text:
# The Football Fest || FIFA One

## Problem Statement

The very famous football club Manchester United decided to popularize football in India by organizing a football fest. The fest had many events for different sections of people.
For the awesome coders of Hacker Earth, there was an event called PASS and BACK. In this event, the coders were given N passes and players having ids between 1 and 1000000.
Initially, some player with a given id had the ball in his possession. The coders had to make a program to display the id of the player who possessed the ball after exactly N passes.

Description of the passes:
There were two kinds of passes:
1. P ID
2.

## Explanation :

for the first kind of pass, the player in possession of the ball passes the ball to player with id=ID while for the second kind of a pass, the player in possession of the ball passes the ball back to the player who had passed the ball to him.

## NOTE:

It is guaranteed that the given order of all the passes will be a valid order.

## INPUT :

The first line of the input contains the number of test cases. For each test case, two space-separated integers Nand ID ( of the player possessing the ball in the very beginning). N lines follow describing the passes. ( for a description of the passes, refer the statement above. )

## OUTPUT :

Output to each test case should be a single line containing the “Player” ID (quotes for clarity) of the player who possesses the ball after N passes.

1≤T≤100
1≤N≤100000

## SAMPLE INPUT

1
10 23
P 86
P 63
P 60
B
P 47
B
P 99
P 9
B
B

## SAMPLE OUTPUT

Player 9

## Explanation

Initially, Player having id=23 posses ball. After pass 1, Player having id=86 posses ball. After pass 2, Player having id=63 posses ball. After pass 3, Player having id=60 posses ball. After pass 4, Player having id=63posses ball. After pass 5, Player having id=47 posses ball. After pass 6, Player having id=63 posses ball. After pass 7, Player having id=99 posses ball. After pass 8, Player having id=9 posses ball. After pass 9, Player having id=99 posses ball. After pass 10, Player having id=9 posses ball.

________________________________________________________________________________

# SOLUTION

#include <bits/stdc++.h>

using namespace std;

stack < int > s;

int main()
{
ios::sync_with_stdio(false);cin.tie(0);          // Decrease Time of cin, cout.

int n , t, d, id;
char c;

cin >> t;

while(t--){

cin >> n >> d;

while(!s.empty()) s.pop();      // clear the stack in each test case

s.push(d);

for(int i = 0; i < n; i++){

cin >> c;

if(c == 'B'){

int second = s.top();   // pop 2 elements
s.pop();

int first = s.top();
s.pop();

s.push(second);     // push the 2 elements in reverse order
s.push(first);
continue;
}

cin >> id;
s.push(id);         // normal pushing in the stack
}

cout << "Player " << s.top() << "\n";
}

return 0;
}

## Brief Description:

Given two space separated integers N and ID ( of the player possessing the ball in the very beginning).
N lines follow describing the passes.

1. P ID
2. B

For the first kind of pass, the player in possession of the ball passes the ball to player with id=ID while for the second kind of a pass, the player in possession of the ball passes the ball back to the player who had passed the ball to him.

Print the “Player” ID (quotes for clarity) of the player who possesses the ball after N passes.
(The top value on the Stack of the passes).

## Detailed Editorial:

This Question depends on your data structure knowledge as you can easily solve this
question using suitable data structure which is a stack in our question.

You have 2 types of instructions in this question:

1) PID:

So you can easily push the new ID in the stack to keep track of the latest player
who has the ball.

2) B:

Now you should pass the ball to the previous player so you can easily pop the last
player from the stack, but you must first push the current player, then the previous player
to make the previous player at the top of the stack.

Note: In the B instruction: you should also push the current player before pushing the previous
player because you want also to update the current player to be the penultimate player.

## Memory Space Complexity:O(N)O(N).

{Code With Codeater}


text:
This website was created for the predictions for the upcoming Congressional Elections made by the Political Statistics class at Montgomery Blair High School in Silver Spring, Maryland. Under the guidance of Mr. David Stein, this model (which we named the Overall Results of an Analytical Consideration of the Looming Elections a.k.a. ORACLE of Blair) was developed by a group of around 70 high school seniors, working diligently since the start of September. Apart from the youth and enthusiasm that went into making it, the advantage our model has over professionally developed models is transparency. Unlike professionals, we need not have any secrets in regards to how our predictions are generated. In fact, the sections that follow attempt to detail exactly how we come up with all of the numbers involved in our model. If you are interested by politics, statistics, education, or just agree with our predictions, please tell your friends and social media followers about the work that we’ve done.

# Preliminaries

First, some conventions used throughout the explanation of our methodology. All calculations made are based on the two party vote, meaning that any votes for a third-party or independent candidate do not count. For example, in Oregon's 3rd Congressional District, we say that in 2014 the Democratic vote percentage was $$78.7\%$$, even though the Democratic candidate only got $$72.3\%$$ of the actual vote. The $$78.7\%$$ represents the ratio of votes cast for the Democrat to the total votes cast for either the Republican or the Democrat, hence it’s a two-party vote percentage. The two-party vote percentage differs from the actual vote percentage as some votes are not counted due to the votes being cast for a third-party or an independent candidate.

Another convention used is that, when using a metric centered at $$0$$, positive numbers mean a Democratic advantage and negative numbers mean a Republican advantage. For example, when talking about margin in the elections, we usually look at the Democratic margin, as opposed to the Republican Margin. A margin of $$10\%$$ means that the Democratic candidate received $$10\%$$ more of the two-party vote than the Republican candidate. This means that the Democrat got $$55\%$$ of the two-party vote. Going in the other direction, a Democratic margin of $$-20\%$$ means that the Democratic candidate received $$20\%$$ less of the two-party vote than the Republican Candidate. This means that the Republican candidate prevailed with $$60\%$$ of the two-party vote.

### BPI and SEER

The Blair Partisan Index (BPI) is a metric that measures a district’s partisan voting tendencies relative to the nation. In order to calculate this, we subtract the national Democratic vote percentage from a district’s Democratic vote percentage for four recent elections: the 2012 and 2016 Presidential elections and the 2014 and 2016 House elections. The subtraction leads to the relative Democratic vote percentages of a district. For example, in AR-02 during the 2016 House election, the district cast $$39\%$$ of their two-party vote for the Democratic nominee. In comparison, the nation as a whole cast $$50.57\%$$ of their two-party vote for a Democratic nominee.

To be consistent with our predicted 2018 incumbency advantage (weights for SEER), we adjusted 2014 and 2016 House vote proportions based on incumbency in those years. Specifically, if there was a Democratic incumbent running we subtracted $$3.45\%$$ ($$6.90\%/2$$) from the Democratic vote proportion and if there was a Republican incumbent running we added $$4.09\%$$ ($$8.17\%/2$$) to the Democratic two-party vote proportion. So in 2016, there was a Republican incumbent running for the House for AR-02. Therefore, we added $$4.09\%$$ to the district's Democratic vote percentage leading to an adjusted vote percentage of $$39\% + 4.09\% = 43.09\%$$ for the 2016 House election. We then take the adjusted vote percentage and subtract it from the national average of $$50.57\%$$ to get a relative 2016 House vote percentage of $$43.09\% - 50.57\% = -7.48\%$$. The BPI is a weighted average of the relative vote percentages, weighted by the table below:

 Relative 2012 Presidential Vote Weight: 0.133 Relative 2014 House Vote Weight: 0.278 Relative 2016 Presidential Weight: 0.244 Relative 2016 House Weight: 0.345

The BPI of a district with relative vote percentages $$v_1, v_2, v_3, v_4$$ is $$0.133v_{1}+0.278v_{2}+0.244v_{3}+0.345v_{4}$$. So for AR-02, the BPI would be $$0.133(-8.01\%) + 0.278(-1.35\%) + 0.244(-6.79\%)+0.345(-7.48\%)$$, which is $$-5.68\%$$.

Districts in which the 2014 and/or 2016 House elections did not have both a Democratic candidate and a Republican candidate pose a problem, since only one major party has a vote total, and that vote total is not meaningful. Therefore, if a past race was unopposed we did not use it in calculating that district's BPI. That is to say, we calculated the BPI as the weighted average of only the races with candidates from both major parties, using the above weights for each election and dividing the weighted sum by the sum of the weight for the elections used. As an example, consider AR-04, where there was no Democratic candidate for the House seat in 2016. AR-04 has relative vote percentages of $$-15.22\%$$ for Obama in 2012, $$-2.85\%$$ for the Democratic House candidate in 2014, and $$-18.33\%$$ for Clinton in 2016. There was no incumbent running for the House seat in 2014, so no adjustment is needed. The BPI of AR-04 is therefore $$0.133(-15.22\%)+0.278(-2.85\%)+0.244(-18.33\%)/(0.133+0.278+0.244)=-14.39%$$

The Synthesized using Earlier Elections as Rationale (SEER) percentage is the first forecast given by the ORACLE of Blair process. SEER begins with a prediction of the Democratic margin, or by how much of the two-party vote the Democratic nominee will win or lose. The Democratic margin prediction is a scale of the BPI shifted by incumbency, incumbency being whether or not a Democratic or a Republican incumbent is running. The scale of the BPI and the shift based on incumbency are listed below:

 Partisanship Weight: 2 Dem Incumbency Weight: 6.90% Rep Incumbency Weight: 8.17%

By our positive-negative convention, we subtract $$8.17\%$$ if a Republican incumbent is running and add $$6.90\%$$ if a Democratic incumbent is running. So, as AR-02 has a BPI of $$-5.68\%$$ and a Republican incumbent running, SEER predicts a Democratic margin of $$2 (-5.68\%) - 8.17\% = -19.63\%$$, meaning SEER predicts that the Democratic nominee will lose by $$19.63\%$$ of the two-party vote. This leads to the final SEER prediction of the Democratic two-party vote, which is calculated by $$50\% + 0.5 (margin)$$. So for AR-02 SEER predicts that $$50\% + 0.5(-19.63\%) = 40.24\%$$ of the two-party vote will go to the Democratic nominee.

We also assigned a standard deviation to each SEER prediction. Based on some tests of the predictive accuracy of past election results, we used a standard deviation of $$0.066$$ if there is an incumbent (of either party) running in 2018 and $$0.074$$ if there is not. These standard deviations are for the two-party vote proportion predictions.

## National Mood Shift (bigmood)

To calculate and adjustment for the national mood, we started with an estimation of the expected major-party voter turnout in 2018. To get this, we started with the sum of Democratic and Republican voters in each Congressional election in 2014. In cases where the 2014 turnout in a district was problematic due to their not being a contested House race (leading to low vote totals) or due to redistricting, we used the 2016 turnout in that district scaled down by the ratio of 2014 average contested district turnout to 2016 average contested district turnout (which is less than 1). In cases where the 2016 turnout is also problematic, we resorted to using the average 2014 turnout for those districts. We then determined, based on our SEER predictions for each district and the number of expected major-party voters in each district, the predicted total number of Democratic and Republican votes on the national level, from which we get the predicted national two-party Democratic vote.

We then compare this to current generic ballot polls. Generic ballot polls ask respondents nationwide whether they plan to vote for the Democratic Congressional candidate or for the Republican Congressional candidate in their districts without naming the candidate. We use a weighted average of generic ballot polls, averaged using the same method as used for polls in an individual district (discussed below); the weighted average Democratic two-party vote is referred to here as the current National Mood. We then find the difference between the current National Mood and the National fundamental prediction discussed in the first paragraph. Let us call this difference the National Mood Shift.

Each district has an elasticity (taken from 538), which quantifies how much more or less than the national average a district is affected by shifts in the national mood. The average elasticity is $$1.00$$ (affected the same as the country as a whole). For example, the Democratic vote percentage of a hypothetical district with elasticity $$0.90$$ is expected to shift by $$0.9$$ points for every 1-point shift in the National Mood, in the same direction as the national mood.

To combine the National Mood with the SEER predictions, we add the product of the National Mood Shift and the elasticity for each district to the SEER prediction for the district to get the bigmood prediction for each district.

There are two sources of variation of the shift. The first the generic ballot polling average. The standard deviation of the polling average ($$\sigma_{p}$$) isn’t calculated from the standard deviation of the individual polls, but rather from looking at the relationship between generic ballot average and national popular House vote from 2002 to 2016. This yields a ($$\sigma_{p}$$) of $$1.38$$ percentage points. The second source of variation is the variation of the SEER predictions, which persists when they are applied to the 2014 voter turnout. The standard deviation ($$\sigma_{q}$$) of the SEER-based projected national vote is therefore:

$$\sigma_{q} = \frac{\sqrt{\sum\limits_{i=1}^{435} t_{i}^{2}+\sigma_{i}^{2}}}{\sum \limits_{i=1}^{435} t_{i}}$$

Where the $$i$$th district (from $$1$$ to $$435$$) has 2014 two-party turnout $$t_{i}$$ and the SEER standard deviation for the ith district has standard deviation $$\sigma_{i}$$. The overall standard deviation of the national mood shift is given by $$\sqrt{\sigma_{p}^{2}+\sigma_{q}^{2}}$$.

## Averaging polls

For districts that have polls for their congressional races, we construct a weighted average of those polls. We start by getting for each poll: the two-party Democratic vote prediction $$(0 < p < 1)$$, the number of days before the election that the poll was finished ($$t$$), and the sample size of the poll ($$n$$). The polls are weighted by their ages, based on the relative values of this function for the different ages of the polls:

$$f(t)=e^{-t/30}$$

Therefore, if there are $$m$$ polls in a district, and the $$a$$th poll is from $$t_{a}$$ days before the election, it will have weight $$w_{a}$$:

$$w_{a}=\frac{e^{\frac{-t_{a}}{30}}}{\sum\limits_{i=1}^m e^{\frac{-t_{i}}{30}}}$$

This ensures that old polls are weighted less than newer polls, that the marginal penalty for being one day older decreases as the polls get older, and that the weights sum to $$1$$. We then get the weighted average of the predictions by summing the products of each poll’s predicted Democratic two-party vote proportion and that poll’s weight. However, there is some variation in polls which we include as uncertainty in our model. One source of this variation is sampling error. Given a poll with Democratic vote share $$p$$, Republican vote share $$(1-p)=q$$, and sample size $$n$$, the average sampling error $$se$$ of such a poll is given by:

$$se=\sqrt{\frac{pq}{n}}$$

For now, we will say that the standard deviation $$\sigma$$ of each poll (i.e. the average error) is its sampling error.

However, there are other sources of variation that we are unable to empirically calculate. We divide the polls into four grades (A, B, C, and D; taken from 538) based on the quality of the pollster’s methodology. For each grade of polls, we took polls from previous House elections (starting in 2012) and found the average distance around the line of best fit for actual election results vs. poll results (i.e. the line that best predicted the actual election result given a poll result). Let’s call this average distance the past poll standard deviation for each grade.

We then find the average sampling error of all polls of each grade. For each grade, we then add ((past poll standard deviation) - (average sampling error)) to the standard deviation of each poll with that grade. Although the numbers will change slightly as we add more polls with slightly different average sampling errors, we found that the average sampling error is generally between $$0.02$$ and $$0.025$$ regardless of grade. Here are approximate standard deviation increases for each grade:

 Grade Standard Deviation Around Regression Line Est. Standard Deviation Increase A 0.0565 0.034 B 0.0662 0.044 C 0.0767 0.054 D 0.1303 0.105

Given two polls $$P_{1}$$ and $$P_{2}$$ with standard deviations $$\sigma_{2}$$ and weights $$w_{1}$$ and $$w_{2}$$, we can find the standard deviation of the weighted sum of $$P_{1}$$ and $$P_{2}$$ with:

$$\sigma_{1}+\sigma_{2}=\sqrt{w_{1}^{2}\sigma_{1}^{2}+w_{2}^{2}\sigma_{2}^{2}}$$

This can be generalized to more than two polls to get one ‘weighted average’ standard deviation for the weighted average of the polls in a district.

## Weighting Polls vs Bigmood

For districts with polls, we take a weighted average of the bigmood prediction and aggregate poll prediction to get to our final prediction. The weight ($$0 < weight < 1$$) of the polls depends on 1) the number of polls, 2) the grade of each poll, and 3) the age (in days before the election) of each poll. We first calculate the Grade Point Sum (GPS) of all of the polls. If a district has $$m$$ polls, and the $$i$$th poll has grade $$g_{i}$$ and is from $$t_{i}$$ days before the election, the GPS is:

$$GPS=\sum\limits_{i=1}^m G(g_{i})e^{\frac{-t_{i}}{167}}$$

Where $$G(g_{i})$$ is $$0.177$$ if $$g_{i}$$ is A, $$0.151$$ if $$g_{i}$$ is B, $$0.130$$ if $$g_{i}$$ is C, and $$0.077$$ if $$g_{i}$$ is D.

The weighting of the aggregate poll prediction ($$w_{p}$$) is then given by:

$$w_{p}=\frac{1.9}{\pi} \cdot \arctan(6.12 \cdot GPS)$$

And the weighting of the bigmood prediction is ($$1-w_{p}$$).

The constants of the arctan function were chosen so that it has an asymptote of $$0.95$$ (a district with a very large GPS has a poll weight very close to $$0.95$$) and so that a district with two B polls, each finished the day of the election, will have poll weight $$0.6$$.

## Blairvoyance

Blairvoyance provides a hypothesized poll result for districts without polls by interpolating on demographics of the districts that do have polls. First, Blairvoyance creates linear relationships between districts using select demographics for that district. Figure 1 is a simplified example where the three districts are plotted against two axes representing two demographics, which is effectively a representation of a relationship between districts.

Figure 1. Blairvoyance demographics mapping

The closer they are in Figure 1, the more similar the districts by some metric formed by the demographic data. The red dots represent the districts with polling data. Blairvoyance then moves the points that represent district with polls to another dimension, which represent the poll result. Figure 2 is a representation of Figure 1 with the third axes of poll result is added, and Figure 3 shows the points being moved into the third dimension representing polls.

Figure 2. Blairvoyance mapping in 3-space

Figure 3. Blairvoyance adding in poll results

Keep in mind that, our model contains more than two demographics, therefore Blairvoyance utilizes more than 3-dimensions. Blairvoyance then tries to fit a curve relating demographics and poll results. The blue line in Figure 4 represents a potential fitted curve Blairvoyance might create in that simplified case. Then, in order to create a hypothesized poll result for a district without poll, it will use the poll result of the closest poll in the fitted curve, as can be seen in Figure 5.

Figure 4. Blairvoyance curve fitting

Figure 5. Blairvoyance poll hypothesis

Blairvoyance takes in polls and returns poll hypotheses. Typically, polls are only done on the districts with close races. Therefore, Blairvoyance should make poll hypotheses closer than actual polls would be; the weight of Blairvoyance should decrease as partisanship increases. Hence, the weight of Blairvoyance should be at most the best poll we can have, which is an A graded poll taken one day before the election. Let the weight of one A graded poll taken one day before election be $$w_a$$. Then the weight of Blairvoyance would be $$w_a (1 + 2|bigmood - 0.5|)$$. This satisfies both properties that it is at most $$w_{a}$$ and decreases as partisanship increases. If some hypothetical district is completely partisan, Blairvoyance would not count toward the prediction of that district.

## Calculating AUSPICE

The Agglomeration Utilizing Statistical Predictions and Inquiries Concerning Elections (AUSPICE) is the final election forecast that the ORACLE of Blair gives a district every run. AUSPICE is a combination of the distribution provided by Bigmood and the averaged polls, or Blairvoyance in the absence of polls, weighted as mentioned above. Suppose the weight of bigmood is $$w_b$$ and the weight of the averaged polls for a district (or Blairvoyance) to be $$w_l$$. And suppose that bigmood gives a mean of $$\mu_b$$ and a standard deviation of $$\sigma_b$$ for the proportion of the two-party vote going to the Democratic nominee, and the polls give a mean of $$\mu_l$$ and a standard deviation of $$\sigma_l$$. The AUSPICE for that district will return a mean of $$w_l \mu_l + w_b \mu_b$$ and a standard deviation of $$\sqrt{(w_l \sigma_l)^2+(w_b \sigma_b)^2}$$.

## National Predictions

At this point, we have predicted vote shares and standard deviations of vote shares, so it might seem that calculating the overall chance that a party gets a majority of seats is rather simple. However, this is not the case.

In order to avoid excessive computations or calculating an exact number, we simulate the entire House election $$10,000,000$$ times each time we run our model. The probability that each party wins a majority of seats in the House is then about the number of simulations in which this happens divided by $$10,000,000$$.

A naïve approach to use here would be to simulate each district separately, using the numbers we already have. However, this approach has the implicit assumption that the district vote shares are all independent, which is certainly not the case. For one, the systematic bias of polls can often be caused by the same factors from district to district. Also, since our model is not perfect, it may be consistently off in one direction for most districts. Thus, our simulation must introduce some correlation between districts.

The way we chose to implement this is to, for each simulation, choose a number $$s$$ from a normal distribution with mean $$0$$ and variance $$\sigma^2$$ (we will explain where $$\sigma$$ comes from later). Then, for every district $$d$$ with mean vote share $$\mu_d$$ and variance of the vote share $$\sigma_d^2$$, we choose a number $$v_d$$ from a normal distribution with mean vote share $$\mu_d$$ and variance $$\sigma_d^2 - \sigma^2$$ and let the simulated vote share be $$v_d + s$$. Since $$v_d$$ and $$s$$ are independent, this means that the distribution of $$v_d + s$$ has mean $$\mu_d$$ and variance $$\sigma_d^2$$, matching our earlier computations The important part is that $$s$$ is the same for all districts, meaning that a portion of the variance of the districts is shared.

It suffices to find a suitable value for $$\sigma$$. There is no clear way to do this, but we decided on the following: the outcome of the House is very correlated with the House popular vote, so a good choice for $$\sigma$$ is one such that the resulting distribution of the House popular vote matches our uncertainty about it. Since the individual district variation (on the order of $$\sigma \sim 5\%$$) is all independent, the contribution to the House popular vote from these sources of variation is on the order of $$5\%/\sqrt{435} \sim 0.25\%$$ which is far lower than the actual uncertainty. So we should have $$\sigma$$ be our uncertainty about the House popular vote.

There are many ways to forecast the House popular vote, but using the standard deviation of the national voter turnout prediction is close to the best one can do. Therefore, we use the historical standard deviation of the national generic ballot polls versus the national popular vote, which corresponds to $$\sigma \approx 1.38\%$$.

## Example: California’s 25th District

Please note that in this example calculations may be shown with intermediate rounding. However, the actual computations used to produce results did not use intermediate rounding. This district is currently represented by Republican Steve Knight. He is being challenged by Democrat Katie Hill. This section was updated on October 26th.

## BPI

Obama received $$49.1\%$$ of the two-party vote in this district in 2012. As a consequence of California’s jungle primary system, there was no Democrat on the 2014 House ballot. We therefore used the aggregate two-party Democratic performance in the jungle primary, which was $$32.8\%$$; there was no incumbent running in 2014. Clinton received $$53.6\%$$ of the two-party vote in this district in 2016. There was a Democratic candidate for the House seat in 2016 and he got $$46.9\%$$ of the vote; there was a Republican incumbent. Since there was a Republican incumbent in 2016, we adjust the Democrat’s vote share in 2016 to be $$51.0\%$$.

This district thus has a BPI of $$0.133(49.1-50) + 0.278(32.8-50) + 0.244(53.6-50)+0.345(51.0-50) = -3.59\%$$. Therefore, as there is a Republican incumbent running in 2018, this district has a SEER prediction of $$50\% + 0.5(2(-3.59\%) - 8.17\%) = 42.4\%$$. Since there is an incumbent, we will use a standard deviation of $$0.066$$ for the SEER prediction.

## Bigmood

Using the SEER predictions and procedure described above in the National Mood Shift section, our nationwide fundamental prediction with a 2014 turnout model is that Democrats would get $$49.5\%$$ of the two-party vote. The current (at the time or writing) generic ballot average is that Democrats will get $$54.3\%$$ of the two-party vote. This implies an average shift of $$+4.8\%$$ to the Democratic two-party votes in each district to get the big mood predictions. This number will change as more generic ballot polls are conducted, but it is suitable for this example. CA-25 has an elasticity of $$0.97$$, so it has a bigmood prediction of $$42.4\% + 0.97 \cdot 4.8\% = 47.1\%$$.

## Polling

This district has six polls:

 Poll # Grade Age (days) Poll result Sample size 1 D 44 0.521 650 2 A 48 0.489 500 3 C 117 0.500 400 4 C 138 0.421 400 5 B 264 0.556 283 6 B 282 0.570 650

The standard deviation and weight of each poll are:

 Poll # Standard Deviation Poll weight function Poll weight 1 0.125 0.231 0.498 2 0.056 0.202 0.436 3 0.079 0.020 0.044 4 0.079 0.010 0.022 5 0.074 0.000 0.000 6 0.063 0.000 0.000

These variances include the sampling error and the grade-based standard deviation increase discussed above in the Averaging Polls section. The weighted average of the polls is approximately $$0.498 \cdot 0.521+0.436 \cdot 0.489+0.044 \cdot 0.500+0.022 \cdot 0.421+0.00 \cdot 0.556+0.00 \cdot 0.570 = 0.504 = 50.4\%$$. The weighted average has a standard deviation of approximately $$\sqrt{0.498^{2} \cdot 0.125^{2} + 0.436^{2} \cdot 0.056^{2} + 0.044^{2} \cdot 0.079^{2} + 0.022^{2} \cdot 0.079^{2} \cdot 0.000^{2} \cdot 0.074^{2} + 0.000^{2} \cdot 0.063^{2}} = 0.067$$.

## Poll Weighting

This district’s polls have $$GPS = 0.077e^{\frac{-44}{167}}+0.177e^{\frac{-48}{167}}+0.130e^{\frac{-117}{167}}+0.130e^{\frac{-138}{167}}+0.151e^{\frac{-264}{167}}+0.151e^{\frac{-282}{167}} = 0.372$$. Therefore, the polls in this district have weight $$\frac{1.9}{π} \cdot \arctan(6.12 \cdot 0.372) = 0.70$$. The bigmood prediction consequently has weight $$1-0.70 = 0.30$$.

## Final Prediction

The final prediction for CA-25 is that Hill will get on average $$0.70 \cdot 50.4\% + 0.30 \cdot 47.1\% = 49.4\%$$ of the two-party vote. This has a standard deviation of $$\sqrt{0.70^{2} \cdot 0.067^{2} + 0.30^{2} \cdot 0.066^{2}} = 0.051$$. Therefore, Hill getting at least $$50\%$$ of the two-party vote has a $$z$$-score of $$z = \frac{0.50-0.494}{0.051} = 0.11$$. This implies a win probability of $$45.4\%$$. In other words, if this particular general election were held a very large number of times, we would expect Hill to win $$45.4\%$$ of the time and Knight to win $$54.6\%$$ of the time.

format: plain


text:
Algebra

Challenge Quizzes

If $\alpha,\beta$ are the roots of the quadratic equation

$x^2-\left( 3+2^{\sqrt{\log_23}}-3^{\sqrt{\log_32}} \right)x-2\left( 3^{\log_32}-2^{\log_23} \right) =0$,

then find the value of $\left( \alpha^2+\alpha\beta+\beta^2 \right)$.

Find the number of quadratic equations which are unchanged by squaring their roots.

Positive integers $a, b$ and $c$ satisfy the condition that the quadratic equation $ax^2 - bx +c =0$ has 2 distinct real roots that are strictly between 0 and 1. What is the minimum value of $a$?

How many integral values of $a$ lie in the interval $[-1000,1000]$ for which the inequality ${25}^x+(a+2).5^x-(a+3)<0$ is satisfied for at least one real $x$.

If the McDonald's logo were stored as a set of pixels, enlargement would quickly result in distorted or pixelated images, which are an eyesore. As such, companies often make vector images of their logos, in which the information is stored as mathematical formulae. Such vector images are easily scaled while maintaining sharp, crisp images.

As a first approximation, the logo is deconstructed and approximated as 2 parabolic curves of the form $y = -A(x-5)^2$ and $y = - A (x+5)^2$. The McDonald's logo has a height to length ratio of 1.05. What is $A$?

×


text:
We started this module by discussing large, ancient buildings and wondering how their structure had mostly remained the same for hundreds of years. It is clear now that Friction plays a big role here. Due to static friction, the rocks that are on top of other rocks don't move even if there are forces acting on them.

One other thing: notice that the maximum frictional force is given by $$F_f = \mu N$$. Here, $$N$$, the normal force, depends on the mass of the object. So, the heavier the rock, the more the frictional force is. The more the frictional force is, the more stable the structure will be.

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I somewhat advanced since 6j symbols trouble when irrep multiplicity >1 (because I found the decades old papers where the Racah algebra is dealt with, e.g. "Coupling Coefficients and Tensor Operators for Chains of Groups", P. H. Butler) but now it's even worse: I just don't call my finding "antinomy" because I reserve the possibility I'm too flamin' stupid :-)
So, I need just one irrep with multiplicity >=2 (take for example the "fusion mini" $A\bigotimes{A}=1\bigoplus{2A}\bigoplus{B},A\bigotimes{B}=A,B\bigotimes{B}=1$ and just use $A$ for "coloring") and asking the innocent question: Is >1-2< equal to >2-1< ?
Excuses for the ASCII art. These are two connected cubic nodes where both nodes are labeled with different multiplicities (I don't know what's the standard way to label with multiplicity, but "coloring" the node seems logical).
A: They are not equal. But "mutant logic" says they are equal anyway, expressing them as linear combinations of 4-tangles that ARE mutant-symmetric (like >1-1<) makes them obviously equal (weasel out: the set of mutant-symmetric 4-tangles is not a complete vector space basis) and doing a 6j recoupling to the rotated projectors makes them obviously equal too (a multiplicity flip in a 6j symbol built from just one irrep has no effect).
B: They are equal. But then the vector basis gets the wrong dimension (in my example, e.g. 6 projectors, but then two of them are the same) and even worse, "multiply" the equation from the right and reduce the loop by Schur: >1-2<>1- = >2-1<>1- . Complete nonsense results.

If someone would have the complete 6j set for a fusion ring with multiplicity, I might find out where the bug sits and how to encode the 6j formulas (Biedenharn-Elliott, Racah) correctly in case of multiplicity (and non-self-duality, which also annoys :-). Or can you tell me directly what I am doing wrong?

The fusion ring you consider is known as (even part of) $E_6-$subfactor. The corresponding 6j symbols (also called associativity matrices) are given in the paper by Hagge and Hong, see Appendix.


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## Equivalent Diameter when Flow in Rectangular Duct Solution

STEP 0: Pre-Calculation Summary
Formula Used
Equivalent Diameter = (4*Length of Rectangular Section*Breadth of Rectangle)/(2*(Length of Rectangular Section+Breadth of Rectangle))
De = (4*L*B)/(2*(L+B))
This formula uses 3 Variables
Variables Used
Equivalent Diameter - (Measured in Meter) - Equivalent Diameter of an irregularly shaped object is the diameter of a sphere of equivalent volume.
Length of Rectangular Section - (Measured in Meter) - Length of Rectangular Section is the total distance from one end to other end, length is the longest side of rectangle.
Breadth of Rectangle - (Measured in Meter) - Breadth of Rectangle is the distance from one end to the other which is perpendicular to the length.
STEP 1: Convert Input(s) to Base Unit
Length of Rectangular Section: 750 Meter --> 750 Meter No Conversion Required
Breadth of Rectangle: 4 Meter --> 4 Meter No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
De = (4*L*B)/(2*(L+B)) --> (4*750*4)/(2*(750+4))
Evaluating ... ...
De = 7.95755968169761
STEP 3: Convert Result to Output's Unit
7.95755968169761 Meter --> No Conversion Required
7.95755968169761 Meter <-- Equivalent Diameter
(Calculation completed in 00.016 seconds)
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## < 10+ Basics of Heat Transfer Calculators

Log Mean Temperature Difference for Counter Current Flow

## Log Mean Temperature Difference for Counter Current Flow

Formula
"LMTD" = (("T"_{"ho"}-"T"_{"ci"})-("T"_{"hi"}-"T"_{"co"}))/ln(("T"_{"ho"}-"T"_{"ci"})/("T"_{"hi"}-"T"_{"co"}))

Example
"19.57615K"=(("20K"-"5K")-("35K"-"10K"))/ln(("20K"-"5K")/("35K"-"10K"))

Calculator
LaTeX
Log Mean Temperature Difference = ((Outlet Temperature of Hot Fluid-Inlet Temperature of Cold Fluid)-(Inlet Temperature of Hot Fluid-Outlet Temperature of Cold Fluid))/ln((Outlet Temperature of Hot Fluid-Inlet Temperature of Cold Fluid)/(Inlet Temperature of Hot Fluid-Outlet Temperature of Cold Fluid))
Log Mean Temperature Difference for CoCurrent Flow

## Log Mean Temperature Difference for CoCurrent Flow

Formula
"LMTD" = (("T"_{"ho"}-"T"_{"co"})-("T"_{"hi"}-"T"_{"ci"}))/ln(("T"_{"ho"}-"T"_{"co"})/("T"_{"hi"}-"T"_{"ci"}))

Example
"18.20478K"=(("20K"-"10K")-("35K"-"5K"))/ln(("20K"-"10K")/("35K"-"5K"))

Calculator
LaTeX
Log Mean Temperature Difference = ((Outlet Temperature of Hot Fluid-Outlet Temperature of Cold Fluid)-(Inlet Temperature of Hot Fluid-Inlet Temperature of Cold Fluid))/ln((Outlet Temperature of Hot Fluid-Outlet Temperature of Cold Fluid)/(Inlet Temperature of Hot Fluid-Inlet Temperature of Cold Fluid))
Logarithmic Mean Area of Cylinder

## Logarithmic Mean Area of Cylinder

Formula
"A"_{""mean""} = ("A"_{"o"}-"A"_{"i"})/ln("A"_{"o"}/"A"_{"i"})

Example
"9.865214m²"=("12m²"-"8m²")/ln("12m²"/"8m²")

Calculator
LaTeX
Logarithmic Mean Area = (Outer Area of the Cylinder-Inner Area of the Cylinder)/ln(Outer Area of the Cylinder/Inner Area of the Cylinder)
Equivalent Diameter when Flow in Rectangular Duct

## Equivalent Diameter when Flow in Rectangular Duct

Formula
"D"_{"e"} = (4*"L"*"B")/(2*("L"+"B"))

Example
"7.95756m"=(4*"750m"*"4m")/(2*("750m"+"4m"))

Calculator
LaTeX
Equivalent Diameter = (4*Length of Rectangular Section*Breadth of Rectangle)/(2*(Length of Rectangular Section+Breadth of Rectangle))
Heat Transfer from Stream of Gas flowing in Turbulent Motion

## Heat Transfer from Stream of Gas flowing in Turbulent Motion

Formula
"h"_{"transfer"} = (16.6*"c"_{"p"}*("G")^0.8)/("D"^0.2)

Example
"86.84618W/m²*K"=(16.6*"0.0002kcal(IT)/kg*°C"*("13kg/s/m²")^0.8)/("3m"^0.2)

Calculator
LaTeX
Heat Transfer Coefficient = (16.6*Specific Heat Capacity*(Mass Velocity)^0.8)/(Internal Diameter of Pipe^0.2)
Colburn Factor using Chilton Colburn Analogy

## Colburn Factor using Chilton Colburn Analogy

Formula
"j"_{"H"} = "Nu"/("Re")*("Pr")^(1/3)

Example
"0.001243"="7"/("5000")*("0.7")^(1/3)

Calculator
LaTeX
Colburn's j-factor = Nusselt Number/(Reynolds Number)*(Prandtl Number)^(1/3)
Heat Transfer Coefficient based on Temperature Difference

## Heat Transfer Coefficient based on Temperature Difference

Formula
"h"_{"transfer"} = "q"/"ΔT"_{"Overall"}

Example
"0.312727W/m²*K"="17.2W"/"55K"

Calculator
LaTeX
Heat Transfer Coefficient = Heat Transfer/Overall Temperature Difference
Equivalent Diameter of Non-Circular Duct

## Equivalent Diameter of Non-Circular Duct

Formula
"D"_{"e"} = (4*"A"_{"cs"})/"P"

Example
"1.25m"=(4*"25m²")/"80m"

Calculator
LaTeX
Equivalent Diameter = (4*Cross Sectional Area of Flow)/Wetted Perimeter

Formula
"r"_{"H"} = "A"_{"cs"}/"P"

Example
"0.3125m"="25m²"/"80m"

Calculator
LaTeX
Hydraulic Radius = Cross Sectional Area of Flow/Wetted Perimeter
Colburn J-Factor given Fanning Friction Factor

## Colburn J-Factor given Fanning Friction Factor

Formula
"j"_{"H"} = "f"/2

Example
"0.245"="0.49"/2

Calculator
LaTeX
Colburn's j-factor = Fanning Friction Factor/2

## Equivalent Diameter when Flow in Rectangular Duct Formula

Equivalent Diameter = (4*Length of Rectangular Section*Breadth of Rectangle)/(2*(Length of Rectangular Section+Breadth of Rectangle))
De = (4*L*B)/(2*(L+B))

## What is Heat Transfer?

Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy between physical systems. Heat transfer is classified into various mechanisms, such as thermal conduction, thermal convection, thermal radiation, and transfer of energy by phase changes.

## Define Thermal Conductivity & Factors affecting it?

Thermal conductivity is defined as the ability of a substance to conduct heat. Factors Affecting The Thermal Conductivity are: Moisture, Density of material, Pressure, Temperature & Structure of material.

## How to Calculate Equivalent Diameter when Flow in Rectangular Duct?

Equivalent Diameter when Flow in Rectangular Duct calculator uses Equivalent Diameter = (4*Length of Rectangular Section*Breadth of Rectangle)/(2*(Length of Rectangular Section+Breadth of Rectangle)) to calculate the Equivalent Diameter, The Equivalent Diameter when Flow in Rectangular Duct formula is defined as the ratio of 4 multiply by length and breadth of rectangle to the 2 multiply by summation of length and breadth of rectangle. Equivalent Diameter is denoted by De symbol.

How to calculate Equivalent Diameter when Flow in Rectangular Duct using this online calculator? To use this online calculator for Equivalent Diameter when Flow in Rectangular Duct, enter Length of Rectangular Section (L) & Breadth of Rectangle (B) and hit the calculate button. Here is how the Equivalent Diameter when Flow in Rectangular Duct calculation can be explained with given input values -> 7.95756 = (4*750*4)/(2*(750+4)).

### FAQ

What is Equivalent Diameter when Flow in Rectangular Duct?
The Equivalent Diameter when Flow in Rectangular Duct formula is defined as the ratio of 4 multiply by length and breadth of rectangle to the 2 multiply by summation of length and breadth of rectangle and is represented as De = (4*L*B)/(2*(L+B)) or Equivalent Diameter = (4*Length of Rectangular Section*Breadth of Rectangle)/(2*(Length of Rectangular Section+Breadth of Rectangle)). Length of Rectangular Section is the total distance from one end to other end, length is the longest side of rectangle & Breadth of Rectangle is the distance from one end to the other which is perpendicular to the length.
How to calculate Equivalent Diameter when Flow in Rectangular Duct?
The Equivalent Diameter when Flow in Rectangular Duct formula is defined as the ratio of 4 multiply by length and breadth of rectangle to the 2 multiply by summation of length and breadth of rectangle is calculated using Equivalent Diameter = (4*Length of Rectangular Section*Breadth of Rectangle)/(2*(Length of Rectangular Section+Breadth of Rectangle)). To calculate Equivalent Diameter when Flow in Rectangular Duct, you need Length of Rectangular Section (L) & Breadth of Rectangle (B). With our tool, you need to enter the respective value for Length of Rectangular Section & Breadth of Rectangle and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Equivalent Diameter?
In this formula, Equivalent Diameter uses Length of Rectangular Section & Breadth of Rectangle. We can use 1 other way(s) to calculate the same, which is/are as follows -
• Equivalent Diameter = (4*Cross Sectional Area of Flow)/Wetted Perimeter
Let Others Know


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Toxic gas and less dense than air. Ball-and-stick model, geometric structure and   Feb 17, 2019 Workout the chemical formula of carbon monoxide by cross over multiple their valency Get the answers you need, now! Hitta stockbilder i HD på What Chemical Formula Carbon Monoxide Chalkboard och miljontals andra royaltyfria stockbilder, illustrationer och vektorer i  Chlorine monoxide is a chemical radical with the chemical formula ClO. It plays an important role in the process of ozone depletion.3d illustration.The molecule  Carbon monoxide. Regulatory process names 4 630-08-0. Mol. formula: CO. formula. Hazard classification & labelling Help Hazard classification and labelling. en chemical of the formula CO. + 3 definitioner.

Carbon monoxide, with the chemical formula CO, is a colourless, odourless and tasteless gas. It is the product of the incomplete combustion of carbon-containing compounds, notably in Carbon monoxide formula is one of the most simple and easy formulae which students can remember. In this article, we will talk about carbon monoxide and some key things about it to know. This chemical compound has no odor or taste, but having the potential to become toxic. This article will explain the Carbon monoxide formula with examples. 2015-06-12 2018-05-11 2021-04-06 The formula for Dinitrogen Monoxide: Its chemical formula is; $$N_2O$$ Its chemical equation is: $$2 N_2 + O_2 \rightarrow 2 N_2O$$ Properties: Its melting point is -88.48 degree C. And its boiling point is -90.86 degree C. Its density is 1.98 kg per cubic cm.

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# A Numerical Method for the Elliptic Monge-Ampère Equation with Transport Boundary Conditions

@article{Froese2012ANM,
title={A Numerical Method for the Elliptic Monge-Amp{\e}re Equation with Transport Boundary Conditions},
author={Brittany D. Froese},
journal={SIAM J. Sci. Comput.},
year={2012},
volume={34}
}
The problem of optimal mass transport arises in numerous applications including image registration, mesh generation, reflector design, and astrophysics. One approach to solving this problem is via the Monge-Amp\ere equation. While recent years have seen much work in the development of numerical methods for solving this equation, very little has been done on the implementation of the transport boundary condition. In this paper, we propose a method for solving the transport problem by…
87 Citations

## Figures and Tables from this paper

### An efficient numerical algorithm for the L2 optimal transport problem with periodic densities

• Mathematics, Computer Science
• 2015
An extension of the numerical method of Loeper and Rapetti (2005) for the Monge–Ampere equation by a Newton’s algorithm that converges for sufficiently large damping coefficients and is used to solve the optimal transport problem with quadratic cost.

### Convergent Finite Difference Solvers for Viscosity Solutions of the Elliptic Monge-Ampère Equation in Dimensions Two and Higher

• Mathematics
SIAM J. Numer. Anal.
• 2011
This article builds a wide stencil finite difference discretization for the Monge-Ampere equation and proves convergence of Newton's method and provides a systematic method to determine a starting point for the Newton iteration.

### Numerical Solution of the Optimal Transportation Problem via viscosity solutions for the Monge-Ampere equation

• Mathematics
• 2012
A numerical method for the solution of the elliptic Monge-Ampere Partial Differential Equation, with boundary conditions corresponding to the Optimal Transportation (OT) problem is presented. A local

### A Finite Element/Operator-Splitting Method for the Numerical Solution of the Three Dimensional Monge-Ampère Equation

• Mathematics, Computer Science
J. Sci. Comput.
• 2019
The present article extends to the three-dimensional elliptic Monge–Ampère equation the method discussed in Glowinski et al. ( 2019) for the numerical solution of its two-dimensional variant, taking advantage of an equivalent divergence formulation of the Monge-Ampere equation, involving the cofactor matrix of the Hessian of the solution.

### A Finite Element/Operator-Splitting Method for the Numerical Solution of the Two Dimensional Elliptic Monge–Ampère Equation

• Mathematics, Computer Science
J. Sci. Comput.
• 2019
The methodology discussed in this article is able to handle domains with curved boundaries and unstructured meshes, using piecewise affine continuous approximations, while preserving optimal, or nearly optimal, convergence orders for the approximation error.

### Solving the Monge–Ampère equations for the inverse reflector problem

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• 2015
The inverse reflector problem arises in geometrical nonimaging optics: given a light source and a target, the question is how to design a reflecting free-form surface such that a desired light

### Numerical Methods for Hamilton-Jacobi-Bellman Equations with Applications

This thesis proposes a deep neural network framework for the HJB equations emerging from the study of American options in high dimensions, which addresses the curse of dimensionality issue that state-of-the-art approaches suffer.

### Two-scale method for the Monge-Ampère equation: Convergence to the viscosity solution

• Mathematics
Math. Comput.
• 2019
We propose a two-scale finite element method for the Monge-Amp\ere equation with Dirichlet boundary condition in dimension $d\ge2$ and prove that it converges to the viscosity solution uniformly.

### Viscosity subsolutions of the second boundary value problem for the Monge-Amp\ere equation

It is well known that the quadratic-cost optimal transportation problem is formally equivalent to the second boundary value problem for the Monge-Amp\`ere equation. Viscosity solutions are a powerful

## References

SHOWING 1-10 OF 68 REFERENCES

### Convergent Finite Difference Solvers for Viscosity Solutions of the Elliptic Monge-Ampère Equation in Dimensions Two and Higher

• Mathematics
SIAM J. Numer. Anal.
• 2011
This article builds a wide stencil finite difference discretization for the Monge-Ampere equation and proves convergence of Newton's method and provides a systematic method to determine a starting point for the Newton iteration.

### Moving Mesh Generation Using the Parabolic Monge--Amp[e-grave]re Equation

• Computer Science
SIAM J. Sci. Comput.
• 2009

### An Efficient Numerical Method for the Solution of the L2 Optimal Mass Transfer Problem

• Computer Science
SIAM J. Sci. Comput.
• 2010
A new computationally efficient numerical scheme for the minimizing flow approach for the computation of the optimal L(2) mass transport mapping is presented, employing a direct variational method.

### Grid Generation and Adaptation by Monge-Kantorovich Optimization in Two and Three Dimensions

• Physics
IMR
• 2008
It is shown that grids which are generated by the Monge-Ampere (MA) equation also satisfy equations of an alternate variational principle minimizing grid distortion, indicating robustness to grid tangling.

### Two Numerical Methods for the elliptic Monge-Ampère equation

• Mathematics
• 2010
The numerical solution of the elliptic Monge-Ampere Partial Differential Equation has been a subject of increasing interest recently [Glowinski, in 6th International Congress on Industrial and


text:
# Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.5

Students can Download Maths Chapter 2 Algebra Ex 2.5 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.5

Miscellaneous Practice Problems

Question 1.
The sum of three numbers is 58. The second numberis three times of two-fifth of the first number and the third number is 6 less than the first number. Find the three numbers.
Solution:
Here what we know
a + b + c = 58 (sum of three numbers is 58)
Let the first number be ‘x’
b = a + 3 (the second number is three times of $$\frac{1}{2}$$ of the first number)
b = 3 x $$\frac{2}{5}$$x = $$\frac{6}{5}$$x
Third number = x – 6
Sum of the numbers is given as 58.
∴ x + $$\frac{6}{5}$$x + (x – 6) = 58
Multiplying by 5 throughout, we get
5 × x + 6x + 5 × (x – 6) = 58 x 5
5x + 6x + 5x – 30 = 290
∴ 16x = 290 + 30
∴ 16x = 320
∴ x = $$\frac{320}{16}$$x = 20
1st number =20
2nd number = 3 x $$\frac{2}{5}$$ x 20 = 24
3rd number = 24 – 6 = 14

Question 2.
In triangle ABC, the measure of ∠B is two third the measure of ∠A. The measure of ∠C is 20° more than the measure of ∠A. Find the measures of the three angles.
Solution:
Let angle ∠A be a°
Given that ∠B = $$\frac{2}{3}$$ x ∠A = $$\frac{2}{3}$$a
& given ∠C = ∠A + 20 = a + 20
Since A, B & C are angles of a triangle, they add up to 180° (∆ property)
∠A + ∠B + ∠C = 180°
a + $$\frac{2}{3}$$ a + a + 20 = 180°
$$\frac{3a+2a+3a}{3}$$ + 20 = 180°
$$\frac{8a}{3}$$ = 180 – 20 = 160°

∠C = 80°

Question 3.
Two equal sides of an isosceles triangle are 5y – 2 and 4y + 9 units. The third side is 2y + 5 units. Find y and the perimeter of the triangle.
Solution:
Given that 5y – 2 & 4y + 9 are the equal sides of an isosceles triangle.
The 2 sides are equal

∴ 5y – 4y = 9 + 2 (by transposing)
∴ y = 11
∴ 1st side = 5y – 2 = 5 x 11-2 = 55 – 2 = 53
2nd side = 53
3rd side = 2y + 5 = 2 x 11 + 5 = 22 + 5 = 27
Perimeter is the sum of all 3 sides
∴ P = 53 + 53 + 27 = 133 units

Question 4.
In the given figure, angle XOZ and angle ZOY form a linear pair. Find the value of x.

Solution:
Since ∠XOZ & ∠ZOY form a linear pair,
by property, we have their sum to be 180°
∴ ∠XOZ + ∠ZOY = 180°
3x – 2 + 5x + 6 = 180°
8x + 4 = 180 = 8x = 180 – 4
∴ 8x = 176 ⇒ x = $$\frac{176}{8}$$ ⇒ x = 22°
XOZ = 3x – 2 = 3 x 22 – 2 = 66 – 2 = 64°
YOZ = 5x + 6 = 5 x 22 + 6
= 110 + 6 = 116

Question 5.
Draw a graph for the following data:

Does the graph represent a linear relation?
Solution:
Graph between side of square & area

When we plot the graph, we observe that it is not a linear relation.

Challenging Problems

Question 6.
Three consecutive integers, when taken in increasing order and multiplied by 2, 3 and 4 respectively, total up to 74. Find the three numbers.
Solution:
Let the 3 consecutive integers be ‘x’, ’x + 1 & ‘x + 2’
Given that when multiplied by 2, 3 & 4 respectively & added up, we get 74

Simplifying the equation, we get
2x + 3x + 3 + 4x + 8 = 74
9x + 11 = 74
9x = 63 ⇒ x = $$\frac{63}{9}$$ = 7
First number = 7
Second numbers = x + 1
Third numbers = x + 2
∴ The numbers are 7, 8 & 9

Question 7.
331 students went on a field trip. Six buses were filled to capacity and 7 students had to travel in a van. How many students were there in each bus?
Solution:
Let the number of students in each bus be ‘x’
number of students in 6 buses = 6 × x = 6x
Apart from 6 buses, 7 students went in van
A total number of students is 331
∴ 6x + 7 = 331
∴ 6x = 331 – 7 = 324
∴ x = $$\frac{324}{6}$$ = 54
∴ There are 54 students in each bus.

Question 8.
A mobile vendor has 22 items, some which are pencils and others are ball pens. On a particular day, he is able to sell the pencils and ball pens. Pencils are sold for ₹ 15 each and ball pens are sold at ₹ 20 each. If the total sale amount with the vendor is ₹ 380, how many pencils did he sell?
Solution:
Let vendor have ‘p’ number of pencils & ‘b’ number of ball pens
Given that total number of items is 22
∴ p + b = 22
Pencils are sold for ₹ 15 each & ball pens for ₹ 20 each
total sale amount = 15 x p + 20 x b
= 15p + 20b which is given to be 380.
∴ 15p + 20b = 380
Dividing by 5 throughout,
$$\frac{15p}{5}$$ + $$\frac{20b}{5}$$ = $$\frac{380}{5}$$
⇒ 3p + 4b = 76
Multiplying equation (1) by 3 we get
3 x 7 + 3 x b = 22 x 3
⇒ 3p + 3b = 66
Equation (2) – (3) gives

∴ b = 10
∴ p = 12
He sold 12 pencils

Question 9.
Draw the graph of the lines y = x, y = 2x, y = 3x and y = 5x on the same graph sheet. Is there anything special that you find in these graphs?
Solution:

1. y = x
2. y = 2x
3. y = 3x
4. y = 5x

1. y = x
When x = 1, y = 1
x = 2, y = 2
x = 3, y = 2

2. y = 2x
When x = 1, y = 2
X = 2, y = 4
X = 3, y = 6

3. y = 3x
when x = 1, y = 3
x = 2, y = 6
x = 3, y = 9

4. y = 5x
When x = 1, y = 5
x = 2, y = 10
x = 3, y = 15

When we plot the alijpve points & join the points to form line, we notice that the lines become progressively steeper. In other words, the slope keeps increasing.

Question 10.
Consider the number of angles of a convex polygon and the number of sides of that polygon. Tabulate as follows:

Use this draw a graph illustrating the relations hip between the number of angles and the number of sides of a polygon.
Solution:
Angles:


text:
# American Institute of Mathematical Sciences

February  2017, 37(2): 983-1012. doi: 10.3934/dcds.2017041

## A stationary core-shell assembly in a ternary inhibitory system

 Department of Mathematics, The George Washington University, Washington, DC 20052, USA

* Corresponding author: Xiaofeng Ren

Received  March 2015 Revised  January 2016 Published  November 2016

A ternary inhibitory system motivated by the triblock copolymer theoryis studied as a nonlocal geometric variational problem. The free energyof the system is the sum of two terms: the total size of the interfacesseparating the three constituents, and a longer ranging interaction energythat inhibits micro-domains from unlimited growth. In a particular parameterrange there is an assembly of many core-shells that exists as a stationaryset of the free energy functional. The cores form regions occupied by thefirst constituent of the ternary system, the shells form regionsoccupied by the second constituent, and the background is taken by thethird constituent. The constructive proof of the existence theorem revealsmuch information about the core-shell stationary assembly: asymptoticallyone can determine the sizes and locations of all the core-shells in theassembly. The proof also implies a kind of stability for the stationaryassembly.

Citation: Xiaofeng Ren, Chong Wang. A stationary core-shell assembly in a ternary inhibitory system. Discrete & Continuous Dynamical Systems - A, 2017, 37 (2) : 983-1012. doi: 10.3934/dcds.2017041
##### References:

show all references

##### References:
A double bubble assembly on the left and a core-shell assembly on the right
 [1] Xiaofeng Ren. Shell structure as solution to a free boundary problem from block copolymer morphology. Discrete & Continuous Dynamical Systems - A, 2009, 24 (3) : 979-1003. doi: 10.3934/dcds.2009.24.979 [2] Tamara Fastovska. Long-time behaviour of a radially symmetric fluid-shell interaction system. Discrete & Continuous Dynamical Systems - A, 2018, 38 (3) : 1315-1348. doi: 10.3934/dcds.2018054 [3] Stéphanie Portet, Julien Arino. An in vivo intermediate filament assembly model. Mathematical Biosciences & Engineering, 2009, 6 (1) : 117-134. doi: 10.3934/mbe.2009.6.117 [4] Peng-Fei Yao. On shallow shell equations. Discrete & Continuous Dynamical Systems - S, 2009, 2 (3) : 697-722. doi: 10.3934/dcdss.2009.2.697 [5] Murray R. Bremner. Polynomial identities for ternary intermolecular recombination. Discrete & Continuous Dynamical Systems - S, 2011, 4 (6) : 1387-1399. doi: 10.3934/dcdss.2011.4.1387 [6] Michael Helmers, Barbara Niethammer, Xiaofeng Ren. Evolution in off-critical diblock copolymer melts. Networks & Heterogeneous Media, 2008, 3 (3) : 615-632. doi: 10.3934/nhm.2008.3.615 [7] Jean-Philippe Lessard, Evelyn Sander, Thomas Wanner. Rigorous continuation of bifurcation points in the diblock copolymer equation. Journal of Computational Dynamics, 2017, 4 (1&2) : 71-118. doi: 10.3934/jcd.2017003 [8] Tatsuya Maruta, Yusuke Oya. On optimal ternary linear codes of dimension 6. Advances in Mathematics of Communications, 2011, 5 (3) : 505-520. doi: 10.3934/amc.2011.5.505 [9] Anish Ghosh, Dubi Kelmer. A quantitative Oppenheim theorem for generic ternary quadratic forms. Journal of Modern Dynamics, 2018, 12: 1-8. doi: 10.3934/jmd.2018001 [10] Marek Bodnar, Urszula Foryś. Time Delay In Necrotic Core Formation. Mathematical Biosciences & Engineering, 2005, 2 (3) : 461-472. doi: 10.3934/mbe.2005.2.461 [11] G.W. Hunt, Gabriel J. Lord, Mark A. Peletier. Cylindrical shell buckling: a characterization of localization and periodicity. Discrete & Continuous Dynamical Systems - B, 2003, 3 (4) : 505-518. doi: 10.3934/dcdsb.2003.3.505 [12] Thomas Wanner. Computer-assisted equilibrium validation for the diblock copolymer model. Discrete & Continuous Dynamical Systems - A, 2017, 37 (2) : 1075-1107. doi: 10.3934/dcds.2017045 [13] Michael Baur, Marco Gaertler, Robert Görke, Marcus Krug, Dorothea Wagner. Augmenting $k$-core generation with preferential attachment. Networks & Heterogeneous Media, 2008, 3 (2) : 277-294. doi: 10.3934/nhm.2008.3.277 [14] François Alouges, Sylvain Faure, Jutta Steiner. The vortex core structure inside spherical ferromagnetic particles. Discrete & Continuous Dynamical Systems - A, 2010, 27 (4) : 1259-1282. doi: 10.3934/dcds.2010.27.1259 [15] Domingo González, Gamaliel Blé. Core entropy of polynomials with a critical point of maximal order. Discrete & Continuous Dynamical Systems - A, 2019, 39 (1) : 115-130. doi: 10.3934/dcds.2019005 [16] Jason Chao-Hsien Pan, Ku-Kuang Chang, Yu-Cheng Hsiao. Optimal inventory policies for serial-type and assembly-type supply chains with equal sized batch. Journal of Industrial & Management Optimization, 2015, 11 (3) : 1021-1040. doi: 10.3934/jimo.2015.11.1021 [17] Tania Biswas, Sheetal Dharmatti. Control problems and invariant subspaces for sabra shell model of turbulence. Evolution Equations & Control Theory, 2018, 7 (3) : 417-445. doi: 10.3934/eect.2018021 [18] Inger Daniels, Catherine Lebiedzik. Existence and uniqueness of a structural acoustic model involving a nonlinear shell. Discrete & Continuous Dynamical Systems - S, 2008, 1 (2) : 243-252. doi: 10.3934/dcdss.2008.1.243 [19] Shumin Li, Masahiro Yamamoto, Bernadette Miara. A Carleman estimate for the linear shallow shell equation and an inverse source problem. Discrete & Continuous Dynamical Systems - A, 2009, 23 (1&2) : 367-380. doi: 10.3934/dcds.2009.23.367 [20] Sylvia Anicic. Existence theorem for a first-order Koiter nonlinear shell model. Discrete & Continuous Dynamical Systems - S, 2019, 12 (6) : 1535-1545. doi: 10.3934/dcdss.2019106

2018 Impact Factor: 1.143


text:
# Talk:Convolution power

WikiProject Mathematics (Rated Stub-class, Low-priority)
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Field:  Analysis

## Tidy up

The following things need to be addressed:

• The whole of this article depends on the definition of negative convolution power, which in turn depends on the inverse Fourier transform. It must be made abundantly clear that for the majority of functions, this does not exist. I'm sure similar limitations apply to the exponentiation and logarithm stuff.
err.. what do you mean by the 'majority of functions' ? If you mean that the cardinality of the set of all functions from R->R (or C->C) on which the inverse transform operator may be applied is less than the cardinality of the set all functions from R->R (C->C), well, that limitation applies to pretty much all the calculus/analysis articles on wikipedia. So, I suppose, it would be good to be abundantly clear what you mean by 'the majority of functions'. Perhaps the best way would be to give some examples of these convolution operators applied to some functions, and to give some other examples where the operators fail. mike40033 (talk) 09:13, 14 August 2008 (UTC)
• The notation is highly inconsistent. If ${\displaystyle \ k}$ is a constant, then ${\displaystyle \ x^{*(k)}}$, ${\displaystyle \ x^{*(y)}}$ and ${\displaystyle \ k^{*(y)}}$ are all defined differently.
Partially fixed.
• The series expansion stuff is woefully under-explained -- what precisely are the "replacement" rules? This is probably because it's of very limited applicability (how exactly does it hold for "e.g. Fourier series"?). It's really just a limited case of the convolution theorem, which is much more general. Therefore, I'm not quite sure what the point of this section is.
Removed.
• The series expansion relationship involving ${\displaystyle e^{*(x)}=\sum _{k=0}^{\infty }{\frac {x^{*(k)}}{k!}}}$ is just a repetition of the previous definition.
Removed.
• Need references for definitions of the following (Google, etc. brings up practically nothing):
• convolution power for anything other than the natural numbers
• convolution exponentiation and logarithm section - where else are these definitions used??

Oli Filth 18:48, 5 July 2007 (UTC)

Certainly a bit of material on convolution exponential and logarithmic functions is found in Daniel Stroock's book Probability Theory: An Analytic View, in the section on infinite divisibility. Michael Hardy 13:15, 19 July 2007 (UTC)

## Central limit theorem

Shouldn't there be some mention of the Central limit theorem as an application of convolution powers? -- Spireguy (talk) 22:10, 15 September 2009 (UTC)

Yes. I will add something. Sławomir Biały (talk) 11:41, 18 September 2009 (UTC)


text:
# How to use a string to form the name of an existing command?

In my project, I already have a file containing commands for names of my documents :

\newcommand{\docOne}{Title of my first document}
\newcommand{\docTwo}{Title of my second document}
...

I've also created a function to obtain the word associated with a number between 1 and 9:

\newcommand{\getIntToString}{wrongInput}
\newcommand{\setIntToString}[1]{%
\ifthenelse{\equal{#1}{1}}{\renewcommand{\getIntToString}{One}}{}%
\ifthenelse{\equal{#1}{2}}{\renewcommand{\getIntToString}{Two}}{}%
...
}

(where \ifthenelse and \equal comes from the ifthen package)

This function produce the desired output, for instance when compiling \setIntToString{0}\getIntToString\ \setIntToString{1}\getIntToString in the document, I obtain "wrongInput One" in the pdf.

I would like to write a function like the following:

\newcommand{\docTitle}[1]{%
\doc\setIntToString{#1}\getIntToString%
}

So that \doc\setIntToString{#1}\getIntToString becomes \docOne if I write \docTitle{1}, \docTwo if I write \docTitle{2}, etc.

I guess this has something to do with \csname, \expandafter, \csuse, \@namedef ... but I don't know how to use them.

Here is my MWE:

\documentclass{report}
\usepackage{ifthen}

\newcommand{\docOne}{Title of my first document}
\newcommand{\docTwo}{Title of my second document}

\newcommand{\getIntToString}{wrongInput}
\newcommand{\setIntToString}[1]{%
\ifthenelse{\equal{#1}{1}}{\renewcommand{\getIntToString}{One}}{}%
\ifthenelse{\equal{#1}{2}}{\renewcommand{\getIntToString}{Two}}{}%
\ifthenelse{\equal{#1}{3}}{\renewcommand{\getIntToString}{Three}}{}%
\ifthenelse{\equal{#1}{4}}{\renewcommand{\getIntToString}{Four}}{}%
\ifthenelse{\equal{#1}{5}}{\renewcommand{\getIntToString}{Five}}{}%
\ifthenelse{\equal{#1}{6}}{\renewcommand{\getIntToString}{Six}}{}%
\ifthenelse{\equal{#1}{7}}{\renewcommand{\getIntToString}{Seven}}{}%
\ifthenelse{\equal{#1}{8}}{\renewcommand{\getIntToString}{Eight}}{}%
\ifthenelse{\equal{#1}{9}}{\renewcommand{\getIntToString}{Nine}}{}%
}

\newcommand{\docTitle}[1]{%
\doc\setIntToString{#1}\getIntToString%
}

\begin{document}

\setIntToString{0}\getIntToString\ \setIntToString{1}\getIntToString

%\docTitle{1}, \docTitle{2}

\end{document}

• You need \csname ... \endcsname. – user156344 May 7 at 15:03
• I guess, yes, but how? Especially that the desired command names already exist... – Apeiron May 7 at 15:08

You might do

\documentclass{report}
\usepackage{ifthen}

\newcommand{\docOne}{Title of my first document}
\newcommand{\docTwo}{Title of my second document}

\newcommand{\getIntToString}{wrongInput}
\newcommand{\setIntToString}[1]{%
\ifthenelse{\equal{#1}{1}}{\renewcommand{\getIntToString}{One}}{}%
\ifthenelse{\equal{#1}{2}}{\renewcommand{\getIntToString}{Two}}{}%
\ifthenelse{\equal{#1}{3}}{\renewcommand{\getIntToString}{Three}}{}%
\ifthenelse{\equal{#1}{4}}{\renewcommand{\getIntToString}{Four}}{}%
\ifthenelse{\equal{#1}{5}}{\renewcommand{\getIntToString}{Five}}{}%
\ifthenelse{\equal{#1}{6}}{\renewcommand{\getIntToString}{Six}}{}%
\ifthenelse{\equal{#1}{7}}{\renewcommand{\getIntToString}{Seven}}{}%
\ifthenelse{\equal{#1}{8}}{\renewcommand{\getIntToString}{Eight}}{}%
\ifthenelse{\equal{#1}{9}}{\renewcommand{\getIntToString}{Nine}}{}%
}

\newcommand{\docTitle}[1]{%
\begingroup\setIntToString{#1}%
\expandafter\endgroup
\csname doc\getIntToString\endcsname
}

\begin{document}

\setIntToString{0}\getIntToString\ \setIntToString{1}\getIntToString

\docTitle{1}, \docTitle{2}

\end{document}

The group ensures that the redefinition of \getIntToString is forgotten.

Much better with higher level functions.

\documentclass{report}
\usepackage{xparse}

\ExplSyntaxOn
\cs_new:Nn \apeiron_int_to_string:n
{
\int_case:nnF { #1 }
{
{1}{One}
{2}{Two}
{3}{Three}
{4}{Four}
{5}{Five}
{6}{Six}
{7}{Seven}
{8}{Eight}
{9}{Nine}
}
{WrongInput}
}

\NewDocumentCommand{\docTitle}{m}
{
\use:c { doc \apeiron_int_to_string:n { #1 } }
}
\ExplSyntaxOff

\newcommand{\docOne}{Title of my first document}
\newcommand{\docTwo}{Title of my second document}

\newcommand{\docWrongInput}{WRONG}

\begin{document}

\docTitle{1}, \docTitle{2}, \docTitle{42}

\end{document}

Using the \name-macro (see in the example below) you can easily both define and call control-sequence-tokens/macros whose names may also contain digits. These digits then may as well come from the values of counters. Using this technique you are not bound to using digits.

The gist of the \name-macro is: It takes material before the first following opening brace for its first argument and material nested into the first following opening brace and the matching closing brace for its second argument and applies \csname..\endcsname to that second argument while leaving the first argument in place. Hereby the first argument my be empty.

You can do things like
\name\global\long\def{foo}...\global\long\def\foo...
\name{foo}...\foo...
\name\string{foo}...\string\foo...
\name\expandafter{foo}...\expandafter\foo...
\name\name\expandafter{foo}{bar}\expandafter\foo\bar
\name\name\let{foo}={bar}\let\foo=\bar

\documentclass{article}

\newcommand\exchange[2]{#2#1}%
\newcommand\name{}\long\def\name#1#{\romannumeral0\innername{#1}}%
\newcommand\innername[2]{\expandafter\exchange\expandafter{\csname#2\endcsname}{ #1}}%
\newcommand\checkwhethernameundefined{}\name\global\let\checkwhethernameundefined={@ifundefined}%

% let's define the macros \doc1, \doc2, ... , \doc12, and \docWeird:
\name\newcommand{doc1}{Title of my first document}
\name\newcommand{doc2}{Title of my second document}
\name\newcommand{doc3}{Title of my third document}
\name\newcommand{doc4}{Title of my fourth document}
\name\newcommand{doc5}{Title of my fifth document}
\name\newcommand{doc6}{Title of my sixth document}
\name\newcommand{doc7}{Title of my seventh document}
\name\newcommand{doc8}{Title of my eighth document}
\name\newcommand{doc9}{Title of my nineth document}
\name\newcommand{doc10}{Title of my tenth document}
\name\newcommand{doc11}{Title of my eleventh document}
\name\newcommand{doc12}{Title of my twelveth document}
\name\newcommand{docWeird}{Title of my weird document}

% let's define a macro for calling these macros:
\newcommand{\docTitle}[1]{%
\checkwhethernameundefined{doc#1}{%
\begingroup
\def\mytempa{#1}\name{@onelevel@sanitize}\mytempa
\name{@latex@error}%
{\string\docTitle{\mytempa} not available:\MessageBreak
Command \name\string{doc\mytempa} undefined}%
{Use \string\docTitle\space with number in argument in range from 1 to 12 or with\MessageBreak
the string "Weird".}%
\endgroup
}{%
\name{doc#1}%
}%
}%

\begin{document}

\noindent
\verb|\docTitle{1}|: \docTitle{1}\\
\verb|\docTitle{2}|: \docTitle{2}\\
\verb|\docTitle{3}|: \docTitle{3}\\
\verb|\docTitle{4}|: \docTitle{4}\\
\verb|\docTitle{5}|: \docTitle{5}\\
\verb|\docTitle{6}|: \docTitle{6}\\
\verb|\docTitle{7}|: \docTitle{7}\\
\verb|\docTitle{8}|: \docTitle{8}\\
\verb|\docTitle{9}|: \docTitle{9}\\
\verb|\docTitle{10}|: \docTitle{10}\\
\verb|\docTitle{11}|: \docTitle{11}\\
%
% You can also obtain the thing depending on the value stored in a count-register/a LaTeX counter:
\newcounter{docnumber}%
\setcounter{docnumber}{12}%
Counter \verb|docnumber|'s current value is: \number\value{docnumber}\\
\verb|\docTitle{\number\value{docnumber}}|: \docTitle{\number\value{docnumber}}\\
%
% You are not bound to numbers:
% You can define and use macros with almost arbitrary strings in their names:
\verb|\docTitle{Weird}|: \docTitle{Weird}\\
%
% This yields an error-message as \doc13 is undefined:
% \verb|\docTitle{13}|: \docTitle{13}\\

\end{document}

If you insist in control sequences \docOne, \docTwo, ... instead of \doc1, \doc2, ... , you can still use the \name-macro for creating a generic macro which "maps" from digit sequences to these control-sequences.

In the example below the macro \map has the syntax

\map{⟨macro-name-part before the string which is to be mapped to another string⟩}%
{⟨string which is to be mapped to another string⟩}%
{⟨macro-name-part behind the string which is to be mapped to another string⟩}

E.g.,
\map{Foo}{3}{Bar}
checks whether a mapping from the string 3 to some other string is defined.
If not so, an error-message is raised.
If so, and if 3 is mapped to the string Three, it is checked whether the control-sequence \FooThreeBar is defined.
If not so, an error-message is raised.
If so, that control-sequence is called.

The macro \definemapping{⟨string⟩}{⟨replacement⟩} is used for defining the mappings.

E.g., for the above to work, you would need to apply the command \definemapping{3}{Three}% first.

\documentclass{article}

\newcommand\exchange[2]{#2#1}%
\newcommand\name{}\long\def\name#1#{\romannumeral0\innername{#1}}%
\newcommand\innername[2]{\expandafter\exchange\expandafter{\csname#2\endcsname}{ #1}}%
\newcommand\checkwhethernameundefined{}\name\global\let\checkwhethernameundefined={@ifundefined}%

\newcommand*\Mappinglist{}%

\newcommand\map[3]{%
\checkwhethernameundefined{my-map-phrase-#2}{%
\begingroup
\def\mytempa{#2}\name{@onelevel@sanitize}\mytempa
\name{@latex@error}{%
Cannot map from string \mytempa' to textual phrase:\MessageBreak
Macro \name\string{my-map-phrase-\mytempa} undefined
}{Mapping is defined for the string(s) \MessageBreak\Mappinglist.\MessageBreak
You can use \string\definemapping\space for defining more mappings.
}%
\endgroup
}{%
\checkwhethernameundefined{#1\name{my-map-phrase-#2}#3}{%
\begingroup
\def\mytempa{#1}\name{@onelevel@sanitize}\mytempa
\def\mytempb{#3}\name{@onelevel@sanitize}\mytempb
\name{@latex@error}{%
Control sequence \name\string{\mytempa\name{my-map-phrase-#2}\mytempb} undefined%
}{%
Seems the control sequence \name\string{\mytempa\name{my-map-phrase-#2}\mytempb} needs to be defined.%
}%
\endgroup
}{%
\name{#1\name{my-map-phrase-#2}#3}%
}%
}%
}%

\newcommand\definemapping[2]{%
\checkwhethernameundefined{my-map-phrase-#1}{%
\name\newcommand{my-map-phrase-#1}{#2}%
\name\name\global\let{my-map-phrase-#1}={my-map-phrase-#1}%
\begingroup
\def\mytempa{#1'}\name{@onelevel@sanitize}\mytempa
}{%
\begingroup
\def\mytempa{#1}\name{@onelevel@sanitize}\mytempa
\name{@latex@error}{%
Mapping for string \mytempa'\space already defined%
}{%
Each string can have only one mapping.\MessageBreak
Mappings are already defined for the following strings:\MessageBreak
\Mappinglist.%
}%
\endgroup
}%
}%

\definemapping{1}{One}%
%\definemapping{1}{Uno}% <- This will raise an error-message.
\definemapping{2}{Two}%
\definemapping{3}{Three}%
\definemapping{4}{Four}%
\definemapping{5}{Five}%
\definemapping{6}{Six}%
\definemapping{7}{Seven}%
\definemapping{8}{Eight}%
\definemapping{9}{Nine}%
\definemapping{10}{Ten}%
\definemapping{11}{Eleven}%
\definemapping{12}{Twelve}%
\definemapping{13}{Thirteen}%
\definemapping{Weird}{Weird}%

% let's define the macros \docOne, \docTwo, ... , \docTwelve and \docWeird:
\newcommand{\docOne}{Title of my first document}
\newcommand{\docTwo}{Title of my second document}
\newcommand{\docThree}{Title of my third document}
\newcommand{\docFour}{Title of my fourth document}
\newcommand{\docFive}{Title of my fifth document}
\newcommand{\docSix}{Title of my sixth document}
\newcommand{\docSeven}{Title of my seventh document}
\newcommand{\docEight}{Title of my eighth document}
\newcommand{\docNine}{Title of my nineth document}
\newcommand{\docTen}{Title of my tenth document}
\newcommand{\docEleven}{Title of my eleventh document}
\newcommand{\docTwelve}{Title of my twelveth document}
\newcommand{\docWeird}{Title of my weird document}

% let's define a macro for calling these macros:
\newcommand{\docTitle}[1]{\map{doc}{#1}{}}%

\begin{document}

\noindent
\verb|\docTitle{1}|: \docTitle{1}\\
\verb|\docTitle{2}|: \docTitle{2}\\
\verb|\docTitle{3}|: \docTitle{3}\\
\verb|\docTitle{4}|: \docTitle{4}\\
\verb|\docTitle{5}|: \docTitle{5}\\
\verb|\docTitle{6}|: \docTitle{6}\\
\verb|\docTitle{7}|: \docTitle{7}\\
\verb|\docTitle{8}|: \docTitle{8}\\
\verb|\docTitle{9}|: \docTitle{9}\\
\verb|\docTitle{10}|: \docTitle{10}\\
\verb|\docTitle{11}|: \docTitle{11}\\
%
% You can also obtain the thing depending on the value stored in a count-register/a LaTeX counter:
\newcounter{docnumber}%
\setcounter{docnumber}{12}%
Counter \verb|docnumber|'s current value is: \number\value{docnumber}\\
\verb|\docTitle{\number\value{docnumber}}|: \docTitle{\number\value{docnumber}}\\
%
% You are not bound to numbers:
% You can define and use macros with almost arbitrary strings in their names:
\verb|\docTitle{Weird}|: \docTitle{Weird}\\
%
% This yields an error-message as \docThirteen is undefined:
%\verb|\docTitle{13}|: \docTitle{13}\\
%
% This yields an error-message as mapping from 14 to phrase is undefined:
%\verb|\docTitle{14}|: \docTitle{14}\\

\end{document}

Another common strategy is using macro-names which have roman numerals as parts of their names:

\documentclass{article}

% let's define the macros \doci, \docii, ... , \docxii:
\newcommand{\doci}{Title of my first document}
\newcommand{\docii}{Title of my second document}
\newcommand{\dociii}{Title of my third document}
\newcommand{\dociv}{Title of my fourth document}
\newcommand{\docv}{Title of my fifth document}
\newcommand{\docvi}{Title of my sixth document}
\newcommand{\docvii}{Title of my seventh document}
\newcommand{\docviii}{Title of my eighth document}
\newcommand{\docix}{Title of my nineth document}
\newcommand{\docx}{Title of my tenth document}
\newcommand{\docxi}{Title of my eleventh document}
\newcommand{\docxii}{Title of my twelveth document}

% let's define a macro for calling these macros:
\makeatletter
\newcommand{\docTitle}[1]{%
\@ifundefined{doc\romannumeral0#1}{%
\begingroup
\def\mytempa{#1}\@onelevel@sanitize\mytempa
\expandafter\def\expandafter\mytempb\expandafter{\romannumeral0#1}%
\@onelevel@sanitize\mytempb
\@latex@error
{\string\docTitle{\mytempa} not available:\MessageBreak
Command \expandafter\string\csname doc\mytempb\endcsname\space undefined}%
{Use \string\docTitle\space with number in argument in range from 1 to 12.}%
\endgroup
}{%
\csname doc\romannumeral0#1\endcsname
}%
}%
\makeatother

\begin{document}

\noindent
\verb|\docTitle{1}|: \docTitle{1}\\
\verb|\docTitle{2}|: \docTitle{2}\\
\verb|\docTitle{3}|: \docTitle{3}\\
\verb|\docTitle{4}|: \docTitle{4}\\
\verb|\docTitle{5}|: \docTitle{5}\\
\verb|\docTitle{6}|: \docTitle{6}\\
\verb|\docTitle{7}|: \docTitle{7}\\
\verb|\docTitle{8}|: \docTitle{8}\\
\verb|\docTitle{9}|: \docTitle{9}\\
\verb|\docTitle{10}|: \docTitle{10}\\
\verb|\docTitle{11}|: \docTitle{11}\\
%
% You can also obtain the thing depending on the value stored in a count-register/a LaTeX counter:
\newcounter{docnumber}%
\setcounter{docnumber}{12}%
Counter \verb|docnumber|'s current value is: \number\value{docnumber}\\
\verb|\docTitle{\number\value{docnumber}}|: \docTitle{\number\value{docnumber}}\\
%
% This yields an error-message as \docxiii is undefined:
%\verb|\docTitle{13}|: \docTitle{13}\\

\end{document}
`
• This was very enlightning, thank you very much :) – Apeiron May 9 at 6:39


text:
# CERN Preprints

2015-09-01
16:24
 Search for a high-mass Higgs boson decaying to a $W$ boson pair in $pp$ collisions at $\sqrt{s} = 8$ TeV with the ATLAS detector / , A search for a high-mass Higgs boson $H$ is performed in the $H\to WW\to\ell\nu\ell\nu$ and $H\to WW\to \ell\nu{qq}$ decay channels using $pp$ collision data corresponding to an integrated luminosity of 20.3 fb$^{-1}$ collected at $\sqrt{s}=$ 8 TeV by the ATLAS detector at the Large Hadron Collider. [...] CERN-PH-EP-2015-185 ; arXiv:1509.00389. - 2015. - 51 p. Previous draft version - Full text - Full text

2015-08-31
20:14
 Search for single top-quark production via flavour changing neutral currents at 8 TeV with the ATLAS detector / , A search for single top-quark production via flavour changing neutral current processes from gluon plus up- or charm--quark initial states in proton--proton collisions at the LHC is presented. [...] CERN-PH-EP-2015-172 ; arXiv:1509.00294. - 2015. - 29 p. Previous draft version - Full text - Full text

2015-08-31
16:35
 Search for invisible decays of a Higgs boson using vector-boson fusion in $pp$ collisions at $\sqrt{s}=8$ TeV with the ATLAS detector / , A search for a Higgs boson produced via vector-boson fusion and decaying into invisible particles is presented, using 20.3 fb$^{-1}$ of proton--proton collision data at a centre-of-mass energy of 8 TeV recorded by the ATLAS detector at the LHC. [...] CERN-PH-EP-2015-186 ; arXiv:1508.07869. - 2015. - 26 p. Previous draft version - Full text - Full text

2015-08-31
15:06
 CMS inner detector: the Run 1 to Run 2 transition, and first experience of Run 2 / Lange, Clemens (Zurich U.) /CMS Collaboration The CMS inner detector is the largest silicon tracker ever built. Located around the center of CMS where the LHC particle beams are brought into collision, it consists of a hybrid pixel detector with 66 million channels and a 200~m$^2$ silicon strip detector with about 10~million readout channels. [...] CMS-CR-2015-158.- Geneva : CERN, 2015 - 8 p. Fulltext: PDF; In : 24th International Workshop on Vertex Detectors, Santa Fe, New Mexico, United States, 1 - 5 Jun 2015

2015-08-31
14:48
 ROOT - A C++ Framework for Petabyte Data Storage, Statistical Analysis and Visualization / Naumann, Axel (CERN) ; Antcheva, Ilka ; Ballintijn, Maarten ; Bellenot, Bertrand (CERN) ; Biskup, Marek ; Brun, Rene (CERN) ; Buncic, Nenad ; Canal, Philippe (Fermilab) ; Casadei, Diego (New York University, NY, USA) ; Couet, Olivier (CERN) et al. ROOT is an object-oriented C++ framework conceived in the high-energy physics (HEP) community, designed for storing and analyzing petabytes of data in an efficient way. [...] CERN-PH-SFT-2015-001. - 2009. - 33 p.

2015-08-31
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 Search for a charged Higgs boson in pp collisions at $\sqrt{s} =$ 8 TeV / , A search for a charged Higgs boson is performed with a data sample corresponding to an integrated luminosity of 19.7 $\pm$ 0.5 fb$^{-1}$ collected with the CMS detector in proton-proton collisions at $\sqrt{s} =$ 8 TeV. [...] CERN-PH-EP-2015-221 ; CMS-HIG-14-023-003 ; arXiv:1508.07774. - 2015. Additional information for the analysis - CMS AuthorList - Full text - Full text

2015-08-31
07:20
 Searches for CP Violation in charm at LHCb / Naik, Paras P (Bristol U.) LHCb has collected the world's largest sample of charmed hadrons. [...] arXiv:1508.07200. - 2015. - 10 p. Preprint

2015-08-30
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 Search for supersymmetry in the vector-boson fusion topology in proton-proton collisions at $\sqrt{s} =$ 8 TeV / , The first search for supersymmetry in the vector-boson fusion topology is presented. [...] CERN-PH-EP-2015-213 ; CMS-SUS-14-005-003 ; arXiv:1508.07628. - 2015. Additional information for the analysis - CMS AuthorList - Full text - Full text

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 Measurement of differential cross sections for Higgs boson production in the diphoton decay channel in pp collisions at $\sqrt{s}=$ 8 TeV / , A measurement is presented of differential cross sections for the Higgs boson (H) production in pp collisions at $\sqrt{s}=$ 8 TeV. [...] CERN-PH-EP-2015-195 ; CMS-HIG-14-016-003 ; arXiv:1508.07819. - 2015. Additional information for the analysis - CMS AuthorList - Full text - Full text

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text:
ISSN 0439-755X
CN 11-1911/B

中国科学院心理研究所

Acta Psychologica Sinica ›› 2014, Vol. 46 ›› Issue (5): 666-676.

### Effect of Interpersonal Distance on Neural Basis of Self- and Other Representation: Evidence from the oFRN Component

YANG Shuai;HUANG Xiting;CHEN Youguo;FU Yuling;LIU Mengchao

1. (1 Key Laboratory of Cognition and Personality (Ministry of Education) and School of Psychology, Southwest University, Chongqing 400715, China) (2 Teaching and Research Institute of Political Theory, Chongqing University of Posts and Telecommunications, Chongqing 400065, China) (3 Chengdu University of Technology, Chengdu 610059, China)
• Received:2013-01-21 Online:2014-05-24 Published:2014-05-24
• Contact: HUANG Xiting

Abstract:

Since the concept of inclusion of other in the self (IOS) appeared, the topic of interpersonal distance has become an important research framework in social cognition. Similar to the self-reference effect, the other-reference effect was found when information related to close others (e.g. mother) was processed. These findings suggest that the representations of self and others are similar. Interestingly, studies on overcome evaluation indicated the same phenomenon. When participants monitor errors made by others, a similar ERP waveform could be detected by the observers just as they committed errors themselves. It is so-called observer feedback related negativity (oFRN). Although interpersonal distance is considered to play an important role, it is still unclear that how self- and other representation are modulated by interpersonal distance, since there are no consistent findings in previous studies. Thus, the purpose of this article was to confirm the effect of interpersonal distance on the oFRN, and to reveal the similarity of brain mechanism between self and others. Seventeen participants (9 males and 8 females) completed a simplified monetary gambling task in self-execution, friend-observation, and stranger-observation conditions, respectively. Actual observation paradigm was used in observation conditions, in which the electroencephalogram (EEG) participant and the partner were sitting side by side in the gambling task. The data were recorded from 64 scalp sites using tin electrodes mounted in an elastic cap. ERPs time-locked to the onset of feedback stimuli were averaged for epochs of 700ms starting 100ms prior to the stimulus (baseline). The ERPs were statistically evaluated by SPSS 15.0 with repeated measure analysis of variance (ANOVA). Results showed that there were more negative ERP responses, as indexed by the FRN, for loss trials compare to gain trials in monitoring both self and friend gambling (p<0.01, respectively). No significant difference was detected between loss and gain trials in stranger-observation condition (p>0.05), but a negative deflection was found in loss feedback. As for the oFRN, a marginally significant difference was found between friend- and stranger-observation conditions when distinct waves were statistically evaluated (p = 0.062). A correlation analysis was made to further explore the agency effect on the oFRN, and the results showed that interpersonal distance (evaluated by a subjective assessment) and the ERPs were significantly related (p<0.05). It indicates that a friend (who was rated closer) might be represented slightly more important than a stranger in “mirror” performance-monitoring system. That is, the oFRN is sensitive to interpersonal distance. The negative deflection across three agency conditions in loss trials suggested that similar monitoring mechanisms for both self- and other generated errors may be partly based on the mirror neuron system. However, the agency difference on the oFRN between friend and stranger observation conditions suggested that the vicarious experiences of others were modulated by interpersonal distance.


text:
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# Find parametric equations for the line segment from $(-2, 18, 31)$ to $(11, -4, 48)$.

## $x=-2+13 t, \quad y=18-22 t, \quad z=31+17 t, \quad 0 \leq t \leq 1$

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### Video Transcript

So again, we want to find the parametric equation for the line segment minus two, 18, 31 two, 11 -4 and 48. Again, this is my there's no doubting that I'll start here And then my normal vector and is just 11 last two -4 -18. And then 48 when is 31? It was just 13 minus 22 Than seven in 17. So they question is a Last three times and was his -2 18, 31. Last three times 13 -20- 17. This is my anguish. Yeah. And then we wanted to start at minus 2 18 and 31. So he must start from zero And then must change at 11 -4 and 48 13 must end at one. That's it. That stands

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This page contains a combination of traditional lecture materials (slides) and code demonstrating the relevant methods. The short course will proceed by working through both. We will use several recent packages in our examples; see the About page for information about the package versions.

library(tidyverse)
library(grpreg)
library(splines)
library(refund)
library(refund.shiny)

In this section we’ll use the NHANES dataset to …

## Scalar-on-function regression

Slides below introduce scalar-on-function regression.

PLACEHOLDER FOR SPEAKERDECK SLIDES

## Practical example

PLACEHOLDER FOR CODE


text:
## Lists

Lists are a very common structure in mathematical papers. For example, a list may represent different parts of a theorem, different parts of a proof, a list of properties, etc. Latex has several pre-defined environments (itemize, enumerate, description) for typesetting lists. Alternatively, a list can also be typeset without any special coding other than inserting paragraph breaks (i.e., one or more blank lines) between the list items, or inline, with commas or semicolons separating the list items.

The main considerations in typesetting a list are (i) the choice of environment (if any) for the list; (ii) the style of labels; and (iii) the formatting (if any) of the labels.

My advice on these issues is the following:

### Use labels with balanced parentheses, e.g., "(i)", "(a)", "(P)".

Labels with unbalanced parentheses, such as "1)", look ugly, and they can mess with the parenthesis matching of an editor. Labels with a period at the end (e.g. "1.") make references to these labels tricky. For example, "By 1. we have" doesn't look good, but the alternative "By 1 we have" has its own problem. Using labels with balanced parentheses avoids these problems.

### Use lower case Roman letters (e.g., "(i)"), rather than numerals, as labels for parts of a theorem

This is the most common practice in mathematical writing. Its main advantage is that it avoids conflicts with numbered equations. Even if one doesn't have numbered equations in a paper, using numbered labels still may cause confusion, since upon seeing a reference like "by (1) we have ..." most readers would instinctively assume that (1) refers to an equation and hunt for an equation numbered (1).

### Avoid using explicit fonts or special formatting in conjunction with labels

I often see papers in which the authors try to achieve a particular "look" of a label, by placing the label, or a reference to it, inside a pair of dollar signs ("$(i)$"), or within a font instruction ("\textit{(i)}"). This is unnecessary, and it is contrary to standard typesetting practice. Moreover, the dollar sign version violates the basic rule of using math mode (i.e., $...$) only for mathematical material.

(There is one situation in which explicit formatting may be appropriate: In a list inside theorem, it may be appropriate to enclose "(i)" in a "\textup{...}" to prevent it from being italicized; i.e., instead of \item[(i)] use \item[\textup{(i)}].)

### Use the itemize or enumerate environment for lists in theorems, or other lists with short items (no more than a few lines each). Use no special environment, other than paragraph breaks separating the items, for lists in proofs, or "long" lists

The list environments cause the entire list to be indented. This is the desired look for a short list (e.g., in a theorem, or a list of properties), but it is inappropriate for longer lists, especially if they extend over more than one page.

To generate the item labels, put the desired label in brackets after "\item", e.g., "\item[(i)]", etc.

If explicit labels are given, there is no essential difference between the itemize and enumerate environments, so either one can be used. Without such explicit labels, an enumerate environment generates numbered labels (though usually not in the desired style), whereas an itemize environment produces a bullet list.

### Lists at the beginning of a theorem statement. The \mbox{} trick.

Anyone who has had to typeset theorems with lists probably has run into the problem that if the list occurs at the beginning of the statement (i.e., with no text preceding it inside the theorem), it gets typeset right after the theorem name, and not on a new line and indented in the same way as the other items, and putting spacing commands or explicit newlines before the list doesn't seem to help. The reason for this behavior is that the theorem environment "eats up" (nearly) any space at the beginning of the theorem, so commands like \hfill or \newline or \par have no effect. The remedy is to trick TeX into thinking that there is material before the list, by putting an "\mbox{}" before the list. See the example below.

### An example. A three part theorem, with proof

Here is an example that illustrates a typical situation.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{theorem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\mbox{}
\begin{itemize}

\item[(i)] $QP\cap \Omega \subset Q$.

\item[(ii)] If $(X,T)$ is minimal, then $QP=PQ$.

\item[(iii)]  Let $\pi:X \to Y$ be a homomorphism of minimal
flows. Then $\pi(Q_XP_X)=Q_YP_Y$.

\end{itemize}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{theorem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{proof}

(i) Let $(x,z)\in QP\cap \Omega$. Let $y\in X$ such that
$(x,y)\in Q$ and $(y,z)\in P$. Let $I$ be a minimal right ideal
in $E(X)$ such that $yr=zr$ for all $r\in I$, and let $u$ be an
idempotent in $I$ such that $(x,z)u=(x,z)$. Then $(x,z)=(x,z)u= (x,y)u \in Qu\subset Q$.

(ii)  Note that if the relation $R$ is symmetric, so is
$C(R)$. Therefore $C(Q)$ is symmetric. Now $C(Q)=PQ$ (\cite{AGl}), so
$QP=(PQ) ^{-1}=C(Q)^{-1}=C(Q)=PQ$.

(iii)  Let $(y_1,y_2)\in Q_Y$ and $(y_2,y_3)\in P_Y$. Let
$v$ be a minimal idempotent in $E(Y)$ such that $y_3=y_2v$, and
let $w$ be a minimal idempotent in $E(X)$ such that $\theta (w)=v$. Let $(x_1,x_2)\in Q(X)$ such that $\pi(x_1,x_2)= (y_1,y_2)$. Then $\pi(x_1,x_2w)=(y_1,y_2v)=(y_1,y_3)$ and $(x_1, x_2w)\in Q_XP_X$.
\end{proof}