text:
## Pages

### Bayesian Aggregation

This page provides some technical background on the Bayesian poll aggregation models used on this site. The page has been updated for the 2016 Federal election.

### General overview

The aggregation or data fusion models I use are known as hidden Markov models. They are are sometimes referred to as state space models or latent process models.

I model the national voting intention (which cannot be observed directly; it is "hidden") for each and every day of the period under analysis. The only time the national voting intention is not hidden, is at an election. In some models (known as anchored models), we use the election result to anchor the daily model we use.

In the language of modelling, our estimates of the national voting intention for each day being modeled are known as states. These "states" link together to form a Markov process, where each state is directly dependent on the previous state and a probability distribution linking the states. In plain English, the models assume that the national voting intention today is much like it was yesterday. The simplest models assume the voting intention today is normally distributed around the voting intention yesterday.

The model is informed by irregular and noisy data from the selected polling houses. The challenge for the model is to ignore the noise and find the underlying signal. In effect, the model is solved by finding the the day-to-day pathway with the maximum likelihood given the known poll results.

To improve the robustness of the model, we make provision for the long-run tendency of each polling house to systematically favour either the Coalition or Labor. We call this small tendency to favour one side or the other a "house effect". The model assumes that the results from each pollster diverge (on average) from the from real population voting intention by a small, constant number of percentage points. We use the calculated house effect to adjust the raw polling data from each polling house.

In estimating the house effects, we can take one of a number of approaches. We could:

• anchor the model to an election result on a particular day, and use that anchoring to establish the house effects.
• anchor the model to a particular polling house or houses; or
• assume that collectively the polling houses are unbiased, and that collectively their house effects sum to zero.

Typically, I use the first and third approaches in my models. Some models assume that the house effects sum to zero. Other models assume that the house effects can be determined absolutely by anchoring the hidden model for a particular day or week to a known election outcome.

There are issues with both approaches. The problem with anchoring the model to an election outcome (or to a particular polling house), is that pollsters are constantly reviewing and, from time to time, changing their polling practice. Over time these changes affect the reliability of the model. On the other hand, the sum-to-zero assumption is rarely correct. Of the two approaches, anchoring tends to perform better.

Another adjustment we make is to allow discontinuities in the hidden process when either party changes its leadership. Leadership changes can see immediate changes in the voting preferences. For the anchored models, we allow for discontinuities with the Gillard to Rudd and Abbott to Turnbull leadership changes. The unanchored models have a single discontinuity with the Abbott to Turnbull leadership change.

Solving a model necessitates integration over a series of complex multidimensional probability distributions. The definite integral is typically impossible to solve algebraically. But it can be solved using a numerical method based on Markov chains and random numbers known as Markov Chain Monte Carlo (MCMC) integration. I use a free software product called JAGS to solve these models.

### The dynamic linear model of TPP voting intention with house effects summed to zero

This is the simplest model. It has three parts:

1. The observational part of the model assumes two factors explain the difference between published poll results (what we observe) and the national voting intention on a particular day (which, with the exception of elections, is hidden):

1. The first factor is the margin of error from classical statistics. This is the random error associated with selecting a sample; and
2. The second factor is the systemic biases (house effects) that affect each pollster's published estimate of the population voting intention.

2. The temporal part of the model assumes that the actual population voting intention on any day is much the same as it was on the previous day (with the exception of discontinuities). The model estimates the (hidden) population voting intention for every day under analysis.

3. The house effects part of the model assumes that house effects are distributed around zero and sum to zero.

This model builds on original work by Professor Simon Jackman. It is encoded in JAGS as follows:

model {
## -- draws on models developed by Simon Jackman

## -- observational model
for(poll in 1:n_polls) {
yhat[poll] <- houseEffect[house[poll]] + hidden_voting_intention[day[poll]]
y[poll] ~ dnorm(yhat[poll], samplePrecision[poll]) # distribution
}

## -- temporal model - with one discontinuity
hidden_voting_intention[1] ~ dunif(0.3, 0.7) # contextually uninformative
hidden_voting_intention[discontinuity] ~ dunif(0.3, 0.7) # ditto

for(i in 2:(discontinuity-1)) {
hidden_voting_intention[i] ~ dnorm(hidden_voting_intention[i-1], walkPrecision)
}
for (j in (discontinuity+1):n_span) {
hidden_voting_intention[j] ~ dnorm(hidden_voting_intention[j-1], walkPrecision)
}
sigmaWalk ~ dunif(0, 0.01)
walkPrecision <- pow(sigmaWalk, -2)

## -- house effects model
for(i in 2:n_houses) {
houseEffect[i] ~ dunif(-0.15, 0.15) # contextually uninformative
}
houseEffect[1] <- -sum( houseEffect[2:n_houses] )
}

Professor Jackman's original JAGS code can be found in the file kalman.bug, in the zip file link on this page, under the heading Pooling the Polls Over an Election Campaign.

### The anchored dynamic linear model of TPP voting intention

This model is much the same as the previous model. However, it is run with data from prior to the 2013 election to anchor poll performance. It includes two discontinuities, with the ascensions of Rudd and Turnbull. And, because it is anchored, the house effects are not constrained to sum to zero.

model {

## -- observational model
for(poll in 1:n_polls) {
yhat[poll] <- houseEffect[house[poll]] + hidden_voting_intention[day[poll]]
y[poll] ~ dnorm(yhat[poll], samplePrecision[poll]) # distribution
}

## -- temporal model - with two or more discontinuities
# priors ...
hidden_voting_intention[1] ~ dunif(0.3, 0.7) # fairly uninformative
for(j in 1:n_discontinuities) {
hidden_voting_intention[discontinuities[j]] ~ dunif(0.3, 0.7) # fairly uninformative
}
sigmaWalk ~ dunif(0, 0.01)
walkPrecision <- pow(sigmaWalk, -2)

# Up until the first discontinuity ...
for(k in 2:(discontinuities[1]-1)) {
hidden_voting_intention[k] ~ dnorm(hidden_voting_intention[k-1], walkPrecision)
}

# Between the discontinuities ... assumes 2 or more discontinuities ...
for( disc in 1:(n_discontinuities-1) ) {
for(k in (discontinuities[disc]+1):(discontinuities[(disc+1)]-1)) {
hidden_voting_intention[k] ~ dnorm(hidden_voting_intention[k-1], walkPrecision)
}
}

# after the last discontinuity
for(k in (discontinuities[n_discontinuities]+1):n_span) {
hidden_voting_intention[k] ~ dnorm(hidden_voting_intention[k-1], walkPrecision)
}

## -- house effects model
for(i in 1:n_houses) {
houseEffect[i] ~ dnorm(0, pow(0.05, -2))
}
}

### The latent Dirichlet process for primary voting intention

This model is more complex. It takes advantage of the Dirichlet (pronounced dirik-lay) distribution, which always sums to 1, just as the primary votes for all parties would sum to 100 per cent of voters at an election. A weakness is a transmission mechanism from day-to-day that uses a "tightness of fit" parameter, which has been arbitrarily selected.

The model is set up a little differently to the previous models. Rather than pass the vote share as a number between 0 and 1; we pass the size of the sample that indicated a preference for each party. For example, if the poll is of 1000 voters, with 40 per cent for the Coalition, 40 per cent for Labor, 11 per cent for the Greens and and 9 per cent for Other parties, the multinomial we would pass across for this poll is [400, 400, 110, 90].

More broadly, this model is conceptually very similar to the sum-to-zero TPP model: with an observational component, a dynamic walk of primary voting proportions (modeled as a hierarchical Dirichlet process), a discontinuity for Turnbull's ascension, and a set of house effects that sum to zero across polling houses and across the parties.

data {
zero <- 0.0
}
model {

#### -- observational model

for(poll in 1:NUMPOLLS) { # for each poll result - rows
adjusted_poll[poll, 1:PARTIES] <- walk[pollDay[poll], 1:PARTIES] +
houseEffect[house[poll], 1:PARTIES]
}

#### -- temporal model with one discontinuity

tightness <- 50000 # kludge - today very much like yesterday
# - before discontinuity
for(day in 2:(discontinuity-1)) {
# Note: use math not a distribution to generate the multinomial ...
multinomial[day, 1:PARTIES] <- walk[day-1,  1:PARTIES] * tightness
walk[day, 1:PARTIES] ~ ddirch(multinomial[day, 1:PARTIES])
}
# - after discontinuity
for(day in discontinuity+1:PERIOD) {
# Note: use math not a distribution to generate the multinomial ...
multinomial[day, 1:PARTIES] <- walk[day-1,  1:PARTIES] * tightness
walk[day, 1:PARTIES] ~ ddirch(multinomial[day, 1:PARTIES])
}

## -- weakly informative priors for first and discontinuity days
for (party in 1:2) { # for each major party
alpha[party] ~ dunif(250, 600) # majors between 25% and 60%
beta[party] ~ dunif(250, 600) # majors between 25% and 60%
}
for (party in 3:PARTIES) { # for each minor party
alpha[party] ~ dunif(10, 250) # minors between 1% and 25%
beta[party] ~ dunif(10, 250) # minors between 1% and 25%
}
walk[1, 1:PARTIES] ~ ddirch(alpha[])
walk[discontinuity, 1:PARTIES] ~ ddirch(beta[])

## -- estimate a Coalition TPP from the primary votes
for(day in 1:PERIOD) {
CoalitionTPP[day] <- sum(walk[day, 1:PARTIES] *
preference_flows[1:PARTIES])
CoalitionTPP2010[day] <- sum(walk[day, 1:PARTIES] *
preference_flows_2010[1:PARTIES])
}

#### -- house effects model with two-way, sum-to-zero constraints

## -- vague priors ...
for (h in 2:HOUSECOUNT) {
for (p in 2:PARTIES) {
houseEffect[h, p] ~ dunif(-0.1, 0.1)
}
}

## -- sum to zero - but only in one direction for houseEffect[1, 1]
for (p in 2:PARTIES) {
houseEffect[1, p] <- 0 - sum( houseEffect[2:HOUSECOUNT, p] )
}
for(h in 1:HOUSECOUNT) {
# includes constraint for houseEffect[1, 1], but only in one direction
houseEffect[h, 1] <- 0 - sum( houseEffect[h, 2:PARTIES] )
}

## -- the other direction constraint on houseEffect[1, 1]
zero ~ dsum( houseEffect[1, 1], sum( houseEffect[2:HOUSECOUNT, 1] ) )
}

### The anchored Dirichlet primary vote model

The anchored Dirichlet model follows. It draws on elements from the anchored TPP model and Dirichlet model above. It is the most complex of these models. It also takes the longest time to produce reliable results. For this model, I run 460,000 samples, taking every 23rd sample for analysis. On my aging Apple iMac and JAGS 4.0.1 it takes around 100 minutes to run.

model {

#### -- observational model
for(poll in 1:NUMPOLLS) { # for each poll result - rows
adjusted_poll[poll, 1:PARTIES] <- walk[pollDay[poll], 1:PARTIES] +
houseEffect[house[poll], 1:PARTIES]
}

#### -- temporal model with multiple discontinuities
# - tightness of fit parameters
tightness <- 50000 # kludge - today very much like yesterday

# Up until the first discontinuity ...
for(day in 2:(discontinuities[1]-1)) {
multinomial[day, 1:PARTIES] <- walk[day-1,  1:PARTIES] * tightness
walk[day, 1:PARTIES] ~ ddirch(multinomial[day, 1:PARTIES])
}

# Between the discontinuities ... assumes 2 or more discontinuities ...
for( disc in 1:(n_discontinuities-1) ) {
for(day in (discontinuities[disc]+1):(discontinuities[(disc+1)]-1)) {
multinomial[day, 1:PARTIES] <- walk[day-1,  1:PARTIES] * tightness
walk[day, 1:PARTIES] ~ ddirch(multinomial[day, 1:PARTIES])
}
}

# After the last discontinuity
for(day in (discontinuities[n_discontinuities]+1):PERIOD) {
multinomial[day, 1:PARTIES] <- walk[day-1,  1:PARTIES] * tightness
walk[day, 1:PARTIES] ~ ddirch(multinomial[day, 1:PARTIES])
}

# weakly informative priors for first day and discontinutity days ...
for (party in 1:2) { # for each minor party
alpha[party] ~ dunif(250, 600) # minors between 25% and 60%
}
for (party in 3:PARTIES) { # for each minor party
alpha[party] ~ dunif(10, 200) # minors between 1% and 20%
}
walk[1, 1:PARTIES] ~ ddirch(alpha[])

for(j in 1:n_discontinuities) {
for (party in 1:2) { # for each minor party
beta[j, party] ~ dunif(250, 600) # minors between 25% and 60%
}
for (party in 3:PARTIES) { # for each minor party
beta[j, party] ~ dunif(10, 200) # minors between 1% and 20%
}
walk[discontinuities[j], 1:PARTIES] ~ ddirch(beta[j, 1:PARTIES])
}

## -- estimate a Coalition TPP from the primary votes
for(day in 1:PERIOD) {
CoalitionTPP[day] <- sum(walk[day, 1:PARTIES] *
preference_flows[1:PARTIES])
}

#### -- sum-to-zero constraints on house effects
for(h in 1:HOUSECOUNT) {
for (p in 2:PARTIES) {
houseEffect[h, p] ~ dnorm(0, pow(0.05, -2))
}
}
# need to lock in ... but only in one dimension
for(h in 1:HOUSECOUNT) { # for each house ...
houseEffect[h, 1] <- -sum( houseEffect[h, 2:PARTIES] )
}
}

### Beta model of primary vote share for Palmer United

A simplification of the Dirichlet distribution is the Beta distribution. I use the Beta distribution (in a similar model to the Dirichlet model above), to track the primary vote share of the Palmer United Party. This model does not provide for a discontinuity in voting associated with the change of Prime Minister.

model {

#### -- observational model
for(poll in 1:NUMPOLLS) { # for each poll result - rows
}

#### -- temporal model (a daily walk where today is much like yesterday)
tightness <- 50000 # KLUDGE - tightness of fit parameter selected by hand
for(day in 2:PERIOD) { # rows
binomial[day] <- walk[day-1] * tightness
walk[day] ~ dbeta(binomial[day], tightness - binomial[day])
}

## -- weakly informative priors for first day in the temporal model
alpha ~ dunif(1, 1500)
walk[1] ~ dbeta(alpha, 10000-alpha)

#### -- sum-to-zero constraints on house effects
for(h in 2:HOUSECOUNT) { # for each house ...
houseEffect[h] ~ dnorm(0, pow(0.1, -2))
}
houseEffect[1] <- -sum(houseEffect[2:HOUSECOUNT])
}

### Code and data

I have made most of my data and code base available on Google Drive. Please note, this is my live code base, which I play with quite a bit. So, there will be times when it is broken or in some stage of being edited.

What I have not made available is the Excel spreadsheets into which I initially place my data. These live in the (hidden) raw-data directory. However, the collated data for the Bayesian model lives in the intermediate directory, visible from the above link.


text:
## ionizing collision

in mass spectrometry
https://doi.org/10.1351/goldbook.I03214
An ion/neutral species interaction in which an electron or electrons are stripped from the ion and/or the neutral species in the collision. Generally, this term has come to be used to describe collisions of fast-moving ions with a neutral species in which the neutral species is ionized with no change in the number of charges carried by the ion. Care should be taken when this term is used to emphasize if charge @[email protected] of the ion has taken place.
Source:
PAC, 1991, 63, 1541. (Recommendations for nomenclature and symbolism for mass spectroscopy (including an appendix of terms used in vacuum technology). (Recommendations 1991)) on page 1555 [Terms] [Paper]

format: plain


text:
mersenneforum.org How to properly put a MacBook to Rust in Pieces
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 2012-04-29, 01:56 #1 ewmayer ∂2ω=0     Sep 2002 República de California 265008 Posts How to properly put a MacBook to Rust in Pieces Like many laptops, my vintage-2009 MacBook is undercooled. Its fine for normal daily tasks, but I also run my Mlucas code (similar thermal properties as Prime95, big-FFT code which hammers the SIMD unit) on it. Even single-threaded is sufficient to get the fan maxed out and the CPU hot enough that it frequently goes into throttle-down mode. I have no complaints about the MacBook fan - it sounds as quietly powerful as the day I first used it, and is the best fan I've ever had in a laptop. These things are just not intended to run heavy-duty code 24/7. Based on timings and 'top'-based profiling, the only time the code really runs at 100% is for a few minutes after startup, and if I set the lid-open unit next to an open window on a cool night. With the nights now being warmer as summer nears, even the latter method is not enough to keep the CPU cool enough to prevent auto-throttling, so I've been trying a few things, and finally hit upon what appears to be an optimum configuration for cooling when the unit is running DC-style code but not being actively used by me. I call it the "A-frame": simply tip the lid-open notebook onto the 2 long edges created by its opened lid, with the fan vents thus pointing up toward the ceiling. This aids cooling in 2 major ways: 1. The plastic-shielded underside of the unit, which normally gets quite hot since it is sealed off from airflow and has the plastic housing interfering with heat dissipation, is now fully exposed to the air. 2. One creates a "chimney" effect which maximizes natural convection in aiding the fan: air freely enters the fan intakes (mainly via the porously-designed keyboard) through the open ends of the A-frame, and blows out the top of the thus-created chimney, in the natural direction heated air wants to go. It looks a bit odd, but works beautifully. My results-file timings show full speed ahead for the entire night, meaning 10-20% more throughput than sitting on my desk in "orthodox" clam-style mode. I will try running stuff on both CPUs tonight and see if any throttling occurs. Here is a high-tech schematic showing a side view of the A-frame configuration, complete with a Schlieren photograph of the lovely convective whorls of hot air exiting the top of the chimney: Code:  ? ? ? /\ / \ / \ / \` Last fiddled with by Batalov on 2012-07-09 at 22:16 Reason: /\
 2012-04-29, 03:24 #2 axn     Jun 2003 2×32×269 Posts When was the last time you opened it up and blew out the dust?
 2012-04-29, 14:01 #3 rogue     "Mark" Apr 2003 Between here and the 52×13×19 Posts I have a stand (mStand), although my MacBook Pro is even more vintage than yours, circa 2006.
2012-04-30, 07:14   #4
debrouxl

Sep 2009

11110100012 Posts

Quote:
 These things are just not intended to run heavy-duty code 24/7.
Indeed. Because they have a strong focus on design (no air intakes all around the bottom) rather than on reliability, MacBooks have, by far, the worst fan system for a laptop I know of...

On the '2006 MacBook Pro I was given in my previous job, it took only one year of 24/7 usage to damage the fan system, despite blowing it multiple times (including once case full open, when I had to change the HDD, the second HDD killed by the computer in merely four years...): it became very noisy. And yet, it was hard to keep bare hands on the metal casing, which means that its temperature was circa 50°C...

Last fiddled with by debrouxl on 2012-04-30 at 07:38

2012-04-30, 12:41   #5
rogue

"Mark"
Apr 2003
Between here and the

52×13×19 Posts

Quote:
 Originally Posted by debrouxl On the '2006 MacBook Pro I was given in my previous job, it took only one year of 24/7 usage to damage the fan system, despite blowing it multiple times (including once case full open, when I had to change the HDD, the second HDD killed by the computer in merely four years...): it became very noisy. And yet, it was hard to keep bare hands on the metal casing, which means that its temperature was circa 50°C...
Interesting. I run pfgw on both cores of mine 24x7. It does get warm, but it has never overheated. The only problem I've had was that the battery died and I had it replaced at the Apple Store at no cost in 2010. The use of the stand seems to make a difference.

Last fiddled with by rogue on 2012-04-30 at 12:42

2012-04-30, 19:45   #6
ewmayer
2ω=0

Sep 2002
República de California

26×181 Posts

Quote:
 Originally Posted by rogue Interesting. I run pfgw on both cores of mine 24x7. It does get warm, but it has never overheated. The only problem I've had was that the battery died and I had it replaced at the Apple Store at no cost in 2010. The use of the stand seems to make a difference.
Mine is a basic white-plastic-case MacBook (non-pro, that is), which embodies the issue debrouxl notes, the bottom is sealed and the plastic is an insulating material.

Last week I installed a piece of freeware called smcFanControl, which allows me to monitor internal temp and fan speed. When nothing significant is running, e.g. as I write this, internal temp shows 50-60C depending on configuration (bottom elevated or not) and ambient temp/airflow. If I fire up some heavy-duty crunching program the temp very quickly gets up toward 80C, though (on my system) auto-throttling is a serious issue only for highly-optimized FPU-based code. Note the 80C is accompanied by the fan revving to 6-7krpm, and there are no odd noises from it, and a robust warm-airflow out the back vent which tells me the fan is working fine.

I notice a significant difference depending on the nature of the code: When running my Mlucas TF code which is highly optimized but 64-integer (as opposed to FPU) based, it gets warm but stays below the auto-throttling temperature threshold. When running Mlucas in LL-test mode it hammers the FPU by way of SSE2, and the temperature according to the smcFan app rises from 50C to 80C in under a minute. Whether auto-throttling kicks in depends on ambient conditions ... but at ambient temps over 70 it starts auto-throttling in no more than 5-10 minutes.

I sometimes will elevate the back of the unit slightly (by e.g. putting it on the edge of a book) which allows me to type and gives better airflow, but again that helps less that it should because it only keeps the plastic shell down there a little cooler - there are no air intakes on the bottom which could really benefit from it.

 2012-07-08, 20:30 #7 ewmayer ∂2ω=0     Sep 2002 República de California 26×181 Posts The cooling fan on the above MacBook died last night - I have been running an app called smcFanControl to help monitor temperature and fan RPM, noticed last night that 'top' showed kernel_task running at 100%, which indicates big-time thermal throttling (I know the "100% CPU" is counterintuitive, but based on now-long experience with this system, kernel_task "CPU usage" really correlates with "throttle-idled cycles"), which normally only happens in a closed warm space, never with the system in a cool-breeze spot as it was last night. Sure enough, smcFanControl showed RMPs stuck at 0, and when I put my ear close to the vent I heard faint ticking noises which I'd never heard before. The system is over 3 years old and out of warranty, so I'd like to do a do-it-yourself fan replacement. Have found plenty of online guides to this, but oddly, have been unable to find a replacement fan unit for this particular Macbook model (5,2) ... for example when I google macbook replacement "cooling fan" amazon.com I get lots of hits, but all are for MacBook Pro models. Adding "13-inch" "5,2" to the query does not help. (I *hate* that apple did not give their non-pro, non-air plain-vanilla MacBook a distinguishing 2nd name, to help differentiate it from the Pro and Air in searches, I ran into similar time-wastage a couple months ago when I needed to find a replacement battery for my system. How about "MacBook Classic", Apple? But, I digress). Any advice for finding a replacement fan unit for non-pro MacBook model family 5,* would be greatly welcome. I am also planning on picking up a micro-toolkit like this one.
2012-07-08, 21:07   #8
Mathew

Nov 2009

1010111102 Posts

Something like this?

Part Number(s): 922-8273
Compatible Model(s):
Quote:
 MacBook 13.3" 2.0GHz Core 2 Duo (MB881LL/A, A1181, MacBook5,2) - Early 2009 MacBook 13.3" 2.13GHz Core 2 Duo (MC240LL/A, A1181, MacBook5,2) - Early 2009

2012-07-09, 01:58   #9
ewmayer
2ω=0

Sep 2002
República de California

26×181 Posts

Quote:
 Originally Posted by Mathew Something like this? Part Number(s): 922-8273 Compatible Model(s):
Yeah, that's the ticket - thanks!

I wonder why that was in none of my first-few-pages of search results ... did you use a search engine or search on Amazon?

-----------------------------

Edit: Now that I know the part #, I also found it on ebay for ~1/3 the price. (eBay was $12.50 + free ship, amazon is$30 + \$5 shipping, not eligible for free super saver shipping). Sweet! I will provide update once the part arrives.

 2012-07-09, 02:50 #10 Prime95 P90 years forever!     Aug 2002 Yeehaw, FL 67×109 Posts I replaced one of my fans 9 months ago and it died again last week :( They are a pain to replace, I'd rather upgrade my laptop than replace fans every year. BTW, you can use CoolBookController to look at the actual CPU speed. My 2.5 GHz Pro is now running at 1.6 - 1.8 GHz on one fan.
2012-07-09, 18:39   #11
ewmayer
2ω=0

Sep 2002
República de California

26·181 Posts

Quote:
 Originally Posted by Prime95 I replaced one of my fans 9 months ago and it died again last week :( They are a pain to replace, I'd rather upgrade my laptop than replace fans every year.
I do intend to buy an AVX-capable macbook toward EOY, but would still like to be able to run DC stuff on the current one, and make it my "beater laptop".

What tools did you need for your fan-swap? (Not being a PC repair geek, I'm still trying to figure out what a "spudger" is ... sounds like something one uses to dig up potatoes :).

Quote:
 BTW, you can use CoolBookController to look at the actual CPU speed. My 2.5 GHz Pro is now running at 1.6 - 1.8 GHz on one fan.
Downloaded, installed into Applications and just tried - the entire right-half of the dialog (e.g. throttling level) is grayed out. Do I need to purchase a license to see that info?

BTW, and with a "do not try this at home" legal disclaimer, using the above "A frame" trick and placing the macbook in my open screened bedroom window last night allowed it run some pure-integer factoring code all night with very little throttling (inferred from kernel_task CPU % and the smcFan temperature display) even with no fan ... but my SSE2-floating-point-heavy code runs much too hot.

 Similar Threads Thread Thread Starter Forum Replies Last Post a1call FactorDB 5 2016-04-17 16:59 Rodrigo Linux 8 2015-12-13 23:22 ixfd64 Software 4 2011-01-01 18:58 ixfd64 Hardware 15 2009-12-16 20:05 Uncwilly Puzzles 35 2006-11-15 01:07

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text:
# Model category of all model categories

Is there some natural model category structure on the category of all small model categories and some specified functors among them (i.e. not necessarily all functors), say Quillen adjunctions? What is the most natural choice to this question: what should the classses of fibrations, cofibrations and weak equivalencies be?

• To define a category, you have to be sure that for every object $X,Y$ $Hom(X,Y)$ is a set. Are you sure that the set of functors between two categories endowed witha Quillen model is a set ? – Tsemo Aristide Dec 1 '18 at 15:41
• Well, we may encounter an illegitimate category $\mathbf{CAT}$. – user122424 Dec 2 '18 at 16:03
• I've edited my question by inserting the word "small". Everything goes smoothly now ? – user122424 Dec 2 '18 at 16:42
• Well, very few model categories of interest are small...Anyway, that's not really the point. A model category structure gives, in particular, homotopy limits and colimits, and it's extremely difficult to construct even these for model categories. Julie Bergner is one author who's written papers on special cases like homotopy fiber products. Much easier is to study similar questions $\infty$-categorically. – Kevin Carlson Dec 2 '18 at 21:13
• @KevinCarlson "Very few model categories of interest are small". Maybe that's true in your universe. I'm in the next one up, and I see lots of interesting small model categories! – Alex Kruckman Dec 3 '18 at 16:30


text:
Mathematics
OpenStudy (anonymous):

Hi can somebody help me please: 4 log x = log 4 x im a bit stuck on this one. :(

OpenStudy (***[isuru]***):

hi, Is it something like this ?$4logx = \log4x$ and u r trying to solve it for x ?

OpenStudy (anonymous):

yes

OpenStudy (***[isuru]***):

Hi, this is a basic rule u need to solve this question $\log a + \log b = \log (ab)$ So... apply it to the log4x and rewrite it as this.... $\log4x = \log4 + logx$ now ur question is modified to this... $4logx = logx + \log 4$$4logx - \log x = \log 4$$3logx = \log4$$logx = \frac{ \log4 }{3}$ Now let's take $logx = \frac{ \log4 }{3 } = y$ then $\log x = y$ $10^{y} = x\ ----(1)$ also $\frac{ \log4 }{3 } = y$$\log4 = 3y$$10^{3y} = 4$$(10^{y})^{3} =4$$10^{y} = \sqrt[3]{4} ----(2)$ hence the left side of (1) = (2) ---> The right hand side also should equal which mean $x = \sqrt[3]{4}$ Hope this will hep ya!!!

OpenStudy (anonymous):

that really helps thanks so much!

OpenStudy (***[isuru]***):

u r welcome!!


text:
# WPF Application Resource, Content, and Data Files

Microsoft Windows applications often depend on files that contain non-executable data, such as Extensible Application Markup Language (XAML), images, video, and audio. Windows Presentation Foundation (WPF) offers special support for configuring, identifying, and using these types of data files, which are called application data files. This support revolves around a specific set of application data file types, including:

• Resource Files: Data files that are compiled into either an executable or library WPF assembly.

• Content Files: Standalone data files that have an explicit association with an executable WPF assembly.

• Site of Origin Files: Standalone data files that have no association with an executable WPF assembly.

One important distinction to make between these three types of files is that resource files and content files are known at build time; an assembly has explicit knowledge of them. For site of origin files, however, an assembly may have no knowledge of them at all, or implicit knowledge through a pack uniform resource identifier (URI) reference; the case of the latter, there is no guarantee that the referenced site of origin file actually exists.

To reference application data files, Windows Presentation Foundation (WPF) uses the Pack uniform resource identifier (URI) Scheme, which is described in detail in Pack URIs in WPF).

This topic describes how to configure and use application data files.

## Resource Files

If an application data file must always be available to an application, the only way to guarantee availability is to compile it into an application's main executable assembly or one of its referenced assemblies. This type of application data file is known as a resource file.

You should use resource files when:

• You don't need to update the resource file's content after it is compiled into an assembly.

• You want to simplify application distribution complexity by reducing the number of file dependencies.

• Your application data file needs to be localizable (see WPF Globalization and Localization Overview).

Note

The resource files described in this section are different than the resource files described in XAML Resources and different than the embedded or linked resources described in Manage Application Resources (.NET).

### Configuring Resource Files

In WPF, a resource file is a file that is included in an Microsoft build engine (MSBuild) project as a Resource item.

<Project xmlns="http://schemas.microsoft.com/developer/msbuild/2003" ... >
...
<ItemGroup>
<Resource Include="ResourceFile.xaml" />
</ItemGroup>
...
</Project>

Note

In Microsoft Visual Studio, you create a resource file by adding a file to a project and setting its Build Action to Resource.

When the project is built, MSBuild compiles the resource into the assembly.

### Using Resource Files

To load a resource file, you can call the GetResourceStream method of the Application class, passing a pack URI that identifies the desired resource file. GetResourceStream returns a StreamResourceInfo object, which exposes the resource file as a Stream and describes its content type.

As an example, the following code shows how to use GetResourceStream to load a Page resource file and set it as the content of a Frame (pageFrame):

// Navigate to xaml page
Uri uri = new Uri("/PageResourceFile.xaml", UriKind.Relative);
StreamResourceInfo info = Application.GetResourceStream(uri);
this.pageFrame.Content = page;

' Navigate to xaml page
Dim uri As New Uri("/PageResourceFile.xaml", UriKind.Relative)
Dim info As StreamResourceInfo = Application.GetResourceStream(uri)
Me.pageFrame.Content = page

While calling GetResourceStream gives you access to the Stream, you need to perform the additional work of converting it to the type of the property that you'll be setting it with. Instead, you can let WPF take care of opening and converting the Stream by loading a resource file directly into the property of a type using code.

The following example shows how to load a Page directly into a Frame (pageFrame) using code.

Uri pageUri = new Uri("/PageResourceFile.xaml", UriKind.Relative);
this.pageFrame.Source = pageUri;

Dim pageUri As New Uri("/PageResourceFile.xaml", UriKind.Relative)
Me.pageFrame.Source = pageUri

The following example is the markup equivalent of the preceding example.

<Frame Name="pageFrame" Source="PageResourceFile.xaml" />

### Application Code Files as Resource Files

A special set of WPF application code files can be referenced using pack URIs, including windows, pages, flow documents, and resource dictionaries. For example, you can set the Application.StartupUri property with a pack URI that references the window or page that you would like to load when an application starts.

<Application
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
StartupUri="SOOPage.xaml" />

You can do this when a XAML file is included in an Microsoft build engine (MSBuild) project as a Page item.

<Project xmlns="http://schemas.microsoft.com/developer/msbuild/2003" ... >
...
<ItemGroup>
<Page Include="MainWindow.xaml" />
</ItemGroup>
...
</Project>

Note

In Visual Studio, you add a new Window, NavigationWindow, Page, FlowDocument, or ResourceDictionary to a project, the Build Action for the markup file will default to Page.

When a project with Page items is compiled, the XAML items are converted to binary format and compiled into the associated assembly. Consequently, these files can be used in the same way as typical resource files.

Note

If a XAML file is configured as a Resource item, and does not have a code-behind file, the raw XAML is compiled into an assembly rather than a binary version of the raw XAML.

## Content Files

A content file is distributed as a loose file alongside an executable assembly. Although they are not compiled into an assembly, assemblies are compiled with metadata that establishes an association with each content file.

You should use content files when your application requires a specific set of application data files that you want to be able to update without recompiling the assembly that consumes them.

### Configuring Content Files

To add a content file to a project, an application data file must be included as a Content item. Furthermore, because a content file is not compiled directly into the assembly, you need to set the MSBuildCopyToOutputDirectory metadata element to specify that the content file is copied to a location that is relative to the built assembly. If you want the resource to be copied to the build output folder every time a project is built, you set the CopyToOutputDirectory metadata element with the Always value. Otherwise, you can ensure that only the newest version of the resource is copied to the build output folder by using the PreserveNewest value.

The following shows a file that is configured as a content file which is copied to the build output folder only when a new version of the resource is added to the project.

<Project xmlns="http://schemas.microsoft.com/developer/msbuild/2003" ... >
...
<ItemGroup>
<Content Include="ContentFile.xaml">
</Content>
</ItemGroup>
...
</Project>

Note

In Visual Studio, you create a content file by adding a file to a project and setting its Build Action to Content, and set its Copy to Output Directory to Copy always (same as Always) and Copy if newer (same as PreserveNewest).

When the project is built, an AssemblyAssociatedContentFileAttribute attribute is compiled into the metadata of the assembly for each content file.

[assembly: AssemblyAssociatedContentFile("ContentFile.xaml")]

The value of the AssemblyAssociatedContentFileAttribute implies the path to the content file relative to its position in the project. For example, if a content file was located in a project subfolder, the additional path information would be incorporated into the AssemblyAssociatedContentFileAttribute value.

[assembly: AssemblyAssociatedContentFile("Resources/ContentFile.xaml")]

The AssemblyAssociatedContentFileAttribute value is also the value of the path to the content file in the build output folder.

### Using Content Files

To load a content file, you can call the GetContentStream method of the Application class, passing a pack URI that identifies the desired content file. GetContentStream returns a StreamResourceInfo object, which exposes the content file as a Stream and describes its content type.

As an example, the following code shows how to use GetContentStream to load a Page content file and set it as the content of a Frame (pageFrame).

// Navigate to xaml page
Uri uri = new Uri("/PageContentFile.xaml", UriKind.Relative);
StreamResourceInfo info = Application.GetContentStream(uri);
this.pageFrame.Content = page;

' Navigate to xaml page
Dim uri As New Uri("/PageContentFile.xaml", UriKind.Relative)
Dim info As StreamResourceInfo = Application.GetContentStream(uri)
Me.pageFrame.Content = page

While calling GetContentStream gives you access to the Stream, you need to perform the additional work of converting it to the type of the property that you'll be setting it with. Instead, you can let WPF take care of opening and converting the Stream by loading a resource file directly into the property of a type using code.

The following example shows how to load a Page directly into a Frame (pageFrame) using code.

Uri pageUri = new Uri("/PageContentFile.xaml", UriKind.Relative);
this.pageFrame.Source = pageUri;

Dim pageUri As New Uri("/PageContentFile.xaml", UriKind.Relative)
Me.pageFrame.Source = pageUri

The following example is the markup equivalent of the preceding example.

<Frame Name="pageFrame" Source="PageContentFile.xaml" />

## Site of Origin Files

Resource files have an explicit relationship with the assemblies that they are distributed alongside, as defined by the AssemblyAssociatedContentFileAttribute. But, there are times when you may want to establish either an implicit or non-existent relationship between an assembly and an application data file, including when:

• A file doesn't exist at compile time.

• You don't know what files your assembly will require until run time.

• You want to be able to update files without recompiling the assembly that they are associated with.

• Your application uses large data files, such as audio and video, and you only want users to download them if they choose to.

It is possible to load these types of files by using traditional URI schemes, such as the file:/// and http:// schemes.

<Image Source="file:///C:/DataFile.bmp" />
<Image Source="http://www.datafilewebsite.com/DataFile.bmp" />

However, the file:/// and http:// schemes require your application to have full trust. If your application is a XAML browser application (XBAP) that was launched from the Internet or intranet, and it requests only the set of permissions that are allowed for applications launched from those locations, loose files can only be loaded from the application's site of origin (launch location). Such files are known as site of origin files.

Site of origin files are the only option for partial trust applications, although are not limited to partial trust applications. Full trust applications may still need to load application data files that they do not know about at build time; while full trust applications could use file:///, it is likely that the application data files will be installed in the same folder as, or a subfolder of, the application assembly. In this case, using site of origin referencing is easier than using file:///, because using file:/// requires you to work out the full path the file.

Note

Site of origin files are not cached with an XAML browser application (XBAP) on a client machine, while content files are. Consequently, they are only downloaded when specifically requested. If an XAML browser application (XBAP) application has large media files, configuring them as site of origin files means the initial application launch is much faster, and the files are only downloaded on demand.

### Configuring Site of Origin Files

If your site of origin files are non-existent or unknown at compile time, you need to use traditional deployment mechanisms for ensuring the required files are available at run time, including using either the XCopy command-line program or the Microsoft Windows Installer.

If you do know at compile time the files that you would like to be located at the site of origin, but still want to avoid an explicit dependency, you can add those files to an Microsoft build engine (MSBuild) project as None item. As with content files, you need to set the MSBuildCopyToOutputDirectory attribute to specify that the site of origin file is copied to a location that is relative to the built assembly, by specifying either the Always value or the PreserveNewest value.

<Project xmlns="http://schemas.microsoft.com/developer/msbuild/2003" ... >
...
<None Include="PageSiteOfOriginFile.xaml">
<CopyToOutputDirectory>Always</CopyToOutputDirectory>
</None>
...
</Project>

Note

In Visual Studio, you create a site of origin file by adding a file to a project and setting its Build Action to None.

When the project is built, MSBuild copies the specified files to the build output folder.

### Using Site of Origin Files

To load a site of origin file, you can call the GetRemoteStream method of the Application class, passing a pack URI that identifies the desired site of origin file. GetRemoteStream returns a StreamResourceInfo object, which exposes the site of origin file as a Stream and describes its content type.

As an example, the following code shows how to use GetRemoteStream to load a Page site of origin file and set it as the content of a Frame (pageFrame).

// Navigate to xaml page
Uri uri = new Uri("/SiteOfOriginFile.xaml", UriKind.Relative);
StreamResourceInfo info = Application.GetRemoteStream(uri);
this.pageFrame.Content = page;

' Navigate to xaml page
Dim uri As New Uri("/SiteOfOriginFile.xaml", UriKind.Relative)
Dim info As StreamResourceInfo = Application.GetRemoteStream(uri)
Me.pageFrame.Content = page

While calling GetRemoteStream gives you access to the Stream, you need to perform the additional work of converting it to the type of the property that you'll be setting it with. Instead, you can let WPF take care of opening and converting the Stream by loading a resource file directly into the property of a type using code.

The following example shows how to load a Page directly into a Frame (pageFrame) using code.

Uri pageUri = new Uri("pack://siteoforigin:,,,/SiteOfOriginFile.xaml", UriKind.Absolute);
this.pageFrame.Source = pageUri;

Dim pageUri As New Uri("pack://siteoforigin:,,,/Subfolder/SiteOfOriginFile.xaml", UriKind.Absolute)
Me.pageFrame.Source = pageUri

The following example is the markup equivalent of the preceding example.

<Frame Name="pageFrame" Source="pack://siteoforigin:,,,/SiteOfOriginFile.xaml" />

## Rebuilding After Changing Build Type

After you change the build type of an application data file, you need to rebuild the entire application to ensure those changes are applied. If you only build the application, the changes are not applied.


text:
# dynamical_matrix command

## Syntax

dynamical_matrix group-ID style gamma args keyword value ...

• group-ID = ID of group of atoms to displace

• style = regular or eskm

• gamma = finite different displacement length (distance units)

• one or more keyword/arg pairs may be appended

keyword = file or binary
file name = name of output file for the dynamical matrix
binary arg = yes or no or gzip

## Examples

dynamical_matrix 1 regular 0.000001
dynamical_matrix 1 eskm 0.000001
dynamical_matrix 3 regular 0.00004 file dynmat.dat
dynamical_matrix 5 eskm 0.00000001 file dynamical.dat binary yes

## Description

Calculate the dynamical matrix of the selected group.

## Restrictions

The command collects the entire dynamical matrix a single MPI rank, so the memory requirements can be very significant for large systems.

This command assumes a periodic system.

This command is part of the USER-PHONON package. It is only enabled if LAMMPS was built with that package. See the Build package doc page for more info.

## Default

The default settings are file = “dynmat.dyn”, binary = no


text:
# Hit the bullseye

This is another puzzle grabbed from a popular Italian magazine. The rules are pretty much the same except that there are more words! While this might look harmlessly easy, unlike my Star of Dice, don't be deceived! It will be a nice mental exercise. Also, there is only one correct order. Multiple associations might be there, at least apparently, but you must choose the ones that will allow you to use up all the words properly.

Objective

Starting from the word indicated by the arrow (in this case MASTER), you must reach the center of the target, progressively associating and eliminating all the words included in it, that can be found in any ring, not necessarily in order, according to the rules below.

Rules

1. The word can be an anagram of the previous one.
2. It can be a synonym or an antonym of the previous one.
3. It can be obtained by adding, removing, or replacing a letter in the previous one.
4. It can be associated to the previous one because of a saying (proverb), metaphor, or word association.
5. It can form, together with the previous one, the name of a celebrity, or a place.
6. It can associated to a work of art (book, theater, poetry, etc.), its author, or its title including any genre.

Hint

urn $\rightarrow$ eternity

For quick reference, the words in the bullseye, going towards the center, are:

MASTER sea nee horse wing entirety force driver rhcp plane fore sting keats leaf row bow stream faction poisoned plan genre gowan opposite fraction stage golf attraction gown finger shore software eternity guitar wagon river swing poseidon arrow flea antidote stake ball bee urn prod program theater string fringe green drop knee HAIR

• Is there any significance to the rings, or is this just a more difficult to read unordered list? Do we have to go from one ring to the next? Can rings spin? I don't understand the purpose of the diagram. – Ian MacDonald Oct 20 '16 at 16:28
• So really what you're asking is for an ordering that includes all of the following words starting at MASTER and ending at HAIR according to the rules you've listed: MASTER, FORE, STING, KEATS, LEAF, ROW, SEA, NEE, HORSE, WING, ENTIRETY, FORCE, DRIVER, RHCP, SCHEDULE, STREAM, FACTION, POISONED, PLAN, GENRE, GOWAN, OPPOSITE, FRACTION, STAGE, GOLF, BOW, GOWN, FLEA, FINGER, SHORE, GROUP, ETERNITY, GUITAR, WAGON, ATTRACTION, POSEIDON, ARROW, ANTIDOTE, STAKE, BALL, RIVER, SWING, THEATER, STRING, BEE, URN, PROD, PROGRAM, FRINGE, GREEN, DROP, KNEE, HAIR – Ian MacDonald Oct 20 '16 at 16:41
• Some of these rules seem very vague to me. Anagrams, synonyms, and antonyms seem fine, as well as adding or removing letters, but word association allows for an awful lot of interpretation. I wouldn't be surprised to see several different "legitimate" answers, based on whatever the poster thinks is a valid word association. – GentlePurpleRain Oct 20 '16 at 16:58
• It seems like Rule 6 could allow for just about anything. Associating to any work of art, author, or title? I could probably pick almost any two words in this list, and find the title of some work of art that includes them. – GentlePurpleRain Oct 20 '16 at 17:01
• Adesso la settimana enigmistica vanta un tentativo di imitazione anche su StackExchange! – Francesco Oct 20 '16 at 20:11

Master

Stream via Anagram
River via Synonym
Wagon via Association ->Wagon Driver
Gowan via Anagram
Gown via Subtracted Letter
Ball via Association ->Ball Gown
Golf via Association ->Golf Ball
Fore via Association -> Yell 'Fore' while playing Golf
Attraction via Association -> Electromagnetic Attraction is a Force
Opposite via Association -> Opposites feel Attraction (Opposites Attract)
Faction via Association -> Factions imply opposed groups
Entirety via Antonym
Eternity via Anogram
Urn via Association -> 'Eternity' funerary urns
Keats via Association -> Wrote 'Ode on a Grecian Urn'
Stake via Anagram
Stage via Letter Replacement
Program via Association ->Stage Program
Software via Association ->Software Program
Plan via Association ->Software execution Plan
Wing via Association ->Plane Wing Swing via Added Letter
Sting via Changing Letter
Antidote via Antonym
Poisoned via Antonym
Poseidon via Anagram
Sea via Association -> God of the Sea (and Horses)
Horse via Association -> Seahorse (these can be inverted for fun and profit)
Shore via Anagram ->(If above is inverted, Association->She sells sea shells by the Sea Shore)
Row via Association ->Dinah, Row your boat to Shore
Bow via Changing Letter
Arrow via Association ->Bow and Arrow Knee via Association ->Arrow to the Knee
Nee via Dropped Letter
Bee via Changing Letter
Drop via Association -> Bee produces honey Drop
Prod via Anagram
Finger via Association ->Prod with Finger
Fringe via Anagram
String via Association ->String Fringe
Guitar via Association ->Guitar String
RHCP via Association ->Guitar Band
Flea via Association ->Flea is a member of RHCP
Leaf via Anagram
Green via Association ->Leaf Green
Genre via Anagram
Theater via Association -> Theater Genres

And we're at Hair!

• Yes, RHCP is what you think it is. I couldn't fit it otherwise. :D – Alenanno Oct 20 '16 at 16:49
• You've done huge progress, but there's a word association that doesn't work in your list. – Alenanno Oct 20 '16 at 17:07
• I think it should go POSEIDON $\rightarrow$ SEA $\rightarrow$ HORSE $\rightarrow$ SHORE. – GentlePurpleRain Oct 20 '16 at 18:44
• Theater seems like it doesn't fit with anything – MMAdams Oct 20 '16 at 19:34
• @MMAdams It does. :P – Alenanno Oct 20 '16 at 20:47

Okay, I feel like I'm very close, but there are a number of words that only seem to connect to one other word, and I can't see any way to connect them to another:

OPPOSITE, ETERNITY, ANTIDOTE, URN, KNEE (among others)

Here are the groups I've come up with so far. Maybe someone else can find the connections I'm missing...

MASTER anagrams to
STREAM is a synonym of
RIVER is one letter short of DRIVER is a club in
GOLF is where people often say
FORE is one letter short of
FORCE is a synonym of
ATTRACTION is in a common phrase with
OPPOSITE

ETERNITY anagrams to
ENTIRETY is an antonym of
FRACTION is one letter more than
FACTION is a synonym of
GENRE anagrams to
GREEN describes a
LEAF anagrams to
FLEA is a singer in
RHCP where they play
GUITAR has a
STRING which is often rolled into a
BALL is in a common phrase with
GOWN is one letter short of
GOWAN anagrams to
WAGON is pulled by a
HORSE anagrams to
SHORE is in a common phrase with
SEA is the domain of
POSEIDON anagrams to
POISONED is (sort of) an antonym of
ANTIDOTE

URN is in the title of "Ode to a Grecian Urn" by
KEATS anagrams to
STAKE is one letter different from
STAGE is commonly found in a
THEATER is usually performed by a
GROUP

WING is one letter short of
SWING is one letter different from
STING is done by a
BEE is one letter different from
NEE is one letter short of
KNEE

ARROW is often paired with a
BOW is one letter different from
ROW

PROGRAM is a synonym of
SCHEDULE is a synonym of
PLAN

DROP is an anagram of
PROD is a synonym of
FINGER is an anagram of
FRINGE is commonly found in
HAIR

• I'm fairly sure 'knee' and 'arrow' are linked via the phrase 'took an arrow to the knee' – MMAdams Oct 20 '16 at 18:33
• I fixed the bullseye. Replaced something and it seems that in one of all the changes I was doing, I missed one word. Now it has been added. Sorry for the inconvenience. – Alenanno Oct 20 '16 at 20:48
• @Alenanno It looks like you removed GROUP and SCHEDULE and added SOFTWARE and PLANE. Is that correct? – GentlePurpleRain Oct 20 '16 at 20:56
• @GentlePurpleRain Yes. I have included the text list in the main question and also added a hint for the hardest association. It will be helpful. Sorry again for the mess! – Alenanno Oct 20 '16 at 20:59

Here's what I came up with and I only traverse inwards. Though, I'm not sure if I misused rule #4:

1. Master -> Gowan using #5 (celebrity; "Gowan" is a popular singer/song-writer)
2. Gowan -> Gown using #3 (removing a letter)
3. Gown -> Ball using #4 (word association; you wear a "gown" to a "ball" or dance)
4. Ball -> String using #4 (word association; a "ball" of "string")
5. String -> Fringe using #4 (word association; decoration of threads or "string" on the edge of clothing is called a "fringe")
6. Fringe -> Hair using #2 ("fringe" is another word for bangs as in the "hair" covering your forehead)

I also wondered if you could make associations like this:

1. Master -> Gowan using #5 (Gowan is a popular singer/song-writer)
2. Gowan -> Group using #4 (word association; Gowan is the lead singer of the "group" Styx)
3. Group -> Stake using #4 (word association; a "stake" is a stick ("Styx"))

Or is that too much of a stretch?

EDIT: Perhaps I misread the puzzle. We have to use every single word?

• Yes, every single word. Of course, the rules seem to allow any association possible, but if you're missing words in your list or if some associations are really stretched, I'd suggest revisiting. – Alenanno Oct 20 '16 at 16:48
• I fixed the bullseye. Replaced something and it seems that in one of all the changes I was doing, I missed one word. Now it has been added. Sorry for the inconvenience. – Alenanno Oct 20 '16 at 20:49

Start:
Master

1: Stream an anagram of Master (@Beastly-Gerbil)
2: River a synonym of Stream
3: Driver is River with a D added
4: Golf a Golf Driver is a kind of golf club
5: Ball, a Golf Ball
6: Gown, a Ball Gown
7: Gowan, Gown with an A added
8: Wagon, an anagram of Gowan
9: Horse, a Horse and Wagon
10: Shore, an anagram of Horse
11: Sea, a Sea shore
12: Poseidon, the god of the Sea
13: Poisoned, an anagram of Poseidon
14: Antidote, a cure for one who is poisoned
15: Drop, a drop of Antidote
16: Prod, anagram of Drop

Some of the middle is definitely:

Guitar, String, Sting, Swing, Wing
Stage, Stake, Keats, Urn
Faction, Fraction, Entirety, Eternity
Bee, Nee, Knee
Leaf, Flea, RHCP
Force, Fore
Genre, Green
Row, Bow, Arrow
Opposite, Attraction

Not sure of all middle steps but the end could be:

Finger, what someone uses to Prod
Fringe, an anagram of Finger
Hair, Fringe is another word for bangs, which are made of Hair

• I fixed the bullseye. Replaced something and it seems that in one of all the changes I was doing, I missed one word. Now it has been added. Sorry for the inconvenience. – Alenanno Oct 20 '16 at 20:48

First one is

MASTER is an anagram of STREAM on the next row

• I fixed the bullseye. Replaced something and it seems that in one of all the changes I was doing, I missed one word. Now it has been added. Sorry for the inconvenience. – Alenanno Oct 20 '16 at 20:49


text:
# Circle Touches both x-axis and y-axis

We will learn how to find the equation of a circle touches both x-axis and y-axis.

The equation of a circle with centre at (h, k) and radius equal to a, is (x - h)$$^{2}$$ + (y - k)$$^{2}$$ = a$$^{2}$$.

When the circle touches both x-axis and y-axis i.e., h = k = a.

Then the equation (x - h)$$^{2}$$ + (y - k)$$^{2}$$ = a$$^{2}$$ becomes (x - a)$$^{2}$$ + (y - a)$$^{2}$$ = a$$^{2}$$

If a circle touches both the co-ordinate axes then the abscissa as well as ordinate of the centre will be equal to the radius of the circle. Hence, the equation of the circle will be of the form:

(x - a)$$^{2}$$ + (y - a)$$^{2}$$ = a$$^{2}$$

⇒ x$$^{2}$$ + y$$^{2}$$ - 2ax - 2ay + a$$^{2}$$ = 0

Solved example on the central form of the equation of a circle touches both x-axis and y-axis:

1. Find the equation of a circle whose radius is 4 units and touches both x-axis and y-axis.

Solution:

Radius of the circle = 4 units.

Since, the circle touches both x-axis and y-axis the centre of the circle is (4, 4).

The required equation of the circle whose radius is 4 units and touches both x-axis and y-axis is

(x - 4)$$^{2}$$ + (y - 4)$$^{2}$$ = 4$$^{2}$$

⇒ x$$^{2}$$ - 8x + 16 + y$$^{2}$$ - 8y + 16 = 16

⇒ x$$^{2}$$ - 8x - 8y + 16 = 0

2. Find the equation of a circle whose radius is 8 units and touches both x-axis and y-axis.

Solution:

Radius of the circle = 8 units.

Since, the circle touches both x-axis and y-axis the centre of the circle is (8, 8).

The required equation of the circle whose radius is 8 units and touches both x-axis and y-axis is

(x - 8)$$^{2}$$ + (y - 8)$$^{2}$$ = 8$$^{2}$$

⇒ x$$^{2}$$ - 16x + 64 + y$$^{2}$$ - 16y + 64 = 64

⇒ x$$^{2}$$ + y$$^{2}$$ - 16x - 16y + 64 = 0

`

The Circle


text:
Last edited by Arashikinos
Sunday, August 2, 2020 | History

2 edition of Maxwell equations. found in the catalog.

Maxwell equations.

Sybren Ruurds de Groot

# Maxwell equations.

## by Sybren Ruurds de Groot

Written in English

Subjects:
• Maxwell equations

• Edition Notes

Bibliography: p. 162-167.

The Physical Object ID Numbers Series Studies in statistical mechanics, v. 4 Pagination 179 p. ; Number of Pages 179 Open Library OL19195905M

Designed for upper division electromagnetism courses or as a reference for electrical engineers and scientists, this book introduces Maxwell’s equations and electromagnetic waves as soon as possible (i.e., in the first third of the book), and then goes on to discuss electrostatics, magnetostatics, induction, etc., in the light of those equations. The book also provides a thorough discussion.   The differential form of Maxwell’s Equations (Equations \ref{m_e1}, \ref{m_e2}, \ref{m_e3}, and \ref{m_e4}) involve operations on the phasor representations of the physical quantities. These equations have the advantage that differentiation with respect to time is replaced by multiplication by $$j\omega$$.

An Essential Guide to Maxwell’s Equations first reviews the Ranada field line solutions of Maxwell’s equations in a vacuum, describing a topologically non-trivial electromagnetic field, as well as their relation with the knot theory. Also, the authors present a generalization of these solutions to the non-linear electrodynamics recently published in the literature. Maxwell equations are solved for spheres or ellipsoids with or without substrate [2, ].However, analytical solutions have not been obtained for other complicated structures such as triangular, tetrahedral particles with/without unsymmetrical protrusions or pit, because isolation of variables in the differential equations is substantially difficult.

Maxwell's Equations. We have so far established that the total flux of electric field out of a closed surface is just the total enclosed charge multiplied by 1 / ε 0, ∫ E → ⋅ d A → = q / ε 0. This is Maxwell’s first equation. It represents completely covering the surface with a large number of tiny patches having areas d A →. We. Maxwell’s equations are paraphrased here in words because their mathematical statement is beyond the level of this text. However, the equations illustrate how apparently simple mathematical statements can elegantly unite and express a multitude of concepts—why mathematics is the language of science.

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### Maxwell equations by Sybren Ruurds de Groot Download PDF EPUB FB2

A Student's Guide to Maxwell's Equations by D. Fleisch- focuses attention on the four Maxwell's equations in pages. This book consists of five chapters, and the first four chapters deal with four Maxwell's equations, respectively.

The final chapter is about inducing electric and magnetic wave equations from Maxwell's by:   The book is highly detailed on most of the math involved with Maxwell's equations.

However, this doesn't seem to be a decent book for most undergraduate students to reference. The math is advanced, it requires previous knowledge of the topic, and on a few parts, the author seems to stumble on about how students fail to understand various Cited by: 9.

The direction of the emf opposes the change. This third of Maxwell’s equations, Equationis Faraday’s law of induction and includes Lenz’s law. The electric field from a changing magnetic field has field lines that form closed loops, without any beginning or end.

Ampère-Maxwell law. Most books have incorrect Equations. Here is the correct “Maxwell's Equations” Here are fields, B=zD=uH=E/c X=[d/dr, Del]=[d/cdt,Del] XE=[d/dr,Del][e,E] XE=[de/dr- Del.E,dE/dr + Del e + DelxE] Maxwell's Equation is Stationary Equation 0=XE=[de/dr.

Maxwell deals with the motion-related aspect of electromagnetic induction, v × B, in equation (77), which is the same as equation (D) in Maxwell's original equations as listed below. It is expressed today as the force law equation, F = q (E + v × B), which sits adjacent to Maxwell's equations and bears the name Lorentz force, even though.

Welcome to the website for A Student’s Guide to Maxwell’s Equations, and thanks for visiting. The purpose of this site is to supplement the material in the book by providing resources that will help you understand Maxwell’s Equations. Maxwell equations.

book On this site, you’ll find: Complete solutions to every problem in the book. Maxwell’s equations describe all (classical) Maxwell equations. book phenomena: 0/n, a relationship that we will use extensively in this book.

More generally, constitutive relations may be inhomogeneous, anisotropic, nonlin-ear, frequency dependent (dispersive). Maxwell’s Equations and the Principles of Electromagnetism. ISBN: The publisher recognizes and respects all marks used by companies, manufacturers, and developers as a means to distinguish their brand names and product names mentioned in this book are trademarks or service marks of their respective companies.

Collectively they are known as Maxwell’s equations. The above equations may also be written in differential forms as 0 0 t t ρ ε µµε ∇⋅ = ∂ ∇× =− ∂ ∇⋅ = ∂ ∇× = + ∂ E B E B E BJ G G G G G GG () where ρand are the free charge and the conduction current densities, respectively.

In the absence of sources. Elementary Electromagnetic Theory Volume 3: Maxwell’s Equations and their Consequences is the third of three volumes that intend to cover electromagnetism and its potential theory. The third volume considers the implications of Maxwell's equations, such as electromagnetic radiation in simple cases, and its relation between Maxwell's equation.

This book was developed at Simon Fraser University for an upper-level physics course. Along with a careful exposition of electricity and magnetism, it devotes a chapter to ferromagnets. According to the course description, the topics covered were “electromagnetics, magnetostatics, waves, transmission lines, wave guides,antennas, and radiating systems.”.

Maxwell's equations are a set of four differential equations that form the theoretical basis for describing classical electromagnetism: Gauss's law: Electric charges produce an electric field. The electric flux across a closed surface is proportional to the charge enclosed. Gauss's law for magnetism: There are no magnetic monopoles.

The magnetic flux across a closed surface is zero. Designed for upper division electromagnetism courses or as a reference for electrical engineers and scientists, this book introduces Maxwell’s equations and electromagnetic waves as soon as possible (i.e., in the first third of the book), and then goes on to discuss electrostatics, magnetostatics, induction, etc., in the light of those by:   Hooke's law at equation (1) in section I became Maxwell's fifth equation in the original list of eight "Maxwell's Equations" in his paper "A Dynamical Theory of.

LEVEL: 🧠🧠⚪⠀ 🕒 understand Maxwell-Equations within 40 minutes. Maxwell equations can be written in a form independent of the coordinates by describing the fields as differential forms.

A very accessible introduction is given in [4, 5]. We give here a succinct introduction to the matter. We will only define them in an affine Euclidean space of dimension n.

Maxwell's Equations - Physics Joke, Men's Tshirt. out of 5 stars 1.  FREE Shipping on your first order shipped by Amazon. Audible Listen to Books & Original Audio Performances: Book Depository Books With Free Delivery Worldwide: Box Office Mojo Find Movie Box Office Data.

I revere Maxwell's Equations. I don't recommend this as a starter book for Maxwell's equations. Just read a typical high school physics textbook that does not shy away from Maxwell (AP text, w/ calculus). E&M culminates with Maxwell. Now that you've met Maxwell's Equations, learn their depth by this book, aimed at the student/5.

The final chapter shows how Maxwell’s equations may be combined to produce the wave equation, the basis for the electromagnetic theory of light. This book is a wonderful resource for undergraduate and graduate courses in electromagnetism and electromagnetics.

Maxwell's Equations. Gauss’s law. The electric flux through any closed surface is equal to the electric charge $$Q_{in}$$ enclosed by the surface. Gauss’s law (Equation \ref{eq1}) describes the relation between an electric charge and the electric field it produces. This book arose from a lecture on Maxwell’s equations given by the authors between??

and The emphasis is put on three topics which are clearly structured into Chapters 2,??, and??. In each of these chapters we study ﬁrst the simpler scalar case where we replace the Maxwell system by the scalar Helmholtz equation.A Student's Guide to Maxwell's Equations by Daniel Fleisch Book Resume: Gauss's law for electric fields, Gauss's law for magnetic fields, Faraday's law, and the Ampere–Maxwell law are four of the most influential equations in science.

In this guide for students, each equation is the subject of an entire chapter, with detailed, plain-language.Maxwell’s Equations A dynamical theory of the electromagnetic field James Clerk Maxwell, F.

R. S. Philosophical Transactions of the Royal Society of London, published 1 January


text:
## International Journal of Differential Equations

### Bounds for Products of Zeros of Solutions to Nonhomogeneous ODE with Polynomial Coefficients

Michael Gil’

#### Abstract

We consider the equation ${u}^{″}=P(z)u+F(z) (z\in \mathbb{C})$, where $P(z)$ is a polynomial and $F(z)$ is an entire function. Let ${z}_{k}(u) (k=\mathrm{1,2},\dots )$ be the zeros of a solution $u(z)$ to that equation. Lower estimates for the products ${\prod }_{k=\mathrm{1}}^{j}|{z}_{k}(u)| (j=\mathrm{1,2},\dots )$ are derived. In particular, they give us a bound for the zero free domain. Applications of the obtained estimates to the counting function of the zeros of solutions are also discussed.

#### Article information

Source
Int. J. Differ. Equ., Volume 2015 (2015), Article ID 690519, 6 pages.

Dates
Revised: 22 October 2015
Accepted: 29 October 2015
First available in Project Euclid: 20 January 2017

https://projecteuclid.org/euclid.ijde/1484881449

Digital Object Identifier
doi:10.1155/2015/690519

Mathematical Reviews number (MathSciNet)
MR3429206

Zentralblatt MATH identifier
1339.34094

#### Citation

Gil’, Michael. Bounds for Products of Zeros of Solutions to Nonhomogeneous ODE with Polynomial Coefficients. Int. J. Differ. Equ. 2015 (2015), Article ID 690519, 6 pages. doi:10.1155/2015/690519. https://projecteuclid.org/euclid.ijde/1484881449

#### References

• I. Laine, Nevanlinna Theory and Complex Differential Equations, Walter de Gruyter, Berlin, Germany, 1993.MR1207139
• B. Belaidi and A. El Farissi, “Differential polynomials generated by some complex linear differential equations with meromorphic coefficients,” Glasnik Matematicki, vol. 43, no. 2, pp. 363–373, 2008.MR2460705
• T. B. Cao and H. X. Yi, “On the complex oscillation theory of ${f}^{''}+A(z)f=0$ where $A(z)$ is analytic in the unit disc,” Mathematische Nachrichten, vol. 282, no. 6, pp. 820–831, 2009.
• T. B. Cao and H. X. Yi, “Complex oscillation theory of linear differential equations with analytic coefficients in the unit disc,” Acta Mathematica Scientia Series A, vol. 28, no. 6, pp. 1046–1057, 2008.MR2517355
• T.-B. Cao, K. Liu, and H.-Y. Xu, “Bounds for the sums of zeros of solutions of $u(m)=P(z)u$ where P is a polynomial,” Electronic Journal of Qualitative Theory of Differential Equations, no. 60, pp. 1–10, 2011.
• A. E. Farissi and B. Belaïdi, “On oscillation theorems for differential polynomials,” Electronic Journal of Qualitative Theory of Differential Equations, no. 22, pp. 1–10, 2009.
• M. Gaudenzi, “On the number of the zeros of solutions of a linear differential equation,” Journal of Mathematical Analysis and Applications, vol. 221, no. 1, pp. 306–325, 1998.
• C. Z. Huang, “Real zeros of solutions of second order linear differential equations,” Kodai Mathematical Journal, vol. 14, no. 1, pp. 113–122, 1991.MR1099319
• Z. Latreuch, B. Belaïdi, and A. El Farissi, “Complex oscillation of differential polynomials in the unit disc,” Periodica Mathematica Hungarica, vol. 66, no. 1, pp. 45–60, 2013.
• C. H. Lin, Y. Sibuya, and T. Tabara, “Zeros of solutions of a second order linear differential equation with polynomial coefficients,” Funkcialaj Ekvacioj, vol. 36, no. 2, pp. 375–384, 1993.MR1245345
• G. M. Muminov, “On the zeros of solutions of the differential equation ${\omega }^{(2m)}+p(z)\omega =0$,” Demonstratio Mathematica, vol. 35, no. 1, pp. 41–48, 2002.
• J. Tu and Z.-X. Chen, “Zeros of solutions of certain second order linear differential equation,” Journal of Mathematical Analysis and Applications, vol. 332, no. 1, pp. 279–291, 2007.
• J. F. Xu and H. X. Yi, “Solutions of higher order linear differential equations in an angle,” Applied Mathematics Letters, vol. 22, no. 4, pp. 484–489, 2009.MR2502241
• M. I. Gil', “Bounds for zeros of solutions of second order differential equations with polynomial coefficients,” Results in Mathematics, vol. 59, no. 1-2, pp. 115–124, 2011.MR2772181
• M. I. Gil, “Sums of zeros of solutions to second order ODE with non polynomial coefficients,” Electronic Journal of Differential Equations, vol. 2012, no. 107, pp. 1–8, 2012.
• M. I. Gil, “Sums of zeros of solutions to non-homogeneous ODE with polynomial coefficients,” Journal of Mathematical Analysis and Applications, vol. 421, no. 2, pp. 1917–1924, 2015.
• Y. L. Daleckii and M. G. Krein, Stability of Solutions of Differential Equations in Banach Space, American Mathematical Society, Providence, RI, USA, 1971.
• M. I. Gil, “Inequalities for products of zeros of polynomials and entire functions,” Mathematical Inequalities and Applications, vol. 15, no. 4, Article ID 15-70, pp. 827–833, 2012.
• M. I. Gil', “Inequalities for zeros of entire functions,” Journal of Inequalities, vol. 6, no. 4, pp. 463–471, 2001.MR1888437
• M. Gil', Localization and Perturbation of Zeros of Entire Functions, CRC Press, Taylor and Francis Group, New York, NY, USA, 2010.MR2589816 \endinput


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## anonymous one year ago In the coordinate plane, three vertices of trapezoid STUV are S(0, 0), T(a, c), and V(d, 0). If point U is halfway between points S and V, what are the coordinates of point U?

• This Question is Open
1. Mehek14

first we find the distance between S and V do you know the distance formula @Erion ?

2. anonymous

$\huge~\rm~\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$


text:
# Tag Info

2

There are ${^{10}C_{5}}$ ways to choose 5 of 10 people. That equals $252$. $${^{10}C_{5}} = \dfrac{10!}{5!\,5!} = 252$$ You want to count the ways to choose 2 of 4 women and 3 of 6 men. $${^{\Box}C_{\Box}}\cdot{^{\Box}C_{\Box}} = \dfrac{\Box!}{\Box!\,\Box!}\dfrac{\Box!}{\Box!\,\Box!} = \Box$$ Can you not fill in the boxes? Thank you, I can fill in ...

0

Bayes' Theorem states that: \begin{align} \mathsf P(A\mid B, C) & = \dfrac{P(B\mid A, C)\;P(A\mid C)}{P(B\mid C)} \\[2ex] \therefore p(\theta \mid \mathbf X,\alpha) & = \dfrac{p(\mathbf X\mid \theta, \alpha)\; p(\theta\mid \alpha)}{p(\mathbf X\mid \alpha)} \end{align} Now $\mathbf X$ is a vector of data points $x_i$ each from a distribution ...

2

There are $7$ students, and the number of project assignments is $4+5+6=15$. Since every student did at least $2$ (presumably different!) projects, one overburdened student did $3$, and the others did $2$ each. The student who did $3$ can be chosen in $\binom{7}{1}$ ways. Now we must distribute the remaining projects, $2$ of each, between the remaining $6$ ...

1

$$D = b^2-4a = (4k)^2 - 4 \cdot 4 \cdot (k+2) = 16(k^2 - k - 2) = 16(k-2)(k+1)$$ We need $D \ge 0$ for real roots, which happens when $k \ge 2$ or $k \le -1$. Can you take it from here?

-1

It is much easier to prove if you consider the intersection of events ($\cap$). This is associative and commutative just like the set intersection operator. For any events A, B: $p(A|B) = \dfrac{p(A \cap B)}{p(B)}$ So we have $p(\theta | X,\alpha) = \dfrac{p(\theta \cap (X \cap \alpha))}{p(X \cap \alpha)} = \dfrac{p((\theta \cap X) \cap \alpha)}{p(X \cap ... 0 In the worst case, the machine does not work$365{\tiny .25}$days in the year. At least one component fails every day, which has probability$(1 - 0.99^3)^{356.25} = 1.5\times 10^{-558}$In the best case, the machine does not work$0$days in the year. No component fails on any day, which has probability$0.99^{3\cdot365.25} = 1.6\times 10^{-5}On the ... 2 What you are doing is correct, except there was a typo where the '2' should be a '1'. \begin{aligned} M(t)&=E(e^{tx})=\int_{-1}^{\infty} e^{tx} e^{-x-1} dx \\ &=e^{-1}\int_{-1}^{\infty} e^{(t-1)x} dx \\ &=e^{-1}\frac{1}{t-1}\left. e^{(t-1)x} \right|_{-1}^\infty \\ &=e^{-1}\frac{e^{1-t}}{1-t} \\ &=\frac{e^{-t}}{1-t} \\ \end{aligned} ... 0 You seem a little confused about terminology in general and the binomial and Poisson distributions in particular. The cure for that is to go back in your text or lecture notes to where these ideas are explained. (You will rarely find an exact formula that is 'tailor made' to answer a particular probability problem so that you can plug into the formula ... 2 As the comments mentioned the worst case scenario is it never works, best case is it works every day. Some information that may be of interest: Let A be the event that A works (thus A^c means it does not work), and M be that the machine works (define similarly for B,C). Assuming A,B, and C are independent, on any given day the chance the ... 0 It's a way of gathering things that sum to a constant. For example, \begin{align} \sum_{n=0}^{\infty}a_{n}z^{n}\sum_{n=0}^{\infty}b_{n}z^{n} & = \sum_{n=0}^{\infty}\left(\sum_{j+k=n}a_{j}b_{k}\right) z^{n} \\ & = \sum_{n=0}^{\infty}\left(\sum_{j=0}^{n}a_{j}b_{n-j}\right)z^{n}. \end{align} A Fourier transform can be viewed as an integral ... 0 In the case of n persons the graph of possible friendships is made of {n \choose 2} pairs. For each edge e =\{a,b\} of this graph we consider a random variable X_e if X_e = 1 this means that they are friends. If X_e=0 they are not friends. (a,b are two individuals of our population). We assume that X_e are independent random variables and ... 3 Hint: E[X|X+Y=1] = E[Y|X+Y=1]. Now use additivity of conditional expectations. Of course this assumes these conditional expectations exist. 1 It seems like you may be referring to the Law of Large Numbers. The Wikipedia page linked to gives a good explanation, from what I can see. See also: the relevant part of this SE answer for more information on the theorem, and why it is important, this SE question for another good, layman oriented explanation of the theorem, the top-rated answer to this SE ... 0 P(A_1\cup A_2 \cup A_3)= P(A_1\cup A_2) + P(A_3) - P((A_1\cup A_2)\cap A_3 ) =P(A_1\cup A_2) + P(A_3) - P(A_1\cup A_2).P(A_3 ) ..since (A_1\cup A_2) and (A_3) are independent Similarly, P(A_1\cup A_2)=P(A_1)+P(A_2)-P(A_1\cap A_2)=P(A_1)+P(A_2) - P(A_1).P(A_2) Now you can just do the calculation. I guess the final answer is 3/4 2 Possibly you're thinking of the Poisson distribution. Suppose you have a one-in-1{,}000{,}000 chance of success on each trial, and there are 3{,}600{,}000 trials. The expected number of successes is then 3.6. If we ask for the probability that there are exactly 5 successes, we get \frac{3.6^5 e^{-3.6}}{5!} = \frac{3.6^5 e^{-3.6}}{120} \approx ... 0 If you want a set of outcomes with equal probability, you need to account somehow for the fact that when the contestant chooses the prize door on the first guess, Monty has a choice of which door to open. This is best modeled as a random variable with two equally likely outcomes. We might as well assume Monty flips a coin and opens the leftmost possible ... 0 The number of subsets of a set withn$elements is$2^n$since each element is either in the subset or it is not. Each of the$2^4 = 16$subsets of$A$except the empty set could be the range of$f: A \to A$. Hence, there are$15$possible ranges for$f$. There are also$15$possible ranges for$g: A \to A$. Hence, there are$15^2 = 225$possible pairs ... 0 I think you might be able to follow this route ($q=1-p$):$\sum_k {N\choose k} (xp)^k (1-p)^{N-k}=(xp+q)^nf(z)=(zp+q)^n(\frac{p}{z}+q)^n$has your sum$\sum_k {N\choose k}^2 p^{2k} (1-p)^{2N-2k}$as the free term (no$z$) Your sum is the integral$\frac{1}{2\pi i}\int\frac{f(z)}{z}dz$over the unit circle.$z=e^{2\pi i t}$reduces the integral to ... 1 "Randomly chosen" isn't clearly defined, but let's take it to mean that we randomly choose the values of$f(1),f(2),f(3), f(4)$and similarly for$g$. We'll work case by case, indexed by the size of the range of$f$. I will not complete the calculation, it's a bit messy and I'll leave some of the arithmetic off. Case I: Range of$f$has exactly 1 element. ... 0$\newcommand{\var}{\operatorname{var}}$The conditional distribution of$X_D-X_{\overline D}$given the event$C=c$is normal with expected value$\Big(\mu+g(c)\Big) - g(c) = \mu$and variance$1+1=2, since \begin{align} \var(X_D - X_{\overline D} \mid C=c) & = \var(X_D \mid C=c) + \var( -X_{\overline D} \mid C=c) \\[10pt] & = \var(X_D \mid C=c) + ... 3 I believe that you are either misinterpreting the video, the video is wrong, or something else is going on because, if\sum_{i = 0}^{n} \binom{n + i}{i} = \binom{2n + 1}{n + 1}$and$\sum_{i = 0}^{n} \binom{n + i}{i} = \binom{n + k + 1}{k}$for$0 \leq k \leq n$, that would imply$\sum_{i = 0}^{n} \binom{n + i}{i} = \binom{2n + 1}{n + 1} = \binom{n + k ...

0

Let $X$ be a random variable with density function $f_X(x)$, $g(x)$ be a monotone increasing function of $x$, and $Y = g(X)$. We seek the density function $f_Y(y)$ of $Y.$ In this instance it can be shown that $$f_Y(y) = f_X(g^{-1}(y)) \frac{dg^{-1}(y)}{dy},$$ where $g^{-1}$ is the inverse function of $g$. In your Question, $X \sim Exp(1),\;$$g(x) = ... 1 One more explanation, given a CS or CS-like background: Consider the set of all n-bit numbers, from 0 through 2^n-1, inclusive; there are thus 2^n of them, by inspection. (For instance, there are 2^3 = 8 numbers from 0 through 2^3-1 = 7: 0, 1, 2, 3, 4, 5, 6, 7.) We could, also, count them as follows: How many of the numbers have no 1 bits? ... 4 Another explanation is to observe that, since \,\dbinom nk is the number of subsets of k elements in a set of n elements, the sum of all these coefficients (n fixed) is but the number of all subsets of a set of n elements – which is known to be 2^n. 7 It's from the Binomial Theorem, expanding (1+1)^n 2 There is no coincidence here. We learn nothing from A about X only if the coin is fair (otherwise the fact that both tosses produced the same outcome enhances our belief that the outcome was the more likely one) or "one-sided" (i.e., P(X=0)=0 or P(X=1)=0). When the coin is not fair, the mutual information of A and X is non-zero: if P(X=0)=p, ... 0 You can use the approximate confidence interval for a proportion. In general it is used for normally distributed variables. But you have 3333 identical and independent distributed (iid) random variables. So you can apply the central limit theorem. The two sided limits are \Large{\left[\hat p-z_{1-\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat p\cdot (1-\hat ... 0 If a and b are independent there is a very simple solution. First convince yourself that$$a + b \leq C \Leftrightarrow \mathrm{max}(a,b) \leq \frac{C}{2}$$Then due to independence, P\{\mathrm{max}(a,b) \leq \frac{C}{2}\} = P\{a \leq \frac{C}{2}\}P\{b \leq \frac{C}{2}\} I am assuming both a and b have uniform distributions. Then, P\{a \leq ... 0 Can be done in \min(O(A\log B),O(B\log A)) using binary serach. 0 If anyone was curious, I was able to solve this using optimization by minimizing the objective function$$ (F(b) - F(a) - 0.95)^2 + (F(a) - (1 - F(b))^2 $$where F is the cdf. The first term requires that P(a \leq X \leq b) = 0.95 and the second requires that P(X < a) = P(X > b), thereby "centering" this 95\% interval on the median. MATLAB's ... 2 The probability is in \Omega\left(n^{-2}\right), and apparently also in O\left(n^{-2}\right) (and thus in \Theta\left(n^{-2}\right)). I'll focus on odd n because it makes things easier, but I believe with some slight complications similar considerations apply to even n. If we add another entry for j=-1 to S, we have S_{-1}=-S_{n-1}=:\sigma. ... 0 No. The following example is due to R. O. Davies, Separate approximate continuity implies measurability, Mathematical Proceedings of the Cambridge Philosophical Society, Vol. 73, Issue 03, May 1973, pp 461. Let \kappa be a real valued measurable cardinal and m: \mathcal{P}(\kappa) \to [0, 1] be a \kappa-additive diffused probability measure. Put ... -1 It suffices to use the following property of integrals$$f \leq g \Rightarrow \int f(y)\, dy \leq \int g(y)\, dy$$and note that$$1_{\{0 \leq y <t\}} y^2 Q(y)\leq t^2 1_{\{0 \leq y <t\}} Q(y)$$the same observation will allow you to solve the other inequalities. 0 Let there be n women. To meet the criteria, men must either be in two blocks of 4 & 2 and positioned in two of the (n+1) gaps between women (including the ends), or in a block of 6 positioned in the (n+1) gaps, i.e. in [(n+1)\cdot n + (n+1)] patterns, having 4!2! and 6! permutations respectively, thus$$\frac{[(n+1)\cdot n!]\cdot(n\cdot 4!\cdot2! ... -1 Let us assume that the men and the women are distinguishable. From the data, you must have EITHER a group of 6 men together, arranged amongst$n$women, OR a group of 4 men and a group of two men arranged so that there is at least one woman between these groups. In the first case, the number of arrangements is $$6!\times(n+1)!$$ In the second case the men ... 1 Suppose there are$x$women. I assume them all to be the same, like you seem to do in the question. Case 1. If there is a block of exactly 4 men, we have two possibilities. Case 1.1. If the block of 4 men is at the end or the beginning of the row, we only need one women enclosing it. So we have the block$B$consisting of 4 men and 1 woman, which can ... 1 Consider the men as$M, M,\ldots,Mto get 4 aligned and no men alone you need to form 2 groups, one with4$men and one with$2$men. let$G(4),G(2)$be the group with$4$and$2$men respectively. A favorable occurrence is of the form $$W(n_1) G(4) W(n_2) G(2) W(n_3)$$ $$W(n_1) G(2) W(n_2) G(4) W(n_3)$$ or $$W(N_1) G(4) G(2) W(N_2)$$ where ... 0 I think that they are almost ok but it depends on what do you exactly need. I see two potential problems among the kernel functions you mentioned (they arise because kernel functions are made for density estimation but you want something little bit different). 1) All of these functions are symetrical around$0$but you have distances that are always ... 2 I'll take the statement "All players are equally skilled." to mean that every player has a$50\%$chance to win against every other player.$7$players have been eliminated by$p_5$; the probability that$p_1$was not among them is$24/31$. If$p_1$has not been eliminated, the identity of the winner of$p_5$'s branch of the tournament is irrelevant to her; ... 0 The prime numbers is the set of numbers derived by removing the set of composite nunbers from the set of natural numbers. The pattern that the natural numbers follow is obvious and easily understood. The set of composite numbers also follow a discernable pattern:it is the union of the factors of 2, the factors of 3 etc. Therefore the prime numbers as ... 0 As Did pointed out, we have to assume that$X$and$Y$are independent. Otherwise, consider a symmetric non-degenerated random variable$X$,$(X_n)$and$(Y_n)$independent identically distributed sequences where$X_1$and$Y_1$are distributed as$X$. Then$X_n\to X$in distribution,$Y_n\to -X$in distribution, but the sequence$(X_n+Y_n)_{n\geqslant 1}$... 2 For both parts you need to calculate conditional probabilities. For a) let$B$denote the event that the biased coin is chosen at the final stage, and let$3H$denote the event of obtaining three heads. We require $$p(B|3H)=\frac{p(B\cap3H)}{p(3H)}=\frac{\frac 16\times\frac{27}{64}}{\frac 16\times\frac{27}{64}+\frac 56\times\frac 18}=\frac{27}{67}$$ For ... 0 Your calculation of the probability distribution of the absolute difference is correct:$\Pr(D=1) = \frac34\Pr(D=3) = \frac14$You could have used these to find the expected absolute difference:$E[D]=1\times \frac34 +3 \times \frac14 = \frac32 .1 \begin{align} \mathsf P(X < Y) &=\int_1^3 \left\{\int_0^y f_X(x) \operatorname dx\right\}g_Y(y) \operatorname dy\\ &\neq \tfrac{1}{4} \int_1^3 \int_0^y \operatorname dx \operatorname dy \end{align} Here's the problem. The inner integral's upper bound should be\min(2, y)$because the support for$X$is$(0;2)$. Watch out for ... 1 Split the inteval and use Baye's Theorem to get $$P(X<Y) = P(X<Y | X<1) P(X<1) + P(X<Y|X\geq1)P(X\geq 1))$$ sind$X$is uniform on$(0,2)$, we know that$P(X<1)=P(X>1) = \frac12$. Furthermore,$P(X<Y|X<1) = 1$since$Y$is uniform on$(1,3)$. This yields $$P(X<Y) = 1 \cdot \frac12 + P(X<Y|X\geq1)\frac12$$ Now$P(X<Y| ...

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If you want to solve the problem using integrals then you should notice that you have wrong upper limit in the inner integral. It should be min(y,2).

9

This sketch might help. You want the red area as a proportion of the red and blue areas.

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We .can draw the rectangle and it's interior $0 \leq x \leq 2$ and $1 \leq y \leq 3$. Then we can draw the line $y=x$. Let's look at our event. So we should draw the line $y=x$. Now, the region delimited is given by the triangle whose vertices are $(1,1)$, $(2,2)$ and $(2,1)$.The probanility is $\int\limits_{1}^{2}\int\limits_{1}^{x} \frac{1}{4}dydx$.

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Answer: Divide the regions of X with respect to Y for the condition $X<Y$. For $0<X<1$, $P(X<Y) = \frac{1}{2}$ For $1<X<2$ and $1<Y<2$ $P(X<Y) = \int_{1}^{2}\int_{x}^{2} \frac{1}{2}\frac{1}{2}dydx = \frac{1}{8}$ For $1<X<2$, and $2<Y<3$ $P(X<Y) = \frac{1}{2}.\frac{1}{2}=\frac{1}{4}$ Thus $P(X<Y) = ... 0 Not a complete answer, but too long for a comment: Technically, the definition of a median is any number$m$such that$\Pr(X\le m)\ge0.5$and$\Pr(X\ge m)\ge0.5$. Note that the first one may also be written as$1-\Pr(X>m)\ge0.5\$, with the inequality being strict. If your random variable is continuous, then the strictness of the inequality doesn't ...

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