text:
See title.

Yes, you read right. SteamCMD is not working on CephFS (and maybe GlusterFS too, I can’t test it on GlusterFS ). The reason for that is a “failing” getdents when on the CephFS. A colleague told me that error may be related to the st_size and the rbytes of CephFS.

strace on xfs filesystem:

 1 2 3 4 5 6 7 8  [...] [pid 123] stat64("./zip0.zip", 0xf71c9c40) = -1 ENOENT (No such file or directory) [pid 123] open(".", O_RDONLY|O_NONBLOCK|O_LARGEFILE|O_DIRECTORY|O_CLOEXEC) = 2 [pid 123] fstat64(2, {st_mode=S_IFDIR|0755, st_size=100, ...}) = 0 [pid 123] getdents(2, /* 7 entries */, 32768) = 152 [pid 123] getdents(2, /* 0 entries */, 32768) = 0 [pid 123] close(2) [...]

strace on CephFS:

 1 2 3 4 5 6 7  [...] [pid 123] stat64("./zip0.zip", 0xf720ac40) = -1 ENOENT (No such file or directory) [pid 123] open(".", O_RDONLY|O_NONBLOCK|O_LARGEFILE|O_DIRECTORY|O_CLOEXEC) = 2 [pid 123] fstat64(2, {st_mode=S_IFDIR|0755, st_size=5, ...}) = 0 [pid 123] getdents(2, 0xf7002114, 65536) = -1 EOVERFLOW (Value too large for defined data type) [pid 123] close(2) [...]

As you can see the st_size on the xfs filesystem reported is higher (100). The third parameter (unsigned int count from getdents man page), comes as a result from the “weird” st_size, given to the getdents() function is lower 32768 than the int 65536 on the CephFS.

In C an unsigned short int can only be 65535 max, but 65536 got returned..

Next I tried to look into the rbytes that a colleague mentioned. And guess what I found looking for a way to disable/fix/modify the st_size from CephFS. I found out that the documented rbytes and for me interesting norbytes mount option. When looking at the docs we see that both mount options are documented: http://docs.ceph.com/docs/hammer/man/8/mount.ceph/#options I used GitHub’s code search but the only occurrences I found are in the man page files.

I have directly created an issue at the Ceph’s Bugtracker to see if it is an issue in my setup or really a missing feature in Ceph. For the interested here is the link to the issue: http://tracker.ceph.com/issues/16738.


text:
Matlab and the 3 Step Adams-Bashforth Method

I am trying to run and plot the solutions to the 3-step Adams-Bashforth method and am unable to understand where my code is wrong. I am very new to Matlab and have been asked to code this without a good prior knowledge of Matlab. Matlab plots my exact solution fine on the interval but I am not having the same luck with my approximated solution. Below is my code and any help would be greatly appreciated.

% To solve y' = -3*y+6*t+5 s.t. y(0) = 2e^3-1 for -1 <= t <= 2 using

clear, clc, clf
f = @(t, y) -3*y+6*t+5; % RHS
h = 0.2; % time step size
T = 3; % This is length of the time interval for which you're solving for, i.e. 2-(-1) = 3.
N = T/h; % total number of times steps

F = @(t) 2*exp(-3*t)+2*t+1; % true solution

% Preallocations:
t = zeros(N+1, 1);
y = zeros(N+1, 1);

% Initializations:
y(1) = 2*exp(3)-1; % Initial value of the ODE IVP

% Compute the solution at t = h by using the true solution:
t(2) = h;
y(2) = F(t(2));

t(3) = 2*h;
y(3) = F(t(3));

% Main loop for marching N steps:
for i = 2:N
t(i+2) = i*h; % time points
y(i+2) = y(i+1) + (h/12)*(23*f(t(i+1), y(i+1)) - 16*f(t(i), y(i)) +5*f(t(i-1),y(i-1))); % 2-step Adams-Bashforth method!!!
end

% Plotting:
plot(t, y), hold on % plot the numerical solution obtained by Adams-Bashforth

% Plot the true solution:
tt = linspace(-1, 2, 1000); % sampling points for true solution
ff = F(tt); % function values at sampling points
plot(tt, ff, 'r') % plot the true solution using the sampling values

Below is the Graph I am currently outputting - which is very off for a 3rd order approximation.

Since that comment got a bit long here a "comment-answer"

• How does your solution look? Does it explode, does it converge to something different?
• Can you include a plot, if there is one where you can actually see something?
• As that is an explicit method: Does a reduction of the step-size help?
• y(0) = 2e^3-1 for -1 <= t <= 2 using What does this mean? Do you want to start at $0$ or at $-1$?

Edit:

If I understand correctly: You want to solve: \begin{align*} y'(t) &= -3y+6t+5 \qquad \text{in }[-1,2]\\ y_{-1} &= 2e^3-1 = y(-1) \end{align*}

But what your code actually solves is: \begin{align*} y'(t) &= -3y+6t+5 \qquad \text{in } [-1,2]\\ y_{0} &= 2e^3-1 \neq y(0) \end{align*}

Matlab will not let you have a negative value for y(t).

Yes, it is normal, that you can't enter negative numbers there. What you use in matlab is an array, and arrays start at index 1. Your solution is supposed to be on the intervall $[-1,2]$, so we have $$t_1 = -1,\qquad t_2 = -1+h \qquad t_3=-1+2h \qquad … \qquad t_i = -1+(i-1)h$$

In Matlab that would be:

 t[1] = -1 % Stores -1 at index 1
t[2] = -1+h % Stores -1+h at index 2
...

So you should fix this line

t(i+2) = i*h; % time points

What you can try to test if the rest of your implementation is correct, is setting the initial value to: $$y_0=y(0) = 3$$ in line y(1) = 2*exp(3)-1; % Initial value of the ODE IVP

This would mean that you solve the correct ODE with the correct starting value at $t=0$.

• Unsure as to how I can show the plot on mathstack, it follows the direction of the curve however is very volatile and doesn't follow the curve of the exact solution well. y(-1) = 2exp^3-1 is the solution given for the original value problem, however Matlab will not let you have a negative value for y(t). So I want to start at -1. I believe the error could lie where I get the 3rd value for y as I am unsure as to what t(3) and y(3) should be set equal to. – user496388 Oct 27 '17 at 14:58
• Simply take a screenshot from the plot-window. Then edit your question (there is an edit button below the question) and click on the image icon in the bar at the top of the edit window. There you can simply include your picture. – P. Siehr Oct 27 '17 at 15:01
• @user496388 Have a look at the edit in my answer. – P. Siehr Oct 27 '17 at 15:20
• I just changed my code and as I make the values of H smaller is works, thank you very much for your help! – user496388 Oct 27 '17 at 15:33
• Did you only change the size of h, or also what I mentioned in my answer? Since you are new: You can mark the answer that helped the most with a green arrow (on the left side of the answer). This will help other people with the same problem. Another indicator of good answers are up-votes. – P. Siehr Oct 27 '17 at 15:41

format: plain


text:
Tennis scores meets Code Golf

In light of the 2014 Australian Open in Melbourne, and the win by Stanislas Wawrinka against Novak Djokovic, I propose the following code golf challenge!

Wawrinka and Djokovic are at it again for the basis of this CGC. Your challenge is to take a string consisting of 1's and 2's and determine the tennis scores based on the tennis scoring system. A "1" indicates that Wawrinka has scored a point, whereas a "2" indicates that Djokovic has scored a point.

For the sake of example: 1211222122111122111 will generate a two-line output:

Wawrinka 1 - 40
Djokovic 1 - 30

The longer the string, the further into the game the score is.

Rules:

• Shortest code wins, pure and simple.
• The scores when displayed must be right aligned and also column aligned; e.g.
• Sample alignment of output:

Wawrinka 7 5 3 -  0
Djokovic 6 7 4 - 15

• If a game is in progress, the score must be displayed after the set scores. Game scores must follow set scores with a space-dash-space separator. If this happens to be the first game, a 0-0 set score must be displayed.

Wawrinka 0 -  0
Djokovic 0 - 15

• A game will go in order of 0, 15, 30, 40, and game point. If a game reaches 40-40, a deuce is declared, and the score must display the deuce on the player row scoring the point:

Wawrinka 7 5 3 - 40
Djokovic 6 7 4 - 40 Deuce

• If a deuce is reached, the next point will be an advantage to the player scoring the point:

Wawrinka 7 5 3 - Ad
Djokovic 6 7 4 - 40

The point after that, if scored by the advantaged player, wins the game, else the game goes back to deuce.

• If more than one deuce is scored in between advantage points, the number of the deuce will be indicated by a number in parentheses:

Wawrinka 7 5 3 - 40
Djokovic 6 7 4 - 40 Deuce (2)

• A set is won if a player reaches 6 games or more and with a two game margin in lead, e.g. 7-5, 6-4... 9-7 (In the case of last night's fifth set)

• Consecutive sets must be separated with a space.

• There are tie breaks in the Australian Open, therefore if a 6-6 set is reached, a tie-break is established by the first player reaching seven points with a two point margin lead. Display of tie breaks are shown within square brackets as follows:

Wawrinka 7 6 [6]
Djokovic 6 6 [4]

The winner of the tie-break is said to have won the previous set 7-6.

• If the fifth set reaches 6-6, play is continued without tie-break until a two game margin is reached with one player, that player being the winner of that set. There are NO tie-breaks in the fifth set.

• If a player has won three sets, a third line will be displayed saying Wawrinka wins or Djokovic wins depending on who has won and the code then terminates.

Wawrinka 2 6 6 3 9
Djokovic 6 4 2 6 7
Wawrinka wins

• If the string ends such that the last point has won a game, set or match, the game score is not displayed...

Wawrinka 7 5 3
Djokovic 6 7 4

• If the string goes beyond a match being won, the rest of the string is ignored, the score displayed and the winner declared.

• I really don't think we need a new tag called sports; game should work just fine. – Justin Jan 30 '14 at 0:48
• What will be the result? Just the end score or every step along the way? – Teun Pronk Jan 30 '14 at 8:17
• @TeunPronk depending on the score, the match will be finished or not. The idea seems to be: Given the string of points, what is the current score. – Tim Seguine Jan 30 '14 at 14:18
• @McKay - you shouldn't be able to reach 9-7 in the third set of the Australian Open; the tiebreak should apply in every set except the fifth. – Chowlett Jan 30 '14 at 16:41
• @WallyWest you were correct with your previous edit; only the fifth set has no tie-breaker. I fixed your example to make this more clear. – primo Jan 31 '14 at 2:58

Perl - 424 bytes

*1=a;*2=b;@1=(Wawrinka,0);@2=(Djokovic,0);
$$_++>{o=S^_}&&$$_>3and$1=$2=0,
$w=$w[$_]+=($$_[.]++>$$o[$.]||(($t=$$_[.]==$$o[$.])&&!$%&&$.<5))&&
$$_[.]>5+%&&!(1[.]=2[.+=!%]=w<2&&0,$$_[$.-1]+=$%,$%=$t)for<>=~/./g;
@s=(' 0',15,30,(40)x($e=$$o<3||$$o-2),Ad);$%and$_="[$_]"for@1[-1],@2[-1];$d[$&]=$1>2&&$1==$2&&' Deuce'.($1>3&&" ($e)"); print"@1",$w<3&&" - $s[$1]$d[1]","\n@2",$w<3&&" - $s[$2]$d[2]",$w>2&&"\n${$&}[0] wins"

I believe this to be a complete solution, according to the scoring of the Australian Open:

• Best of five sets (a.k.a. race to three).
• Sets 1-4 are played as a 6-6 tie-breaker.
• Set 5 is played as an advantage set.

Test Cases

1211222122111122111

Wawrinka 1 - 40
Djokovic 1 - 30

12112221221111221112

Wawrinka 1 - 40
Djokovic 1 - 40 Deuce

121122212211112211122

Wawrinka 1 - 40

1211222122111122111221

Wawrinka 1 - 40 Deuce (2)
Djokovic 1 - 40

22111111212122221122111212212112121221212211221121222222112112221121121122221122221211111222121222122211212122111212112211222121211212211212211122121211112222222212211121122

Wawrinka 6 [5] - 30
Djokovic 6 [6] - 40

221111112121222211221112122121121212212122112211212222221121122211211211222211222212111112221212221222112121221112121122112221212112122112122111221212111122222222122111211222

Wawrinka 6 0 -  0
Djokovic 7 0 -  0

1122222211121211121211111121111211221222212212112221211222211222112212211121122122212122212222122212212211221111121222111221211111211112222212122122112111212121221221212211112122212211111111112111212222221112212121122212121111122111222222111212221121221111222122122222111212111111221121122111122122111222222121122221112221221122221121211212111122111121212112112121222122

Wawrinka 5 4 6 6 5 - 15
Djokovic 7 6 4 4 6 - 40

11222222111212111212111111211112112212222122121122212112222112221122122111211221222121222122221222122122112211111212221112212111112111122222121221221121112121212212212122111121222122111111111121112122222211122121211222121211111221112222221112122211212211112221221222221112121111112211211221111221221112222221211222211122212211222211212112121111221111212121121121212221222

Wawrinka 5 4 6 6 5
Djokovic 7 6 4 4 7
Djokovic wins

222221112112212212222111222211111111121111121112211221221211212121122211222112111112122122212222211112122212221111121111121211212112112112221221121122121121112221221222122122211222212121212112112111221221121112222212122222221111112222222221221122211221121111221121222222111111122221122111211121222112112122212122221121222221222121212111121221221112111212212222122212212212112111112112112121112221111221221221121222122211221212211111111222222121221112221212

Wawrinka 6 7 6 4 7 - 40
Djokovic 3 5 7 6 8 - Ad

2222211121122122122221112222111111111211111211122112212212112121211222112221121111121221222122222111121222122211111211111212112121121121122212211211221211211122212212221221222112222121212121121121112212211211122222121222222211111122222222212211222112211211112211212222221111111222211221112111212221121121222121222211212222212221212121111212212211121112122122221222122122121121111121121121211122211112212212211212221222112212122111111112222221212211122212122

Wawrinka 6 7 6 4 7
Djokovic 3 5 7 6 9
Djokovic wins

All intermediate scores for the last test case can be seen here: http://codepad.org/FzDIcf0W

• Isn't 11-9 only meant to be valid as a fifth set? – Volatility Jan 30 '14 at 23:54
• @Volatility actually, you were right. I'll need to update. – primo Jan 31 '14 at 2:58

ECMAScript 6 - 635 Characters

f=(p,q,C)=>{B='';T=' ';L=(x)=>(B+x).length;M=(x,y)=>x>y?x:y;E=(a)=>{var x=L(a[0]),y=L(a[1]),q=M(x,y);for(;x++<q;)a[0]=T+a[0];for(;y++<q;)a[1]=T+a[1]};E(a=[p,q]);g=[m=n=o=i=t=z=0,0];s=[0,0];S=[0,0];w=2;O=(i)=>a[i]+(n+o>0?T+s[i]:B)+(o>0?(t?' ['+g[i]+']':' - '+[' 0',15,30,40,'Ad'][g[i]]+(z>0&&g[0]+g[1]==6&&i==w?' Deuce'+(z>1?' ('+z+')':B):B)):B)+'\n';while(W=C[i++]){w=--W;++o;if((d=++g[w])>M(t?6:3,(e=g[l=1-w])+1)){g=[o=z=0,0];j=++s[w];k=s[l];t=++n>11;if(j>M(5,m>4||k<6?k+1:6)){E(s);a[0]+=T+s[0];a[1]+=T+s[1];s=[n=0,0];++m;if(++S[w]>2)break}}else if(!t&&d+e>7){--g[w];--g[l];++z}}E(s);E(g);return O(0)+O(1)+(S[w]>2?(w?q:p)+' wins':B)}

// Function f takes arguments:
//   p - Player 1 name
//   q - Player 2 name
//   C - String of 1s and 2s representing points won by players.
f=(p,q,C)=>{
/* Empty String          */ B='';
/* Space                 */ T=' ';
/* String Length Func.   */ L=(x)=>(B+x).length;
/* Max Function          */ M=(x,y)=>x>y?x:y;
/* Equalize Length Func. */ E=(a)=>{var x=L(a[0]),y=L(a[1]),q=M(x,y);for(;x++<q;)a[0]=T+a[0];for(;y++<q;)a[1]=T+a[1]};
/* No. of sets           */ m=0;
/* No. games in set      */ n=0;
/* No. points in game    */ o=0;
/* Input Index           */ i=0;
/* Output String         */ E(a=[p,q]);
/* Current Game's Points */ g=[0,0];
/* Current Set's Games   */ s=[0,0];
/* No. sets won          */ S=[0,0];
/* Is a tiebreaker       */ t=0;
/* No. of deuces         */ z=0;
/* Current match result  */ w=2;
/* Output Fnctn  */ O=(i)=>a[i]+(n+o>0?T+s[i]:B)+(o>0?(t?' ['+g[i]+']':' - '+[' 0',15,30,40,'Ad'][g[i]]+(z>0&&g[0]+g[1]==6&&i==w?' Deuce'+(z>1?' ('+z+')':B):B)):B)+'\n';
while(W=C[i++]){
w=--W;
// w - index of winner of current game
// l - index of loser of current game
// d - winner of current point's game score
// e - loser of current point's game score
++o;
if((d=++g[w])>M(t?6:3,(e=g[l=1-w])+1)){
g=[0,0];  // Reset the game score.
o=z=0;
j=++s[w]; // j = Increment the winner's set score
k=s[l];   // k = Loser's set score
t=++n>11; // Is a tiebreak?
if(j>M(5,m>4||k<6?k+1:6)){
E(s);
s=[n=0,0];  // Reset current set's no. of games & no. of deuces.
++m;        // Increment no. of sets.
if(++S[w]>2)break;     // Increment winners no. sets won and check if match won.
}
}
else if(!t&&d+e>7){--g[w];--g[l];++z}// Check if deuces increased.
}
E(s); // Format sets strings.
E(g); // Format games strings.
return O(0)+O(1)+(S[w]>2?(w?q:p)+' wins':B);
}

Testing

var tests=[
'',
'1',
'2',
'11',
'222',
'1111',
'2222',
'1112221',
'11122212',
'121212121',
'1212121212',
'1211222122111122111',
'12112221221111221112',
'121122212211112211122',
'1211222122111122111221',
'1111222211112222111122221111222211112222111122221212121212121212121',
'11111111111111111111111111111111111111111111111111111111111111111111111',
'111111111111111111111111111111111111111111111111111111111111111111111111',
'1111111111111111111111111111111111111111111111111111111111111111111111111'
];
for(var ex in tests) console.log('Test '+ex+'\n'+tests[ex]+'\n'+f('Wawrinka','Djokovic',tests[ex]));

Test Outputs

Test 0

Wawrinka
Djokovic

Test 1
1
Wawrinka 0 - 15
Djokovic 0 -  0

Test 2
2
Wawrinka 0 -  0
Djokovic 0 - 15

Test 3
11
Wawrinka 0 - 30
Djokovic 0 -  0

Test 4
222
Wawrinka 0 -  0
Djokovic 0 - 40

Test 5
1111
Wawrinka 1
Djokovic 0

Test 6
2222
Wawrinka 0
Djokovic 1

Test 7
1112221
Djokovic 0 - 40

Test 8
11122212
Wawrinka 0 - 40
Djokovic 0 - 40 Deuce

Test 9
121212121
Djokovic 0 - 40

Test 10
1212121212
Wawrinka 0 - 40
Djokovic 0 - 40 Deuce (2)

Test 11
1211222122111122111
Wawrinka 1 - 40
Djokovic 1 - 30

Test 12
12112221221111221112
Wawrinka 1 - 40
Djokovic 1 - 40

Test 13
121122212211112211122
Wawrinka 1 - 40

Test 14
1211222122111122111221
Wawrinka 1 - 40 Deuce
Djokovic 1 - 40

Test 15
1111222211112222111122221111222211112222111122221212121212121212121
Wawrinka 6 [10]
Djokovic 6 [ 9]

Test 16
11111111111111111111111111111111111111111111111111111111111111111111111
Wawrinka 6 6 5 - 40
Djokovic 0 0 0 -  0

Test 17
111111111111111111111111111111111111111111111111111111111111111111111111
Wawrinka 6 6 6
Djokovic 0 0 0
Wawrinka wins

Test 18
1111111111111111111111111111111111111111111111111111111111111111111111111
Wawrinka 6 6 6
Djokovic 0 0 0
Wawrinka wins

Javascript - 743 bytes

var b=process.argv[2],c=[0,0],e=[],g=[0,0],h=[0,0],k=["0","15","30","40"],l=["Wawrinka","Djokovic"];function m(a){var d="",f;for(f in e)d+=e[f][a]+" ";3<=c[0]||3<=c[1]||(d+=g[a]+" - ",f=h[a],a=h[(a+1)%2],d=n()?d+("["+f+"]"):3>f||3>a?d+k[f]:f>a?d+"Ad":d+k[3]);return d}function n(){return 6<=g[0]&&6<=g[1]}function p(){var a=q;g[a]++;h=[0,0];7<=g[a]?(e.push(g),g=[0,0],c[a]++):6<=g[a]&&g[a]>g[(a+1)%2]+1&&(e.push(g),g=[0,0],c[a]++)} for(var r in b){var q=parseInt(b[r])-1,s=++h[q];n()?7<=s&&(4>e.length?p():s>h[(q+1)%2]+1&&p()):4<=s&&s>h[(q+1)%2]+1&&p()}console.log(l[0]," ",m(0));console.log(l[1]," ",m(1),n()?"":2<h[0]&&2<h[1]&&h[0]==h[1]?"Deuce"+(3<h[0]?" ("+(h[0]-2)+")":""):"");console.log(3<=c[0]||3<=c[1]?(3<=c[0]?l[0]:l[1])+" wins":"");

Without closure compiler:

var input = process.argv[2];

var score = [0,0];
var match = [];
var set = [0,0];
var game = [0,0];
var gameScore = [ '0', '15', '30', '40' ];
var names = ['Wawrinka', 'Djokovic'];

function printScores(who)
{
var out = '';
for (var i in match) {
out += match[i][who] + ' ';
}

if (!isDone()) {
out += set[who] + ' - ';

var point = game[who], otherPoint = game[(who+1)%2];

if (isTieBreaker()) { // Tie breaker
out += '['+point+']';
} else {
if (point < 3 || otherPoint < 3) {
out += gameScore[point];
} else if (point > otherPoint) {
} else {
out+= gameScore[3];
}
}
}

return out;
}

function printDeuce()
{
if (isTieBreaker()) {
return '';
} else {
return (game[0] > 2 && game[1] > 2 && game[0] == game[1]) ? ('Deuce' + (game[0] > 3 ? ' (' + (game[0] - 2) + ')' : '')) : '';
}

}

function isDone()
{
return score[0] >= 3 || score[1] >= 3;
}

function isTieBreaker()
{
return set[0] >= 6 && set[1] >= 6;
}

function getOther(who)
{
return (who + 1) % 2;
}

{
var points = ++game[who];

if (isTieBreaker()) {
if (points >= 7) {
if (match.length < 4) {
} else if (points > (game[getOther(who)]+ 1)) {
}
}
} else {
if (points >= 4 && points > (game[getOther(who)] + 1)) {
}
}
}

{
set[who]++;
game = [0,0];

if (set[who] >= 7) { // Won Tiebreaker
} else if (set[who] >= 6 && set[who] > (set[getOther(who)] + 1)) {
}
}

{
match.push(set);
set = [0,0];
score[who]++;
}

// Play game
for(var i in input) {


text:
# Probability- drawing cards from a deck.

by johnsa9
Tags: combinations, probability
 P: 1 I am having some issues with trying to figure out how to go about solving this problem. You have a standard deck of 52 cards and you draw 10 without replacement. What is the probability of drawing four of the same rank? (ex. 4 aces) Order doesn't matter, so I am thinking I have to do some sort of combination over the number of ways to draw ten cards. I could be completely wrong though... Please help? :D
HW Helper
Thanks
P: 26,167
 Quote by johnsa9 You have a standard deck of 52 cards and you draw 10 without replacement. What is the probability of drawing four of the same rank? (ex. 4 aces)
Hi johnsa9! Welcome to PF!

Is this an exam question, or did you just make it up?

Exam questions are usually less complicated than this (in the sense that they don't involve too much number-slugging).

You have to count all the ways of drawing 10 cards (which is easy!), and all the ways in which 4 of them are the same (which is really complicated).
 PF Gold P: 1,059 $$13*1\frac{48!}{42!6!}*\frac{42!10!}{52!}$$=1.01%
P: 403

## Probability- drawing cards from a deck.

 Quote by tiny-tim You have to count all the ways of drawing 10 cards...
$${52\choose 10}$$

 Quote by tiny-tim and all the ways in which 4 of them are the same.
Well, there are aces, kings, queens, jacks, etc...So,

#combinations we are interested in =
#combinations with 4 aces (including combinations with 4 kings, etc... )
PLUS
#combinations with 4 kings (including combinations with 4 aces, etc... )
PLUS
etc...
MINUS
#combinations with 4 aces e 4 kings (because they were counted twice )
MINUS
#combinations with 4 aces e 4 queens (because they were counted twice )
MINUS
etc...

That is:
$${13\choose 1}{48\choose 6} - {13\choose 2}{44\choose 2}$$

So, the probability of getting four of same rank is
$$\frac{{13\choose 1}{48\choose 6}-{13\choose 2}{44\choose 2}}{{52\choose 10}}=\frac{6483}{643195}$$

HW Helper
Thanks
P: 26,167
 Quote by Rogerio $${13\choose 1}{48\choose 6} - {13\choose 2}{44\choose 2}$$ So, the probability of getting four of same rank is $$\frac{{13\choose 1}{48\choose 6}-{13\choose 2}{44\choose 2}}{{52\choose 10}}=\frac{6483}{643195}$$
Hi Rogerio!

(btw, please don't give too much of the answer away before the OP has had a chance to try it him/herself)

two things …

i] you have to add back in the probability of three fours (ok, i know that's pretty small )

ii] you haven't included the factor for eg, the number of ways of laying down 4 aces among 10 cards.
P: 403
 Quote by tiny-tim Hi Rogerio! (btw, please don't give too much of the answer away before the OP has had a chance to try it him/herself) two things … i] you have to add back in the probability of three fours (ok, i know that's pretty small ) ii] you haven't included the factor for eg, the number of ways of laying down 4 aces among 10 cards.
Hi tiny-tim!
the two things...

i] only Chuck Norris (and maybe David Copperfield) could get three fours using just 10 cards.

ii] the "number of ways of laying down 4 aces among 10 cards" is already the factor $${48\choose 6}$$

So, the probability of getting four of same rank is
 Quote by Rogerio $$\frac{{13\choose 1}{48\choose 6}-{13\choose 2}{44\choose 2}}{{52\choose 10}}=\frac{6483}{643195}$$
 Sci Advisor HW Helper Thanks P: 26,167 Hi Rogerio! i] oops! ii] no, 48C6 is only the number of ways if the first four cards are four aces … but the aces could be anywhere
P: 403
 Quote by tiny-tim ... ii] no, 48C6 is only the number of ways if the first four cards are four aces … but the aces could be anywhere
Tiny-tim, the order doesn't matter...
So, you have exactly 48_C_6 combinations.

P: 100
 Quote by Rogerio ... i] only Chuck Norris (and maybe David Copperfield) could get three fours using just 10 cards....

FACT: There is no <CTRL> key on Chuck Norris's keyboard. Chuck Norris is always in control. Believe it.
 PF Gold P: 1,059 Rogerio: Tiny-tim, the order doesn't matter... So, you have exactly 48_C_6 combinations. I thought the simplist way to look at this was to just take out the four cards that are the same--say aces. Well, then there are only 13 such sets of four cards in the whole deck. Now if we add a 5th card, there are 48 ways of adding that fifth card, as Poker players know. So by extension we arrive at $$13{48\choose 6}$$. Order does not matter. But, I did not consider, as did Rogerio did, that duplicate sets could occur after we draw 8 or more cards!
 P: 2 I would like to first stress that I have never taken a statistics course in my life, so my logic may be flawed, but if so it would be helpful if someone explained to me where the flaw is. So as far as I can tell the problem is trying to calculate the probability that 4 aces will be drawn out of a 52 card deck when 10 cards are drawn. I believe this ratio would be equal to: the number of hands possible not counting for order that can be drawn with 4 aces/the number of hands possible not counting for order that can be drawn so, when looking at the numerator I would assume that each of those hands contain 4 aces and 6 other cards, and that each grouping of 6 would have at least one different card from each other grouping. which would be equal to 48!/(6!42!) total hands (as if I'm drawing 6 cards from a deck of 48). The 48! in this being the number of ways the deck can be arranged, the 42! accounting for the number of ways the nonviewed cards could be arranged without changing the 6 cards in the hand and the 6! accountinf for the number of ways the viewed cards could be arranged without changing the 6 cards that will be in hand. Out of similar logic I derive the denominator to be 52!/(10!42!) because the number of ways the deck could be arranged is 52!, the number of ways that the viewed and nonviewed cards could be arranged without changing the hand are 10! and 42! respectively. at this point I come to (48!/(6!42!))/(52!/(10!42!)) =(48!/6!)/(52!/10!) =(10!/6!)/(52!/48!) =(10*9*8*7)/(52*51*50*49) =5040/6,497,400 =.000775694... =.0775694%
 P: 2 Never mind, I misread the problem

 Related Discussions Set Theory, Logic, Probability, Statistics 8 Precalculus Mathematics Homework 6 Introductory Physics Homework 4 Set Theory, Logic, Probability, Statistics 2 General Math 6


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# Thread: Aaaah confused on analytic geometry questions!!

1. ## Aaaah confused on analytic geometry questions!!

Urgent =( I managed to get the other twenty two but I need help on these three!! If anybody is willing to give me an answer and explanation I would be very very grateful!

[size=3]16. Which point on the circle x^2 + y^2 -12x - 4y= 50 is the closest to the origin? Which point is farthest? Explain.

2. 1. 3 diameters + 1 circumference

2. 4 diameters + 1 circumference

(for 1 and 2 ... see why?)

3. complete the square to get the circle equation in the form

$(x-h)^2 + (y-k)^2 = r^2$

then sketch the circle to see what point is closest/farthest to the origin

3. Originally Posted by skeeter
1. 3 diameters + 1 circumference

2. 4 diameters + 1 circumference

(for 1 and 2 ... see why?)

3. complete the square to get the circle equation in the form

$(x-h)^2 + (y-k)^2 = r^2$

then sketch the circle to see what point is closest/farthest to the origin
Thanks you so much! I get it now, thanks SO MUCH.

I'm not sure if I'm supposed to post an entire new thread so I'll post this here, I want to see if I'm off or not.

The question is:

An isosceles triangle has two sides of length p and one of length m. IN terms of these lengths, write calculator-ready formulas for the sizes of the angles of this triangle.

For one of the two base angles, would the formula be something along the lines of:

inverse cos= .5m/p

Is that right or totally off??

4. I put the three circles on a set of coordinate axes.

$(x-6)^{2} + y^{2} = 6^{2}$

$(x+6)^{2} + y^{2} = 6^{2}$

$x^{2} + (y-6\sqrt{3})^{2} = 6^{2}$

No notice a few things.

If you connect the centers, you get an equilateral triangle. Draw this triangle. Your band obviously is longer than this.

What do you suppose you get if you extend the three tangents until they intersect? Draw this triangle. Your band obviously is shorter than this.

Draw all six radii, two from each center, intersecting the tangents on the circle. These radii are perpendicular to the tangents.

As far as the four pipes, is there a unique solution? You'll have to explore that.


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# Introduction

## Installing

Install KrylovKit.jl via the package manager:

using Pkg
Pkg.add("KrylovKit")

KrylovKit.jl is a pure Julia package; no dependencies (aside from the Julia standard library) are required.

## Getting started

After installation, start by loading KrylovKit

using KrylovKit

The help entry of the KrylovKit module states

KrylovKitModule
KrylovKit

A Julia package collecting a number of Krylov-based algorithms for linear problems, singular value and eigenvalue problems and the application of functions of linear maps or operators to vectors.

KrylovKit accepts general functions or callable objects as linear maps, and general Julia objects with vector like behavior as vectors.

The high level interface of KrylovKit is provided by the following functions:

source

## Common interface

The for high-level function linsolve, eigsolve, geneigsolve, svdsolve, exponentiate and expintegrator follow a common interface

results..., info = problemsolver(A, args...; kwargs...)

where problemsolver is one of the functions above. Here, A is the linear map in the problem, which could be an instance of AbstractMatrix, or any function or callable object that encodes the action of the linear map on a vector. In particular, one can write the linear map using Julia's do block syntax as

results..., info = problemsolver(args...; kwargs...) do x
y = # implement linear map on x
return y
end

Read the documentation for problems that require both the linear map and its adjoint to be implemented, e.g. svdsolve, or that require two different linear maps, e.g. geneigsolve.

Furthermore, args is a set of additional arguments to specify the problem. The keyword arguments kwargs contain information about the linear map (issymmetric, ishermitian, isposdef) and about the solution strategy (tol, krylovdim, maxiter). Finally, there is a keyword argument verbosity that determines how much information is printed to STDOUT. The default value verbosity = 0 means that no information will be printed. With verbosity = 1, a single message at the end of the algorithm will be displayed, which is a warning if the algorithm did not succeed in finding the solution, or some information if it did. For verbosity = 2, information about the current state is displayed after every iteration of the algorithm. Finally, for verbosity > 2, information about the individual Krylov expansion steps is displayed.

The return value contains one or more entries that define the solution, and a final entry info of type ConvergeInfo that encodes information about the solution, i.e. whether it has converged, the residual(s) and the norm thereof, the number of operations used:

KrylovKit.ConvergenceInfoType
struct ConvergenceInfo{S,T}
converged::Int
residual::T
normres::S
numiter::Int
numops::Int
end

Used to return information about the solution found by the iterative method.

• converged: the number of solutions that have converged according to an appropriate error measure and requested tolerance for the problem. Its value can be zero or one for linsolve, exponentiate and expintegrator, or any integer >= 0 for eigsolve, schursolve or svdsolve.
• residual: the (list of) residual(s) for the problem, or nothing for problems without the concept of a residual (i.e. exponentiate, expintegrator). This is a single vector (of the same type as the type of vectors used in the problem) for linsolve, or a Vector of such vectors for eigsolve, schursolve or svdsolve.
• normres: the norm of the residual(s) (in the previous field) or the value of any other error measure that is appropriate for the problem. This is a Real for linsolve and exponentiate, and a Vector{<:Real} for eigsolve, schursolve and svdsolve. The number of values in normres that are smaller than a predefined tolerance corresponds to the number converged of solutions that have converged.
• numiter: the number of iterations (sometimes called restarts) used by the algorithm.
• numops: the number of times the linear map or operator was applied
source

There is also an expert interface where the user specifies the algorithm that should be used explicitly, i.e.

results..., info = problemsolver(A, args..., algorithm(; kwargs...))

Most algorithm constructions take the same keyword arguments (tol, krylovdim, maxiter and verbosity) discussed above.

As mentioned before, there are two auxiliary structs that can be used to define new vectors, namely

KrylovKit.RecursiveVecType
v = RecursiveVec(vecs)

Create a new vector v from an existing (homogeneous or heterogeneous) list of vectors vecs with one or more elements, represented as a Tuple or AbstractVector. The elements of vecs can be any type of vectors that are supported by KrylovKit. For a heterogeneous list, it is best to use a tuple for reasons of type stability, while for a homogeneous list, either a Tuple or a Vector can be used. From a mathematical perspectve, v represents the direct sum of the vectors in vecs. Scalar multiplication and addition of vectors v acts simultaneously on all elements of v.vecs. The inner product corresponds to the sum of the inner products of the individual vectors in the list v.vecs.

The vector v also adheres to the iteration syntax, but where it will just produce the individual vectors in v.vecs. Hence, length(v) = length(v.vecs). It can also be indexed, so that v[i] = v.vecs[i], which can be useful in writing a linear map that acts on v.

source
KrylovKit.InnerProductVecType
v = InnerProductVec(vec, dotf)

Create a new vector v from an existing vector dotf with a modified inner product given by inner. The vector vec, which can be any type (not necessarily Vector) that supports the basic vector interface required by KrylovKit, is wrapped in a custom struct v::InnerProductVec. All vector space functionality such as addition and multiplication with scalars (both out of place and in place using mul!, rmul!, axpy! and axpby!) applied to v is simply forwarded to v.vec. The inner product between vectors v1 = InnerProductVec(vec1, dotf) and v2 = InnerProductVec(vec2, dotf) is computed as dot(v1, v2) = dotf(v1.vec, v2.vec) = dotf(vec1, vec2). The inner product between vectors with different dotf functions is not defined. Similarly, The norm of v::InnerProductVec is defined as v = sqrt(real(dot(v, v))) = sqrt(real(dotf(vec, vec))).

In a (linear) map applied to v, the original vector can be obtained as v.vec or simply as v[].

source


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# Hitting Set Problem with non-minimal Greedy Algorithm

The Hitting Set Problem is defined as having a universal set $$\mathfrak{U}$$, and nonempty sets $$S_i \subseteq \mathfrak{U}$$ for $$1 \leq i \leq n$$, and finding a set $$\mathcal{H} \subset \mathfrak{U}$$ such that $$|\mathcal{H} \cap S_i| \geq 1$$ for all $$1 \leq i \leq n$$.

We may ask for the minimal cardinality of $$\mathcal{H}$$, that is, what is the least number of elements needed to "Hit" every $$S_i$$?

Further, we may use a greedy algorithm to ensure we find a hitting set. In this greedy algorithm, we set $$\mathcal{H} = \emptyset$$, and while we still have sets $$S_i$$ that have not been hit, we add to $$\mathcal{H}$$ an element whom appears in the most $$S_i$$ that have not been hit, breaking ties arbitrarily if there are any.

My question is: What is an example of a Universe set $$\mathfrak{U}$$ and subsets $$S_i$$, where $$1 \leq i \leq n$$ for some $$n \in \mathbb{N}$$, such that the greedy algorithm above does not find a minimal Hitting Set $$\mathcal{H} \subset \mathfrak{U}$$?

For a longer (and probably clearer) description, and more info on the Hitting Set problem, see http://theory.stanford.edu/~virgi/cs267/lecture5.pdf, or Prove that Hitting Set is NP-Complete.

$$U = \{1, 2, 3, 4, 5\}\\ \mathcal{F} = \{ \{1, 2, 3\}, \{1, 3, 4\}, \{1, 4, 5\}, \{1, 2, 5\}, \{2, 3\}, \{4, 5\} \},$$ where $$U$$ is the universe and $$\mathcal{F}$$ is the family of sets to be hit. A minimum solution is hence $$\{2, 4\}$$. Any solution including $$1$$ is not minimum, since the set must hit $$\{2, 3\}$$ and $$\{4, 5\}$$ using at least one element of each resulting in a solution of size at least 3. However, the greedy algorithm includes $$1$$ in the first step and hence can not yield a minimum solution.


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Aller au contenu
Author
Divya Choudhary
AbstractComparative study of linear classifiers in the context of text classification


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# Tag Info

I'm not familiar with the ins and outs of self driving cars but I imagine that the action space is not discrete. For instance, the car may want to decide what angle it needs to turn (rather than left or right). The update in Q-Learning involves taking $\max_aQ(s,a)$; this is theoretically possible for a continuous action space but it would itself require ...


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# Math Help - area of the shaded area of the circle.

1. ## area of the shaded area of the circle.

The center C of the larger circle lies on another circle whose diameter coincides with the chord AB of the larger circle. Find the area of the shaded region if the length AB measures 10cm.

2. Originally Posted by ejaykasai
The center C of the larger circle lies on another circle whose diameter coincides with the chord AB of the larger circle. Find the area of the shaded region if the length AB measures 10cm.

What have you tried, where are you stuck?

Post what you have done.

CB

3. i got the area of the half circle which is 1/2 pi r^2 but then i don't know how to get the area of the segment to minus it on the half area of the circle.

4. Since AB, of length 10cm, is a diameter, the radius of the smaller circle is 5 cm. That tells you that the segment from O to P, the center of the smaller circle, is 5 cm. Now, Look at the right triangle formed by OAP. It has legs of 5 and 5 so you can find the length of OA, a radius of the larger circle.

5. Since $\angle AOB$ is inscribed in a semicircle: $\angle AOB = 90^o.$

As HallsofIvy pointed out: . $PA = PO = PB = 5$

You can now find the area of sector $AOB$ of the large circle
and the area of right triangle $AOB$,
then determine the area of the unshaded segment.


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## Friday, November 16, 2007

Heat engines
The second law of thermodynamics puts a fundamental limit on the thermal efficiency of heat engines. Surprisingly, even an ideal, frictionless engine can't convert anywhere near 100% of its input heat into work. The limiting factors are the temperature at which the heat enters the engine, $T_H,$, and the temperature of the environment into which the engine exhausts its waste heat,$T_C,$, measured in the absolute Kelvin or Rankine scale. From Carnot's theorem, for any engine working between these two temperatures:
$eta_{th} le 1 - frac{T_C}{T_H},$
This limiting value is called the Carnot cycle efficiency because it is the efficiency of an unattainable, ideal, lossless (reversible) engine cycle called the Carnot cycle. No heat engine, regardless of its construction, can exceed this efficiency.
Examples of $T_H,$ are the temperature of hot steam entering the turbine of a steam power plant, or the temperature at which the fuel burns in an internal combustion engine. $T_C,$ is usually the ambient temperature where the engine is located, or the temperature of a lake or river that waste heat is discharged into. For example, if an automobile engine burns gasoline at a temperature of $T_H = 1500^circ F = 1089^circ K,$ and the ambient temperature is $T_C = 70^circ F = 294^circ K,$, then its maximum possible efficiency is given by:
$eta_{th} le 1 - frac{294^circ K}{1089^circ K} = 73.0%,$
In practice, because no practical implementation of the Carnot cycle exists coupled with other irreversibilities such as the combustion process itself and friction, real engines fall far short of the Carnot efficiency. Real automobile engines are only around 25% efficient, combined cycle power stations have efficiencies much higher however will still fall at least 15% points short of the Carnot value. As Carnot's theorem only applies to heat engines, devices that convert the fuel's energy directly into work without burning it such as fuel cells, can exceed the Carnot efficiency.

Carnot efficiency
For an energy conversion device like a boiler or furnace, the thermal efficiency is $eta_{th} equiv frac{Q_{out}}{Q_{in}}$. So, for a boiler that produces 210 kW (or 700,000 BTU/h) output for each 300 kW (or 1,000,000 BTU/h) heat-equivalent input, its thermal efficiency is 210/300 = 0.70, or 70%. This means that the 30% of the energy is lost to the environment.
An electric resistance heater has a thermal efficiency of at or very near 100%, so, for example, 1500W of heat are produced for 1500W of electrical input. When comparing heating units, such as a 100% efficient electric resistance heater to an 80% efficient natural gas-fueled furnace, an economic analysis is needed to determine the most cost-effective choice.

Energy conversion
Heat pumps, refrigerators, and air conditioners, for example, move heat, rather than convert it, so other measures are needed to describe their thermal performance. The common measures are the coefficient of performance (COP), energy efficiency ratio (EER), and seasonal energy efficiency ratio (SEER). The Efficiency of a Heat pump (HP) and Refrigerators (R)*: $E_{HP}=frac{|Q_H|}{|W|}$ $E_{R}=frac{|Q_L|}{|W|}$ $displaystyle E_{HP} - E_{R} = 1$ If temperatures at both ends of the Heat Pump or Refrigerator are constant and their processes reversible: $E_{HP}=frac{T_H}{T_H - T_L}$ $E_{R}=frac{T_L}{T_H - T_L}$
Energy efficiency

Annual fuel utilization efficiency (AFUE)
Electrical efficiency
Mechanical efficiency
Figure of merit


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## Precalculus (6th Edition) Blitzer

When the graph of $y=\sin 2x$ is shifted one unit upward, then the graph of $y=\sin 2x+1$ is obtained. For the standard function in the form $y=A\sin \left( Bx+C \right)$ \begin{align} & \text{Amplitude=}\left| A \right| \\ & \text{Period=}\frac{2\pi }{B} \\ & \text{Phase}\,\text{Shift}=\frac{C}{B} \end{align} $\text{Quarter-period}=\frac{2\pi }{4}$ So, Amplitude is 0 The period is given below: \begin{align} & \text{Period = }\frac{2\pi }{B} \\ & =\frac{2\pi }{2} \\ & =\pi \end{align} Phase shift is 0 The quarter-period is as follows: $\text{Quarter-period}=\frac{\pi }{4}$ Now, add quarter periods starting from $x=0$ to generate x-values for the key points. The x-value for the first key point is as follows: $x=0$ ] And the x-value for the second key point is: \begin{align} & x=0+\frac{\pi }{4} \\ & =\frac{\pi }{4} \end{align} And the x-value for the third key point is: \begin{align} & x=\frac{\pi }{4}+\frac{\pi }{4} \\ & =\frac{\pi }{2} \end{align} And the x-value for the fourth key point is: \begin{align} & x=\frac{\pi }{2}+\frac{\pi }{4} \\ & =\frac{3\pi }{4} \end{align} And the x-value for the fifth key point is: \begin{align} & x=\frac{3\pi }{4}+\frac{\pi }{4} \\ & =\pi \end{align}


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# Math Help - Counting using inclusion/exclusion principle.

1. ## Counting using inclusion/exclusion principle.

I am interested in arrangements of MATHEMATICS
with

both T's before both A's or
both A's before both M's or
both M's before the E

I define the following sets

$A_1$ = {arrangements with both T's before both A's}

$A_2$ = {arrangements with both A's before both M's}

$A_3$ = {arrangements with both M's before E}

I am able to find the cardinality of everything else but $A_1 \cap A_3$

i.e. arrangements with both T's before both A's and both M's before E.

I know I want to pick four of the 11 spots (11 choose 4)

and arrange H,I,C,S in those spots (4! ways)

But I dont know whaat to do next, I can't place only the T's in the two left most spots because certainly go T,M,M,E,T.

Thank you for your help.

2. ## Re: Counting using inclusion/exclusion principle.

Originally Posted by Jame
I am interested in arrangements of MATHEMATICS with
both T's before both A's or
both A's before both M's or
both M's before the E
I define the following sets
$A_1$ = {arrangements with both T's before both A's}
$A_2$ = {arrangements with both A's before both M's}
$A_3$ = {arrangements with both M's before E}
I am able to find the cardinality of everything else but $A_1 \cap A_3$
Surely $\|A_1 \cap A_3\|=\binom{11}{4}(4!)\binom{7}{4}$.

i.e. Place the $\{H,I,C,S\}$ and arrange; select the four places to place $\{T,T,A,A\}$(only one way); finally put the $\{M,M,E\}$ is the remaining three places.


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Lesson Objectives
• Demonstrate an understanding of how to solve word problems
• Learn how to set up and solve word problems with quadratic equations

## How to Solve Word Problems with Quadratic Equations

We previously learned how to solve a word problem that involves setting up and solving a linear equation. Here, we will use the same six-step process with problems that involve quadratic equations.

### Six-step method for Applications of Linear Equations in One Variable

1. Read the problem carefully and determine what you are asked to find
• Write down the main objective of the problem
2. Assign a variable to represent the unknown
• If more than one unknown exists, we express the other unknowns in terms of this variable
3. Write out an equation that describes the given situation
4. Solve the equation
5. State the answer using a nice clear sentence
6. Check the result by reading back through the problem
• We need to make sure the answer is reasonable. In other words, if asked how many miles were driven to the store, the answer shouldn't be (-3) as we can't drive a negative amount of miles.
Example 1: Solve each word problem.
The area of a rectangular window is 143 square feet. If the length is 2 feet more than the width, what are the dimensions?
Step 1) Read the problem carefully and determine what you are asked to find:
Here, we are asked to find the dimensions of the window.
Step 2) Assign a variable to represent the unknown:
Here, we don't know the length or the width. We do know that the length is 2 feet more than the width.
let x = width of the window
then x + 2 = length of the window
Step 3) Write out an equation that describes the given situation:
To do this, let's make a little picture, which can help us to visualize the situation: Area of a rectangle: length • width
We can multiply x, our width by (x + 2), our length, and set this equal to 143. $$x(x+2)=143$$ Step 4) Solve the equation: $$x(x+2)=143$$ $$x^2 + 2x=143$$ $$x^2 + 2x - 143=0$$ $$(x - 11)(x + 13)=0$$ $$x=11, -13$$ Since x represents the width of a window, we can throw out -13 since we can't have a width of -13 feet. This tells us that our only valid solution is 11 for the width and 13 (11 + 2) for the length.
Step 5) State the answer using a nice clear sentence:
The width of the window is 11 feet and the length is 13 feet.
Step 6) Check the result by reading back through the problem:
Reading back through the problem, we can see that our window has an area of 143 square feet. $$11 \text{ft}\cdot 13 \text{ft}=143 \text{ft}^2 ✓$$Additionally, we are told that the length is 2 feet more than the width. The length is 13 feet, which is 2 feet more than the width.
Example 2: Solve each word problem.
Two cars left an intersection at the same time. One of the cars traveled directly north, while the other car traveled directly east. Three hours later, they ended up being exactly 175 miles apart from each other. If the car that traveled north traveled 35 miles more than the car that traveled east, how far did each car travel?
Step 1) Read the problem carefully and determine what you are asked to find:
Here, we are asked to find how far each car traveled.
Step 2) Assign a variable to represent the unknown:
Here, we don't know how far the northbound car drove or how far the eastbound car drove.
We are told that the northbound car drove 35 miles more than the eastbound car.
let x = distance traveled by the eastbound car
then x + 35 = distance traveled by the northbound car
Step 3) Write out an equation that describes the given situation:
To do this, let's make a little picture, which can help us to visualize the situation: From our image above, we can see that the Pythagorean formula applies to our problem. $$a^2 + b^2=c^2$$ $$x^2 + (x + 35)^2=175^2$$ Step 4) Solve the equation: $$x^2 + (x + 35)^2=175^2$$ $$x^2 + x^2 + 70x + 1225=30{,}625$$ $$2x^2 + 70x - 29{,}400=0$$ Use the quadratic formula: $$a=2, b=70, c=-29{,}400$$ $$x=\frac{-70 \pm \sqrt{70^2 - 4(2)(-29{,}400)}}{2(2)}$$ $$x=\frac{-70 \pm \sqrt{240{,}100}}{4}$$ $$x=\frac{-70 \pm 490}{4}$$ $$x=\frac{-70 + 490}{4}=\frac{420}{4}=105$$ $$x=\frac{-70 - 490}{4}=-\frac{560}{4}=-140$$ Since x represents the distance traveled by the eastbound car, we can throw out the -140. Recall that we can't have a negative amount for distance traveled. (Driving backwards does not count!)
This tells us that our only valid solution is 105 miles traveled by the eastbound car and 140 miles (105 + 35 = 140) traveled by the northbound car.
Step 5) State the answer using a nice clear sentence:
The eastbound car traveled a total distance of 105 miles, while the northbound car traveled a total distance of 140 miles.
Step 6) Check the result by reading back through the problem:
Reading back through the problem, we can see that our two cars are 175 miles apart at the three-hour mark. Additionally, based on the position of each, we can check using the Pythagorean formula. $$\sqrt{105^2 + 140^2}\stackrel{?}{=}175$$ $$175=175 ✓$$

#### Skills Check:

Example #1

Solve each word problem.

Olivia needs to wash a window in a building that is 16 feet from the ground. To avoid a fence, she decides to rest the ladder against the building. For stability, Olivia decides she should place the ladder 12 feet away from the building. How long of a ladder will Olivia need?

A
20 ft
B
22 ft
C
18 ft
D
29 ft
E
32 ft

Example #2

Solve each word problem.

Two cars left an intersection at the same time. One of the cars traveled directly north, while the other car traveled directly west into heavy traffic. An hour later, they ended up being exactly 41 miles apart from each other. If the car that traveled north traveled 31 miles more than the car that traveled west, how far did each car travel?

Note: NB -> northbound, WB -> westbound

A
WB: 15 mi, NB: 13mi
B
WB: 9 mi, NB: 40mi
C
WB: 27 mi, NB: 40mi
D
WB: 6 mi, NB: 37mi
E
WB: 8 mi, NB: 39mi

Example #3

Solve each word problem.

The length of a garden is 12 feet more than the width. If the area of the garden is 325 square feet, what are the dimensions of the garden? Note: (Area is length times width)

A
L: 22, W: 10
B
L: 25, W: 13
C
L: 27, W: 15
D
L: 30, W: 18
E
L: 33, W: 21


text:
Question

# $$77/90$$ The lines $$2x+y-1=0,ax+3y-3=0$$ and $$3x+2y-2=0$$ are concurrent for

A
All a
B
a=4 only
C
$-1< a<3$
D
a>0 only

Solution

## The correct option is A All $$a$$$$2x + y - 1 = 0 ..... \left( 1 \right)$$$$ax + 3y - 3 = 0 ..... \left( 2 \right)$$$$3x + 2y - 2 = 0 ..... \left( 3 \right)$$These lines are concurrent, i.e., point of intersection of any two lines lies on the third line.Intersection point of $$\left( 1 \right)$$ and $$\left( 3 \right)$$-Multiplying eqiuation $$\left( 1 \right)$$ by $42$$, we get$$4x + 2y - 2 = 0 ..... \left( 4 \right)$$Subtracting equaton$$\left( 3 \right)$$from$$\left( 4 \right)$$, we have$$\left( 4x + 2y - 2 \right) - \left( 3x + 2y - 2 \right) = 0x = 0$$Substituting the value of$$x$$in equation$$\left( 1 \right)$$, we have$$0 + y - 1 = 0y = 1$$Now, the point$$\left( 0, 1 \right)$$must satisfy$$\left( 2 \right)$$.Therefore,$$a \cdot 0 + 3 \cdot 1 - 3 = 00 = 0$$Hence the given lines will be concurrent for all values of$$a$\$.Maths

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#### • Class 11 Physics Demo

Explore Related Concepts

# how to identify gold ore

Question:$5 dollar gold coin Answers:The best advice available here is to check the link below for a coin dealer specializing in US gold. Question:How many cubic centimeters of ore containing 0.22% by mass gold must be processed to obtain$100 worth of gold? The density of the ore is 8.0 g/cm and the price of gold is $418 per troy ounce. (14.6 troy oz. = 1.0 ordinary pound called an avoirdupois pound, and 1 lb = 454 g.) (a) 42 cm (b) 2.7 x 10^4 cm (c) 4.20 x 10 cm (d) 6200 cm (e) 9.3 x 10^-1 cm Answers:C is the answer (4.20 x 10 cm ) First determine how many grams of gold is equivalent to$100. You will need to converting troy ounce to grams. 1 troy ounce is exactly 31.1034768 grams http://en.wikipedia.org/wiki/Troy_ounce ($418.00 /toz) * (1 toz / 31.1035 grams) =$13.44/gram Then ($100) / ($13.44/gram) = 7.44 grams of gold to get $100 The the ore is just 0.22% Au, So divide grams of gold needed by this to get grams of ore needed. (7.44 grams of gold) / (0.0022) = 3382.29 grams of ore needed Finally, multibly by the density to get cm needed: (3382.29 grams of ore) / (8.0 g/ cm ) = 422 cm of ore From Youtube FINDING GOLD II - Concentrating Gold :www.treasuresites.com Location guides & tools. How to find and process your own gold and platinum! Click "More From This User" to get all the films. Gold and platinum are 15-19 times heavier than other streambed materials and concentrate in low pressure areas and cracks that run across rivers and streams. You look for a crack on the bank, and follow it out until you meet the "gold line" and there you suck it out with your dredge. Gold will be on the outside edge of a river gravel bar, at the head of the bar (large gold but usually beneath big boulders), and at the tail end of a bar (vast concentrations due to river bars forming in the shape of an airfoil and sucking fine gold to the tail end) but be small to microscopic at the tail end. Gold will travel down a river or stream in a line, usually off center of the high pressure water. Gold will settle behind a boulder. A good place to fish, can also be an excellent place to find gold. "Black sand" is iron ore that can be readilly identified in gravel bars and is a ready indicator that gold is probably present. The most effective and economical way for the average person to find paying concentrations of gold in a river or stream is with a simple ($80) sluice that you shovel into and the riffles retain gold, platinum, gems and anything heavy for you. Gold can be found up high on the old river channels and recovered with metal detectors, a gold wheel, a highbanker, or simply by identifying the material, shoveling it in your ...


text:
# Preliminary Analysis of PEN Experiment: A Precision Measurement of the Pion Electronic Decay Branching Ratio

Glaser, Charles, Physics - Graduate School of Arts and Sciences, University of Virginia
Pocanic, Dinko, Physics, University of Virginia
Abstract:

Lepton universality is a key hypothesis of the Standard Model of particle physics which predicts that the charged leptons couple equally in electroweak interactions. All electroweak interactions with the charged leptons are therefore identical under the umbrella of lepton universality. This hypothesis may be confirmed through either high energy collisions, or low energy precision experiments. The PEN experiment, performed at the Paul Scherrer Institute, sought to obtain the pion electronic decay branching ratio $\Gamma(\pi\rightarrow \text{e}\nu (\gamma)/\Gamma(\pi \rightarrow \mu\nu(\gamma))$ to a relative precision of $5\times 10^{-4}$. This ratio provides the best test for electron-muon universality. Deviations from Standard Model expectation would be indicative of physics beyond the Standard Model, or new physics''. Confirmation of the Standard Model expectations put limits on new physics. This dissertation describes a blinded cut-based preliminary blinded analysis for the period comprising the 2009 and 2010 data taken by the PEN collaboration. This dissertation includes the reconstruction of key observables from detector response, the construction and implementation of an ultra-realistic Monte Carlo simulation, and the necessary event selection criteria in order to obtain the branching ratio.
Emphasis is placed on the construction of the Monte Carlo simulation needed for the characterization of the low energy tail of the CsI electromagnetic calorimeter as well as determination of the acceptances and cut efficiencies used in the analysis. Characterization of various quantities and uncertainties needed to obtain the branching ratio is also discussed. The partial unblinding of the analysis when including key cut efficiencies is $(1.2534 \pm 0.0013)\times 10^{-4}$ with a relative statistical uncertainty of $8.3\times 10^{-4}$ for the 2009 run period. Full unblinding is expected to lower the branching ratio value on the order of 2\% . The blinded branching ratio for the 2010 run period results in a $(1.1751\pm0.0010)\times 10^{-4}$ result with a relative statistical uncertainty of $6.7\times 10^{-4}$. Combining the two years of the data results in $5.3\times 10^{-4}$ relative statistical uncertainty and an overall relative uncertainty of $7.6\times 10^{-4}$.
With further study on acceptances, event rejection criteria, and inclusion of the 2008 data, the statistical uncertainty will be diminished to produce a more competitive result with the inclusion of higher statistics.

Degree:
PHD (Doctor of Philosophy)
Keywords:
pion decay , Standard Model, Electroweak, Realistic Monte Carlo
Language:
English
Rights:


text:
# Thread: probability - variance

1. ## probability - variance

Hi,

I'm pretty stuck with the following question, any help would be appreciated

A system is composed of 5 subsystems, each of them has a random independent lifetime, exponentially distributed with mean = 1.
The system works only if all the subsystems are working.
What is the variance of the whole system's lifetime?

Thanks

2. Originally Posted by maria clara
The system works only if all the subsystems are working.
This means that the lifetime of the whole system is the minimum of the lifetimes of the subsystems.
Do you know the distribution of the minimum of two independent exponentially distributed random variables?

3. well... I calculated it now.
if X~exp(1) and Y~exp(1) and Z = min(X,Y)
then the probability density function of Z is 2exp(-2z) for z>0 and 0 otherwise.
So it's also exponential with parameter=2.
does that mean that the minimum of the five would be also exponential with a parameter of 5? and then the answer should be 1/25..

but as far as I recall, the sum of two exponentials is an exponential with a parameter that equals the sum of parameters, is it true? does it mean that minimum and sum of exponentially distributed random variables are equivalent?...

thanks again!

4. Your computation is right. More generally, if $X$ and $Y$ are exponential r.v. of parameters $\lambda$ and $\mu$, then $\min(X,Y)$ is an exponential r.v. with parameter $\lambda+\mu$. If you apply this 4 times (first, the min of the lifetimes of the first two subsystems, then the minimum of the first two and the third,...), you get exactly what you guessed: the lifetime is an exponential r.v. of parameter 5.

As for the sum, be careful, it is called a gamma distribution, and not exponential.

5. thanks a lot!

and regarding the sum - I think I confused the exponential distribution with poisson, where the rule does hold...


text:
# “Living by faith”

“Living by faith”
March 30, 2022

Mike and MonaRé Shields

International Ministries to Latin America

Dear Friends,

“Living by faith” is now a stark reality in Latin America. We’ve been responding to many situations with the resources people like you have sent.

Last month as the crisis shut down cities and even entire countries,

we sent funds for food and distribution to a variety of places. This includes Mexico, Cuba, Bolivia, Argentina, Venezuela, and Chile.

We helped a local church working nonstop in Costa Rica. They have been delivering food to scores of families left destitute by the crisis.

We received an urgent appeal to help provide food for an orphanage in northern Chile. It was because of your sacrificial offerings we said yes. Friends in Venezuela have kept us apprised of the unfolding humanitarian tragedy already exacerbated by the cruel Maduro regime.  We sent $750 to help two dozen families on the brink. Your offerings are the breakthrough! Here are some comments from Venezuela: “Brother Mike, we’re so grateful for the offering. Buying food is very difficult. Having some in our house under quarantine is a relief.” — – (Here’s from a Pastor) “Yesterday I gave$16 to one of my leaders for medicine; he’s so sick. Today, God sent us your offering! It fulfills the promise “Give and it shall be given to you.”

We are laser focused on responding as quickly and generously as possible in this time of need and perplexity. Like so many others we cling to the promise of Psalm 91:5-7 “You will not fear the terror of night, nor the arrow that flies by day, nor the pestilence that stalks in the darkness, nor the plague that destroys at midday. A thousand may fall at your side, ten thousand at your right hand, but it will not come near you.”

Your prayers, your words of encouragement, your offerings…they really matter. Thank you for the courage to help us say YES. May the Lord give YOU peace, protection and favor in the coming days as we intercede and intervene on behalf of God’s people in Latin America.

Global Outreach/Mike & Mona Ministries
320 S. Flamingo Road   Pembroke Pines, FL   33027    goye777@gmail.com


text:
## Ad Hoc Compression Methods: RLE (part II)

Yestermorning I presented an ad hoc compression method known as RLE, or run-length encoding and I got quite more reactions than I thought I would. Many suggested doing plenty of other stuff despite the fact that the post was about RLE, not about <insert your favorite compression method>. But something very nice happened: people got interested and many suggestions came forth.

I decided to implement the median-based prediction method as suggested by Dark Shikari here. Let us see the results!

The median prediction, as used in the FFV1 lossless video codec that uses only intra-frame coding, uses the following context:

The prediction is given by:

$\mathrm{pred}=\mathrm{med3}(C_1,C_2,C_1+C_2-C_3)$

Which means that if color images are considered, the predicted color of the pixel ‘?’ is given by the median of the colors. In RGB (or more like some other YUV-like color space), that means that the prediction is applied on each color plane separately. The difference with the actual color and the predicted color is then encoded.

Using “continuous” colors, the residual (the difference) is likely to be smaller in magnitude than the original range of the colors; using an efficient code yields compression.

What happens when you have a palette image? As Dark Shikari suggested, let us use the color in the palette that is the closest to the prediction as the predicted color. But that leaves the problem of encoding the error. Palette images have a problem in this regard: colors are not necessarily sorted in any smart way! Encoding the error pretty much means encoding the correct palette index.

To circumvent this problem, I use the same technique as I used with the simpler context prediction of yesterday; I compute the ranking for each color in the palette at each pixel. The closest to the prediction gets rank 0, the furthest, rank 3.

To measure the difference with the palette colors and the predicted color, I tried two approaches. One is based on the approximate perceived brightness difference, where the difference is given by the formula $299|r_1-r_2|+587|g_1-g_2|+114|b_1-b_2|$. The second is to count the number of components that differ, something like $(r_1\neq r_2)+(g_1\neq g_2)+(b_1\neq b_2)$. I labeled the first “median-luma” and the second “median-diff”.

Results, in increasing compression ratios

The “media-luma” method yields an average compression ratio of 1.77:1, and the “median-diff” method 1.82:1, just wee short of the context-prediction that yields 1.9:1. The median prediction is followed by the same steps as the context predictor; so results compare directly the quality of prediction. Why median-diff does a bit better than the median-luma method, I am not sure; intuitively I would say that the bitmaps exhibit some color structure that makes some color components (not palette entries, but separate components such as R, G, B) more likely to be the same as they are pretty much primary-colored. What’s your take on this?

*
* *

So the important point here is not that RLE is state-of-the-art (because it certainly is not) but that compression algorithms perform better when there are multiple stages of analysis, prediction, and encoding. A good prediction or transform helps the following stages exploit redundancy. In our case, we have median or context prediction that transforms the original pixel stream into a ranking stream, which, it turns out, has more repetitions than the original stream. The repetitions are then exploited by an auto-trigger RLE encoder, whose output is further encoded by a code reminescent of Huffman codes.

And most codecs are structured like that, at the global scale—although a lot more complex.

### 11 Responses to Ad Hoc Compression Methods: RLE (part II)

1. Dark Shikari says:

Here’s a possibly even-better method:

1. Use no prediction at all.
2. Use a very simple arithmetic coder with context choice decided by the value of the top neighbor and the value of the left neighbor.
3. Either write the probabilities into the frame header (similar to optimal huffman coding) or make it context-adaptive to avoid such a header at the cost of adaptation time.

So, for example, let’s say we have 4 colors in our palette, the Left pixel is “1”, the top pixel is “3”, and the current pixel is “0”.

Thus, context = Left + Top * 4 =1 + 3 * 4 = 13

Now, how do we write the symbol itself? You could use a multi-symbol rangecoder, in which case making things adaptive would be way too complex (you’d just want to hardcode the range values in the header). You could also use a binary rangecoder, and do the following:

encode_decision( context*3 + 0, pixel == 0 )
encode_decision( context*3 + 1, pixel == 1 )
encode_decision( context*3 + 2, pixel == 2 )

So we write 3 decisions to the bitstream, three of which are zero (for the values which the pixel isn’t) and one of which is one (for the value the pixel is). If the pixel is 3, nothing gets written, because the last bit would be redundant.

Another scheme would be:

encode_decision( context*3, (pixel&1) )
encode_decision( context*3 + (pixel&1) + 1, pixel>>1 )

I suspect these are probably practically equivalent.

• Dark Shikari says:

Note this is using x264 encode_decision syntax, where encode_decision is defined as encode_decision( context, bit ).

• Arithmetic coding is probably very feasible if you also have to manage really limited precision, like it would be the case here. Probability estimates given contexts are necessarily low precision because you won’t be using zillions of examplars to formulate your estimate of the pdf. If a given context shows up 10 times, you’re pretty sure that one decimal point is quite enough. I don’t know how we can exploit that to build a “lazy” arithmetic coder…

• Dark Shikari says:

Well, with only 16 contexts and a thousand pixels or so per sprite, you’re sure to each one a decent number of times.

What you’re probably looking for is an arithmetic coder that adapts quickly–this can be done pretty trivially. You could also make one that adapts faster for contexts which are used more rarely; that would be interesting.

The easiest way, but perhaps less interesting, would probably be to calculate optimal arithcodes and store them in the frame header.

The typical ways to deal with variable adaptation rates are:

1) Count the number of times a given context decides each way. Current probability estimate is a/(a+b). Simple and good, but involves a division, and the initial estimate is heavily quantized.

2) Use a state machine. Instead of h264’s flat statespace, it can be a tree, where higher levels in the tree increment probability by larger amounts, and only the leaves form a closed cycle. Easy to add to an h264-style coder (and it was in one of the drafts iirc), but initial estimate is still heavily quantized.

3) As 2, except leave statespace flat and switch transition tables instead.

Optimal bayesian updating of pdfs should always compress better than explicitly coding an initial pdf (updating gives you exactly the right amount of less precision in less frequently used contexts). But explicit coding lets you leave the probability estimation out of the standard (and of course makes decoding faster).

2. I’ve been thinking. Another possiblity would be to encode larger tiles progressively. Start with an encoding of the bitmap using 2×2 or larger tiles. Nothing special, just a code for uniform/non-uniform and if uniform, in a separate stream, the uniform color. If non-uniform, read four pixels, possibly predicted from surroundings.

• Dark Shikari says:

A similar method to what you’re describing is used in the (currently prototype-stage) fast lossless video codec FFV2.

Note that this is *not* generally better than a single-pixel arithmetic coding solution; it was chosen here because it reduces waste as a result of VLCs being of limited bit-length-precision and because it improves performance on decoding due to fewer VLCs having to be read.

The way FFV2 works is as follows:

1. Median-filter the image.
2. Divide the frame up into 8×4 blocks.
3. Divide each 8×4 block up into 2×2 blocks. Create a bitmask based on which 2×2 blocks have all-zero residual and which don’t. Write this bitmask to the bitstream; this is called the coded block pattern. This bitmask isn’t written directly, instead, the CBP has its own VLC table for this.
4. For each 2×2 block, write the VLC for the block. Each of the four residuals in the block can have one of 6 states:

ESC, -2, -1, 0, 1, 2

This leads to a total of 6^4 VLCs.

5. If the block has escape codes, after writing the block, write, in order, the escape codes for the escaped coefficients. These have their own VLC table as well.

The format has, of course, massive Huffman tables written at each keyframe.

Personally I’d think the arithmetic coding solution would be better for what you’re doing, though this solution might work if you hardcode the VLC tables into the device instead of each individual image (there’d be far fewer VLCs due there being only 4 possible pixel values, of course).

• Dark Shikari says:

By the way, if you’re interested in more in-depth realtime discussion, feel free to drop by #x264dev or #x264 on Freenode; you’ll probably find more concentrated knowledge about entropy coding there than anywhere else I can think of.

• That actually might be a good idea. Although I am very well informed regarding probability estimation and integer (tree-structured) coding, my knowledge of arithmetic and range coders is rather sketchy.

3. enrileVar says:

great domain name for blog like this)))

4. Custom Building Products says:


text:
# equivalent condition for being a fundamental system of entourages

###### Lemma.

Let $X$ be a set and let $\mathcal{B}$ be a nonempty family of subsets of $X\times X$. Then $\mathcal{B}$ is a fundamental system of entourages of a uniformity on $X$ if and only if it satisfies the following axioms.

• (B1) If $S$, $T\in\mathcal{B}$, then $S\cap T$ contains an element of $\mathcal{B}$.

• (B2) Each element of $\mathcal{B}$ contains the diagonal $\Delta(X)$.

• (B3) For any $S\in\mathcal{B}$, the inverse relation of $S$ contains an element of $\mathcal{B}$.

• (B4) For any $S\in\mathcal{B}$, there is an element $T\in\mathcal{B}$ such that the relational composition $T\circ T$ is contained in $S$.

###### Proof.

Suppose $\mathcal{B}$ is a fundamental system of entourages for a uniformity $\mathcal{U}$. Verification of axiom (B2) is immediate, since $\mathcal{B}\subseteq\mathcal{U}$ and each entourage is already required to contain the diagonal of $X$. We will prove that $\mathcal{B}$ satisfies (B1); the proofs that (B3) and (B4) hold are analogous.

Let $S$, $T$ be entourages in $\mathcal{B}\subseteq\mathcal{U}$. Since $\mathcal{U}$ is closed under binary intersections, $S\cap T\in\mathcal{U}$. By the definition of fundamental system of entourages, since $S\cap T\in\mathcal{U}$, there exists an entourage $B\in\mathcal{B}$ such that $B\subseteq S\cap T$. Thus $\mathcal{B}$ satisfies axioms (B1) through (B4).

To prove the converse, define a family of subsets of $X\times X$ by

 $\mathcal{U}=\{S\subseteq X\times X\colon B\subseteq S\text{\ for some B\in% \mathcal{B}}\}.$

By construction, each element of $\mathcal{U}$ contains an element of $\mathcal{B}$, so all that remains is to show that $\mathcal{U}$ is a uniformity. Suppose $T$ is a subset of $X\times X$ that contains an element $S\in\mathcal{U}$. By the definition of $\mathcal{U}$, there exists some $B\in\mathcal{B}$ such that $B\subseteq S$. Since $S\subseteq T$, it follows that $B\subseteq T$, so $T$ satisfies the requirement for membership in $\mathcal{U}$. Thus $\mathcal{U}$ is closed under taking supersets. The remaining axioms for a uniformity follow directly from the appropriate axioms for the fundamental system of entourages by applying the axiom we have just checked. Hence $\mathcal{B}$ is a fundamental system of entourages for a uniformity on $X$. ∎

Title equivalent condition for being a fundamental system of entourages EquivalentConditionForBeingAFundamentalSystemOfEntourages 2013-03-22 16:30:02 2013-03-22 16:30:02 mps (409) mps (409) 5 mps (409) Derivation msc 54E15


text:
• ### Determination of $\Lambda_{\overline{\textrm{MS}}}^{(n_f=2)}$ and analytic parameterization of the static quark-antiquark potential(1804.10909)

April 29, 2018 hep-ph, hep-lat
While lattice QCD allows for reliable results at small momentum transfers (large quark separations), perturbative QCD is restricted to large momentum transfers (small quark separations). The latter is determined up to a reference momentum scale $\Lambda$, which is to be provided from outside, e.g. from experiment or lattice QCD simulations. In this article, we extract $\Lambda_{\overline{\textrm{MS}}}$ for QCD with $n_f=2$ dynamical quark flavors by matching the perturbative static quark-antiquark potential in momentum space to lattice results in the intermediate momentum regime, where both approaches are expected to be applicable. In a second step, we combine the lattice and the perturbative results to provide a complete analytic parameterization of the static quark-antiquark potential in position space up to the string breaking scale. As an exemplary phenomenological application of our all-distances potential we compute the bottomonium spectrum in the static limit.
• ### Photon-photon scattering at the high-intensity frontier(1712.06450)

The tremendous progress in high-intensity laser technology and the establishment of dedicated high-field laboratories in recent years have paved the way towards a first observation of quantum vacuum nonlinearities at the high-intensity frontier. We advocate a particularly prospective scenario, where three synchronized high-intensity laser pulses are brought into collision, giving rise to signal photons, whose frequency and propagation direction differ from the driving laser pulses, thus providing various means to achieve an excellent signal to background separation. Based on the theoretical concept of vacuum emission, we employ an efficient numerical algorithm which allows us to model the collision of focused high-intensity laser pulses in unprecedented detail. We provide accurate predictions for the numbers of signal photons accessible in experiment. Our study paves the way for a first verification of quantum vacuum nonlinearity in a well-controlled laboratory experiment at one of the many high-intensity laser facilities currently coming online.
• ### High magnetic fields for fundamental physics(1803.07547)

Various fundamental-physics experiments such as measurement of the birefringence of the vacuum, searches for ultralight dark matter (e.g., axions), and precision spectroscopy of complex systems (including exotic atoms containing antimatter constituents) are enabled by high-field magnets. We give an overview of current and future experiments and discuss the state-of-the-art DC- and pulsed-magnet technologies and prospects for future developments.
• ### Photon polarization tensor in circularly polarized Hermite- and Laguerre-Gaussian beams(1712.08898)

Dec. 24, 2017 hep-ph
We derive analytical expressions for the photon polarization tensor in circularly polarized Hermite- and Laguerre-Gaussian beams, complementing the corresponding results for linearly polarized beams obtained recently. As they are based upon a locally constant field approximation of the one-loop Heisenberg-Euler effective Lagrangian for quantum electrodynamics (QED) in constant fields, our results are generically limited to slowly varying electromagnetic fields, varying on spatial (temporal) scales much larger than the Compton wavelength (time) of the electron.
• ### All-optical signatures of Strong-Field QED in the vacuum emission picture(1712.03232)

Dec. 8, 2017 quant-ph, hep-th, hep-ph
We study all-optical signatures of the effective nonlinear couplings among electromagnetic fields in the quantum vacuum, using the collision of two focused high-intensity laser pulses as an example. The experimental signatures of quantum vacuum nonlinearities are encoded in signal photons, whose kinematic and polarization properties differ from the photons constituting the macroscopic laser fields. We implement an efficient numerical algorithm allowing for the theoretical investigation of such signatures in realistic field configurations accessible in experiment. This algorithm is based on a vacuum emission scheme and can readily be adapted to the collision of more laser beams or further involved field configurations. We solve the case of two colliding pulses in full 3+1 dimensional spacetime, and identify experimental geometries and parameter regimes with improved signal-to-noise ratios.
• ### Tadpole diagrams in constant electromagnetic fields(1709.03819)

Oct. 10, 2017 hep-th, hep-ph
We show how all possible one-particle reducible tadpole diagrams in constant electromagnetic fields can be constructed from one-particle irreducible constant-field diagrams. The construction procedure is essentially algebraic and involves differentiations of the latter class of diagrams with respect to the field strength tensor and contractions with derivatives of the one-particle irreducible part of the Heisenberg-Euler effective Lagrangian in constant fields. Specific examples include the two-loop addendum to the Heisenberg-Euler effective action as well as a novel one-loop correction to the charged particle propagator in constant electromagnetic fields discovered recently. As an additional example, the approach devised in the present article is adopted to derive the tadpole contribution to the two-loop photon polarization tensor in constant fields for the first time.
• ### An Addendum to the Heisenberg-Euler effective action beyond one loop(1612.07251)

Sept. 11, 2017 hep-th, hep-ph
We study the effective interactions of external electromagnetic fields induced by fluctuations of virtual particles in the vacuum of quantum electrodynamics. Our main focus is on these interactions at two-loop order. We discuss in detail the emergence of the renowned Heisenberg-Euler effective action from the underlying microscopic theory of quantum electrodynamics, emphasizing its distinction from a standard one-particle irreducible effective action. In our explicit calculations we limit ourselves to constant and slowly varying external fields, allowing us to adopt a locally constant field approximation. One of our main findings is that at two-loop order there is a finite one-particle reducible contribution to the Heisenberg-Euler effective action in constant fields, which was previously assumed to vanish. In addition to their conceptual significance, our results are relevant for high-precision probes of quantum vacuum nonlinearity in strong electromagnetic fields.
• ### The Heisenberg-Euler effective action in slowly varying electric field inhomogeneities of Lorentzian shape(1703.08017)

April 23, 2017 hep-th, hep-ph
We use a locally constant field approximation (LCFA) to study the one-loop Heisenberg-Euler effective action in a particular class of slowly varying inhomogeneous electric fields of Lorentzian shape with $0\leq d\leq 4$ inhomogeneous directions. We show that for these fields, the LCFA of the Heisenberg-Euler effective action can be represented in terms of a single parameter integral, with the constant field effective Lagrangian with rescaled argument as integration kernel. The imaginary part of the Heisenberg-Euler effective action contains information about the instability of the quantum vacuum towards the formation of a state with real electrons and positrons. Here, we in particular focus on the dependence of the instantaneous vacuum decay rate on the dimension $d$ of the field inhomogeneity. Specifically for weak fields, we find an overall parametric suppression of the effect with $(E_0/E_{\rm cr})^{d/2}$, where $E_0$ is the peak field strength of the inhomogeneity and $E_{\rm cr}$ the critical electric field strength.
• ### The quantum vacuum in electromagnetic fields: From the Heisenberg-Euler effective action to vacuum birefringence(1611.09883)

Nov. 23, 2016 hep-th, hep-ph
The focus of these lectures is on the quantum vacuum subjected to classical electromagnetic fields. To this end we explicitly derive the renowned Heisenberg-Euler effective action in constant electromagnetic fields in a rather pedagogical and easy to conceive way. As an application, we use it to study vacuum birefringence constituting one of the most promising optical signatures of quantum vacuum nonlinearity.
• ### Photon propagation in slowly varying electromagnetic fields(1607.01546)

July 6, 2016 hep-ph
We study the effective theory of soft photons in slowly varying electromagnetic background fields at one-loop order in QED. This is of relevance for the study of all-optical signatures of quantum vacuum nonlinearity in realistic electromagnetic background fields as provided by high-intensity lasers. The central result derived in this article is a new analytical expression for the photon polarization tensor in two linearly polarized counter-propagating pulsed Gaussian laser beams. As we treat the peak field strengths of both laser beams as free parameters this field configuration can be considered as interpolating between the limiting cases of a purely right- or left-moving laser beam (if one of the peak field strengths is set to zero) and the standing-wave type scenario with two counter-propagating beams of equal strength.
• ### Probing vacuum birefringence using x-ray free electron and optical high-intensity lasers(1605.09294)

May 30, 2016 quant-ph, hep-ph, physics.optics
Vacuum birefringence is one of the most striking predictions of strong field quantum electrodynamics: Probe photons traversing a strong field region can indirectly sense the applied "pump" electromagnetic field via quantum fluctuations of virtual charged particles which couple to both pump and probe fields. This coupling is sensitive to the field alignment and can effectively result in two different indices of refraction for the probe photon polarization modes giving rise to a birefringence phenomenon. In this article we perform a dedicated theoretical analysis of the proposed discovery experiment of vacuum birefringence at a x-ray free electron laser/optical high-intensity laser facility. Describing both pump and probe laser pulses realistically in terms of their macroscopic electromagnetic fields, we go beyond previous analyses by accounting for various effects not considered before in this context. Our study facilitates stringent quantitative predictions and optimizations of the signal in an actual experiment.
• ### Stimulated photon emission from the vacuum(1412.6050)

March 14, 2016 quant-ph, hep-th, hep-ph
We study the effect of stimulated photon emission from the vacuum in strong space-time-dependent electromagnetic fields. We emphasize the viewpoint that the vacuum subjected to macroscopic electromagnetic fields with at least one nonzero electromagnetic field invariant, as, e.g., attainable by superimposing two laser beams, can represent a source term for outgoing photons. We believe that this view is particularly intuitive and allows for a straightforward and intuitive study of optical signatures of quantum vacuum nonlinearity in realistic experiments involving the collision of high-intensity laser pulses, and exemplify this view for the vacuum subjected to a strong standing electromagnetic wave as generated in the focal spot of two counterpropagating, linearly polarized, high-intensity laser pulses. Focusing on a comparably simple electromagnetic field profile, which should nevertheless capture the essential features of the electromagnetic fields generated in the focal spots of real high-intensity laser beams, we provide estimates for emission characteristics and the numbers of emitted photons attainable with present and near future high-intensity laser facilities.
• ### Photon merging and splitting in electromagnetic field inhomogeneities(1603.00314)

March 1, 2016 quant-ph, hep-th, hep-ph
We investigate photon merging and splitting processes in inhomogeneous, slowly varying electromagnetic fields. Our study is based on the three-photon polarization tensor following from the Heisenberg-Euler effective action. We put special emphasis on deviations from the well-known constant field results, also revisiting the selection rules for these processes. In the context of high-intensity laser facilities, we analytically determine compact expressions for the number of merged/split photons as obtained in the focal spots of intense laser beams. For the parameter range of a typical petawatt class laser system as pump and a terawatt class laser as probe, we provide estimates for the numbers of signal photons attainable in an actual experiment. The combination of frequency upshifting, polarization dependence and scattering off the inhomogeneities renders photon merging an ideal signature for the experimental exploration of nonlinear quantum vacuum properties.
• ### Crystalline Ground States in Polyakov-loop extended Nambu--Jona-Lasinio Models(1510.04012)

Oct. 14, 2015 hep-ph, nucl-th
Nambu--Jona-Lasinio-type models have been used extensively to study the dynamics of the theory of the strong interaction at finite temperature and quark chemical potential on a phenomenological level. In addition to these studies, which are often performed under the assumption that the ground state of the theory is homogeneous, searches for the existence of crystalline phases associated with inhomogeneous ground states have attracted a lot of interest in recent years. In this work, we study the Polyakov-loop extended Nambu--Jona-Lasinio model and find that the existence of a crystalline phase is stable against a variation of the parametrization of the underlying Polyakov loop potential. To this end, we adopt two prominent parametrizations. Moreover, we observe that the existence of a quarkyonic phase depends crucially on the parametrization, in particular in the regime of the phase diagram where inhomogeneous chiral condensation is favored.
• ### Vacuum Birefringence as a Vacuum Emission Process(1510.03178)

Oct. 12, 2015 quant-ph, hep-ph
We argue that the phenomenon of vacuum birefringence in strong inhomogeneous electromagnetic fields can be most efficiently analyzed in terms of a vacuum emission process. In this contribution, we exemplarily stick to the case of vacuum birefringence in a stationary perpendicularly directed, purely magnetic background field extending over a finite spatial extent. Similar field configurations are realized in the BMV and PVLAS experiments. We demonstrate that we can reproduce the conventional constant field result. Our focus is on effects which arise when the probe photons originate in the field free region, are directed towards the magnetic field region, and detected well after the interaction with the magnetic field has taken place, again at zero field.
• ### Vacuum birefringence in strong inhomogeneous electromagnetic fields(1507.01084)

July 4, 2015 quant-ph, hep-th, hep-ph
Birefringence is one of the fascinating properties of the vacuum of quantum electrodynamics (QED) in strong electromagnetic fields. The scattering of linearly polarized incident probe photons into a perpendicularly polarized mode provides a distinct signature of the optical activity of the quantum vacuum and thus offers an excellent opportunity for a precision test of non-linear QED. Precision tests require accurate predictions and thus a theoretical framework that is capable of taking the detailed experimental geometry into account. We derive analytical solutions for vacuum birefringence which include the spatio-temporal field structure of a strong optical pump laser field and an x-ray probe. We show that the angular distribution of the scattered photons depends strongly on the interaction geometry and find that scattering of the perpendicularly polarized scattered photons out of the cone of the incident probe x-ray beam is the key to making the phenomenon experimentally accessible with the current generation of FEL/high-field laser facilities.
• ### Photon propagation in slowly varying inhomogeneous electromagnetic fields(1503.00532)

April 23, 2015 quant-ph, hep-th, hep-ph
Starting from the Heisenberg-Euler effective Lagrangian, we determine the photon current and photon polarization tensor in inhomogeneous, slowly varying electromagnetic fields. To this end, we consider background field configurations varying in both space and time, paying special attention to the tensor structure. As a main result, we obtain compact analytical expressions for the photon polarization tensor in realistic Gaussian laser pulses, as generated in the focal spots of high-intensity lasers. These expressions are of utmost importance for the investigation of quantum vacuum nonlinearities in realistic high-intensity laser experiments.
• ### Quantum reflection of photons off spatio-temporal electromagnetic field inhomogeneities(1412.0951)

Dec. 2, 2014 quant-ph, hep-th, hep-ph
We reconsider the recently proposed nonlinear QED effect of quantum reflection of photons off an inhomogeneous strong-field region. We present new results for strong fields varying both in space and time. While such configurations can give rise to new effects such as frequency mixing, estimated reflection rates based on previous one-dimensional studies are corroborated. On a conceptual level, we critically re-examine the validity regime of the conventional locally-constant-field approximation and identify kinematic configurations which can be treated reliably. Our results further underline the discovery potential of quantum reflection as a new signature of the nonlinearity of the quantum vacuum.
• ### On the Search for Inhomogeneous Phases in Fermionic Models(1410.8181)

Oct. 29, 2014 hep-ph, cond-mat.supr-con, nucl-th
We revisit the Gross-Neveu model with N fermion flavors in 1+1 dimensions and compute its phase diagram at finite temperature and chemical potential in the large-N limit. To this end, we double the number of fermion degrees of freedom in a specific way which allows us to detect inhomogeneous phases in an efficient manner. We show analytically that this "fermion doubling trick" predicts correctly the position of the boundary between the chirally symmetric phase and the phase with broken chiral symmetry. Most importantly, we find that the emergence of an inhomogeneous ground state is predicted correctly. We critically analyze our approach based on this trick and discuss its applicability to other theories, such as fermionic models in higher dimensions, where it may be used to guide the search for inhomogeneous phases.
• ### $\Lambda_{\bar{\textrm{MS}}}^{(n_f=2)}$ from a momentum space analysis of the quark-antiquark static potential(1407.7503)

Sept. 30, 2014 hep-th, hep-ph, hep-lat
We determine $\Lambda_{\bar{\textrm{MS}}}^{(n_f=2)}$ by fitting perturbative expressions for the quark-antiquark static potential to lattice results for QCD with $n_f=2$ dynamical quark flavors. To this end we use the perturbative static potential at the presently best known accuracy, i.e. up to ${\cal O}(\alpha_s^4)$, in momentum space. The lattice potential is computed on a fine lattice with $a \approx 0.042 \, \textrm{fm}$ in position space. To allow for a comparison and matching of both results, the lattice potential is transformed into momentum space by means of a discrete Fourier transform. The value of $\Lambda_{\bar{\textrm{MS}}}^{(n_f=2)}$ is extracted in momentum space. All sources of statistical and systematic errors are discussed. The uncertainty in the value of $\Lambda_{\bar{\textrm{MS}}}^{(n_f=2)}$ is found to be smaller than that obtained in a recent position space analysis of the static potential based on the same lattice data.
• ### Laser photon merging in an electromagnetic field inhomogeneity(1406.2972)

Aug. 13, 2014 quant-ph, hep-ph
We study the effect of laser photon merging, or equivalently high harmonic generation, in the quantum vacuum subject to inhomogeneous electromagnetic fields. Such a process is facilitated by the effective nonlinear couplings arising from charged particle-antiparticle fluctuations in the quantum vacuum subject to strong electromagnetic fields. We derive explicit results for general kinematic and polarization configurations involving optical photons. Concentrating on merged photons in reflected channels which are preferable in experiments for reasons of noise suppression, we demonstrate that photon merging is typically dominated by the competing nonlinear process of quantum reflection, though appropriate polarization and signal filtering could specifically search for the merging process. As a byproduct, we devise a novel systematic expansion of the photon polarization tensor in plane wave fields.
• ### The photon polarization tensor in a homogeneous magnetic or electric field(1308.6184)

Aug. 13, 2014 quant-ph, hep-th, hep-ph
We revisit the photon polarization tensor in a homogeneous external magnetic or electric field. The starting point of our considerations is the momentum space representation of the one-loop photon polarization tensor in the presence of a homogeneous electromagnetic field, known in terms of a double parameter integral. Our focus is on explicit analytical insights for both on- and off-the-light-cone dynamics in a wide range of well-specified physical parameter regimes, ranging from the perturbative to the manifestly nonperturbative strong field regime. The basic ideas underlying well-established approximations to the photon polarization tensor are carefully examined and critically reviewed. In particular, we systematically keep track of all contributions, both the ones to be neglected and those to be taken into account explicitly, to all orders. This allows us to study their ranges of applicability in a much more systematic and rigorous way. We point out the limitations of such approximations and manage to go beyond at several instances.
• ### Quark-antiquark static energy from a restricted Fourier transform(1311.7351)

April 23, 2014 hep-th, hep-ph, hep-lat
We provide a fully analytical determination of the perturbative quark-antiquark static energy in position space as defined by a restricted Fourier transformation from momentum to position space. Such a determination is complicated by the fact that the static energy genuinely decomposes into a strictly perturbative part (made up of contributions $\sim\alpha_s^n$, with $n\in\mathbb{N}$) which is conventionally evaluated in momentum space, and a so-called ultrasoft part (including terms $\sim\alpha_s^{n+m}\ln^m\alpha_s$, with $n\geq3$ and $m\in\mathbb{N}$) which, conversely, is naturally evaluated in position space. Our approach facilitates the explicit determination of the static energy in position space at the accuracy with which the perturbative potential in momentum space is known, i.e., presently up to order $\alpha_s^4$.
• ### Quantum Reflection as a New Signature of Quantum Vacuum Nonlinearity(1305.2320)

Aug. 2, 2013 quant-ph, hep-th, hep-ph
We show that photons subject to a spatially inhomogeneous electromagnetic field can experience quantum reflection. Based on this observation, we propose quantum reflection as a novel means to probe the nonlinearity of the quantum vacuum in the presence of strong electromagnetic fields.
• ### Symmetries of the Three Heavy-Quark System and the Color-Singlet Static Energy at NNLL(1301.3013)

April 15, 2013 hep-ph, nucl-th
We study the symmetries of the three heavy-quark system under exchange of the quark fields within the effective field theory framework of potential non-relativistic QCD. The symmetries constrain the form of the matching coefficients in the effective theory. We then focus on the color-singlet sector and determine the so far unknown leading ultrasoft contribution to the static potential, which is of order alpha_s^4 ln(\mu), and consequently to the static energy, which is of order alpha_s^4 ln(alpha_s). Finally, in the case of an equilateral geometry, we solve the renormalization group equations and resum the leading ultrasoft logarithms for the static potential of three quarks in a color singlet, octet and decuplet representation.


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# The Law of Sines and the Unit Circle

The applet demonstrates the law of sines. It shows a triangle inscribed within the unit circle. You can drag the vertices of the triangle around to change its shape. The size of each angle, its sine, and the length of the side opposite the angle are displayed in a table. The table is updated in real time.
Drag the vertices of the triangle to see how the sizes of the angles, the lengths of the sides, and the sines of the angles vary. Note that the ratio of the length of each side and the sine of the angle opposite that side will not vary (for the unit circle, it will always be two)


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Mathematics
OpenStudy (anonymous):

The exterior angle measure of a regular polygon is 20 degrees. How many sides does the polygon have? 9 16 18 20

Parth (parthkohli):

Interior angle would be 180 - 20 = 160 Now that we know this, let's use the formula. $$\Large \color{MidnightBlue}{\Rightarrow {180n - 360 \over n} = 160 }$$ $$\Large \color{MidnightBlue}{\Rightarrow 160n = 180n - 360 }$$ $$\Large \color{MidnightBlue}{\Rightarrow 0 = 20n - 360 }$$ $$\Large \color{MidnightBlue}{\Rightarrow -20n = -360 }$$ $$\Large \color{MidnightBlue}{\Rightarrow 20n = 360 }$$ $$\Large \color{MidnightBlue}{\Rightarrow n = 18 }$$ 18 sides.

OpenStudy (anonymous):

OH the same formula huh ok

Parth (parthkohli):

Yep :P

OpenStudy (anonymous):

(2n-4)*9=16n => 18n-36=16n => 2n=36=> n=18