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# When Did CD102.5 Book the Summerfest Artists?

CD1025 is an “alternative” radio station here in Columbus. They are one of the few remaining radio stations that are independently owned and they take great pride in it. For data nerds like me, they also put a real time list of recently played songs on their website. The page has the most recent 50 songs played, but you can also click on “Older Tracks” to go back in time. When you do this, the URL ends “now-playing/?start=50”. If you got back again, it says “now-playing/?start=100”.

Using this structure, I decided to see if I could download all of their historical data and see how far it goes back. In the code below, I use the XML package to go to the website and download the 50 songs and then increment the number by 50 to find the previous 50 songs. I am telling the code to keep doing this until I get to January 1, 2012.
library(ggplot2)theme_set(theme_bw())library(XML)library(lubridate)library(sqldf)startNum = 0while (TRUE) {    theurl <- paste0("http://cd1025.com/about/playlists/now-playing/?start=",         startNum)    table <- readHTMLTable(theurl, stringsAsFactors = FALSE)[[1]]    if (startNum == 0) {        playlist = table[, -1]    } else {        playlist = rbind(playlist, table[, -1])    }    dt = mdy(substring(table[1, 4], nchar(table[1, 4]) - 9, nchar(table[1, 4])))    print(dt)    if (dt < mdy("1/1/12")) {        break    }    startNum = startNum + 50}playlist = unique(playlist)  # Remove Dupeswrite.csv(playlist, "CD101Playlist.csv", row.names = FALSE)
This takes a while and is fairly large. My file has over 150,000 songs. If you want just a little data, change the date to last week or so. The first thing I will do is parse the dates and times of the songs, order them, and look at the first few songs. You can see that data only goes back to March of 2012.
dates = mdy(substring(playlist[, 3], nchar(playlist[, 3]) - 9, nchar(playlist[,     3])))times = hm(substring(playlist[, 3], 1, nchar(playlist[, 3]) - 10))playlist$Month = ymd(paste(year(dates), month(dates), "1", sep = "-"))playlist$Day = datesplaylist$Time = timesplaylist = playlist[order(playlist$Day, playlist$Time), ]head(playlist) ## Artist Song Last.Played## 151638 DEATH CAB FOR CUTIE YOU ARE A TOURIST 12:34am03/01/2012## 151637 SLEEPER AGENT GET BURNED 12:38am03/01/2012## 151636 WASHED OUT AMOR FATI 12:41am03/01/2012## 151635 COLDPLAY CHARLIE BROWN 12:45am03/01/2012## 151634 GROUPLOVE TONGUE TIED 12:49am03/01/2012## 151633 SUGAR YOUR FAVORITE THING 12:52am03/01/2012## Month Day Time## 151638 2012-03-01 2012-03-01 34M 0S## 151637 2012-03-01 2012-03-01 38M 0S## 151636 2012-03-01 2012-03-01 41M 0S## 151635 2012-03-01 2012-03-01 45M 0S## 151634 2012-03-01 2012-03-01 49M 0S## 151633 2012-03-01 2012-03-01 52M 0S Using the sqldf package, I can easily see what the most played artists and songs are from the data I scraped. sqldf("Select Artist, Count(Artist) as PlayCount From playlist Group By Artist Order by PlayCount DESC Limit 10") ## Artist PlayCount## 1 SILVERSUN PICKUPS 2340## 2 THE BLACK KEYS 2203## 3 MUSE 1988## 4 THE SHINS 1885## 5 OF MONSTERS AND MEN 1753## 6 PASSION PIT 1552## 7 GROUPLOVE 1544## 8 RED HOT CHILI PEPPERS 1514## 9 METRIC 1495## 10 ATLAS GENIUS 1494 sqldf("Select Artist, Song, Count(Song) as PlayCount From playlist Group By Artist, Song Order by PlayCount DESC Limit 10") ## Artist Song PlayCount## 1 PASSION PIT TAKE A WALK 828## 2 SILVERSUN PICKUPS PIT, THE 825## 3 ATLAS GENIUS TROJANS 819## 4 WALK THE MOON ANNA SUN 742## 5 THE BLACK KEYS LITTLE BLACK SUBMARINES 736## 6 DIVINE FITS WOULD THAT NOT BE NICE 731## 7 THE LUMINEERS HO HEY 722## 8 CAPITAL CITIES SAFE AND SOUND 712## 9 OF MONSTERS AND MEN MOUNTAIN SOUND 711## 10 ALT J BREEZEBLOCKS 691 I am a little surprised that Silversun Pickups are the number one band, but everyone on the list makes sense. Looking at how the plays of the top artists have varied from month to month, you can see a few patterns. Muse has been more popular recently and The Shins and Grouplove have lost some steam. artist.month=sqldf("Select Month, Artist, Count(Song) as Num From playlist Group By Month, Artist Order by Month, Artist")artist=sqldf("Select Artist, Count(Artist) as Num From playlist Group By Artist Order by Num DESC")p=ggplot(subset(artist.month,Artist %in% head(artist$Artist,8)),aes(Month,Num))p+geom_bar(stat="identity",aes(fill=Artist),position='fill',colour="grey")+
 labs(y="Percentage of Plays")

For the play count of the top artists, I see some odd numbers in June and July of 2012. The number of plays went way down.
p + geom_area(aes(fill = Artist), position = "stack", colour = 1) + labs(y = "Number of Plays")

Looking into this further, I plotted the date and time that the song was played by the cumulative number of songs played since the beginning of the list. The plot should be a line with a constant slope, meaning that the plays per day are relatively constant. You can see in June and July of 2012 there are flat spots where there is no playlist history.
qplot(playlist$Day + playlist$Time, 1:length(dates), geom = "path")

There are also smaller flat spots in September and December, but I am going to decide that those are small enough not to affect any further analyses. From this, I am going to use data only from August 2012 to present.
playlist = subset(playlist, Day >= mdy("8/1/12"))
Next up, I am going to use this data to analyze the plays of artists from Summerfest, and try to infer if the play counts varied once they were added to the bill.

# What Is the Probability of a 16 Seed Beating a 1 Seed?

Note: I started this post way back when the NCAA men’s basketball tournament was going on, but didn’t finish it until now.

Since the NCAA Men’s Basketball Tournament has moved to 64 teams, a 16 seed as never upset a 1 seed. You might be tempted to say that the probability of such an event must be 0 then. But we know better than that.

In this post, I am interested in looking at different ways of estimating how the odds of winning a game change as the difference between seeds increases. I was able to download tournament data going back to the 1930s until 2012 from hoopstournament.net/Database.html. The tournament expanded to 64 teams in 1985, which is what I used for this post. I only used match ups in which one of the seeds was higher than the other because this was the easiest way to remove duplicates. (The database has each game listed twice, once with the winner as the first team and once with the loser as the first team. The vast majority (98.9%) of games had one team as a higher seed because an equal seed can only happen at the Final Four or later.)

library(ggplot2); theme_set(theme_bw())brackets=read.csv("NCAAHistory.csv") # use only data from 1985 on in which the first team has the higher seedbrackets=subset(brackets,Seed<Opponent.Seed & Year>=1985 & Round!="Opening Round") brackets$SeedDiff=abs(brackets$Opponent.Seed-brackets$Seed)brackets$HigherSeedWon=ifelse(brackets$Opponent.Seed>brackets$Seed,brackets$Wins,brackets$Losses)brackets$HigherSeedScoreDiff=ifelse(brackets$Opponent.Seed>brackets$Seed,1,-1)*(brackets$Score-brackets$Opponent.Score) Created by Pretty R at inside-R.org ### Use Frequencies The first way is the most simple: look at the historical records when a 16 seed is playing a 1 seed (where the seed difference is 15). As you can see from the plot below, when the seed difference is 15, the higher seeded team has won every time. This is also true when the seed difference is 12, although there have only been 4 games in this scenario. Another oddity is that when the seed difference is 10, the higher seed only has only won 50% of the time. Again, this is largely due to the fact that there have only been 6 games with this seed difference. seed.diffs=sort(unique(brackets$SeedDiff))win.pct=sapply(seed.diffs,function(x) mean(brackets$HigherSeedWon[brackets$SeedDiff==x]))ggplot(data=data.frame(seed.diffs,win.pct),aes(seed.diffs,win.pct))+geom_point()+  geom_hline(yintercept=0.5,linetype=2)+  geom_line()+labs(x="Seed Difference",y="Proportion of Games Won by Higher Seed")
Created by Pretty R at inside-R.org

### Use Score Difference

In many applications, it has been shown that using margin of victory is much more reliable than just wins and losses. For example, in the computer ranking of College Football teams, using score differences is more accurate, but outlawed for fear that teams would run up the score on weaker opponents. So the computer rankings are not as strong as they could be.

We have no such conflict of interest, so we should try to make use of any information available. A simple way to do that is to look at the mean and standard deviation of the margin of victory when the 16 seed is playing the 1 seed. Below is a plot of the mean score difference with a ribbon for the +/- 2 standard deviations.

seed.diffs=sort(unique(brackets$SeedDiff))means=sapply(seed.diffs,function(x) mean(brackets$HigherSeedScoreDiff[brackets$SeedDiff==x]))sds=sapply(seed.diffs,function(x) sd(brackets$HigherSeedScoreDiff[brackets$SeedDiff==x]))ggplot(data=data.frame(seed.diffs,means,sds),aes(seed.diffs,means))+ geom_ribbon(aes(ymin=means-2*sds,ymax=means+2*sds),alpha=.5)+geom_point()+geom_line()+ geom_hline(yintercept=0,linetype=2)+ labs(x="Seed Difference",y="Margin of Victory by Higher Seed") Created by Pretty R at inside-R.org You can see that the ribbon includes zero for all seed differences except 15. If we assume that the score differences are roughly normal, we can calculate the probability that the score difference will be greater than 0. The results are largely the same as before, but we see now that there are no 100% estimates. Also, the 50% win percentage for a seed difference of 10 now looks a little more reasonable, albeit still out of line with the rest. ggplot(data=data.frame(seed.diffs,means,sds),aes(seed.diffs,1-pnorm(0,means,sds)))+ geom_point()+geom_line()+geom_hline(yintercept=0.5,linetype=2)+ labs(x="Seed Difference",y="Probability of Higher Seed Winning Based on Margin of Victory") Created by Pretty R at inside-R.org ### Model Win Percentage as a Function of Seed Difference It is always good to incorporate as much knowledge as possible into an analysis. In this case, we have information on other games besides the 16 versus 1 seed game which help us estimate the 16 versus 1 game. For example, it is reasonable to assume that the larger the difference in seed is, the more likely the higher seed will win. We can build a logistic regression model which looks at all of the outcomes of all of the games and predicts the probability of winning based on the difference in seed. When the two teams have the same seed, I enforced the probability of the higher seed winning to be 0.5 by making the intercept 0. In the plot below, you can see that the logistic model predicts that the probability of winning increases throughout until reaching about 90% for the 16 versus 1. I also included a non-linear generalized additive model (GAM) model for comparison. The GAM believes that being a big favorite (16 vs 1 or 15 vs 2) gives an little boost in win probability. An advantage of modeling is that we can make predictions for match-ups that have never occurred (like a seed difference of 14). ggplot(data=brackets,aes(SeedDiff,HigherSeedWon))+ stat_smooth(method="gam",family="binomial",se=F,formula=y~0+x,aes(colour="Logistic"),size=1)+ stat_smooth(method="gam",family="binomial",se=F,formula=y~s(x),aes(colour="GAM"),size=1)+ geom_jitter(alpha=.15,position = position_jitter(height = .025,width=.25))+ labs(x="Seed Difference",y="Game Won by Higher Seed",colour="Model") Created by Pretty R at inside-R.org ### Model Score Difference as a Function of Seed Difference We can also do the same thing with margin of victory. Here, I constrain the linear model to have an intercept of 0, meaning that two teams with the same seed should be evenly matched. Again, I included the GAM fit for comparison. The interpretations are similar to before, in that it seems that there is an increase in margin of victory for the heavily favored teams. ggplot(data=brackets,aes(SeedDiff,HigherSeedScoreDiff))+ stat_smooth(method="lm",se=F,formula=y~0+x,aes(colour="Linear"),size=1)+ stat_smooth(method="gam",se=F,formula=y~s(x),aes(colour="GAM"),size=1)+ geom_jitter(alpha=.25,position = position_jitter(height = 0,width=.25))+ labs(x="Seed Difference",y="Margin of Victory by Higher Seed",colour="Model") Created by Pretty R at inside-R.org From these models of margin of victory we can infer the probability of the higher seed winning (again, assuming normality). library(gam)lm.seed=lm(HigherSeedScoreDiff~0+SeedDiff,data=brackets)gam.seed=gam(HigherSeedScoreDiff~s(SeedDiff),data=brackets) pred.lm.seed=predict(lm.seed,data.frame(SeedDiff=0:15),se.fit=TRUE)pred.gam.seed=predict(gam.seed,data.frame(SeedDiff=0:15),se.fit=TRUE)se.lm=sqrt(mean(lm.seed$residuals^2))se.gam=sqrt(mean(gam.seed$residuals^2)) df1=data.frame(SeedDiff=0:15,ProbLM=1-pnorm(0,pred.lm.seed$fit,sqrt(se.lm^2+pred.lm.seed$se.fit^2)), ProbGAM=1-pnorm(0,pred.gam.seed$fit,sqrt(se.gam^2+pred.gam.seed$se.fit^2)))ggplot(df1)+geom_hline(yintercept=0.5,linetype=2)+ geom_line(aes(SeedDiff,ProbLM,colour="Linear"),size=1)+ geom_line(aes(SeedDiff,ProbGAM,colour="GAM"),size=1)+ labs(x="Seed Difference",y="Probability of Higher Seed Winning",colour="Model") Created by Pretty R at inside-R.org ### Summary Putting all of the estimates together, you can easily spot the differences between the models. The two assumptions that just used the data between specific seeds look pretty similar. It looks like using score differential is a little more reasonable of the two. The two GAMs have a similar trend and so did the linear and logistic models. If someone asks you what the probability that a 16 seed beats a 1 seed, you have at least 6 different answers. This post highlights the many different ways someone can analyze the same data. Simply statistics talked a bit about this in a recent podcast. In this case, the differences are not huge, but there are noticeable changes. So the next time you read about an analysis that someone did, keep in mind all the decisions that they had to make and what type a sensitivity they would have on the results. logit.seed=glm(HigherSeedWon~0+SeedDiff,data=brackets,family=binomial(logit))logit.seed.gam=gam(HigherSeedWon~s(SeedDiff),data=brackets,family=binomial(logit)) df2=data.frame(SeedDiff=0:15, ProbLM=1-pnorm(0,pred.lm.seed$fit,sqrt(se.lm^2+pred.lm.seed$se.fit^2)), ProbGAM=1-pnorm(0,pred.gam.seed$fit,sqrt(se.gam^2+pred.gam.seed\$se.fit^2)),               ProbLogit=predict(logit.seed,data.frame(SeedDiff=0:15),type="response"),               ProbLogitGAM=predict(logit.seed.gam,data.frame(SeedDiff=0:15),type="response"))df2=merge(df2,data.frame(SeedDiff=seed.diffs,ProbFreq=win.pct),all.x=T)df2=merge(df2,data.frame(SeedDiff=seed.diffs,ProbScore=1-pnorm(0,means,sds)),all.x=T)ggplot(df2,aes(SeedDiff))+geom_hline(yintercept=0.5,linetype=2)+  geom_line(aes(y=ProbLM,colour="Linear"),size=1)+  geom_line(aes(y=ProbGAM,colour="GAM"),size=1)+  geom_line(aes(y=ProbLogit,colour="Logistic"),size=1)+  geom_line(aes(y=ProbLogitGAM,colour="Logistic GAM"),size=1)+  geom_line(aes(y=ProbFreq,colour="Frequencies"),size=1)+  geom_line(aes(y=ProbScore,colour="Score Diff"),size=1)+  geom_point(aes(y=ProbFreq,colour="Frequencies"),size=3)+  geom_point(aes(y=ProbScore,colour="Score Diff"),size=3)+  labs(x="Seed Difference",y="Probability of Higher Seed Winning",colour="Model") ggplot(df2)+geom_hline(yintercept=0.5,linetype=2)+  geom_point(aes(x=SeedDiff,y=ProbFreq,colour="Frequencies"),size=1)
Created by Pretty R at inside-R.org

Note that the GAM functions did not have a way to easily restrict the win probability be equal to exactly 0.5 when the seed difference is 0. That is why you may notice the GAM model is a bit above 0.5 at 0.

# Easily Access Academic Journals Off Campus With a Firefox Bookmark

As a grad student, I do lots of searches for research related to my own. When I am off campus, a lot of the relevant results are not open access. In that case, I have to log onto my school’s library website and search for the journal or article there. It is quite a hassle. Luckily, I recently noticed that the website is predictably modified after I log into the library. I go to Ohio State University, and before and after logging in the websites will be http://www.jstor.org/… and http://www.jstor.org.proxy.lib.ohio-state.edu/…, for example. It doesn’t matter which journal or website I am looking at. If OSU has access, it will show up with .proxy.lib.ohio-state.edu after the .com/.org/etc.

So I was able to save myself a few steps by typing that in instead of going through the library’s website. I should also be sure to note that typing this in brings you to an OSU log in screen. Only those with valid access can use this shortcut, but I assume other schools have the same setup.

This is still more busy work than I would like, so I searched for ways to automatically do this. Lifehacker has some address bar shortcuts, but nothing that quite works. I came across a  Firefox add-on called URL Swap that seems like it could do the trick. However, when setting it up, I happened to find this post on Stack Overflow. The second answer has exactly what I am looking for, changing .ExtraPart to .proxy.lib.ohio-state.edu.

I created a bookmark on my bookmarks toolbar with the javascript:
javascript:(function(){ var curloc = window.location.toString(); var newloc = curloc.substring(0, curloc.indexOf("/", 8)) + ".proxy.lib.ohio-state.edu" + curloc.substring(curloc.indexOf("/", 8)); location.href=newloc; })()

and it works perfectly. This trick is so useful I wanted to share it to save others some time.

# Copying Data From Excel to R and Back

A lot of times we are given a data set in Excel format and we want to run a quick analysis using R’s functionality to look at advanced statistics or make better visualizations. There are packages for importing/exporting data from/to Excel, but I have found them to be hard to work with or only work with old versions of Excel (*.xls, not *.xlsx). So for a one time analysis, I usually save the file as a csv and import it into R.

This can be a little burdensome if you are trying to do something quick and creates a file that needs to be cleaned up later. An easier option is to copy and paste the data directly into R. This can be done by using “clipboard” as the file and specifying that it is tab delimited, since that is how Excel’s clipboard stores the data.

For example, say you have a table in excel you want to copy into R. First, copy it in Excel.

Then go into R and use this function.

read.excel <- function(header=TRUE,...) {  read.table("clipboard",sep="\t",header=header,...)} dat=read.excel()

This function specifies that you are reading data from the clipboard, that it is tab delimited, and that it has a header.

Similarly, you can copy from R to Excel using the same logic. Here I also make row.name=FALSE as default since I rarely have meaningful row names and they mess up the header alignment.

write.excel <- function(x,row.names=FALSE,col.names=TRUE,...) {  write.table(x,"clipboard",sep="\t",row.names=row.names,col.names=col.names,...)} write.excel(dat)
Created by Pretty R at inside-R.org

These functions can be added to you .RProfile so that they are always ready for a quick analysis!

Obviously, this technique does not encourage reproducible research. It is meant to be used for quick, ad hoc analysis and plotting; not something you would use for an analysis that needs to be done on a regular basis.

I'm using this function in R for Windows. When I view the data numerically, it's fine. But when I go to plot the data I get nonsense. How to describe it…the window has 0 to 250 on the x-axis (regardless of my data series) and white boxes across the middle.
I create similar functions and got few tips. For reading from excel following settings can be useful:
na.strings = "" # to prevent replacing NA string to missing value
comment.char = "" # to not loose everything after # sign
quote = "" # or ' or " could mess with data
check.names = FALSE # if you want column names as in excel (spaces, special characters, etc.). You need to use column name in R to reference such columns.

For writing na="" replace missing values by empty string and not "NA" as on default.

Second thing is that you can increase size of clipboard by using e.g. "clipboard-10240" instead of "clipboard" (it's a size in Kb, so it's around 10Mb; see help for connection, section Clipboard) which allow to copy and paste larger tables.
rkward (http://rkward.sourceforge.net/) has a very nifty feature (Edit -> Paste Special…), that allows you to paste the copied data directly into your R source code, already formatted as a single string, vector or matrix.
Excellent article. Real people are always busy and this is just the kind of article that helps us all. I have been doing scan() but when you start entering a few thousand rows that way, it gets a bit slow.
scan() allows you to just paste…

y <- scan()
Thank you for these!!
Thank you for the tip, this will help me a lot.
Also, it seems that, libreOffice also uses clipboard to store copied things. This function also works for libreOffice
I can't resist: you could just use

(The "clipboard" parameter is 'doze only for the foreseeable future)


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# `f(x)=sinhx` Prove that the Maclaurin series for the function converges to the function for all x

Maclaurin series is a special case of Taylor series that is centered at `c=0.` The expansion of the function about `0` follows the formula:

`f(x)=sum_(n=0)^oo (f^n(0))/(n!) x^n`

or

`f(x)= f(0)+(f'(0))/(1!)x+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4 +...`

To determine the Maclaurin series for the given function` f(x)=sinh(x)` , we may apply the formula for Maclaurin series.

For...

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Maclaurin series is a special case of Taylor series that is centered at `c=0.` The expansion of the function about `0` follows the formula:

`f(x)=sum_(n=0)^oo (f^n(0))/(n!) x^n`

or

`f(x)= f(0)+(f'(0))/(1!)x+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4 +...`

To determine the Maclaurin series for the given function` f(x)=sinh(x)` , we may apply the formula for Maclaurin series.

For the list of `f^n(x)` , we may apply the derivative formula for hyperbolic trigonometric functions: `d/(dx) sinh(x) = cosh(x)`  and `d/(dx) cosh(x) = sinh(x)` .

`f(x) =sinh(x)`

`f'(x) = d/(dx) sinh(x)= cosh(x)`

`f^2(x) = d/(dx) cosh(x)= sinh(x)`

`f^3(x) = d/(dx) sinh(x)=cosh(x)`

`f^4(x) = d/(dx) cosh(x)= sinh(x)`

`f^5(x) = d/(dx) sinh(x)= cosh(x)`

Plug-in `x=0` on each `f^n(x)` , we get:

`f(0) =sinh(0)=0`

`f'(0) = cosh(0)=1`

`f^2(0) = sinh(0)=0`

`f^3(0) = cosh(0)=1`

`f^4(0) = sinh(0)=0`

`f^5(0) = cosh(0)=1`

Plug-in the values on the formula for Maclaurin series, we get:

`sum_(n=0)^oo (f^n(0))/(n!) x^n`

`= 0+1/(1!)x+0/(2!)x^2+1/(3!)x^3+0/(4!)x^4+1/(5!)x^5+...`

`= 1/(1!)x+1/(3!)x^3+1/(5!)x^5+...`

`=sum_(n=0)^oo x^(2n+1)/((2n+1)!)`

The Maclaurin series is `sum_(n=0)^oo x^(2n+1)/((2n+1)!)` for the function `f(x)=sinh(x)` .

To determine the interval of convergence for the Maclaurin series: `sum_(n=0)^oo x^(2n+1)/((2n+1)!)` , we may apply Ratio Test.

In Ratio test, we determine the limit as: `lim_(n-gtoo)|a_(n+1)/a_n| = L` .

The series converges absolutely when it satisfies `Llt1` .

In the  Maclaurin series: `sum_(n=0)^oo x^(2n+1)/((2n+1)!)` , we have:

`a_n=x^(2n+1)/((2n+1)!)`

Then,

`1/a_n=((2n+1)!)/x^(2n+1)`

`a_(n+1)=x^(2(n+1)+1)/((2(n+1)+1)!)`

`=x^(2n+2+1)/((2n+2+1)!)`

`=x^((2n+1)+2)/((2n+3)!)`

`=(x^(2n+1)*x^2)/((2n+3)(2n+2)((2n+1)!))`

Applying the Ratio test, we set-up the limit as:

`lim_(n-gtoo)|a_(n+1)/a_n|=lim_(n-gtoo)|a_(n+1)*1/a_n|`

`=lim_(n-gtoo)|(x^(2n+1)*x^2)/((2n+3)(2n+2)((2n+1)!))*((2n+1)!)/x^(2n+1)|`

Cancel out common factors: `x^(2n+1) and ((2n+1)!)` .

`lim_(n-gtoo)|x^2/((2n+3)(2n+2))|`

Evaluate the limit.

`lim_(n-gtoo)|x^2/((2n+3)(2n+2))| = |x^2|lim_(n-gtoo)|1/((2n+3)(2n+2))|`

`=|x^2|*1/oo`

`= |x^2|*0`

`=0`

The `L=0` satisfies` Llt1` for all `x` .

Thus, the Maclaurin series: `sum_(n=0)^oo x^(2n+1)/((2n+1)!)` is absolutely converges for all `x` .

Interval of convergence: `-ooltxltoo` .

Approved by eNotes Editorial Team

format: plain


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# Worksheet on Set Theory

In worksheet on set theory we will solve 10 different types of questions. The questions on sets are basically related on sets and their representation and also methods of representing a set.

1. Which of the following are well-defined sets?

(a) All the colors in the rainbow.

(b) All the points that lie on a straight line.

(c) All the honest members in the family.

(d) All the consonants of the English alphabet.

(e) All the tall boys of the school.

(f) All the efficient doctors of the hospital.

(g) All the hardworking teachers in a school.

(h) All the prime numbers less than 100.

(i) All the letters in the word GEOMETRY.

2. Let A = (a, b, c, d, e, f}. Insert the appropriate symbol ∈ or ∉ in the blank space.

(a) d __ A

(b) y __ A

(c) m __ A

(d) a __ A

(e) e __ A

(f) x __ A

Worksheet on Set Theory
3. Write the following sets in the set builder form.

(a) A = {2, 4, 6, 8}

(b) B = {3, 9, 27, 81}

(c) C = {1, 4, 9, 16, 25}

(d) D = {1, 3, 5, ......}

(e) E = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, ......., 52}

(f) F = {-10, ......, -3, -2, -1, 0, 1, 2, …..., 5}

(g) G = {O}

(h) P = { }

(i) H = {-5, 5}

(j) Q = {V, I, B, G, Y, 0, R}

4. Write the following sets in the roster form.

(a) A = {x : x ∈ W, x ≤ 5}

(b) B = {x : x ∈ I, -3 < x < 3)

(c) C = {x : x is divisible by 12}

(d) D = {x : x = 3p, p ∈ W, p ≤ 3}

(e) E = {x : x = a2, a ∈ N, 3 < a < 7}

(f) F = {x : x = n/(n + 1), n ∈ N and n ≤ 4}

(g) G = {x : x ∈ N, 3x - 2 < 5}

(h) J = {x : x ∈ N, x2 < 16}

(i) K = {x : x is a prime number which is a divisor of 42}

(j) H = {x : x is a 2-digit natural number such that the sum of its digits is 5}

5. Which of the following are the examples of an empty set?

(a) The set of even natural numbers divisible by 3.

(b) The set of all prime numbers divisible by 2.

(c) {x : x ∈ N, 5 < x < 6}

(d) The set of odd natural numbers divisible by 2.

(e) B = {O}

(f) C = { }

(g) D = {x : x ∈ R, x2 = -1}

(h) E = {x : x ∈ W, 3x + 1 = 2}

(i) P = {x : x is a prime number, 54 < x < 58}

(j) Q = {x : x = 2n + 3, n ∈ W, n ≤ 5}

Worksheet on Set Theory
6. Classify the following as finite and infinite sets.

(a) The set of days in a week

(b) A = {x : x ∈ N x > 1}

(c) B = {x : x is an even prime number}

(d) C = {x : x is a multiple of 5}

(e) D = {x : x is a factor of 30}

(f) P = {x : x ∈ Z, x < -1}

(g) The set of all letters in the English alphabet

(h) The set of all real numbers

7. From the sets given below, identify the equal sets.

A = {3, 5, 9, 11}         Q = {m, s, t}

B = {8, 9, 1, 13}         R = {o, p, a, z}

C = {-3, 3}           T = {1, 8, 9, 13}

D = {s, t, m}           M = {3, -3}

P = {9, 3, 5, 11}

X = {a, o, z, p}

8. Are the following pairs of sets equal?

(a) A = {2}           B = {x : x ∈ N, x is an even prime number}.

(b) P = {1, 4, 9}         Q = {x : x = n2, n ∈ N, n ≤ 3)

(c) X = {x : x ∈ W, x < 5}       Y = {x : x ∈ N, x ≤ 5}

(d) M = {a, b, c, d}         N = {p, q, r, s}

(e) D = {x : x is a multiple of 30}     E = {x : x is a factor of 10}

Worksheet on Set Theory
9. Which of the following are equivalent sets?

(a) A = {1, 2, 3}           B = {4, 5}

(b) P = {q, s, m}           Q = {6, 9, 12}

(c) X = {x : x is a prime number less than 10}       Y = {x : x ∈ N, x ≤ 4}

(d) R = {x : x = 2n + 3, n < 4, n ∈ N}      S = {x : x = n/(n + 1), n ∈ R, n ≤ 4}

(e) The set of vowels in the English alphabet

(f) The set of consonants in the English alphabet

10. Find the cardinal number of the following sets.

(a) A = {x : x ∈ I, 2 < x < 7}

(b) B = {x : n ∈ N, x = n2, n < 3}

(c) The set of months in a year

(d) C = {x : x ∈ Z+, x < 100}

(e) D = {x : x = n3, n ∈ W, n < 5}

(f) The set of letters in the word MALAYALAM

Answers for worksheet on set theory are given below to make sure that the answers are correct.

1. (a), (b), (d), (h), (i)

2. (a) ∈

(b) ∉

(c) ∉

(d) ∈

(e) ∈

(f) ∉

3. (a) {x : x is even and x ≤ 8}

(b) {x : x = 3n, n ∈ N, n ≤ 4}

(c) {x : x = n2, n ≤ 5, n ∈ N}

(d) {x : x is odd}

(e) {x : x is even, 4 ≤ x ≤ 52 and y : y = 3(2m + 1), 1 ≤ m ≤ 8}

(f)

Worksheet on Set Theory

Sets and Venn-diagrams - Worksheets

Worksheet on Set Theory

Worksheet on Elements of a Set

Worksheet on Representation on Set

Worksheet on Set Operations

Worksheet to Find the Cardinal Number of the Sets

Worksheet on Cardinal Properties of Sets

Worksheet on Sets using Venn Diagram

Worksheet on Union and Intersection using Venn Diagram


text:
# Posts Tagged ‘quick-sort’

In this short lesson we will discuss how to parallelize a simple and rather inefficient (because this is not an in-place version) implementation of quick-sort using asynchronous tasks and futures.

We will perform some benchmarking and performance analysis and we will try to understand how we can further improve our implementation.

## Quick sort

In this section, I will briefly refresh your memory on quick-sort. I will do so by showing you a simple and self-explicative Haskell version first. We will also write a C++ (serial) version of the same code implementation C++ that we will use as a basis for our parallelization.

Here it goes the Haskell version:

quicksort [] = []
quicksort (p:xs) = (quicksort lesser) ++ [p] ++ (quicksort greater)
where
lesser = filter (< p) xs
greater = filter (>= p) xs

It is beautifully simple: in order to sort a list with at least, p one element is only necessary to partition sort the rest of the elements xs into two sublists:

• lesser: containing all the elements in xs smaller than p
• greater: containing all the elements in xs greater than p
Once both sublists are sorted we can finally return the whole sorted list using by simply returning gluing lesser, p and greater together in this order.

If you still have trouble understanding the quick-sort algorithm please refer to Wikipedia.

## Quick-sort serial version

The following is the serial C++ implementation of the same idea described above. It should be pretty easy to map the following implementation to the Haskell one. Run it on Wandbox

template <typename T>
void quick_sort_serial(vector<T>& v) {
if (v.size() <= 1) return;
auto start_it = v.begin();
auto end_it = v.end();

const T pivot = *start_it;

//partition the list
vector<T> lesser;
copy_if(start_it, end_it, std::back_inserter(lesser),
[&](const T& el) { return el < pivot; });

vector<T> greater;
copy_if(start_it + 1, end_it, std::back_inserter(greater),
[&](const T& el) { return el >= pivot; });

//solve subproblems
quick_sort_serial(lesser);
quick_sort_serial(greater);

//merge
std::copy(lesser.begin(), lesser.end(), v.begin());
v[lesser.size()] = pivot;
std::copy(greater.begin(), greater.end(),
v.begin() + lesser.size() + 1);
}

## Parallelizing Quick-sort using std::future

In order to speed-up things we are going to use the fact that quick-sort is a divide and conquer algorithm. Each subproblem can be solved independently:
creating and sorting lesser and greater are two independent tasks. We can easily perform both on different threads.

The following is the first parallel version of the quick_sort_serial() above.
Run it on Wandbox

template <typename T>
void filter_less_than(const vector<T>& v, vector<T>& lesser, const int pivot) {
for (const auto el : v) {
if (el < pivot) lesser.push_back(el);
}
}

template <typename T>
void quick_sort_parallel1(vector<T>& v) {
if (v.size() <= 1) return;
auto start_it = v.begin();
auto end_it = v.end();

const T pivot = *start_it;
vector<T> lesser;
auto fut1 = std::async(
std::launch::async,
[&]() {
filter_less_than<T>(std::ref(v), std::ref(lesser), pivot);
quick_sort_parallel1<T>(std::ref(lesser));
});

vector<T> greater;
copy_if(start_it + 1, end_it, std::back_inserter(greater),
[&](const T& el) { return el >= pivot; });

quick_sort_parallel1(greater);

fut1.wait();

std::copy(lesser.begin(), lesser.end(), v.begin());
v[lesser.size()] = pivot;
std::copy(greater.begin(), greater.end(),
v.begin() + lesser.size() + 1);
}

As you can notice, the creation and sorting of lesser and are performed in parallel. Each thread running an instance of quick_sort_parallel1() will create another thread running quick-sort on one of the two sub-problems while the other subproblem is solved by the current thread.

This is exactly what we are doing when we spawn the following async task:
we are creating a task that will populate lesser with all the elements from v less than pivot and, once ready, it will sort it.
Please note that everything we need to have modified by reference need to be wrapped in a std::ref as we discussed in the previous lessons.

The following picture shows how the execution unfolds for the unsorted list: [2,7,1,6,9,5,8,3,4,10]:

The following code shows hot to spawn an async thread solving the lesser subproblem:

  vector<T> lesser;
auto fut1 = std::async([&]() {
filter<T>(std::ref(v), std::ref(lesser), pivot);
quick_sort<T>(std::ref(lesser));
});

While this task is running on the newly created thread, we can solve greater on the current thread.

The asynchronous task will recursively spawn other async tasks until a list of size <=1 is created, which is of course already sorted. There is nothing to do in this case.

Once the main thread is done with sorting the greater list, it waits for the asynchronous task to be ready using the std:.future::wait() function.
Once wait returns, both lists are sorted, and we can proceed with merging the result and finally, here it is, we have a sorted list.

## Performance analysis

Let's quickly analyze our implementation. We will compare execution time for the single-thread and async-parallel versions above.

Let's start our analysis by taking looking at this graph depicting the execution time (average of 10 runs) for both versions above:

It might be a surprising result to see that the Async parallel version is way slower than the single threaded version, ~55x slower!
Why is that? The reason is that, the parallel version creates a new thread for every single subproblem, even for the ones that are quite small.
Threads are costly to manage by the OS, they use resources and need to be scheduled. For smaller tasks the overhead caused by the additional thread is larger than the gain in performance that we might get by processing the sublist in parallel. This is exactly what is happening.

In order to solve this issue, we want to modify the async code above so that a new thread is spawned only when the input list v is larger than a certain threshold. The code below implements the aforementioned idea:

template <typename T>
void quick_sort_async_lim(vector<T>& v) {
if (v.size() <= 1) return;
auto start_it = v.begin();
auto end_it = v.end();

const T pivot = *start_it;
vector<T> lesser;

vector<T> greater;
copy_if(start_it + 1, end_it, std::back_inserter(greater),
[&](const T& el) { return el >= pivot; });

if (v.size() >= THRESHOLD) {
auto fut1 = std::async([&]() {
filter<T>(std::ref(v), std::ref(lesser), pivot);
quick_sort_async_lim<T>(std::ref(lesser));
});

quick_sort_async_lim(greater);
fut1.wait();

} else {
//problem is too small.
//Do not create new threads
copy_if(start_it, end_it, std::back_inserter(lesser),
[&](const T& el) { return el < pivot; });
quick_sort_async_lim(lesser);
quick_sort_async_lim(greater);
}

std::copy(lesser.begin(), lesser.end(), v.begin());
v[lesser.size()] = pivot;
std::copy(greater.begin(), greater.end(),
v.begin() + lesser.size() + 1);
}

As you can notice, the only addition that this optimized version has is that a new thread is spawned only when the size of the input list is larger than THRESHOLD If the list is too small, then we fall back on the classic single-thread version.
The following pictures show the result for the optimized version above with value of THRESHOLD=4000. As you can notice the execution time drops sensibly w.r.t single thread version. We have achieved ~4x speedup with a minimal programming effort.

We have introduced a new parameter in our code, and we need to figure out what is the best value of THRESHOLD. In order to do so, let's analyze the performance of the code above for various values of the threshold.
The following graph depict the execution time for various values of THRESHOLD. Note that the y-axis is in log scale. The execution time drops quite abruptly from 0 to 300.

## Conclusion

We have used std::future to parallelize the quick-sort algorithm. The async code differs slightly from the serial single thread implementation, but runs 4x faster. On the other hand, we have learned that running too many threads it is definitely not a good idea because each thread comes with an overhead: the OS needs to allocate resources and time to manage them.


text:
[S0199-6231(00)00204-5]

Stirling engine based units are considered among the most effective alternatives for future solar applications at low power range. In order to analyze and to improve the performance of the three main subsystems of those units, namely, the solar receiver, the thermodynamic gas circuit, and the drive mechanism, simulations codes are under development worldwide. Therefore, the authors must be congratulated not only on the quality of their paper, but also on its current interest.

The authors claim that there is a good agreement between the calculated performance and the experimental results. Unfortunately, however, there is hardly any reference to the influence of velocity on either indicated power or mechanical losses. Yet, I have verified that widely accepted simulation codes can present very different degrees of accuracy depending on the operating point considered.

Organ 1 has introduced similarity criteria that evidence interrelations between the engine speed and other indicated performance parameters. Independently, Prieto and co-workers have derived the complete set of dimensionless parameters influencing the indicated performance and have proposed a method to apply similarity to analysis and design. A wide variety of prototypes has been analyzed, and a general model of performance has recently been proposed to cover wide temperature, pressure, size, and power ranges (see, for example, 2).

The influence of the Mach number, $NMA=nVe1/3/RTc,$ a kind of dimensionless engine speed, on the dimensionless indicated power, $ζ=Ni/PV¯en,$ can be expressed by means of:
$ζ=ζ0−ΦNMA−ΨNMA2$
(1)
where Φ and Ψ are dimensionless factors of the indicated power losses, and the coefficient $ζ0,$ known as quasi-static simulation, represents the theoretical thermodynamic prediction assuming an ideal cycle without any thermal or mechanical irreversibilities. Approximately, $ζ0$ can be computed by the well-known Schmidt’s model.
The values $ζmax$ and $NMA,max$ at the maximum indicated power point fulfill the following equations:
$12ζ0⩽ζmax⩽23ζ0$
(2)

$NMA,max=Φ2+3ζ0Ψ−Φ3Ψ$
(3)

$NMA,max$ can be considered as an index of engine development level, i.e., it is not an independent design parameter but a function which depends on the same parameters influencing the indicated power.

If the authors could obtain experimental measurements of $ζmax$ and $NMA,max,$ coefficients Φ and Ψ could be derived from the following equations:
$Φ=2ζ0−3ζmaxNMA,max$
(4)

$Ψ=2ζmax−ζ0NMA,max2$
(5)

On the other hand, I have recently verified that the dimensionless power of the mechanical losses, $ζmech=Nmech/PV¯en,$ is correlated to a series of dimensionless variables among which the Stirling number, $NSG=P¯/μn,$ where μ is the working fluid viscosity, stands out. $NSG$ evidences the influence of forces, viscosity, and velocity on mechanical power losses and it could be considered as a variation of a classical parameter in Tribology, usually known as the Sommerfeld number (see, for example, 3).

$ζmech$ and $NSG$ could be correlated by combining experimental brake power measurements and indicated power maps based on Eq. (1), which could confirm the accuracy of both the assumptions $n=1200 min−1$ and $μg=μs=μsh=0.4$ for helium at the operating point specified.

1.
Organ
,
A. J.
,
1991
, “
Intimate Thermodynamic Design of the Stirling Engine Gas Circuit Without the Computer
,”
Proc. ImechE (Part C): J. Mech. Engng. Sci.
,
205
, pp.
421
430
.
2.
Prieto
,
J. I.
,
Gonzalez
,
M. A.
,
Gonzalez
,
C.
, and
Fano
,
J.
,
2000
, “
A New Equation Representing the Performance of Kinematic Stirling Engines
,”
Proc. ImechE (Part C): J. Mech. Engng. Sci.
,
214
, pp.
449
464
.
3.
Orlov, P., 1975, Ingenieria de Disen˜o (Spanish edition), Mir. Moscow.


text:
# Supplementary Angles

If the measures of two angles sum up to 180^\circ, they are called supplementary angles. You’ll notice that when this pair of angles are adjacent, they form a straight angle. Each angle is called a supplement of the other.

Take for instance the diagram above, \angle AXN and \angle NXF are supplementary. If we add their angle measures (120^\circ + 60^\circ ), we get 180^\circ.

But the angles don’t have to be adjacent nor share a common side and vertex to be considered as supplementary angles.

\angle H and \angle S are supplementary. Why? Because even though they are non-adjacent angles, the sum of their measures is 180^\circ.

35^\circ + 145^\circ = 180^\circ

## Example Problems Involving Supplementary Angles

Let’s delve more into the relationship of this angle pair by going through some examples.

Example 1: Are \angle ERW and \angle WRQ supplementary?

We have to add the angle measures of both angles in order to find out if they sum up to 180^\circ.

And they do! Therefore, \angle ERW and \angle WRQ are supplementary angles.

Example 2: If \angle CYM and \angle LKG are supplementary, what is the measure of \angle CYM?

We are only given the measure of \angle LKG. However, since we know that supplementary angles add up to 180^\circ, we can simply use subtraction in order to find the measure of \angle CYM.

Thus, the measure of \angle CYM is \textbf{127}^\circ.

Example 3: \angle JBU and \angle UBT are supplementary. Find the missing angle measure.

This problem is similar to our previous example. The only difference is that the two angles are adjacent to each other. However, the concept stays the same. We can find the missing measure by subtracting the given measure of \angle UBT from 180^\circ.

The missing angle measure or the measure of \angle JBU is \textbf{42}^\circ.

This makes sense because if we add both angle measures, we get 180^\circ.

138^\circ - 42^\circ = 180^\circ

This proves that both angles are indeed supplementary.

Example 4: What is the value of x?

Just by looking at the diagram, we can tell that \angle PVH and \angle HVA are supplementary. Together, the angle pair form a straight angle while adjacent to each other. A straight angle measures 180^\circ and so are supplementary angles.

Both of the angle measures are given but one is expressed in the form of an algebraic expression. It may look challenging but it’s really not. Since we know that they are supplementary, we will set up our equation such that the sum of the angle measures is 180^\circ. Then we solve for x.

So, the value of x is \textbf{9}.

To check if we got the correct answer, let’s plug in the value of x into our original equation. If both sides of the equation equal to 180, then we got the correct value for x.

Perfect! 9 indeed is the correct value for x. While checking, we also found out that the measure of \angle HVA is 61^\circ

Example 5: Suppose \angle Q and \angle F are supplementary. Find the measures of the two angles.

Here we are given two supplementary angles whose measures are expressed in algebraic expressions. Let’s go ahead and set up our equation then solve for the variable x.

Now that we know the value of x, we can use this to find the measure of each angle. We’ll simply replace x with 7 on each of the algebraic expressions then simplify.

So the measure of \angle Q is \textbf{156}^\circ and the measure of \angle F is \textbf{24}^\circ.

If we add both angle measures, we get 180^\circ which means our answers are correct.

156^\circ + 24^\circ = 180^\circ

Example 6: Two supplementary angles are such that the measure of one angle is 3 times the measure of the other. Determine the measure of each angle.

Let x^\circ be the measure of the first angle. Since the second angle measures 3 times than the first, then it will be 3x ^\circ. Keep in mind that the angles are supplementary so the right side of the equation must be 180 ^\circ.

Using the value of x, the measure of the second angle will be 3x = 3\left( {45} \right) = 135.

Therefore, the measures of the angles are \textbf{45} ^\circ and \textbf{135} ^\circ which when added sum up to 180 ^\circ .


text:
# Hugh has to earn at least $400 to meet his fundraising goal. He has only 100 books that he plans to sell at$8 each. Which inequality shows the number of books, x, Hugh can sell to meet his goal?

Jul 20, 2018

$8 x \setminus \ge 400$

#### Explanation:

Hugh needs to earn ($y$) at least 400 dollars. $y \setminus \ge 400$
Given $x$ books sold at 8 dollars each,
$8 x$

Hugh's earnings $y$...
$8 x = y \setminus \ge 400$
$8 x \setminus \ge 400$

Jul 20, 2018

$x \ge 50$

#### Explanation:

Total funds:
$T = x \cdot 8$

to meet his goal:

$T \ge 400$

so $x \cdot 8 \ge 400$

multiply both sides by 8

$x \ge \frac{400}{8}$

simplify the fraction

$x \ge 50$


text:
# Blog Archives

## IMO 1983 – Problem 3

Let $N$ and $k$ be positive integers and  let $S$ be a set of  $n$ points in the plane such that

$(i)$ no three points of $S$ are collinear, and
$(ii)$  for any point $P$ of  $S$, there are at least $k$ points of $S$ equidistant from $P$

Prove that  $k$  $<$  $\frac{1}{2}$  $+$  $\sqrt{2n}$

Try the question …
Solution will be updated soon

## IMO 2012 problems

This year IMO problems !!!!

Problem 1 :

Given triangle  the point  is the centre of the excircle opposite the vertex  This excircle is tangent to the side  at , and to the lines  and  at  and , respectively. The lines  and  meet at , and the lines  and  meet at  Let  be the point of intersection of the lines  and , and let  be the point of intersection of the lines  and  Prove that  is the midpoint of .

Problem 2 :

Let ${n\ge 3}$ be an integer, and let ${a_2,a_3,\ldots ,a_n}$ be positive real numbers such that  ${a_{2}a_{3}\cdots a_{n}=1}$  Prove that

$\displaystyle \left(a_{2}+1\right)^{2}\left(a_{3}+1\right)^{3}\dots\left(a_{n}+1\right)^{n}>n^{n}.$

Problem 3 :

The liar’s guessing game is a game played between two players ${A}$ and ${B}$. The rules of the game depend on two positive integers ${k}$ and ${n}$ which are known to both players.

At the start of the game ${A}$ chooses integers ${x}$ and ${N}$ with ${1 \le x \le N.}$ Player ${A}$ keeps ${x}$secret, and truthfully tells ${N}$ to player ${B}$. Player ${B}$ now tries to obtain information about ${x}$ by asking player ${A}$ questions as follows: each question consists of ${B}$ specifying an arbitrary set ${S}$ of positive integers (possibly one specified in some previous question), and asking ${A}$whether ${x}$ belongs to ${S}$. Player ${B}$ may ask as many questions as he wishes. After each question, player ${A}$ must immediately answer it with [i]yes[/i] or [i]no[/i], but is allowed to lie as many times as she wants; the only restriction is that, among any ${k+1}$ consecutive answers, at least one answer must be truthful.

After ${B}$ has asked as many questions as he wants, he must specify a set ${X}$ of at most ${n}$positive integers. If ${x}$ belongs to ${X}$, then ${B}$ wins; otherwise, he loses. Prove that:

1. If ${n \ge 2^k,}$ then ${B}$ can guarantee a win. 2. For all sufficiently large ${k}$, there exists an integer ${n \ge (1.99)^k}$ such that ${B}$ cannot guarantee a win.

Problem 4 :

Find all functions ${f:\mathbb Z\rightarrow \mathbb Z}$ such that, for all integers ${a,b,c}$ that satisfy ${a+b+c=0}$, the following equality holds:

$\displaystyle f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).$

Problem 5 :

Let ${ABC}$ be a triangle with ${\angle BCA=90^{\circ}}$, and let ${D}$ be the foot of the altitude from ${C}$. Let ${X}$ be a point in the interior of the segment ${CD}$. Let ${K}$ be the point on the segment ${AX}$such that ${BK=BC}$. Similarly, let ${L}$ be the point on the segment ${BX}$ such that ${AL=AC}$. Let ${M}$ be the point of intersection of ${AL}$ and ${BK}$.

Show that ${MK=ML}$.

Problem 6 :

Find all positive integers ${n}$ for which there exist non-negative integers ${a_1, a_2, \ldots, a_n}$ such that

$\displaystyle \frac{1}{2^{a_1}} + \frac{1}{2^{a_2}} + \cdots + \frac{1}{2^{a_n}} = \frac{1}{3^{a_1}} + \frac{2}{3^{a_2}} + \cdots + \frac{n}{3^{a_n}} = 1.$

IMO 2012

Thank you and good luck for these delicious problems

## Bumper Problems

These are the set of 5 problems  each of  5 marks

The person who get more than 20 points will get a mathematics book on any topic he want…
So try these beautiful problems to test your mathematics abilities and getting new things…

Ways by which you can answer :
I am adding a form in the end of this post  , you can answer there ….

Problem 1)

The bisectors of the angles $A$ and $B$ of the $\bigtriangleup ABC$  meet the sides
$BC$ and $CA$ at the points $D$ and $E$ , respectively.
Assuming that $AE+BD =AB$, determine the angle $C$

Problem 2)

Given a  $\bigtriangleup ABC$ and  $D$ be point on side  $AC$ such that  $AB = DC$,
$\angle BAC= 60-2X$ ,   $\angle DBC= 5X$ and  $\angle BCA= 3X$
Find the value of  $X$

Problem 3)

If p and q are natural numbers so that

Prove that p is divisible by 1979 .

Problem 4)

Find highest degree n of 1991 for which 1991ⁿ  divides the number :

Problem 5)

Let ƒ(n) denote the sum of the digits of n. Let N = 4444⁴⁴⁴⁴
Find ƒ(ƒ(ƒ(n))))

You can use these symbols to write solutions more conveniently

 Mathematical Operators ∀ ∁ ∂ ∃ ∄ ∅ ∆ ∇ ∈ ∉ ∊ ∋ ∌ ∍ ∎ ∏ ∐ ∑ − ∓ ∔ ∕ ∖ ∗ ∘ ∙ √ ∛ ∜ ∝ ∞ ∟ ∠ ∡ ∢ ∣ ∤ ∥ ∦ ∧ ∨ ∩ ∪ ∫ ∬ ∭ ∮ ∯ ∰ ∱ ∲ ∳ ∴ ∵ ∶ ∷ ∸ ∹ ∺ ∻ ∼ ∽ ∾ ∿ ≀ ≁ ≂ ≃ ≄ ≅ ≆ ≇ ≈ ≉ ≊ ≋ ≌ ≍ ≎ ≏ ≐ ≑ ≒ ≓ ≔ ≕ ≖ ≗ ≘ ≙ ≚ ≛ ≜ ≝ ≞ ≟ ≠ ≡ ≢ ≣ ≤ ≥ ≦ ≧ ≨ ≩ ≪ ≫ ≬ ≭ ≮ ≯ ≰ ≱ ≲ ≳ ≴ ≵ ≶ ≷ ≸ ≹ ≺ ≻ ≼ ≽ ≾ ≿ ⊀ ⊁ ⊂ ⊃ ⊄ ⊅ ⊆ ⊇ ⊈ ⊉ ⊊ ⊋ ⊌ ⊍ ⊎ ⊏ ⊐ ⊑ ⊒ ⊓ ⊔ ⊕ ⊖ ⊗ ⊘ ⊙ ⊚ ⊛ ⊜ ⊝ ⊞ ⊟ ⊠ ⊡ ⊢ ⊣ ⊤ ⊥ ⊦ ⊧ ⊨ ⊩ ⊪ ⊫ ⊬ ⊭ ⊮ ⊯ ⊰ ⊱ ⊲ ⊳ ⊴ ⊵ ⊶ ⊷ ⊸ ⊹ ⊺ ⊻ ⊼ ⊽ ⊾ ⊿ ⋀ ⋁ ⋂ ⋃ ⋄ ⋅ ⋆ ⋇ ⋈ ⋉ ⋊ ⋋ ⋌ ⋍ ⋎ ⋏ ⋐ ⋑ ⋒ ⋓ ⋔ ⋕ ⋖ ⋗ ⋘ ⋙ ⋚ ⋛ ⋜ ⋝ ⋞ ⋟ ⋠ ⋡ ⋢ ⋣ ⋤ ⋥ ⋦ ⋧ ⋨ ⋩ ⋪ ⋫ ⋬ ⋭ ⋮ ⋯ ⋰ ⋱ ⋲ ⋳ ⋴ ⋵ ⋶ ⋷ ⋸ ⋹ ⋺ ⋻ ⋼ ⋽ ⋾ ⋿ Exponents  :   ⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁼ ⁽ ⁾ ₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋ ₌ ₍ ₎

## Important theorems of Geometry(Triangles) – 1

Today we will be discussing about  2 important Triangle theorems which have very High applications in field of Geometry

1) Angle Bisector theorem
2) Stewart’s Theorem

Angle bisector theorem :

Introduction :

The Angle Bisector Theorem states that given triangle  and angle bisector AD, where D is on side BC, then c/m= b/n . Likewise, the converse of this theorem holds as well.

Proof :

Because of the ratios and equal angles in the theorem, we think of similar triangles. There are not any similar triangles in the figure as it now stands, however. So, we think to draw in a carefully chosen line or two. Extending AD until it hits the line through C parallel to AB does just the trick:

Since AB and CE are parallel, we know that  and . Triangle ACE is isosceles, with AC = CE.

By AA similarity. By the properties of similar triangles, we arrive at our desired result:

c/m = b/n

Stewarts Theorem :

Introduction :

Given a triangle  with sides of length  opposite vertices , respectively. If cevian  is drawn so that  and , we have that . (This is also often written , a form which invites mnemonic memorization, e.g. “A man and his dad put a bomb in the sink.”)

Proof :

Applying the Law of Cosines in triangle  at angle  and in triangle  at angle , we get the equations

Because angles  and  are supplementary. We can therefore solve both equations for the cosine term. Using the trigonometric identity   gives us

Setting the two left-hand sides equal and clearing denominators, we arrive at the equation: . However,  so and we can rewrite this as  (A man and his dad put a bomb in the sink).
Thank you

Source :
Wikepedia
Art of problem solving


text:
#### Results from LHCb web retrieved at 11:51 (GMT)

LHCb Computing Model This page is a historical account of the changes developments to the LHCb computing model. Computing Model Documents The latest resource...
LHCb Core Software Meeting Date and Location 05. July 2006 10:00 11:20 CERN (2 R 030) Attendees Florence, Hubert, Joel, Juan, Marco Cl. (minutes), Markus...
Minutes of the Data Quality Meeting of Monday 7 July 2008 Offline Monitoring Muon PID Discussion about the best place to run the high level muon monitoring using...
How to run on stripped DC06 data Introduction This page explains all you need to know to run on DC06 stripped data. If theres something missing, just add it! It...
Streaming of data in LHCb COMP 24 Jan 2007 Ulrik Egede (Imperial College London) Marco Cattaneo (CERN) Patrick Koppenburg (Imperial College...
Last Published Publisher TWikiGuest Date 18 Aug 2008 16:50 {PublishContrib}{Dir} /afs/cern.ch/project/twiki/production/twiki/export/ {PublishContrib}{URL} https...
Correlation between selections Introduction The correlations between the selections run in the DC04v2 stripping (using DaVinci v12r16) all available. Because of the...
Use case describing how to recover from a bug in the stripping Outline of use case A given analysis use as its incoming dataset data from two different processings...
Use case for calibration of the detector Outline of use case The RICH needs to be calibrated before RAW data can be processed for physics. The RICH will also be used...
Use case for the development of a new selection Outline of use case For a new selection access to unstripped data is a requirement. The sample has to be large enough...
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Use case for a physics analysis with a high branching ratio Outline of use case This is high branching ratio channel which means that including sidebands the final...
Use case to perform a detector calibration analysis with data from the 2007 pilot run. Outline of use case The 2007 pilot run will be used to calibrate the detector...
Use case for sparse analysis Outline of use case The main point for the %$B s \rightarrow \mu^ \mu^$% is to measure the branching ratio. Due to the clean signature...
Streaming Task Force remit The task force should examine the needs and requirements of streaming at the trigger/DAQ level as well as in the stripping process. The...
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text:
## 19.9 Embedding abelian categories

In this section we show that an abelian category embeds in the category of abelian sheaves on a site having enough points. The site will be the one described in the following lemma.

Lemma 19.9.1. Let $\mathcal{A}$ be an abelian category. Let

$\text{Cov} = \{ \{ f : V \to U\} \mid f\text{ is surjective}\} .$

Then $(\mathcal{A}, \text{Cov})$ is a site, see Sites, Definition 7.6.2.

Proof. Note that $\mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ is a set by our conventions about categories. An isomorphism is a surjective morphism. The composition of surjective morphisms is surjective. And the base change of a surjective morphism in $\mathcal{A}$ is surjective, see Homology, Lemma 12.5.14. $\square$

Let $\mathcal{A}$ be a pre-additive category. In this case the Yoneda embedding $\mathcal{A} \to \textit{PSh}(\mathcal{A})$, $X \mapsto h_ X$ factors through a functor $\mathcal{A} \to \textit{PAb}(\mathcal{A})$.

Lemma 19.9.2. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{C} = (\mathcal{A}, \text{Cov})$ be the site defined in Lemma 19.9.1. Then $X \mapsto h_ X$ defines a fully faithful, exact functor

$\mathcal{A} \longrightarrow \textit{Ab}(\mathcal{C}).$

Moreover, the site $\mathcal{C}$ has enough points.

Proof. Suppose that $f : V \to U$ is a surjective morphism of $\mathcal{A}$. Let $K = \mathop{\mathrm{Ker}}(f)$. Recall that $V \times _ U V = \mathop{\mathrm{Ker}}((f, -f) : V \oplus V \to U)$, see Homology, Example 12.5.6. In particular there exists an injection $K \oplus K \to V \times _ U V$. Let $p, q : V \times _ U V \to V$ be the two projection morphisms. Note that $p - q : V \times _ U V \to V$ is a morphism such that $f \circ (p - q) = 0$. Hence $p - q$ factors through $K \to V$. Let us denote this morphism by $c : V \times _ U V \to K$. And since the composition $K \oplus K \to V \times _ U V \to K$ is surjective, we conclude that $c$ is surjective. It follows that

$V \times _ U V \xrightarrow {p - q} V \to U \to 0$

is an exact sequence of $\mathcal{A}$. Hence for an object $X$ of $\mathcal{A}$ the sequence

$0 \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(U, X) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(V, X) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(V \times _ U V, X)$

is an exact sequence of abelian groups, see Homology, Lemma 12.5.8. This means that $h_ X$ satisfies the sheaf condition on $\mathcal{C}$.

The functor is fully faithful by Categories, Lemma 4.3.5. The functor is a left exact functor between abelian categories by Homology, Lemma 12.5.8. To show that it is right exact, let $X \to Y$ be a surjective morphism of $\mathcal{A}$. Let $U$ be an object of $\mathcal{A}$, and let $s \in h_ Y(U) = \mathop{Mor}\nolimits _\mathcal {A}(U, Y)$ be a section of $h_ Y$ over $U$. By Homology, Lemma 12.5.14 the projection $U \times _ Y X \to U$ is surjective. Hence $\{ V = U \times _ Y X \to U\}$ is a covering of $U$ such that $s|_ V$ lifts to a section of $h_ X$. This proves that $h_ X \to h_ Y$ is a surjection of abelian sheaves, see Sites, Lemma 7.11.2.

The site $\mathcal{C}$ has enough points by Sites, Proposition 7.39.3. $\square$

Remark 19.9.3. The Freyd-Mitchell embedding theorem says there exists a fully faithful exact functor from any abelian category $\mathcal{A}$ to the category of modules over a ring. Lemma 19.9.2 is not quite as strong. But the result is suitable for the Stacks project as we have to understand sheaves of abelian groups on sites in detail anyway. Moreover, “diagram chasing” works in the category of abelian sheaves on $\mathcal{C}$, for example by working with sections over objects, or by working on the level of stalks using that $\mathcal{C}$ has enough points. To see how to deduce the Freyd-Mitchell embedding theorem from Lemma 19.9.2 see Remark 19.9.5.

Remark 19.9.4. If $\mathcal{A}$ is a “big” abelian category, i.e., if $\mathcal{A}$ has a class of objects, then Lemma 19.9.2 does not work. In this case, given any set of objects $E \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ there exists an abelian full subcategory $\mathcal{A}' \subset \mathcal{A}$ such that $\mathop{\mathrm{Ob}}\nolimits (\mathcal{A}')$ is a set and $E \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}')$. Then one can apply Lemma 19.9.2 to $\mathcal{A}'$. One can use this to prove that results depending on a diagram chase hold in $\mathcal{A}$.

Remark 19.9.5. Let $\mathcal{C}$ be a site. Note that $\textit{Ab}(\mathcal{C})$ has enough injectives, see Theorem 19.7.4. (In the case that $\mathcal{C}$ has enough points this is straightforward because $p_*I$ is an injective sheaf if $I$ is an injective $\mathbf{Z}$-module and $p$ is a point.) Also, $\textit{Ab}(\mathcal{C})$ has a cogenerator (details omitted). Hence Lemma 19.9.2 proves that we have a fully faithful, exact embedding $\mathcal{A} \to \mathcal{B}$ where $\mathcal{B}$ has a cogenerator and enough injectives. We can apply this to $\mathcal{A}^{opp}$ and we get a fully faithful exact functor $i : \mathcal{A} \to \mathcal{D} = \mathcal{B}^{opp}$ where $\mathcal{D}$ has enough projectives and a generator. Hence $\mathcal{D}$ has a projective generator $P$. Set $R = \mathop{Mor}\nolimits _\mathcal {D}(P, P)$. Then

$\mathcal{A} \longrightarrow \text{Mod}_ R, \quad X \longmapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(P, X).$

One can check this is a fully faithful, exact functor. In other words, one retrieves the Freyd-Mitchell theorem mentioned in Remark 19.9.3 above.

Remark 19.9.6. The arguments proving Lemmas 19.9.1 and 19.9.2 work also for exact categories, see [Appendix A, Buhler] and [1.1.4, BBD]. We quickly review this here and we add more details if we ever need it in the Stacks project.

Let $\mathcal{A}$ be an additive category. A kernel-cokernel pair is a pair $(i, p)$ of morphisms of $\mathcal{A}$ with $i : A \to B$, $p : B \to C$ such that $i$ is the kernel of $p$ and $p$ is the cokernel of $i$. Given a set $\mathcal{E}$ of kernel-cokernel pairs we say $i : A \to B$ is an admissible monomorphism if $(i, p) \in \mathcal{E}$ for some morphism $p$. Similarly we say a morphism $p : B \to C$ is an admissible epimorphism if $(i, p) \in \mathcal{E}$ for some morphism $i$. The pair $(\mathcal{A}, \mathcal{E})$ is said to be an exact category if the following axioms hold

1. $\mathcal{E}$ is closed under isomorphisms of kernel-cokernel pairs,

2. for any object $A$ the morphism $1_ A$ is both an admissible epimorphism and an admissible monomorphism,

3. admissible monomorphisms are stable under composition,

4. admissible epimorphisms are stable under composition,

5. the push-out of an admissible monomorphism $i : A \to B$ via any morphism $A \to A'$ exist and the induced morphism $i' : A' \to B'$ is an admissible monomorphism, and

6. the base change of an admissible epimorphism $p : B \to C$ via any morphism $C' \to C$ exist and the induced morphism $p' : B' \to C'$ is an admissible epimorphism.

Given such a structure let $\mathcal{C} = (\mathcal{A}, \text{Cov})$ where coverings (i.e., elements of $\text{Cov}$) are given by admissible epimorphisms. The axioms listed above immediately imply that this is a site. Consider the functor

$F : \mathcal{A} \longrightarrow \textit{Ab}(\mathcal{C}), \quad X \longmapsto h_ X$

exactly as in Lemma 19.9.2. It turns out that this functor is fully faithful, exact, and reflects exactness. Moreover, any extension of objects in the essential image of $F$ is in the essential image of $F$.

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# The identity theroem complex analysis

1. Apr 1, 2008

### Scousergirl

1. The problem statement, all variables and given/known data
Prove that there is no holomorphic function f in the open unit disk such that f(1/n)=((-1)^n)/(n^2) for n=2,3,4....

2. Relevant equations
The identity theorem: Let f and g be holomorphic functions in the connected open subset of C, G. If f(z)=g(z) for all z in a subset of G that has a limit point in G, then f=g.

3. The attempt at a solution

We proved in an earlier example that there is not holomorphic function f in the open unit disk such that f(1/n)=2^(-n) n=2,3,4... Is the above function identical to this? I thought this was the route to go but i am not sure anymore. I know we have to somehow use the identity theorem.

2. Apr 1, 2008

### Pere Callahan

Maybe it helps to consider the function $g(z)=(-1)^{\frac{1}{z}}z^2$


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# How do you factor 3x^3-x^2-2x?

##### 1 Answer
May 31, 2016

$x \left(3 x + 2\right) \left(x - 1\right)$

#### Explanation:

First note that every term has $x$ in it so the expression can be rewritten as

$x \left(3 {x}^{2} - x - 2\right)$

The term in brackets is then a standard quadratic which can be factored as $\left(3 x + 2\right) \left(x - 1\right)$

So the final expression is $x \left(3 x + 2\right) \left(x - 1\right)$


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# Solving for intersection of line and a vector function.

How would one approach a problem of finding intersection points between a line $\vec l = \vec S + d \vec t$ and vector of the form $$\vec v = \begin{pmatrix} x \\ y \\f(x, y) \end{pmatrix}$$

I am mainly looking for fast numerical methods. If anyone would take the time to explain through example of $f(x, y) = sin(x) + sin(y)$

If there is a more general form to solve intersections between the line and vectors of form $\vec v'= \begin{pmatrix} f_1(t) \\ f_2(t) \\f_3(t) \end{pmatrix}$ then I would love to know that as well. To be exact I am looking for the first intersection (intersection with the lowest distance d in direction $\vec t$) Hope the question is not too vague. If it is please tell me about the constraints needed. Any reference or keyword helps. Thank you.

EDIT:

I am thinking that we can write $\vec l$ as a combination of its parts and then look at it as set of equations. Meaning: $$\vec l = \begin{pmatrix} \vec S_x + d \vec t_x \\ \vec S_y + d \vec t_y \\ \vec S_z + d \vec t_z\end{pmatrix} \rightarrow \vec l = \vec v \rightarrow \begin{matrix} \vec S_x + d \vec t_x = x \\ \vec S_y + d \vec t_y = y \\ \vec S_z + d \vec t_z = f(x, y) \end{matrix}$$

• First question: what is $\vec S$ vs. $S$? And do you want $d$ in terms of $x$ and $y$ such that $\vec v = \vec l$? (Or something else?) In general, such a $\vec v$ is a curve parametrized by $x$ and $y$. It is possible that your line and your curve do not intersect. (For example, if $\vec v$ parametrizes a line parallel to $\vec l$). Furthermore, $\vec v'$ is not necessarily more general than $\vec v$, since it is only parametrized by one variable, $t$. – artificial_moonlet Sep 23 '15 at 14:15
• Hopefully erased the confusion with $S$ and $\vec S$. They were meant to be components of $\vec S$. I wish to find d given line $\vec l$ and curve $\vec v$ defined above as a function of one or two variables in Cartesian room. It is possible, that they do not have an intersection point in case I would like to know that there exists no solution if possible. Furthermore it is possible to restrict the d to be non-negative if it is any help. – Joonatan Samuel Sep 24 '15 at 10:42


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Previous issue ·  Next issue ·  Most recent issue · All issues

# Journal of Operator Theory

Volume 67, Issue 1, Winter 2012  pp. 73-100.

The higher-dimensional amenability of tensor products of Banach algebras

Authors Zinaida A. Lykova
Author institution: School of Mathematics and Statistics, Newcastle University, Newcastle upon Tyne, NE1 7RU, U.K.

Summary:  We investigate the higher-dimensional amenability of tensor\mathcal{B}reak products $\mathcal{A} \widehat{\otimes} \mathcal{B}$ of Banach algebras $\mathcal{A}$ and $\mathcal{B}$. We prove that the weak bidimension $db_{\mathrm w}$ of the tensor product $\mathcal{A} \widehat{\otimes} \mathcal{B}$ of Banach algebras $\mathcal{A}$ and $\mathcal{B}$ with bounded approximate identities satisfies $db_{\mathrm w} \mathcal{A} \widehat{\otimes} \mathcal{B} = db_{\mathrm w} \mathcal{A} + db_{\mathrm w} \mathcal{B}.$ We show that it cannot be extended to arbitrary Banach algebras. For example, for a biflat Banach algebra $\mathcal{A}$ which has a left or right, but not two-sided, bounded approximate identity, we have $db_{\mathrm w} \mathcal{A} \widehat{\otimes} \mathcal{A} \leqslant 1$ and $db_{\mathrm w} \mathcal{A} + db_{\mathrm w} \mathcal{A} =2.$ We describe explicitly the continuous Hochschild cohomology $\H^n(\mathcal{A} \widehat{\otimes} \mathcal{B}, (X \widehat{\otimes} Y)^*)$ and the cyclic cohomology $\H\C^n(\mathcal{A} \widehat{\otimes} \mathcal{B})$ of certain tensor products $\mathcal{A} \widehat{\otimes} \mathcal{B}$ of Banach algebras $\mathcal{A}$ and $\mathcal{B}$ with bounded approximate identities; here $(X \widehat{\otimes} Y)^*$ is the dual bimodule of the tensor product of essential Banach bimodules $X$ and $Y$ over $\mathcal{A}$ and $\mathcal{B}$ respectively.

Contents    Full-Text PDF


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# 16.9: Properties of Ideal Solutions

We have found the chemical potential of any component in an ideal solution. Now let us find some other thermodynamic properties of an ideal solution. The value of an extensive thermodynamic property of the solution will be the sum of the values of that property for the separate pure components plus the change that occurs when these components are mixed. (The initial state of the system comprises the pure, separate components at a particular temperature and pressure. The mixed state is a homogeneous liquid solution at the same temperature and pressure.) If the solution contains $$n_A$$ moles of component $$A$$ and $$n_B$$ moles of component $$B$$, the Gibbs free energy is

$G_{\mathrm{solution}}\left(n_A,n_B\right)=n_A{\mu }_A+n_B{\mu }_B\nonumber$

Dividing through by $$n_A+n_B$$ to find the Gibbs free energy of the mixture per mole of solution, we have

\begin{align*} \overline{G}_{\mathrm{solution}} \left(y_A,y_B\right) &=y_A\mu_A + y_B\mu_B \\[4pt] &=y_A \Delta_f G^o\left(A\right)+y_ART \ln y_A + y_B \Delta _fG^o \left(B\right)+y_BRT \ln y_B \end{align*}

To make this mixture, we need $$y_A$$ moles of $$A$$ and $$y_B$$ moles of $$B$$. The Gibbs free energy of these amounts of unmixed pure $$A$$ and $$B$$, each in its standard state, is

${\overline{G}}_{\mathrm{initial}}=y_A{\Delta }_fG^o\left(A\right)+y_B{\Delta }_fG^o\left(B\right)\nonumber$

For the process of mixing pure $$A$$ and pure $$B$$, each originally in its standard state, to form one mole of an ideal solution, the Gibbs free energy change is ${\Delta }_{\mathrm{mix}}\overline{G}={\overline{G}}_{\mathrm{solution}}-{\overline{G}}_{\mathrm{initial}}=y_ART{\ln y_A\ }+y_BRT{\ln y_B\ }\nonumber$

In Section 13.3, we found this same relationship for mixing ideal gases:

${\Delta }_{\mathrm{mix}}\overline{G}\left(\mathrm{gas}\right)=x_ART{\ln x_A\ }+x_BRT{\ln x_B\ }\nonumber$

From $${\left({\partial {\Delta }_{\mathrm{mix}}G}/{\partial T}\right)}_P=-{\Delta }_{\mathrm{mix}}S$$, we find

${\Delta }_{\mathrm{mix}}\overline{S}=-y_AR{\ln y_A\ }-y_BR{\ln y_B\ }\nonumber$

and the entropy of the liquid solution is

${\overline{S}}_{\mathrm{solution}}=y_AS^o_A+y_BS^o_B\ -y_AR{\ln y_A\ }-y_BR{\ln y_B\ }\nonumber$

From $${\left({\partial {\Delta }_{\mathrm{mix}}G}/{\partial P}\right)}_T={\Delta }_{\mathrm{mix}}V$$, we find

${\Delta }_{\mathrm{mix}}\overline{V}=0\nonumber$

and from $${\Delta }_{\mathrm{mix}}H={\Delta }_{\mathrm{mix}}G+T{\Delta }_{\mathrm{mix}}S$$, we find

${\Delta }_{\mathrm{mix}}\overline{H}=0\nonumber$

Thus, $${\Delta }_{\mathrm{mix}}\overline{S}$$, $${\Delta }_{\mathrm{mix}}\overline{V}$$, and $${\Delta }_{\mathrm{mix}}\overline{H}$$ for forming an ideal solution are identical also to the relationships we found for mixing ideal gases.

These results have an important physical interpretation. That $${\Delta }_{\mathrm{mix}}\overline{V}=0$$ implies that the molecules of $$A$$ and the molecules of $$B$$ occupy the same volume in the mixture as they do in the pure state. From $${\Delta }_{\mathrm{mix}}\overline{V}=0$$ and $${\Delta }_{\mathrm{mix}}\overline{H}=0$$, it follows that $${\Delta }_{\mathrm{mix}}\overline{E}=0$$ at constant pressure. In turn, this implies that the forces between an $$A$$ molecule and a $$B$$ molecule are the same as the forces between two $$A$$ molecules or between two $$B$$ molecules. If the force of attraction between an $$A$$ molecule and a $$B$$ molecule were stronger than that between two $$A$$ molecules or between two$$\ B$$ molecules, molecules in the mixture would be—on average—closer together in the mixture than in the separate components; we would find $${\Delta }_{\mathrm{mix}}\overline{V}<0$$. Moreover, the potential energy of the mixed state would be lower than that of the separate components; the mixing process would evolve heat at constant temperature; we would find $${\Delta }_{\mathrm{mix}}\overline{E}<0$$.

Conversely, if the repulsive force between an $$A$$ molecule and a $$B$$ molecule were stronger than the repulsive forces between two $$A$$ molecules or between two $$B$$ molecules, the average separation would be greater in the mixture; we would find $${\Delta }_{\mathrm{mix}}\overline{V}>0$$. The potential energy of the mixed state would be greater than that of the separate components; the mixing process would consume heat at constant temperature; we would find $${\Delta }_{\mathrm{mix}}\overline{E}>0$$.

In an ideal gas, molecules do not interact at all. In an ideal solution, the molecules must interact, because only their mutual attraction can keep them in the liquid state. The ideal solution behaves ideally not because the intermolecular interactions are zero but rather because the intermolecular interactions are the same for all of the kinds of molecules present in the mixture. This interpretation implies that the vapor pressures of the pure components of an ideal solution should be equal. Even for solutions that follow Raoult’s law quite closely, this expected equality is often imperfectly realized. Not surprisingly, ideal-solution behavior is best exhibited when the components are isotopically-substituted versions of the same compound.

In an ideal solution, the activities of the components are equal to their mole fractions. The activity of the solvent depends only on the solvent mole fraction. The properties of the solvent in an ideal solution are independent of the specific substance that comprises the solute; they depend only on the concentration of solute particles present. Systems in which this is a useful approximation are sufficiently common that their properties are given a special name: A colligative property of a solution is a property that depends only on the concentration of solute particles and not on the specific chemical properties of the solute. We expect this approximation to become better as the solute concentration approaches zero. When a solute obeys Raoult’s law or Henry’s law, its effects on the thermodynamic properties of the solvent depend only on the concentration of the solute. Consequently, Raoult’s law and Henry’s law prove to be useful when we seek to model colligative properties.

In Sections 16.10 through 16.14, we evaluate five colligative properties: boiling-point elevation, freezing-point depression, osmotic pressure, solid-solute solubility and gas-solute solubility. We derive the first three of these properties from the perspective that they enable us to determine the molar mass of solutes. However, boiling-point elevation, freezing-point depression, and osmotic pressure are important methods for the measurement of activity coefficients in non-ideal solutions. To illustrate the measurement of activity coefficients, we develop a more detailed analysis of freezing-point depression in Section 16.15.


text:
Algebra Level 4

$\large \dfrac{1}{a^{2}+9} + \dfrac{1}{b^{2}+9} + \dfrac{1}{c^{2}+9}$

Given that $$a,b$$ and $$c$$ are positive reals with a sum of 1. If the maximum value of the above expression can be expressed as $$\dfrac{x}{y}$$ where $$x$$ and $$y$$ are coprime integers, find the value of $$x+y$$.

×


text:
Multiple Plots on the Same Graph

## Description

Here is an example of how to display multiple plots on one set of axes. We will plot the following polynomials:

(1)
\begin{align} P_1 &= (x - 1)(x - 2)(x - 3)\\ P_2 &= (x - 1)(x - 2)(x - 4)\\ P_3 &= (x - 1)(x - 2)(x - 5)\\ P_4 &= (x - 1)(x - 2)(x - 6)(x - 10) \end{align}

Setting each of the plots to a different variable, then adding them together to another variable and using the .show() method allows us to show multiple plots on the same set of axes.

## Sage Cell

#### Code

P1 = plot((x - 1) * (x - 2) * (x - 3), 1, 6)
P2 = plot((x - 1) * (x - 2) * (x - 4), 1, 6)
P3 = plot((x - 1) * (x - 2) * (x - 5), 1, 6)
P4 = plot((x - 1) * (x - 2) * (x - 6) * (x - 10), 1, 6)
P = P1 + P2 + P3 + P4
P.show()

## Options

#### Option

Plotting the functions in different colors is possible. A full list of built in colors can be found by evaluating sorted(colors) in a cell. To set a plot to a certain color, set the color property, with the color you want in single or double quotes after the equal sign.

#### Code

P1 = plot((x - 1) * (x - 2) * (x - 3), 1, 6, color = 'red')
P2 = plot((x - 1) * (x - 2) * (x - 4), 1, 6, color = 'blue')
P3 = plot((x - 1) * (x - 2) * (x - 5), 1, 6, color = 'purple')
P4 = plot((x - 1) * (x - 2) * (x - 6) * (x - 10), 1, 6, color = 'purple')
P = P1 + P2 + P3 + P4
P.show()

## Tags

Primary Tags—Plotting: Two-dimensional plots

Secondary Tags—Two-dimensional plots: Rectangular plots


text:
Volume 334 - The 36th Annual International Symposium on Lattice Field Theory (LATTICE2018) - Weak Decays and Matrix Elements
$B_s\to D^{(*)}_s l\nu$ form factors using heavy HISQ quarks
E. McLean,* C. Davies, A. Lytle, J. Koponen on behalf of HPQCD Collaboration
*corresponding author
Full text: pdf
Published on: 2019 May 29
Abstract
We present progress on an ongoing calculation of the $B_s\to D_s^{(*)} l \nu$ form factors calculated on the $n_f=2+1+1$ MILC ensembles and using the Highly Improved Staggered Quark action for all valence quarks. We perform the calculation at a range of $b$ quark masses (and lattice spacings) so that we can extrapolate to the physical $b$-quark mass.
DOI: https://doi.org/10.22323/1.334.0281
Open Access


text:
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# International Conference on Technology and Instrumentation in Particle Physics 2017(TIPP2017)

May 21 – 26, 2017
Beijing International Convention Center
Asia/Shanghai timezone

## Signal response of diamond detectors in intense proton beams and at high temperature

Not scheduled
15m
Beijing International Convention Center

#### Beijing International Convention Center

No.8 Beichen Dong Road, Chaoyang District, Beijing P. R. China 100101
oral Interface and beam instrumentation

### Speaker

Pradeep Sarin (Indian Institute of Technology, Bombay)

### Description

In extreme radiation environments like monitoring intense proton beams, or at high temperature in a nuclear reactor, semiconductor charged particle detectors made of synthetic diamond provide a promising solution due to their high radiation tolerance and thermal properties. We present new test results on the charge collection performance of diamond detectors tested in intense proton beams at JPARC with bunched 8 GeV beams at the level of $10^{12}$ protons/bunch with 1 microsec pulse repetition rateas expected in the COMET experiment. In addition, for monitoring applications in neutron rich reactor environment where the temperature can be very high, we have studied the charge collection properties at temperature up to 600K, and found surprising trends for devices made with poly and single crystal diamond. The experiment methodology and results will be presented.

### Summary

We are developing diamond detectors for proton beam monitoring in the COMET experiment at JPARC. The aim of the experiment is to detect charged lepton flavor violation. To achieve this goal, the input 8 GeV proton beam must be monitored both for integrated flux and bunch wise extinction. The unprecedented challenge of
measuring extinction of $10^{12}$ protons/bunch at a repetion rate of 1 $\mu sec$ can only be met by diamond detectors. First results from ptototype detectors will be presented.
In parallel, monitoring a neutron rich environment inside a nuclear reactor poses a challenge because of the high temperature in addition to the intense radiation fluence. We have studied the properties of charge collection in diamond at temperature up to 600K - prelimnary results from the high temperature measurement will be presented.

### Primary author

Pradeep Sarin (Indian Institute of Technology, Bombay)

### Presentation materials

There are no materials yet.


text:
1. ## integral Cauchy theorem

Hi,
I dont understand the cauchy theorem.
I have this problem and I would like to have some help for it.
The question is:
Use the Cauchy Integral Theorem to prove that:
$\displaystyle \int_{-\infty}^{+\infty}\frac{1}{x^2-2x+5}dx=\frac{\pi}{2}$

When I read the documentation on it , it is told to have a close surface . but here it is not specify...Or I just don't get it.

B

Hi,
I dont understand the cauchy theorem.
I have this problem and I would like to have some help for it.
The question is:
Use the Cauchy Integral Theorem to prove that:
$\displaystyle \int_{-\infty}^{+\infty}\frac{1}{x^2-2x+5}dx=\frac{\pi}{2}$

When I read the documentation on it , it is told to have a close surface . but here it is not specify...Or I just don't get it.

B
Pretend this is an integration over a complex variable z:
$\displaystyle \int_{-\infty}^{+\infty}\frac{1}{z^2-2z+5}dz$.

Now, the Cauchy theorem says that if you integrate a complex integrand (counterclock-wise) over a closed loop that the integral is $\displaystyle 2\pi i$ times the sum of the residues.

So what closed loop to use? Well, two leap to mind:
1) Start at $\displaystyle -\infty$ on the real axis and go to $\displaystyle \infty$, then close the loop by traveling a half-circle (of radius $\displaystyle \infty$ ) upward and to the left, finishing at $\displaystyle -\infty$ on the real axis. (Technically we go from -R to R on the real axis, and take the limit of the integrals as R goes to $\displaystyle \infty$.)

2) Same idea as the first, but close the loop by doing a half-circle downward and to the left.

In each case the ideal result is that the integral over the half-circle ends up being 0. This is the case here. (This is Jordan's Lemma.)

Which contour you choose depends on the integral and location of the poles. In this case you have one pole in the upper half-plane and one in the lower half-plane. I see no reason to choose one contour over the other here. So pick one and do it. (Or if you want some practice, pick both and do them to see that you get the same answer.)

-Dan

3. Thank you. I will try that now!!

Hi,
I dont understand the cauchy theorem.
I have this problem and I would like to have some help for it.
The question is:
Use the Cauchy Integral Theorem to prove that:
$\displaystyle \int_{-\infty}^{+\infty}\frac{1}{x^2-2x+5}dx=\frac{\pi}{2}$

When I read the documentation on it , it is told to have a close surface . but here it is not specify...Or I just don't get it.

B
Define $\displaystyle f(z) = \frac{1}{z^2-2z+5} = -4i \left( \frac{1}{z-1-2i} - \frac{1}{z-1-2i} \right)$

Integrate over a large semicircle,
$\displaystyle -4i \int_{-R}^R \frac{1}{z-1-2i} - \frac{1}{z-1-2i} dz +\mbox{ error }$$\displaystyle = 2 \pi i(-4i)[f'(1+2i) - f'(1-2i) ]$

By Cauchy integral theorem.

The error term goes to zero as $\displaystyle R\to \infty$.

5. Thank you ThePerfectHacker!

Thank you ThePerfectHacker!
Would you like me do explain what I mean by "error" and why it goes to zero if you do not understand? (I was just sick yesterday and it was hard for me to respond fully).

7. Yes please explain me this point,

I have not noticed the "error" before and I was about to ask you this question. I was suscpecting a term with a e^(-R* "something") that lead to zero when R-> inf. but I couls not find it!

8. I am just delving into CA and don't want to appear anal, but I believe PH means:

$\displaystyle \frac{-i}{4}\left[\frac{1}{z-(1+2i)}-\frac{1}{z-(1-2i)}\right]$

This can be broken into two integrals, giving:

Which gives $\displaystyle \frac{-i}{4}(2{\pi}i)=\frac{\pi}{2}$


text:
×

# Is $$0^0=1$$ and can be applied everywhere in mathematics?

4 years, 3 months ago

Sort by:

exact zero raise to the power exact zero is NOT DEFINED. · 4 years, 3 months ago

Read this funny yet informative article http://www.askamathematician.com/2010/12/q-what-does-00-zero-raised-to-the-zeroth-power-equal-why-do-mathematicians-and-high-school-teachers-disagree/ · 4 years, 3 months ago

Watch this video · 4 years, 3 months ago

http://www.wolframalpha.com/input/?i=lim+x+is+approaching+to+0+x%5Ex · 4 years, 3 months ago

In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form. The indeterminate forms include 0^0, 0/0, 1^∞, ∞−∞,∞/∞, 0 × ∞, and ∞/0.........so it's also a indeterminate form you have given.... · 4 years, 3 months ago

0^0 = 0^(m-m) = 0^(m) . 0^(-m) .... now as, 0^(-m) = 1 / 0 which is undefined... so 0^0 = undefined .....well first I tried to prove it like this but then I found that it wasn't a good proof as it had contradictions . :( · 4 years, 3 months ago

it is indeterminate form in chapter limits.... · 4 years, 3 months ago

but mathematicians prove through limits that it it comes to 1? · 4 years, 3 months ago

ok......can u xplain me how? · 4 years, 3 months ago

read the article posted by Saurabh. · 4 years, 3 months ago

ok thanks....i'll do that... · 4 years, 3 months ago

limits tell us that n/0 = inf but we know that n/0 is undefined · 4 years, 3 months ago

well someone can give me the clear proof of any number raised to 0 is 1. · 4 years, 3 months ago

yes we can prove it by log theorem? take log and calculate. · 4 years, 3 months ago

0^0=0/0-------------not defined · 4 years, 3 months ago

look at this.

4^2 = 16

4^1 = 4

4^0 = 1

why 4^0 = 1?

easy to proof. imagine that the exponent is absolutely like an element of carbon molecule. exponent is like a box that contain something, copy the box and paste then imagine that they will combine and has more energy, it will be two thing that are same like x*x in mathematic.

so easy. it means that there are no element, like an empty box. because the box still present in universe, it mean there must be a symbol to represent it, that is the 1 or one. · 4 years, 3 months ago

If 0^0=1 what's the proof and if 0^0= undefined what's the proof ??? · 4 years, 3 months ago

really nice question....... i think so it is undefined. · 4 years, 3 months ago

I agree with that. You can easily prove that it is an indeterminate form in mathematics, · 4 years, 3 months ago

in laws of exponent they never mention that base a is not equal to 0 · 4 years, 3 months ago

undefined. bt nt equal 2 1. · 4 years, 3 months ago

The way I was taught, even though 0^0 is indeterminate, mathematicians will consider it to be 1. · 4 years, 3 months ago

considering is different thing...sometimes in physics also we consider that 1/0 = infinity...but that is wrong.. 0^0 is undefined · 4 years, 3 months ago

it should be zero.. · 4 years, 3 months ago

But many take it to be 1. · 4 years, 3 months ago

Yes. · 4 years, 3 months ago

only google calculator is showing 1! but other scientific calculators finds math error. · 4 years, 3 months ago

I know that. · 4 years, 3 months ago

0^0 is neither 1 nor indeterminate, it is undefined... there is a difference between indeterminate and undefined. e.g. app.0/app.0 is indeterminate where as 0/0 is undefined similarly, 0^0 is undefined as 0^0 = e^(0log0) where log 0 is undefined, but (app. 0)^app.0 is indeterminate whose value is 1...if both functions which are approaching 0 used in (app0)^app0 are the same. if both functions are x. lim (x app0) x^x can be found by equating it to e^xlogx where if xapp. 0 xlogx approaches 0[this can be found by using LH rule] and hence x^x approaches 1(as e^0 = 1) · 4 years, 3 months ago


text:
Recent advances in the analysis of pointwise state-constrained elliptic optimal control problems
ESAIM: Control, Optimisation and Calculus of Variations, Volume 16 (2010) no. 3, p. 581-600

Optimal control problems for semilinear elliptic equations with control constraints and pointwise state constraints are studied. Several theoretical results are derived, which are necessary to carry out a numerical analysis for this class of control problems. In particular, sufficient second-order optimality conditions, some new regularity results on optimal controls and a sufficient condition for the uniqueness of the Lagrange multiplier associated with the state constraints are presented.

DOI : https://doi.org/10.1051/cocv/2009010
Classification:  49J20,  49K20,  35J65
Keywords: optimal control, pointwise state constraints, first and second order optimality conditions, Lagrange multipliers, Borel measures
@article{COCV_2010__16_3_581_0,
author = {Casas, Eduardo and Tr\"oltzsch, Fredi},
title = {Recent advances in the analysis of pointwise state-constrained elliptic optimal control problems},
journal = {ESAIM: Control, Optimisation and Calculus of Variations},
publisher = {EDP-Sciences},
volume = {16},
number = {3},
year = {2010},
pages = {581-600},
doi = {10.1051/cocv/2009010},
zbl = {1201.49004},
mrnumber = {2674627},
language = {en},
url = {http://www.numdam.org/item/COCV_2010__16_3_581_0}
}

Casas, Eduardo; Tröltzsch, Fredi. Recent advances in the analysis of pointwise state-constrained elliptic optimal control problems. ESAIM: Control, Optimisation and Calculus of Variations, Volume 16 (2010) no. 3, pp. 581-600. doi : 10.1051/cocv/2009010. http://www.numdam.org/item/COCV_2010__16_3_581_0/

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