) [inst : ] [inst : ] :
star 1 = 1
@[simp]
theorem star_pow {R : Type u} [inst : ] [inst : ] (x : R) (n : ) :
star (x ^ n) = star x ^ n
@[simp]
theorem star_inv {R : Type u} [inst : ] [inst : ] (x : R) :
@[simp]
theorem star_zpow {R : Type u} [inst : ] [inst : ] (x : R) (z : ) :
star (x ^ z) = star x ^ z
@[simp]
theorem star_div {R : Type u} [inst : ] [inst : ] (x : R) (y : R) :
star (x / y) = star x / star y

When multiplication is commutative, star preserves division.

def starSemigroupOfComm {R : Type u_1} [inst : ] :

Any commutative monoid admits the trivial *-structure.

See note [reducible non-instances].

Equations
theorem star_id_of_comm {R : Type u_1} [inst : ] {x : R} :
star x = x

Note that since starSemigroupOfComm is reducible, simp can already prove this.

class StarAddMonoid (R : Type u) [inst : ] extends :
• star commutes with addition

star_add : ∀ (r s : R), star (r + s) = star r + star s

A *-additive monoid R is an additive monoid with an involutive star operation which preserves addition.

Instances
@[simp]
theorem starAddEquiv_apply {R : Type u} [inst : ] [inst : ] :
∀ (a : R), starAddEquiv a = star a
def starAddEquiv {R : Type u} [inst : ] [inst : ] :
R ≃+ R

star as an AddEquiv

Equations
• One or more equations did not get rendered due to their size.
@[simp]
theorem star_zero (R : Type u) [inst : ] [inst : ] :
star 0 = 0
@[simp]
theorem star_eq_zero {R : Type u} [inst : ] [inst : ] {x : R} :
star x = 0 x = 0
theorem star_ne_zero {R : Type u} [inst : ] [inst : ] {x : R} :
star x 0 x 0
@[simp]
theorem star_neg {R : Type u} [inst : ] [inst : ] (r : R) :
star (-r) = -star r
@[simp]
theorem star_sub {R : Type u} [inst : ] [inst : ] (r : R) (s : R) :
star (r - s) = star r - star s
@[simp]
theorem star_nsmul {R : Type u} [inst : ] [inst : ] (x : R) (n : ) :
star (n x) = n star x
@[simp]
theorem star_zsmul {R : Type u} [inst : ] [inst : ] (x : R) (n : ) :
star (n x) = n star x
class StarRing (R : Type u) [inst : ] extends :
• star commutes with addition

star_add : ∀ (r s : R), star (r + s) = star r + star s

A *-ring R is a (semi)ring with an involutive star operation which is additive which makes R with its multiplicative structure into a *-semigroup (i.e. star (r * s) = star s * star r).

Instances
instance StarRing.toStarAddMonoid {R : Type u} [inst : ] [inst : ] :
Equations
@[simp]
theorem starRingEquiv_apply {R : Type u} [inst : ] [inst : ] (x : R) :
starRingEquiv x = { unop := star x }
def starRingEquiv {R : Type u} [inst : ] [inst : ] :

star as an RingEquiv from R to Rᵐᵒᵖ

Equations
• One or more equations did not get rendered due to their size.
@[simp]
theorem star_natCast {R : Type u} [inst : ] [inst : ] (n : ) :
star n = n
@[simp]
theorem star_intCast {R : Type u} [inst : Ring R] [inst : ] (z : ) :
star z = z
@[simp]
theorem star_ratCast {R : Type u} [inst : ] [inst : ] (r : ) :
star r = r
@[simp]
theorem starRingAut_apply {R : Type u} [inst : ] [inst : ] :
∀ (a : R), starRingAut a = star a
def starRingAut {R : Type u} [inst : ] [inst : ] :

star as a ring automorphism, for commutative R.

Equations
• One or more equations did not get rendered due to their size.
def starRingEnd (R : Type u) [inst : ] [inst : ] :
R →+* R

star as a ring endomorphism, for commutative R. This is used to denote complex conjugation, and is available under the notation conj in the locale ComplexConjugate.

Note that this is the preferred form (over starRingAut, available under the same hypotheses) because the notation E →ₗ⋆[R] F→ₗ⋆[R] F⋆[R] F for an R-conjugate-linear map (short for E →ₛₗ[starRingEnd R] F→ₛₗ[starRingEnd R] F) does not pretty-print if there is a coercion involved, as would be the case for (↑starRingAut : R →* R)↑starRingAut : R →* R)→* R).

Equations
• = starRingAut

star as a ring endomorphism, for commutative R. This is used to denote complex conjugation, and is available under the notation conj in the locale ComplexConjugate.

Note that this is the preferred form (over starRingAut, available under the same hypotheses) because the notation E →ₗ⋆[R] F→ₗ⋆[R] F⋆[R] F for an R-conjugate-linear map (short for E →ₛₗ[starRingEnd R] F→ₛₗ[starRingEnd R] F) does not pretty-print if there is a coercion involved, as would be the case for (↑starRingAut : R →* R)↑starRingAut : R →* R)→* R).

Equations
theorem starRingEnd_apply {R : Type u} [inst : ] [inst : ] {x : R} :
↑() x = star x

This is not a simp lemma, since we usually want simp to keep starRingEnd bundled. For example, for complex conjugation, we don't want simp to turn conj x into the bare function star x automatically since most lemmas are about conj x.

theorem starRingEnd_self_apply {R : Type u} [inst : ] [inst : ] (x : R) :
↑() (↑() x) = x
instance RingHom.involutiveStar {R : Type u} {S : Type u_1} [inst : ] [inst : ] [inst : ] :
Equations
theorem RingHom.star_def {R : Type u} {S : Type u_1} [inst : ] [inst : ] [inst : ] (f : S →+* R) :
star f =
theorem RingHom.star_apply {R : Type u} {S : Type u_1} [inst : ] [inst : ] [inst : ] (f : S →+* R) (s : S) :
↑(star f) s = star (f s)
theorem Complex.conj_conj {R : Type u} [inst : ] [inst : ] (x : R) :
↑() (↑() x) = x

Alias of starRingEnd_self_apply.

theorem IsROrC.conj_conj {R : Type u} [inst : ] [inst : ] (x : R) :
↑() (↑() x) = x

Alias of starRingEnd_self_apply.

@[simp]
theorem star_inv' {R : Type u} [inst : ] [inst : ] (x : R) :
@[simp]
theorem star_zpow₀ {R : Type u} [inst : ] [inst : ] (x : R) (z : ) :
star (x ^ z) = star x ^ z
@[simp]
theorem star_div' {R : Type u} [inst : ] [inst : ] (x : R) (y : R) :
star (x / y) = star x / star y

When multiplication is commutative, star preserves division.

@[simp]
theorem star_bit0 {R : Type u} [inst : ] [inst : ] (r : R) :
star (bit0 r) = bit0 (star r)
@[simp]
theorem star_bit1 {R : Type u} [inst : ] [inst : ] (r : R) :
star (bit1 r) = bit1 (star r)
def starRingOfComm {R : Type u_1} [inst : ] :

Any commutative semiring admits the trivial *-structure.

See note [reducible non-instances].

Equations
• starRingOfComm = let src := starSemigroupOfComm; StarRing.mk (_ : ∀ (x x_1 : R), star (x + x_1) = star (x + x_1))
class StarOrderedRing (R : Type u) [inst : ] [inst : ] extends :
• addition commutes with ≤≤

add_le_add_left : ∀ (a b : R), a b∀ (c : R), c + a c + b
• characterization of non-negativity

nonneg_iff : ∀ (r : R), 0 r s, r = star s * s

An ordered *-ring is a ring which is both an OrderedAddCommGroup and a *-ring, and 0 ≤ r ↔ ∃ s, r = star s * s≤ r ↔ ∃ s, r = star s * s↔ ∃ s, r = star s * s∃ s, r = star s * s.

Instances
instance StarOrderedRing.instOrderedAddCommGroup {R : Type u} [inst : ] [inst : ] [inst : ] :
Equations
• One or more equations did not get rendered due to their size.
theorem star_mul_self_nonneg {R : Type u} [inst : ] [inst : ] [inst : ] {r : R} :
0 star r * r
theorem star_mul_self_nonneg' {R : Type u} [inst : ] [inst : ] [inst : ] {r : R} :
0 r * star r
theorem conjugate_nonneg {R : Type u} [inst : ] [inst : ] [inst : ] {a : R} (ha : 0 a) (c : R) :
0 star c * a * c
theorem conjugate_nonneg' {R : Type u} [inst : ] [inst : ] [inst : ] {a : R} (ha : 0 a) (c : R) :
0 c * a * star c
theorem conjugate_le_conjugate {R : Type u} [inst : ] [inst : ] [inst : ] {a : R} {b : R} (hab : a b) (c : R) :
star c * a * c star c * b * c
theorem conjugate_le_conjugate' {R : Type u} [inst : ] [inst : ] [inst : ] {a : R} {b : R} (hab : a b) (c : R) :
c * a * star c c * b * star c
class StarModule (R : Type u) (A : Type v) [inst : Star R] [inst : Star A] [inst : SMul R A] :
• star commutes with scalar multiplication

star_smul : ∀ (r : R) (a : A), star (r a) = star r star a

A star module A over a star ring R is a module which is a star add monoid, and the two star structures are compatible in the sense star (r • a) = star r • star a.

Note that it is up to the user of this typeclass to enforce [Semiring R] [StarRing R] [AddCommMonoid A] [StarAddMonoid A] [Module R A], and that the statement only requires [Star R] [Star A] [SMul R A].

If used as [CommRing R] [StarRing R] [Semiring A] [StarRing A] [Algebra R A], this represents a star algebra.

Instances
instance StarSemigroup.to_starModule {R : Type u} [inst : ] [inst : ] :

A commutative star monoid is a star module over itself via Monoid.toMulAction.

Equations

Instance needed to define star-linear maps over a commutative star ring (ex: conjugate-linear maps when R = ℂ).

Equations
class StarHomClass (F : Type u_1) (R : outParam (Type u_2)) (S : outParam (Type u_3)) [inst : Star R] [inst : Star S] extends :
Type (max(maxu_1u_2)u_3)
• the maps preserve star

map_star : ∀ (f : F) (r : R), f (star r) = star (f r)

StarHomClass F R S states that F is a type of star-preserving maps from R to S.

Instances

### Instances #

Equations
@[simp]
theorem Units.coe_star {R : Type u} [inst : ] [inst : ] (u : Rˣ) :
↑(star u) = star u
@[simp]
theorem Units.coe_star_inv {R : Type u} [inst : ] [inst : ] (u : Rˣ) :
(star u)⁻¹ = star u⁻¹
Equations
• One or more equations did not get rendered due to their size.
theorem IsUnit.star {R : Type u} [inst : ] [inst : ] {a : R} :
IsUnit (star a)
@[simp]
theorem isUnit_star {R : Type u} [inst : ] [inst : ] {a : R} :
theorem Ring.inverse_star {R : Type u} [inst : ] [inst : ] (a : R) :
instance Invertible.star {R : Type u_1} [inst : ] [inst : ] (r : R) [inst : ] :
Equations
• = { invOf := , invOf_mul_self := (_ : * star r = 1), mul_invOf_self := (_ : star r * = 1) }
theorem star_invOf {R : Type u_1} [inst : ] [inst : ] (r : R) [inst : ] [inst : Invertible (star r)] :
= (star r)
instance MulOpposite.instStarMulOpposite {R : Type u} [inst : Star R] :

The opposite type carries the same star operation.

Equations
• MulOpposite.instStarMulOpposite = { star := fun r => { unop := star r.unop } }
@[simp]
theorem MulOpposite.unop_star {R : Type u} [inst : Star R] (r : Rᵐᵒᵖ) :
(star r).unop = star r.unop
@[simp]
theorem MulOpposite.op_star {R : Type u} [inst : Star R] (r : R) :
{ unop := star r } = star { unop := r }
instance MulOpposite.instInvolutiveStarMulOpposite {R : Type u} [inst : ] :
Equations
Equations
Equations
Equations
instance StarSemigroup.toOpposite_starModule {R : Type u} [inst : ] [inst : ] :

A commutative star monoid is a star module over its opposite via Monoid.toOppositeMulAction.

Equations


text:
# 4: Ensembles

After the last chapter’s dip into thermodynamics, we return to the discussion of the principles of statistical mechanics begun in chapter 2.

This page titled 4: Ensembles is shared under a CC BY-SA license and was authored, remixed, and/or curated by Daniel F. Styer.


text:
|

Ramanujam !

Blazing goIITian

Posted on
14 Mar 2012 17:41:01 IST
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14 Mar 2012 17:41:01 IST
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1115
Ramanujam !
Engineering Entrance , JEE Main , BITSAT , JEE Main & Advanced , Mathematics , Algebra , Trigonometry , academic , Motivational

Ramanujam:

By age 11, he had exhausted the mathematical knowledge of two college students who were lodgers at his home. He was later lent a book on advanced trigonometry written by S. L. Loney

He met deputy collector V. Ramaswamy Aiyer, who had recently founded the Indian Mathematical Society. Ramanujan, wishing for a job at the revenue department where Ramaswamy Aiyer worked, showed him his mathematics notebooks. As Ramaswamy Aiyer later recalled:

One of the first problems he posed in the journal was: $sqrt{1+2sqrt{1+3 sqrt{1+cdots}}}.$

He waited for a solution to be offered in three issues, over six months, but failed to receive any. At the end, Ramanujan supplied the solution to the problem himself. On page 105 of his first notebook, he formulated an equation that could be used to solve the infinitely nested radicals problem.

$x+n+a = sqrt{ax+(n+a)^2 +xsqrt{a(x+n)+(n+a)^2+(x+n) sqrt{cdots}}}$

Using this equation, the answer to the question posed in the Journal was simply 3. Ramanujan wrote his first formal paper for the Journal on the properties of Bernoulli numbers. One property he discovered was that the denominators (sequence A027642 in OEIS) of the fractions of Bernoulli numbers were always divisible by six. He also devised a method of calculating Bn based on previous Bernoulli numbers. One of these methods went as follows:

It will be observed that if n is even but not equal to zero,

(i) Bn is a fraction and the numerator of ${B_n over n}$ in its lowest terms is a prime number,

(ii) the denominator of Bn contains each of the factors 2 and 3 once and only once,

(iii) $2^n(2^n-1){b_n over n}$ is an integer and $2(2^n-1)B_n,$ consequently is an odd integer.

In his 17-page paper, "Some Properties of Bernoulli's Numbers", Ramanujan gave three proofs, two corollaries and three conjectures.Ramanujan's writing initially had many flaws. As Journal editor M. T. Narayana Iyengar noted.

Mr. Ramanujan's methods were so terse and novel and his presentation so lacking in clearness and precision, that the ordinary [mathematical reader], unaccustomed to such intellectual gymnastics, could hardly follow him.

Some more:

In mathematics, there is a distinction between having an insight and having a proof. Ramanujan's talent suggested a plethora of formulae that could then be investigated in depth later.

It is said that Ramanujan's discoveries are unusually rich and that there is often more to them than initially meets the eye.As a by-product, new directions of research were opened up.

Examples of the most interesting of these formulae include the intriguing infinite series for π, one of which is given below: $rac{1}{pi} = rac{2sqrt{2}}{9801} sum^infty_{k=0} rac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}.$

This result is based on the negative fundamental discriminant d = −4×58 with class number h(d) = 2 (note that 5×7×13×58 = 26390 and that 9801=99×99; 396=4×99) and is related to the fact that

$e^{pi sqrt{58}} = 396^4 - 104.000000177dots.$

Compare to Heegner numbers, which have class number 1 and yield similar formulae. Ramanujan's series for π converges extraordinarily rapidly (exponentially) and forms the basis of some of the fastest algorithms currently used to calculate π. Truncating the sum to the first term also gives the approximation $9801sqrt{2}/4412$ for π, which is correct to six decimal places.

One of his remarkable capabilities was the rapid solution for problems. He was sharing a room with P. C. Mahalanobis who had a problem,

"Imagine that you are on a street with houses marked 1 through n. There is a house in between (x) such that the sum of the house numbers to left of it equals the sum of the house numbers to its right. If n is between 50 and 500, what are n and x?"

This is a bivariate problem with multiple solutions. Ramanujan thought about it and gave the answer with a twist: He gave a continued fraction. The unusual part was that it was the solution to the whole class of problems. Mahalanobis was astounded and asked how he did it.

"It is simple. The minute I heard the problem, I knew that the answer was a continued fraction. Which continued fraction, I asked myself. Then the answer came to my mind", Ramanujan replied.

His intuition also led him to derive some previously unknown identities, such as:

$left [ 1+2sum_{n=1}^infty rac{cos(n heta)}{cosh(npi)}ight ]^{-2} + left [1+2sum_{n=1}^infty rac{cosh(n heta)}{cosh(npi)}ight ]^{-2} = rac {2 Gamma^4 left ( rac{3}{4}ight )}{pi}$

For all $heta$, where $Gamma(z)$ is the gamma function. Expanding into series of

powers and equating coefficients of $heta^0$$heta^4$, and $heta^8$ gives some deep identities for the hyperbolic secant.

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ramanujam's number

sum of 2 cubes

Cool goIITian

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15 Mar 2012 22:31:46 IST
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very nice

Cool goIITian

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15 Mar 2012 22:32:15 IST
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nice one

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Rate it if u find it useful!

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hai! my name is hanani iam giving aquestion please answer this if you can. how many times does 9 greater than 8?

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# The first is geographically dependent and the other is fi

The

first is geographically dependent and the other is fixed. The first component is sensitive to local and temporal variations such as flow, precipitation, evaporation and withdrawal. The intensity depends on water acquisition Ralimetinib mouse technology from surface reservoirs or underground aquifers, brackish water or seawater desalination. This component involves conveyance, which depends on distance and elevation difference between source and use. The second component is fixed. It includes filtration and storage as well as wastewater collection and treatment. In order to develop quantitative intuition, ATM Kinase Inhibitor datasheet we use the following approximations, ordered by water energy intensity, I W: surface water withdrawal (0.4 kWh/kgal), waste water reuse (1.6 kWh/kgal),

ground water pumping (2 kWh/kgal), imported water (3.5 kWh/kgal), brackish water desalination (5 kWh/kgal), deep groundwater withdrawal (6 kWh/kgal) and seawater desalination (13 kWh/kgal). We add a value of 4 kWh/kgal for the fixed component (Gellings 2009) and take into account the overall water losses, which range typically from 0.1 to 25 %. To establish a benchmark, let us calculate water efficiency EPGW in kgal/EP at two extreme cases. Low efficiency case: water from desalination using electricity generated from coal and incurring 25 % conveyance losses resulting in EPGW = 0.35 kgal/EP. High efficiency case: using surface water with only 10 % losses using electricity from combined cycle natural gas resulting in A-1210477 datasheet EPGW = 5.5 kgal/EP. We see that technology use, dictated by local conditions, imply an order of magnitude variation in EPGW. Consolidated monthly energy budget We now consolidate the sustainability of the household’s activities using EP in a manner similar to how multinational businesses consolidate global P&Ls across multiple currency regimes. For example, a household with electricity use, car travel, and water use can convert these disparate activities to energy points in the following way: $$Verteporfin \textEP = \frac\textkWh\textEPG + \frac\textmiles\textMPG + \frac\textkgal\textEPG_\textW$$ (3) Let us demonstrate our approach where disaggregated energy budgets are presented for two hypothetical families in reference to the US average. For pedagogical simplicity we limit our attention to four categories of consumption: electricity, heating,6 car miles, and water. Family A resides in a cold climate in an urban setting. They use natural gas for heating and purchase electricity generated from coal. Family B lives in a suburban house in a warm climate where air conditioning needs are high, water is scarce, and natural gas is used only for cooking. They participate in a utility program that allows most of the electricity to be purchased from solar energy, leading to a high effective EPG = 40 kWh/EP.

Comments are closed.


text:
# Raven DB: How to create “UniqueVisitorCount by date” index

I have an application to track the page visits for a website. Here's my model:

public class VisitSession {
public string SessionId { get; set; }
public DateTime StartTime { get; set; }
public string UniqueVisitorId { get; set; }
public IList<PageVisit> PageVisits { get; set; }
}

When a visitor go to the website, a visit session starts. One visit session has many page visits. The tracker will write a UniqueVisitorId (GUID) cookie when the first time a visitor go to the website. So we are able to know if a visitor is returning visitor.

Now, I want to know how many unique visitors visited the website in a date range. That is, I want to display a table in our webpage like this;

Date        | Unique Visitors Count
------------+-----------------------
2012-05-01  | 100
2012-05-02  | 1000
2012-05-03  | 120

I want to create an index to do this in RavenDB. But I don't know how to write the Map/Reduce query. I though it can be like this:

public class UniqueVisitor_ByDate : AbstractIndexCreationTask<VisitSession, UniqueVisitorByDate>
{
public UniqueVisitor_ByDate()
{
Map = sessions => from s in sessions
select new
{
s.StartTime.Date,
s.UniqueVisitorId
};

Reduce = results => from result in results
group result by result.Date into g
select new
{
Date = g.Key,
UniqueVisitorCount = g.Distinct()
};
}
}

But it's not working. In Ayende's e-book, I know that the result of Map function should be same as the result of Reduce function. So how can I write the correct map/reduce functions?

-

This index should do what you want:

public class UniqueVisitor_ByDate : AbstractIndexCreationTask<VisitSession, UniqueVisitorByDate>
{
public UniqueVisitor_ByDate()
{
Map = sessions => from s in sessions
select new
{
s.StartTime.Date,
s.UniqueVisitorId,
Count = 1,
};

Reduce = results => from result in results
group result by result.Date
into g
select new UniqueVisitorByDate
{
Date = g.Key,
Count = g.Select(x => x.UniqueVisitorId).Distinct().Count(),
UniqueVisitorId = g.FirstOrDefault().UniqueVisitorId,
};
}
}

Note that it requires the extra 'UniqueVisitorId' property in the 'reduce' and the 'count' property in the map, but you can just ignore those.

-
But adding "UniqueVisitorId" to UniqueVisitorByDate class looks strange, is there a better solution? –  Mouhong Lin May 15 '12 at 9:32
Not that I'm aware of. You could change the name of the property to something like 'TempId_DO_NOT_USE' to indicate that it shouldn't be used in your application. –  Simon May 15 '12 at 9:38
@DylanLin UniqueVisitorId has to be part of UniqueVisitorByDate because Map and Reduce must produce objects of the same shape. –  Thomas Freudenberg May 15 '12 at 10:03
But still made the UniqueVisitorByDate class not elegant :( –  Mouhong Lin May 15 '12 at 10:03
Elegance or pragmatism... in this case, as that class is only used to display the results and isn't part of your domain or business logic I think it's a small price to pay for an index that actually works! :-) –  Simon May 15 '12 at 10:07


text:
# Cubic Foot To Cubic Meter Calculator

Cubic Foot (ft3):
Cubic Meter (mt3):

The Cubic Foot to Cubic Meter Calculator is an online tool that is used to find the value of equivalent cubic meter from the given cubic foot. BYJU’S Calculator is a tool that makes calculations simple and interesting. It is a quick way of finding the required value within a fraction of seconds.

## How to use Cubic Foot to Cubic Meter Calculator?

Find the below steps to use the cubic foot to cubic meter calculator.

Step 1: Write the value in the cubic foot in the given input field

Step 2: Click on the “Solve” button

Step 3: The equal value in the cubic meter will appear in the output field

## What is Cubic Foot to Cubic Meter Conversion?

Cubic foot and cubic meter both are the measurements of the volume of the three-dimensional object but in a different medium. A cubic foot is used for linear measurement and a cubic meter is used to express the concentrations of a chemical in a volume of air.

One cubic foot is approximately equal to 0.0283168466 cubic meters.

1 cubic foot (ft3) = 0.0283168466 cubic meter (m3)

## How to Convert Cubic Feet to Cubic Meters?

To convert cubic feet to cubic meters, multiply the given feet value by 0.0283168466 cubic meters. For example, to convert 5 cubic feet into cubic meters, multiply 5 by 0.0283168466 cubic meters.

5 cubic feet = 5 x 0.0283168466 cubic meters.

5 ft3  = 0.141584233 m3.

Thus, 5 cubic feet to cubic meters is approximately equal to 0.141584233 m3.

### Cubic Feet to Cubic Meters Conversion

The conversion from cubic foot to cubic meters for certain values are provided below:

 Cubic Foot (ft3) Cubic Meter (m3) 0.001 0.0000283168 0.01 0.0002831685 0.1 0.0028316847 1 0.0283168466 2 0.0566336932 3 0.0849505398 4 0.1132673864 5 0.141584233 6 0.1699010796 7 0.1982179262 8 0.2265347727 9 0.2548516193 10 0.2831684659 50 1.4158423297 100 2.8316846593

### Cubic Feet to Cubic Meters Examples

Example 1:

Convert 0.06 cubic feet to cubic meters.

Solution:

We know that one cubic foot is approximately equal to 0.0283168466 cubic meters.

Therefore,

0.06 ft3 = 0.06 x 0.0283168466 m3.

0.06 ft3= 0.0016990108 m3.

Hence, 0.06 cubic feet is approximately equal to 0.0016990108 cubic meters.

Example 2:

What is the value of 8 cubic feet in cubic meters?

Solution:

To find: 8 cubic feet in cubic meters.

As, 1 ft3 = 0.0283168466 m3

8 ft3= 8 x 0.0283168466 m3

8 ft3= 0.2265347727 m3

Therefore, 8 cubic feet is approximately equal to 0.2265347727 cubic meters

Example 3:

Convert 120 cubic feet to cubic meters.

Solution:

We know that, 1 ft3 = 0.0283168466 m3

Thus, 120 ft3 = 120 x 0.0283168466 m3.

120 ft3 = 3.3980215912 m3

Therefore, 120 cubic feet to cubic meters is 3.3980215912 m3

## Frequently Asked Questions on Cubic Foot to Cubic Meter

### How to convert cubic feet to cubic meters?

To convert cubic feet to cubic meters, multiply the given cubic feet value by 0.0283168466 cubic meters.

### What is the relation between cubic feet and cubic meter?

The relationship between cubic feet and the cubic meter is given as follows:
1 ft3 = 0.0283168466 m3 (or)
1 m3 = 35.31466672 m3

### What is the value of 15 cubic feet in cubic meters?

We know that 1 cubic foot is equal to 0.0283168466 cubic meters, multiply 15 by 0.0283168466 m3.
Therefore,
15 ft3 = 15 x 0.0283168466 m3
15 ft3 = 0.4247526989 m3.
Hence, the value of 15 cubic feet in cubic meters is 0.4247526989 m3.


text:
# Multi-point boundary value problems for one-dimensionalp-Laplacian at resonance

Research paper by Youyu Wang, Guosheng Zhang, Weigao Ge

Indexed on: 01 Sep '06Published on: 01 Sep '06Published in: Journal of Applied Mathematics and Computing

#### Abstract

In this paper, we consider the multi-point boundary value problems for one-dimensional p-Laplacian at resonance:$$(\phi _p (x'(t)))' = f(t,x(t),x'(t))$$ subject to the boundary value conditions:$$(\phi _p (x'(0)) = \sum\limits_{i = 1}^{n - 2} {\alpha _i \phi _p (x'(\xi _i ))}$$,$$(\phi _p (x'(1)) = \sum\limits_{j = 1}^{m - 2} {\beta _j \phi _p (x'(\eta _i ))}$$ where ϕp(s)=|s|p-2s, p>1,αi(1≤i≤n-2)∈R,β{jit}(1≤j≤m-2)∈R, 0<ξ1<ξ2<...<ξn-2<1, 0<η1<η2<...<ηm-2<1, By applying the extension of Mawhin’s continuation theorem, we prove the existence of at least one solution. Our result is new.


text:
# Upside down a latex

Oct 18, - The upside-down capital A (∀) as in “for all” or “for any” element(s) in a set, can be typeset in LaTeX using the TeX macro in math mode: \forall.

The Comprehensive Symbol Guide is your flexible friend concerning lookup of thousands of symbols: \documentclass{article} \begin{document} $\bot\bot\bot$. Feb 16, - The symbol you are looking for is widecheck and can be found in the mathabx font package. Please consult pages 10 and 16 of the mathabx graphics - Is there an upside-down version of the symbol for \$\pi.

## You the storyteller upside down a latex were visited

So here is a way of doing it in LaTeX: \rotatebox{}{Upside down} Upside down \\ \rotatebox{}{\reflectbox{Upside down}} Upside down and mirrored. Relevant parts of the latex code are reproduced under each of the pictures. This text is found in the latex code, mostly stating that they are parts of some.

Dec 18, - The upside-down A symbol is the universal quantifier from predicate logic. (Also see the more complete discussion of the first-order predicate. If you wanted something upside down mirror-image style, rather than only rotated degrees, you might try a combination of that with.


text:
MedicalVisualization

## Marching Hexahedra

The Marching Hexahedra Algorithm is the extension of the marching quads algorithm from 2D to 3D.

Since each egde of the hexahedron can have an intersection point there are $2^8$ cases to consider. A well established method is the original algorithm published by Lorensen & Cline in 1987.

A couple of cases have shown not to match the original scalar function, so that the method has to be considered to be inconsistent because of ambiguous cases in 3D. It is working very well in practice, though.

For this reason, we consider the Marching Tetrahedra algorithm in the following, which does not exhibit ambiguities.


text:
-rw-------  1 10025 10025  58000 Aug 28 15:20 filename.pdf

(user 10025 is also listed as the owner of the K9 process). In particular, I cannot open them with Sailfish applications (which run as the user nemo) after saving them. This did not happen with my Jolla1 in the past; I could point a Sailfish OS file manager to the file and open them.


text:
# How far back can you trace a photon?

You have a photomultiplier tube pointed at a distant star, exactly 100 light years away. It's perfectly set up so that nothing can get into the tube unless it came from that star. Every hour or so, you get a click in the detector.

Each time you hear a click, are you justified in saying that exactly 100 years ago, a photon was emitted from that star?

Or to put it more precisely: let's say your photomultiplier tube is pre-filtered so for all intents and purposes it only responds to a certain arbitrary atomic transition, say something in the potassium spectrum. The next time you hear a click, are you justified in saying that exactly 100 years ago, a particular potassium atom on that star made that particular transition from higher to lower energy?

Obviously, if I'm asking the question, I don't think you are allowed to make that connection. But I wonder what people think?

EDIT: We can sharpen up the question a little more if we look, for example, at the sun's corona. As Wikipedia explains,

The Fraunhofer lines are typical spectral absorption lines. These dark lines are produced whenever a cold gas is between a broad spectrum photon source and the detector. In this case, a decrease in the intensity of light in the frequency of the incident photon is seen as the photons are absorbed, then re-emitted in random directions, which are mostly in directions different from the original one. This results in an absorption line, since the narrow frequency band of light initially traveling toward the detector, has been turned into heat or re-emitted in other directions. By contrast, if the detector sees photons emitted directly from a glowing gas, then the detector often sees photons emitted in a narrow frequency range by quantum emission processes in atoms in the hot gas, resulting in an emission line. In the Sun, Fraunhofer lines are seen from gas in the outer regions of the Sun, which are too cold to directly produce emission lines of the elements they represent.

If we look at the sun directly, we see absorption lines: but I want to look for emission lines. So let's look at the sun's corona; and let's set up our photomultiplier with appropriate filters and whatever so that we can say, with 90% confidence, that a click in the detector came from a particular potassium-atom transition in the corona. Then the question is perfectly clear: when we hear a click, can we say (with 90% confidence) that exactly 8 minutes ago, a particular potassium atom made a transition from a higher to a lower energy level,thereby emitting a photon...the exact same photon that was just captured in our photomultiplier?

I think the question is pretty clear, and my answer is of course: No. And I think the people who say yes are the same people who would say that in the two-slit experiment, the photon must have passed through either one slit or the other one. Or the people who say that in the Stern-Gerlach experiment, the silver atom jumps into either the spin-up or the spin-down states at the moment when it passed through the magnetic field.

• A) Why every hour? Every $10^{-\text{something very big}}$ seconds is more like it. B) It will be impossible to eliminate outside noise. C) Particles don't necessarily travel in straight lines - a straight line is simply the most likely path. See this. I'd put this into some sort of an answer, but I think someone more qualified than I can organize it much better than I can, and can explain it better. Dec 6 '14 at 22:51
• Can you explain why you are unsure that this is possible? The way you have set up the question it sounds as if the only logical conclusion would be that 100 years ago an atomic transition of potatssium took place at that star. Depending on where abouts on the start it happened there would be a little bit of uncertainty about the timing.
– tom
Dec 6 '14 at 23:02
• How do you propose ensuring that your tube can discriminate between photons emitted from that particular star, and not any point in between (say, by reflection off of a speck of dust)? Purely from an optics perspective, I think you would need an imaging system the size of our solar system in order to determine the distance from which the photon originated (via defocusing or a rangefinder type apparatus) Dec 7 '14 at 0:49

This is a perfect example of the collapse of the wave function. The "potassium" atom makes a transition from higher energy to lower energy, sending out a speherical wavefront into the universe. One hundred years later, that spherical wavefront crosses the photomultiplier tube. Since the energy of that wave is spread over hundreds of square light years, there is only one chance in a gazillion that it is captured and caused a click to be heard. But there are jillions of potassium atoms in that star, emitting trillions of photons every hour or so. Since (1 jillion) x (1 trillion) = 1 gazillion, the result is that every hour or so you get a click in the detector. THAT is the collapse of the wave function, and it happens at the moment of detection.

Or does it? It seems that the doubters out there want the wave function to collapse twice...once at the point of emission and again at the point of detection. In the wave picture...and this is the quantum-mechanical wave picture, mind you...the emission of photon by a potassium transition has a characteistic linewidth associated with a finite emission time - in other words, a wave train. That wave train overlaps with the wave trains of the jillions of other potassium atoms in the star, creating a continuous output.The wave at the photodetector contains contributions from all those jillions of potassium emissions. THAT is the wave that "collapses" into a photon, causing a click in the photodetector.

But that wave cannot be traced back and associated with any particular emission event on the star. Or can it? If you think it can, then you are really saying that the wave function collapses twice...once at the point of emission, and once at the point of detection.

I don't think that's right.

• This is wrong: " The "potassium" atom makes a transition from higher energy to lower energy, sending out a speherical wavefront into the universe. One hundred years later, that spherical wavefront crosses the photomultiplier tube. Since the energy of that wave is spread over hundreds of square light years,". The quantum mechanical solution of the transition amplitude gives a PROBABILITY of finding the photon at the (x,y,z) at time t of the detector. The energy is not spread out. It comes in a quantum of energy that will excite a potassium atom to the analogous level it left. Dec 7 '14 at 15:19
• It seems like I'm saying it didn't become a photon until it entered the detector, and you're saying it was a photon from the moment it left the potassium atom. Dec 7 '14 at 15:28
• I am using the standard quantum mechanical interpretation, which assigns probabilities of emission and detection and certainly no energy is spread out all over the universe in any decay. Dec 7 '14 at 15:36
• If there is no energy spread out, then the energy must be concentrated. And in your "standard quantum mechanics", we have local conservation of energy, which means the energy must be somewhere. So you're saying the photon existed from the moment it left the potassium atom. I suppose you're saying if it encountered a double slit along the way, it must have passed through either one or the other of the slits. Dec 7 '14 at 15:45
• Yes for the double slit. Look at these cosmic gamma rays mpa-garching.mpg.de/lectures/ADSEM/WS0203_Obergaulinger.pdf . gamma rays are photons too. Dec 7 '14 at 16:25

There exists the KISS rule ( keep it simple stupid), and I will.

Let us take an excited potassium atom and solve the Schrodinger equation for the transitions available. This will give for this atom a wavefunction, in principle from -infinity to +infinity in space and time. The square of this wavefunction will give the probability for a photon to fall on a specific area centered at a specific (x,y,z) at time t. A probability.

The transition energy is quantized and it comes as an elementary particle in the standard model of particle physics, with the appropriate frequency to excite another potassium atom to that energy level.

If two potassium atoms existed 100 light years apart with nothing else in between there still will be a probability of identifying the photon coming from the direction of the sender potassium atom, within the aperture of the detector.

If a star existed in the location to have the coordinates and the star were not moving with respect to earth and there were no expansion of space and no gravitational wells on the direction the photon would have to take, the probability would be very very small, but the star has zillions of potassium atoms and we would know that that potassium photon came from that star by detecting the excitation, taking 100 years to arrive and knowing the location of the star from the general light of the star. That is the way x-rays and gamma-rays are identified to specific locations in the sky.

So my answer is that you would know that "a" potassium atom deexcited and sent a photon this way. Photons are elementary particles , as real as the cosmic ray protons or any other particles arriving from space.

Now spectral lines from 100 light years away are redshifted or blue shifted depending on the relative motion of the emitting body, that is how we measure star velocities, so in real life the spectrum will be shifted and the potassium atoms in the detector will not be excited, a diffraction grating would be used to see the frequencies of the arriving light and if there exist potassium lines ( although I have the impression that with higher atomic numbers than hydrogen it is absorption lines one sees).

• You start by saying you are going to "keep it simple" but then you throw in so many extraneous complications that your answer is all but incomprehensible. Let me list the things you bring up that have nothing to do with the question: the motion of the star relative to the earth, the expansion of space, existence of gravitational "wells" (?), the redshift, the blueshift, and the standard model. And why do you have a second potassium atom...and why do you think there are potassium atoms inside the detector? And how do you "see" an absorption line with a photomultiplier? Dec 7 '14 at 16:22
• I am answering the question. The questioner has set potassium acceptors in the detector. For optical light you do not need a photomultiplier. Accumulation of photons will show absorption and emission lines. emission values (as in question) change due to redshift or blue shift and cannot match the potassium the questioner wants in the detector coolcosmos.ipac.caltech.edu/cosmic_classroom/cosmic_reference/… . The standard model comes in to emphasize that the photon is a particle. Dec 7 '14 at 19:29
• Oh come on now. Does your potassium atom make an audible click when it absorbs a photon from the distant star? Dec 7 '14 at 20:01
• It could make a visible point spot on a film, as with single photon patterns , if I were clever enough and knew enough physical chemistry to design such a film with a potassium content. This was not the point of the question. I assumed I could. Dec 8 '14 at 4:08


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## anonymous one year ago Solve for B. A = 3(B + C) B = ((A-C) / (3)) B = ((3A-C) / (3)) B = ((A-3C) / (3))

1. anonymous

@Nnesha How do you make things a fraction on this website?

2. Nnesha

yep

3. Nnesha
4. anonymous

thx

5. anonymous

A

6. anonymous

the first choice? ""((A-C) / (3))""

7. Nnesha

` $$\large\rm\frac{2}{2}$$

8. Nnesha

FIRST STEp!!! distribute parentheses by outside term

9. anonymous

3B + 3C

10. Nnesha

yes right A = 3B +3C move 3C to the left side and then solve for b

11. anonymous

3C + A = 3b $\frac{ 3C }{ 3 } + \frac{ A }{ 3 } = \frac{ 3B }{ 3 }$

12. Nnesha

how would you cancel out 3C frm the right side ?

13. anonymous

|dw:1433940357819:dw|

14. Nnesha

nope how did you get 3C at left side ?

15. anonymous

|dw:1433940467896:dw|

16. anonymous

computer is laggy, getting hard to draw

17. Nnesha

yeS SUBTRACT 3c both sides

18. Nnesha

|dw:1433940641351:dw| now divide both sides by 3

19. anonymous

so the answer is: B= ((A-3C) / (3))

20. Nnesha

yep right and REMEMBER parentheses = jail and outside term = bodyguard or a gunman you CAN"T move anything from parentheses with out his permission (distributive property )

21. Nnesha

so it can't be A bec 3(B+C) you can't subtract C first step is distribute

22. anonymous

thats a interesting way to learn that, and a little unique... never heard of it that way before.

23. Nnesha

wait wait nope

24. Nnesha

okay i just noticed something first choice is it $\large\rm \frac{ a-c }{ 3 }$ or $\large\rm \frac{ a }{ 3}-C$? both are different

25. anonymous

the first one

26. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @YouDontKnowMeOo so the answer is: B= ((A-3C) / (3)) $$\color{blue}{\text{End of Quote}}$$ okay so yes this one is right


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# zbMATH — the first resource for mathematics

One case of appearance of positive solutions of delayed discrete equations. (English) Zbl 1099.39001
The authors study the existence of a positive solution of the nonlinear delay discrete equation $\Delta u(k+n)=f\big (k,u(k),u(k+1),\dots ,u(k+n)\big ), \quad k\in {\mathbb Z},\;k\geq a.$ They derive a sharp sufficient condition for the existence of such solution. The result is a special case with $$b(k)\equiv 0$$ and $$c(k):=\sqrt {k}\, \big ( \frac {n}{n+1} \big )^k$$ of the existence and boundedness result.

##### MSC:
 39A10 Additive difference equations 39A11 Stability of difference equations (MSC2000)
##### Keywords:
positive solution; nonlinear delay difference equation
Full Text:
##### References:
 [1] R. P. Agarwal: Difference Equations and Inequalities, Theory, Methods, and Applications, 2nd ed. Marcel Dekker, New York, 2000. [2] J. Baštinec, J. Diblík, B. Zhang: Existence of bounded solutions of discrete delayed equations. Proceedings of the Sixth International Conference on Difference Equations and Applications, Augsburg 2001, Taylor $$\&$$ Francis (eds.), In print. [3] J. Diblík: A criterion for existence of positive solutions of systems of retarded functional differential equations. Nonlinear Anal. 38 (1999), 327-339. · Zbl 0935.34061 [4] J. Diblík, M. Růžičková: Existence of positive solutions of a singular initial problem for nonlinear system of differential equations. Rocky Mountain J. Math, In print. [5] Y. Domshlak, I. P. Stavroulakis: Oscillation of first-order delay differential equations in a critical case. Appl. Anal. 61 (1996), 359-371. · Zbl 0882.34069 [6] S. N. Elaydi: An Introduction to Difference Equations, 2nd ed. Springer, New York, 1999. · Zbl 0930.39001 [7] Á. Elbert, I. P. Stavroulakis: Oscillation and non-oscillation criteria for delay differential equations. Proc. Amer. Math. Soc. 123 (1995), 1503-1510. · Zbl 0828.34057 [8] L. H. Erbe, Q. Kong, B. G. Zhang: Oscillation Theory for Functional Differential Equations. Marcel Dekker, New York, 1995. [9] K. Gopalsamy: Stability and Oscillations in Delay Differential Equations of Population Dynamics. Kluwer Academic Publishers, Dordrecht, 1992. · Zbl 0752.34039 [10] I. Györi, G. Ladas: Oscillation Theory of Delay Differential Equations. Clarendon Press, Oxford, 1991. · Zbl 0780.34048 [11] A. Tineao: Asympotic classification of the positive solutions of the nonautonomous two-competing species problem. J. Math. Anal. Appl. 214 (1997), 324-348. [12] Xue-Zhong He, K. Gopalsamy: Persistence, attractivity, and delay in facultative mutualism. J. Math. Anal. Appl. 215 (1997), 154-173. · Zbl 0893.34036
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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SEARCH HOME
 Math Central Quandaries & Queries
 Question from Torrie, a student: I need to solve this problem in terms of Sin, but I'm having trouble. 1 + cot x divided by sin x + cos x

Hi Torrie,

The numerator is

$1 + \cot(x) = 1 + \frac{\cos(x)}{\sin(x)}.$

Write the two terms in the numerator with a common denominator and then divide the numerator by $\sin(x) + \cos(x)$ to obtain the original expression and solve.

Penny

Write $\cot x$ as $\frac{\cos x}{\sin x}$ (this should be a conditioned reflex!)

Write 1 as $\frac{\sin x}{ \sin x}$ (this is a standard trick)

Tidy up.

Good Hunting!
RD

Math Central is supported by the University of Regina and the Imperial Oil Foundation.


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location:  Publications → journals → CJM
Abstract view

# Characterisation Results for Steiner Triple Systems and Their Application to Edge-Colourings of Cubic Graphs

Published:2009-12-04
Printed: Apr 2010
• Daniel Král'
• Edita Máčajová
• Attila Pór
• Jean-Sébastien Sereni
 Format: HTML LaTeX MathJax

## Abstract

It is known that a Steiner triple system is projective if and only if it does not contain the four-triple configuration $C_{14}$. We find three configurations such that a Steiner triple system is affine if and only if it does not contain one of these configurations. Similarly, we characterise Hall triple systems using two forbidden configurations. Our characterisations have several interesting corollaries in the area of edge-colourings of graphs. A cubic graph G is S-edge-colourable for a Steiner triple system S if its edges can be coloured with points of S in such a way that the points assigned to three edges sharing a vertex form a triple in S. Among others, we show that all cubic graphs are S-edge-colourable for every non-projective non-affine point-transitive Steiner triple system S.
 MSC Classifications: 05B07 - Triple systems 05C15 - Coloring of graphs and hypergraphs


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\texttt{foo} vs \verb#foo#

The LaTeX commands `\texttt` and `\verb` are very similar, but not the same:

• `\verb` uses the next character as the delimiter, not braces
• `\verb` will not break over a line (and will happily encroach into the margin)
• `\verb` (and the `verbatim` environment) in a Beamer slide needs the `fragile` attribute

I tend to prefer `\verb`, as I never want monospaced text to be broken over a line.


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# I'm not able to leave a certain selection tool

I did, somehow (think I pressed a hotkey), end up with this circle around the mouse cursor, which could select multiple vertices. I do not know how to leave this mode/tool, and I'm not able to do anything but to use this certain tool. I can't use shortcuts, for example cmd+S to save. I cant open any menu, I cant rotate the view, I cant do anything but to change the size of the circle (using the scroll wheel on the mouse) and to select vertices using it. No view hotkeys works (1-9)

This is what the tool looks like:

Would love to at least get to know the name of the tool I'm stuck with, so that I could ask a more accurate question. Since I can't save the file, I don't want to close it. Therefore, if I could get some help soon, that would be very appreciated.

• That looks like circle select. Just press escape or right click Apr 8 '17 at 14:00
• right mouse button to escape circle select mode Feb 6 at 22:00

Try pressing "W", it will change the pointer type from select circle to select box, hope that helps.

i accidentally pressed it and had the same issue, pressing it again fixed it for me.

• this was the answer for my problem. It had gotten stuck in circle select mode, which seems to be different than the mode that you get when pressing "C" Aug 10 '19 at 20:09

That is the likely the circle select tool. Quoting from the page:

To cancel the selection use RMB or key Esc or Enter.

• Nah, it's a different circle select mode. It's not the same as the one you get when you press C (Escape and Enter were the first things I tried). -- Instead to get out of it you have to press W as the other answer suggests. Nov 10 '20 at 8:52

in 2.8 press and hold on the selector in the top left until the options appear

For anyone who still has this issue you need to go to the top where it says "Drag" next to "Drag" should be a drop down box, simply pick "box tool" or "Select" or whatever you need to use to go back to the normal tool selection. Hope this helps! :))

Press Esc key to exit, the C select mode


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1. ## Discrete maths

this question is confusing i dont understand it

2. Hello, sabina_19!

It seems clear enough . . . What part is confusing?

By using the appropriate truth tables, decide whethr
each of the following is tautology, absurdity, or contingency.

A tautology is always true.
An absurdity is always false.
A contingency is sometime true, sometimes false. .("It depends.")

$(a)\;\;\begin{array}{|c||ccc|}\hline
p & p & \to & p \\ \hline \hline
T & T & {\color{blue}T} & T \\ \hline
F & F & {\color{blue}T} & F \\ \hline \end{array}\qquad \text{Tautology}$

$(b)\;\;\begin{array}{|c||ccc|}\hline
p & p & \to & \sim\!p \\ \hline \hline
T & T & {\color{blue}F} & F \\ F & F &{\color{blue}T} & T \\ \hline
$(c)\;\;\begin{array}{|c||ccc|} \hline