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July 1

mathematical outcome to simple problem

I need to know if you were having a quiz with 10 participants - how many questions above 10 would you need to ask to ensure that you cannot have a draw?

mauj—Preceding unsigned comment added by 166.120.202.205 (talk) 05:35, 1 July 2010 (UTC) (email address removed. -- SGBailey (talk) 10:22, 1 July 2010 (UTC))

No matter how many questions you have, there is always a possibility of a draw. However, with some assumptions, you can choose the number of questions to make the probability of a draw small. -- Meni Rosenfeld (talk) 06:57, 1 July 2010 (UTC)

If the questions in the quiz are so easy that everybody answer correctly, then draws will occur, no matter how many questions there are in the quiz. So in order to ensure that you cannot have a draw, you need to assume that the probability p for answering correctly satisfy 0<p<1. If p depend on the participant, then there will be fewer draws than if p is the same number for all participants, so let us for the sake of the argument assume that p is the same number for all participants. Let the number of questions in the quiz be called N, and the number of correct answers from participant no i be called N_i such that 0 ≤ N₁ ≤ N₂ ≤ ··· ≤ N₁₀ ≤ N. You want to be ensured that N₁ < N₂ < ···< N₁₀. A necessary but insufficient condition for that to happen is that N≥9, (but not necessarily ≥10). As the probability for a participant to get n correct answers is $\scriptstyle {\binom {N}{n}}p^{n}(1-p)^{N-n}$, the mean number of participants getting n correct answers is $\scriptstyle \lambda =10{\binom {N}{n}}p^{n}(1-p)^{N-n}$, and the actual number is approximately poisson distributed, such that the probability that n is not a draw is $\scriptstyle e^{-\lambda }(1+\lambda )$, and the probability that none of the numbers n are draws is $\scriptstyle P=\prod _{n=0}^{N}e^{-\lambda }(1+\lambda )$. Bo Jacoby (talk) 13:30, 1 July 2010 (UTC).

Taking the logarithm of $\scriptstyle P$ gives

$\log P=\sum _{n=0}^{N}(-\lambda +\log(1+\lambda ))$.

Expanding $\scriptstyle \log(1+\lambda )=\lambda -{\frac {\lambda ^{2}}{2}}+\cdots$ gives

$\log P=\sum _{n=0}^{N}(-\lambda +\lambda -{\frac {\lambda ^{2}}{2}}+\cdots )=\sum _{n=0}^{N}(-{\frac {\lambda ^{2}}{2}}+\cdots )=-50\sum _{n=0}^{N}{\binom {N}{n}}^{2}p^{2n}(1-p)^{2(N-n)}+\cdots$.

The special case $\scriptstyle p={\frac {1}{2}}$ gives the simpler formula

$\log P\approx -50\cdot 4^{-N}\sum _{n=0}^{N}{\binom {N}{n}}^{2}$.

I do not know how to simplify this. Bo Jacoby (talk) 20:53, 1 July 2010 (UTC).

I don't know how to simplify this either (I'm sure I saw it somewhere but forgot). But thanks to my silicon master, I know what to simplify it to:

$\sum _{n=0}^{N}{\binom {N}{n}}^{2}={\binom {2N}{N}}.$

And with Stirling's formula, I believe you get something like

$4^{-N}{\binom {2N}{N}}\approx {\frac {1}{\sqrt {\pi N}}}.$

-- Meni Rosenfeld (talk) 05:03, 2 July 2010 (UTC)

That's great, Meni! (Of course you mean $\scriptstyle {\binom {2N}{N}}$ rather than $\scriptstyle {\binom {2N}{n}}$. I permitted myself to fix it). If you find the reference you should include the formula in the article on binomial coefficient. Now we should be close to answering the OP's question.

$P\approx e^{-{\frac {50}{\sqrt {\pi N}}}}$.

But there is something rotten. For big values of $\scriptstyle N$ the probability $\scriptstyle P$ approaches one rather than zero. I have made an error. Bo Jacoby (talk) 07:35, 2 July 2010 (UTC).

$\scriptstyle P$ is the probability that there are no draws. $\scriptstyle -{\frac {50}{\sqrt {\pi N}}}\approx \log(P)\approx P-1$ implies the final result:

$N\approx {\frac {2500}{\pi (1-P)^{2}}}$

If you require $\scriptstyle P$ > 0.95 you must have $\scriptstyle N$ > 318310. So the answer is that you cannot be ensured that there will be no draws. Bo Jacoby (talk) 12:14, 2 July 2010 (UTC).

A clarification. In a "winner take all" sort of contest, "no draw" is a weaker condition than a complete, non-degenerate ranking. That is, N₁ ≤ N₂ ≤ ··· ≤ N₉ < N₁₀, rather than N₁ < N₂ < ··· < N₉ < N₁₀. (Because you still have a clear winner if one person gets all the questions right, and everyone else ties with no questions right.) If you want no draws in a Olympics-style first/second/third contest, the condition would be N₁ ≤ N₂ ≤ ···≤ N₇ < N₈ < N₉ < N₁₀. The probabilities of each condition will be different. -- 174.24.195.56 (talk) 18:07, 3 July 2010 (UTC)

Only nine, if you had a knock-out tournament with quickest buzzer answers first and wrong answer means opponent wins. 92.28.245.229 (talk) 21:48, 3 July 2010 (UTC)

searching for a solution

Hello, my todays question is as follows. Let n be a natural number and [1,n] stand for the set {1,2,3...n}. Let b be a natural number and consider the equation x+y-z=b, where b is a fixed natural number > 1. Let $I_{1}=[1,\left\lceil {\frac {b}{5}}\right\rceil ],I_{2}=[\left\lfloor {\frac {b+\left\lceil {\frac {b}{5}}\right\rceil }{2}}+1\right\rfloor ,b-\left\lceil {\frac {b}{5}}\right\rceil ]$.

We need to show that it is not possible to find x,y,z such that they solve the equation and x,y is in I₂, and z in I₁. What I am trying to do is to show that the maximum value of b+z is strictly less then the minimum value of x+y. But all I can show is that $x+y>{\frac {6b}{5}}$ and $b+z<{\frac {6b}{5}}+1$. This does not solve the problem. Can someone advise me what am I doing wrong. Thanks--Shahab (talk) 18:30, 1 July 2010 (UTC)

What you wrote looks good to me; but as you see, just using the identities $x\leq \lceil x\rceil <x+1$ and $x-1<\lfloor x\rfloor \leq x$ may not be sufficient.

Try leaving the expressions unsimplified, i.e. $b+z\leq b+\left\lceil {\frac {b}{5}}\right\rceil$ and $x+y\geq 2\left\lfloor {\frac {b+\left\lceil {\frac {b}{5}}\right\rceil }{2}}+1\right\rfloor$, and then checking cases individually for b modulo 10. 98.235.80.144 (talk) 20:55, 1 July 2010 (UTC)

I do not get what you are asking me to do. Can you please elaborate a little. And from where did you get 10 in the picture. (Thanks for replying)-Shahab (talk) 02:05, 2 July 2010 (UTC)

The phrase "b modulo 10" refers to modular arithmetic or the modulo operation (used in computer programming). Those articles have a lot more detail than the basic idea you need for this problem, though.

Here's the idea: Since b is an integer, we have the following 10 cases-- b=10n+0, b=10n+1, b=10n+2, b=10n+3, ..., up to b=10n+9. We can plug each option for b into the formulas above, compute the floor and ceiling functions exactly, and then see that x+y>b+z in each case.

For example, let's look at the case b=10n+7. Then

$b+z\leq b+\left\lceil {\frac {b}{5}}\right\rceil =10n+7+\left\lceil {\frac {10n+7}{5}}\right\rceil =10n+7+\left\lceil 2n+{\frac {7}{5}}\right\rceil =10n+7+(2n+2)=12n+9$.

On the other hand,

$x+y\geq 2\left\lfloor {\frac {b+\left\lceil {\frac {b}{5}}\right\rceil }{2}}+1\right\rfloor =2\left\lfloor {\frac {10n+7+\left\lceil {\frac {10n+7}{5}}\right\rceil }{2}}+1\right\rfloor =2\left\lfloor {\frac {10n+7+\left\lceil 2n+{\frac {7}{5}}\right\rceil }{2}}+1\right\rfloor$

$=2\left\lfloor {\frac {10n+7+(2n+2)}{2}}+1\right\rfloor =2\left\lfloor {\frac {12n+9}{2}}+1\right\rfloor =2\left\lfloor 6n+{\frac {11}{2}}\right\rfloor =2(6n+5)=12n+10$

This proves that there are no solutions if b=7, 17, 27, 37, or any other positive integer ending in 7. If you check the other 9 cases yourself, you'll have the complete proof for all possible values of b. (Or maybe you'll find a counterexample-- I haven't checked them myself.) The reason to check cases of b modulo 10 is that it lets us cancel the denominators 5 and 2. If the problem had denominators of 7 and 3, then you would want to check all cases for b modulo 21 to use this method. 98.235.80.144 (talk) 02:47, 2 July 2010 (UTC)

Actually, it looks like checking b modulo 5 is sufficient. Using b=5n+0, b=5n+1, b=5n+2, b=5n+3, and b=5n+4 you will always be able to cancel the denominator 2. I should have noticed that from the start, since each expression has order 6b/5. Of course, it still works to check all choices of b modulo 10; it's just twice as much writing.98.235.80.144 (talk) 04:25, 2 July 2010 (UTC)

Thanks for all the help.---Shahab (talk) 13:40, 2 July 2010 (UTC)

No need for checking cases, just use the observation that $\lceil n/2\rceil =\lfloor (n+1)/2\rfloor$ for integer n:

${x+y\geq 2\left\lfloor {\frac {b+\lceil b/5\rceil }{2}}+1\right\rfloor =2\left\lceil {\frac {b+\lceil b/5\rceil +1}{2}}\right\rceil \geq b+\lceil b/5\rceil +1>b+\lceil b/5\rceil \geq b+z}.$

—Emil J. 13:55, 2 July 2010 (UTC)

July 2

probability inequalities

what is bennett's inequality? —Preceding unsigned comment added by 122.106.3.247 (talk) 06:24, 2 July 2010 (UTC)

There is some information about it here. -- Meni Rosenfeld (talk) 08:42, 2 July 2010 (UTC)

Examples of linear, nonlinear. or chaotic time series

While I understand the formulas that could generate these different kinds of time series, could anyone provide real-life examples of the different time series please? Particularly as relating to economic or financial series. What caterory would stock markets or currencies fall into, for example? Thanks 92.28.247.183 (talk) 20:14, 2 July 2010 (UTC)

Linear - any regular fixed amount bill, eg monthly phone plan with inclusive call charges - $=kt. Non-linear - savings versus time with compounding interest - $now = $initial*(1+i)^t. Chaotic - share prices - $=random(t). -- SGBailey (talk) 08:24, 5 July 2010 (UTC)

July 3

Notation for a non-derivative with prime (')

Let's say ${}_{\color {white}.}\varepsilon =X\cdot {Y}{\color {white}{\Big |}}\,\!$ and ${}_{\color {white}.}{\mbox{secondary }}\varepsilon =\varepsilon '={\frac {X}{Y^{2}}};{}_{\color {white}.}\,\!$.
What if, in certain cases, you want to identify $X\,\!$ and——therefore——${}_{\color {white}.}\varepsilon {\color {white}|}\,\!$, with respect to $\lambda \,\!$?: ${\color {white}{\Big |}}\varepsilon (\lambda )=X\cdot {Y}(\lambda )\,\!$. But what about ${}_{\color {white}.}\varepsilon '\,\!$? Is there a standard/established modification for the non-derivative $\varepsilon '\,\!$, maybe $\varepsilon (\lambda )'\,\!$ or $\varepsilon '|(\lambda ){}_{\color {white}.}\,\!$? $(\varepsilon ')(\lambda ){}_{\color {white}.}\,\!$ seems awkward. P=|  ~Kaimbridge~ (talk) 18:06, 3 July 2010 (UTC)

It would just be $\varepsilon '(\lambda )$. You just have to tell that it isn't a derivative from context. If the context isn't clear enough, then don't use primes to distinguish between functions. Use ${\tilde {\varepsilon }}$ or something. --Tango (talk) 18:22, 3 July 2010 (UTC)

July 4

What no questions today? Is everyone taking a holiday? —Preceding unsigned comment added by 122.107.192.187 (talk) 13:02, 4 July 2010 (UTC)

I've done an IP lookup on the last 4 anons who posted questions here. Two are from Australia, one from UK and one from Germany. This suggests that a lot less people ascribe special significance to the 4th of July than you'd think. Also, the day is still young, of course. -- Meni Rosenfeld (talk) 14:23, 4 July 2010 (UTC)

Second Order Differential Equations

Hi. I'm currently working through a course on DEs and now have some SODEs to solve, two examples being:

$y''+y=H(t-\pi )-H(t-2\pi )$

subject to y(0)=y'(0)=0 and y(t), y'(t) continuous at $\pi ,2\pi$

and

$y''-4y=\delta (x-a)$

subject to y being bounded as $|x|\to \infty$, with $\delta (x)$ being the Dirac delta function.

My problems are as follows. For each SODE I can find the complementary function and with the first one, I can find the particular integral in the cases $t<\pi$, $\pi <t<2\pi$ and $t>2\pi$ but I don't know how to ensure that the solution is continuous at pi and 2pi. Then, for the second one, I haven't got a clue what to try for the particular integral. Am I supposed to do it in the same fashion as I did for the Heaviside step function and consider what happens for x=a and x≠a separately? Thanks 92.11.130.6 (talk) 15:09, 4 July 2010 (UTC)

Partial Derivatives

A quick one on PDs. If we set $x=e^{-s}\sin t$ and $y=e^{-s}\cos t$ subject to u(x,y)=v(s,t), I have to find ${\frac {\partial v}{\partial s}}$ and ${\frac {\partial v}{\partial t}}$ in terms of x, y, ${\frac {\partial u}{\partial x}}$ and ${\frac {\partial u}{\partial y}}$.

Via the chain rule I get that ${\frac {\partial v}{\partial s}}$ = $-x{\frac {\partial u}{\partial x}}-y{\frac {\partial u}{\partial y}}$ and ${\frac {\partial v}{\partial t}}$ = $y{\frac {\partial v}{\partial x}}-x{\frac {\partial v}{\partial y}}$. Is this correct? I then have to find ${\frac {\partial ^{2}v}{\partial t^{2}}}$. Do I do this by finding ${\frac {\partial }{\partial t}}({\frac {\partial v}{\partial t}})={\frac {\partial }{\partial t}}(y{\frac {\partial v}{\partial x}}-x{\frac {\partial v}{\partial y}})$? It's quite messy when I try it and so I doubt I'm going about this the right way. Thanks 92.11.130.6 (talk) 19:21, 4 July 2010 (UTC)

Hint: y+ix=e^−s+it. Bo Jacoby (talk) 07:04, 5 July 2010 (UTC).

Thanks, I hadn't actually spotted that and it's quite clever, but the final part of the question is about finding a partial differential equation for v, so I'm not allowed to know what the function is yet. 92.11.130.6 (talk) 10:04, 5 July 2010 (UTC)

Never mind, think I've got this one now. 92.11.130.6 (talk) 14:37, 5 July 2010 (UTC)

July 5

Clifford Algebra Question

Is there a way to perform the map

$Ae_{01}+Be_{02}+Ce_{03}+De_{12}+Ee_{13}+Fe_{23}\mapsto GAe_{01}+HBe_{02}+ICe_{03}+JDe_{12}+KEe_{13}+LFe_{23}$

with "standard" functions acting upon the bivector on the left-hand side, where all the capital letters denote unknown, real-values constants? As it happens, the metric signature is $(4,0)$. By "standard functions", I mean all the derived Clifford products (wedge with some other element, commutator, exp etc.) acting upon the bivector on the left.--Leon (talk) 09:45, 5 July 2010 (UTC)

Integration

$\int {{\frac {1}{x{\sqrt {1-x}}}}dx}$?––Wikinv (talk) 23:24, 5 July 2010 (UTC)

Make the substitution $u={\sqrt {1-x}}$ and then use partial fraction decomposition.-Looking for Wisdom and Insight! (talk) 23:48, 5 July 2010 (UTC)

After doing the substitution, rather than immediately differentiating both sides, square both sides first so that from

$u^{2}=1-x\,$

$2u\,du=-dx\,$

And also, you get

$x=1-u^{2}\,$

Michael Hardy (talk) 17:58, 6 July 2010 (UTC)

July 6

6 sided die different from the standard

So I hear there is a way to rearange the numbers on a pair of 6 sided die which are different from the standard die and which have the same probabilities and outcomes when you toll both and add the outcome.... What are the numbers on each face of the 2 dice? —Preceding unsigned comment added by KatRosenbaum (talk • contribs) 04:34, 6 July 2010 (UTC)

If you're required to use the same collection of numbers (1,1,2,2,3,3,4,4,5,5,6,6) then I think it's impossible. Getting the right probability for 2 and 12 (1/36 each) requires the 1s and 6s to be on opposite dice. Then for getting the right probability for 4 (3/36), the chance of getting a 1 and 3 combo is always 2/36 since where ever each 3 is, there's exactly one 1 on the opposite die. To get the final combination that leads to a 4 requires the 2s to be on opposite dice. Similarly the 5s must be on opposite dice. Then to get the right probability for 6 (5/36) the 3s must be on opposite dice, and similarly for the 4s.

If you can use a different collection of numbers then it would be pretty easy. An obvious way would be to put 0,1,2,3,4,5 on one die and 2,3,4,5,6,7 on the other. Rckrone (talk) 06:31, 6 July 2010 (UTC)

See Sicherman dice. --Matthew Auger (talk) 07:42, 6 July 2010 (UTC)

There was an article in the Monthly a few years ago titled Dice With Fair Sums. Try that on Google Scholar. Michael Hardy (talk) 20:47, 6 July 2010 (UTC)

...April 1988, pages 316 through 328. Michael Hardy (talk) 20:49, 6 July 2010 (UTC)

Equating an exponential function to a linear function

$e^{y}=(e-1)y+1$
Solve for y
How do you solve an equation wherein an exponential function is equated to a linear function ?––220.253.104.38 (talk) 07:10, 6 July 2010 (UTC)

There's not an elementary way to solve for y from such an equation, however in this particular case there are some obvious values of y that should jump out that you might guess and check. Rckrone (talk) 07:31, 6 July 2010 (UTC)

Ah, yes, y=0 and y=1 are solutions. Are there any others?––220.253.104.38 (talk) 07:33, 6 July 2010 (UTC)

You might want to try a bit of graph sketching. -- The Anome (talk) 07:55, 6 July 2010 (UTC)

And convex functions, also. --78.13.141.245 (talk) 07:37, 7 July 2010 (UTC)

Sum of exponentials

$2^{x}+2^{1-x}-3=0$
solve for x
How do you solve an equation involving the sum of multiple exponentials?––220.253.104.38 (talk) 07:54, 6 July 2010 (UTC)

This looks very similar to the problem above. This leads me to think this might be a homework assignment: please read the comments at the top of this talk page. -- The Anome (talk) 07:57, 6 July 2010 (UTC)

No, it doesn't look similar to the one above at all. It's much simpler than the one above. Michael Hardy (talk) 17:55, 6 July 2010 (UTC)

This is a different type of question to the one bove and can be solved in a couple of ways. For starters try thinking of 2^x as a unit rather than x itself. Forget to ask, first of all, have you come across complex numbers? Dmcq (talk) 08:03, 6 July 2010 (UTC)

And if you haven't come across complex numbers, or perhaps even if you have, are you certain about the signs in the equation? Dmcq (talk) 08:09, 6 July 2010 (UTC)

should be –3, not +3, oops.––220.253.104.38 (talk) 08:15, 6 July 2010 (UTC)

$2^{x}+2(2^{x})^{-1}-3=0$--220.253.104.38 (talk) 08:17, 6 July 2010 (UTC)

Actually the thinking of the previous section might be more use to you now you have that minus sign in there. Otherwise if you want to go along this line doing something like z=2^x is how the chunking is normally done. Dmcq (talk) 08:28, 6 July 2010 (UTC)

Guess and check is so dodgy, I much prefer to solve it algebraically. Anyway, I've solved it now, thanks for your help.––220.253.104.38 (talk) 08:33, 6 July 2010 (UTC)

Substitution: Let

$u=2^{x}\,$

Then

${\frac {1}{u}}=2^{-x}$

So

$2^{x}+2\cdot 2^{-x}-3=0$

becomes

$u+2{\frac {1}{u}}-3=0$

Multiply both sides by u:

$u^{2}+2-3u=0\,$

That's an ordinary quadratic equation. Once you've solved it for u, you've got the value of 2^x. Take its base-2 logarithm to get the value of x. Michael Hardy (talk) 17:54, 6 July 2010 (UTC)

Connections and Differential Forms

Let $\scriptstyle M$ be an $\scriptstyle n$-dimensional, real differentiable manifold and let $\scriptstyle \nabla$ be a smooth connection on $\scriptstyle M$. So, for a pair of vector fields on $\scriptstyle M$, say $\scriptstyle X$ and $\scriptstyle Y$, we have a new vector field $\scriptstyle \nabla _{X}Y$. Now let $\scriptstyle \omega$ be a differential n-form on $\scriptstyle M$. I've never quite understood how to understand the expression $\scriptstyle \nabla \omega$. I've seen expressions like

$\scriptstyle \nabla _{X}[\omega (Y_{1},\ldots ,Y_{n})]\,=\,[\nabla _{X}\omega ](Y_{1},\ldots ,Y_{n})\,+\,\omega (\nabla _{X}Y_{1},Y_{2},\ldots ,Y_{n})\,+\,\omega (Y_{1},\nabla _{X}Y_{2},Y_{3},\ldots ,Y_{n})\,+\,\cdots \,+\,\omega (Y_{1},\ldots ,Y_{n-1},\nabla _{X}Y_{n}).$

$\scriptstyle \iff [\nabla _{X}\omega ](Y_{1},\ldots ,Y_{n})\,=\,\nabla _{X}[\omega (Y_{1},\ldots ,Y_{n})]\,-\,\omega (\nabla _{X}Y_{1},Y_{2},\ldots ,Y_{n})\,-\,\omega (Y_{1},\nabla _{X}Y_{2},Y_{3},\ldots ,Y_{n})\,-\,\cdots \,-\,\omega (Y_{1},\ldots ,Y_{n-1},\nabla _{X}Y_{n}).$

And this makes sense from a formal point of view. But thinking about it; it doesn't make any real sense to me. For example; $\scriptstyle \omega (Y_{1},\ldots ,Y_{n})$ is a differential form, and not a vector field. But $\scriptstyle \nabla$ is a connection, i.e. it takes a pair of vector fields and gives another vector field. (It's a type of covariant derivative.) Out of the last displayed line, the only term I don't understand is $\scriptstyle \nabla _{X}[\omega (Y_{1},\ldots ,Y_{n})]$. How can I make sense of it? •• Fly by Night (talk) 20:26, 6 July 2010 (UTC)

Using the moment generating function to find E(y)

The p.d.f. is $f(y)={\frac {1}{2}}e^{\left|y\right|},-\infty <y<\infty$. When I find the moment generating functions, I set the two cases where one is when y is positive and the other where y is negative. $m(t)=-{\frac {1}{2(t-1)}}$ When y is negative, and t < 1 for the integral to converge. $m(t)={\frac {1}{2(t+1)}}$ When y is positive, and t > 1 for the integral to converge. Now, since E(y) is found by setting t = 0, I use the first m.g.f. to calculate the E(y) since that is when t < 1. I get the answer of -1/2 for E(Y).

However, when I try to integrate y * p.d.f. over the ranges from negative infinity to infinity, I get 0. I'm wondering where did I go wrong here. 142.244.151.247 (talk) 23:17, 6 July 2010 (UTC)

The moment generating function is an integral over y, not a function of y. There are not two cases. Bo Jacoby (talk) 06:14, 7 July 2010 (UTC).

The integral of ½e^|y| from –∞ to ∞ is infinite, whereas the integral of a pdf over its support should be 1. Qwfp (talk) 07:37, 7 July 2010 (UTC)

The OP probably ment to say ½e^−|y| Bo Jacoby (talk) 07:55, 7 July 2010 (UTC).

Yeah, I meant to say $f(y)={\frac {1}{2}}e^{-\left|y\right|},-\infty <y<\infty$, sorry about that. What do you exactly mean by integral over y, not function of y? 142.244.148.125 (talk) 14:24, 7 July 2010 (UTC)

In that case your mgf is wrong, and once you have the right one you need to differentiate it with respect to t before setting t to zero to find E(y), as 'moment generating function' explains. Qwfp (talk) 14:49, 7 July 2010 (UTC)

I know the mgf is wrong, but I'm wondering shouldn't there be two cases for mgf when y is positive and when y is negative here, since there's an absolute value involved here. 142.244.148.125 (talk) 14:58, 7 July 2010 (UTC)

The mgf m(t) is for any given t defined by $\int _{-\infty }^{+\infty }{\frac {1}{2}}e^{-|y|}e^{ty}\,dy$ which does not depend on y (that's a bound variable), only on t, so there cannot be any cases depending on whether y is positive or negative. When you evaluate the integral properly, you will actually end up with three cases for t ≤ −1, −1 < t < 1, and t ≥ 1.—Emil J. 14:59, 7 July 2010 (UTC)

July 7

Equation to solve

${\frac {dy}{dx}}={\frac {y}{x}}+{\frac {x}{y}}$? How does one solve?
$y=\int {{\frac {y}{x}}dx}+\int {{\frac {x}{y}}dx}$
$y=y\int {{\frac {1}{x}}dx}+{\frac {1}{y}}\int {xdx}$
$y=y\ln x+{\frac {x^{2}}{2y}}$
apparently this is wrong ––203.22.23.9 (talk) 03:31, 7 July 2010 (UTC)

y is not a constant and so $\scriptstyle \int {{\frac {y}{x}}dx}$ is not equal to $\scriptstyle y\int {{\frac {1}{x}}dx}$. Bo Jacoby (talk) 06:01, 7 July 2010 (UTC).

Then how is this type of problem solved? I've never encountered one like it before.--203.22.23.9 —Preceding unsigned comment added by 220.253.104.38 (talk) 06:16, 7 July 2010 (UTC)

Substitute $\scriptstyle y=xz$

${\frac {d(xz)}{dx}}={\frac {xz}{x}}+{\frac {x}{xz}}$

${\frac {xdz+zdx}{dx}}=z+{\frac {1}{z}}$

$x{\frac {dz}{dx}}={\frac {1}{z}}$

$zdz={\frac {dx}{x}}$

${\frac {1}{2}}z^{2}=\log {x}+C$

$z={\sqrt {2\log {x}+C}}$

$y=x{\sqrt {2\log {x}+C}}$

This solution is on the form y=f(x). It is more general to state that

$E=x^{2}e^{-{\frac {y^{2}}{x^{2}}}}$

is constant. Bo Jacoby (talk) 06:38, 7 July 2010 (UTC).

Rank and stuff

Some notes I were reading gave the following two theorems without proof: "Suppose that the matrix U is in row-echelon form. The row vectors containing leading 1’s (so the non-zero row vectors) will form a basis for the row space of U. The column vectors that contain the leading 1’s from the row vectors will form a basis for the column space of U." "Suppose that A and B are two row equivalent matrices (so we got from one to the other by row operations) then a set of column vectors from A will be a basis for the column space of A if and only if the corresponding columns from B will form a basis for the column space of B." Would someone be so kind as to provide me with a simple proof of these, if they exist? Much thanks. 74.15.137.192 (talk) 06:02, 7 July 2010 (UTC)

• The first thm, rows: they obviously span the space, it thus suffices to show independence. Consider a nontrivial linear combination of the rows. If i is the top-most row which appears with a nonzero coefficient in the combination, and j is the column containing the leading 1 from row i, then only row i will contribute to the jth entry of the linear combination, making it nonzero.
• The first thm, columns: the selected columns are independent by much the same argument as above. Since their number is the number of nonzero rows of the matrix, which is an upper bound on the dimension of the column space, they also have to span the whole space.
• As for the second theorem, the obvious strategy is to prove it by induction on the number of operations, hence it suffices to show it for the case where B is obtained by a single row operation from A. That should be straightforward enough, but I didn't think about the details.—Emil J. 11:46, 7 July 2010 (UTC)

Quine's NF Set theory

Does the set $\{\{x,y\}\mid xRy\}$ exist, for every relation R expressible by a stratifiable formula whose free variables are x and y? Eliko (talk) 16:30, 7 July 2010 (UTC)

Yes, you can define it as $\{z\mid \exists x,y\,(x\,R\,y\land x\in z\land y\in z\land \forall u\,(u\in z\to u=x\lor u=y))\}$, and the latter formula is stratifiable (x, y, u can be assigned the same type, one less than the type of z).—Emil J. 16:59, 7 July 2010 (UTC)

Now I realized that you didn't specify whether R is stratifiable in such a way that x and y receive the same type, which I used in the above argument. If this assumption is violated, then your set won't exist in general: for example, $u=\{\{x,y\}\mid x\in y\}$ does not exist, since otherwise we could construct the paradoxical Russell set $r=\{x\mid x\notin x\}$ by using $r=\{x\mid \{x\}\notin u\}$ (this can be easily written as a stratified formula).—Emil J. 17:08, 7 July 2010 (UTC)

Oh, yes, I've really meant to ask about x and y having the same type. Thanks. Eliko (talk) 07:26, 8 July 2010 (UTC)

Calculating asymetric confidence intervals for truncated rating scale data

Hi: (For reference, this question is related to calculated confidence intervals for data found at www.boardgamegeek.com and is almost completely for curiosity's sake. It's not to cure cancer or anything really important, so consider that in how much time you take to answer me.  :-) ).

Situation: People rate games on a Likert-like scale from 1 to 10 (1 being hate it, 10 being love it). They can assign fractional ratings (i.e. 7.5, 3.225), so the data are technically continuous, although few people will assign ratings of other than whole numbers. However, while continuous, the data are clearly bounded; 10 is the highest value and 1 is the lowest.

I do not have access to these data directly, however. What I do have access to is:

• The mean of the ratings (calculated as if the data were continuous and non-bounded).
• The standard deviation of the ratings (again, calculated as if the data were continous and non-bounded).
• The number of raters

Is it possible to calculate, from this information, an asymetric 95% confidence interval for the mean of the ratings? That is, an interval that will not exceed 10 or go lower than 1? I suspect that this might involve recourse to the truncated normal distribution in some way, but I'm not sure how. It might also involve a transformation of the mean (e.g. logit?) but again, I'm not clear if that would be appropriate. Just to make clear, what I want is a confidence interval that is bounded by 1 and 10; I know how to calculate the confidence interval from the data I have without that restriction.

Thanks. EDIT: Sorry, didn't sign it. Skalchemist (talk) 17:47, 7 July 2010 (UTC)

Can you give an example of the data? My initial suspicion is that the mean will be sufficiently removed from the boundaries, and the standard error sufficiently small, that the symmetric confidence interval will already be within the bounds. -- Meni Rosenfeld (talk) 05:13, 8 July 2010 (UTC)

I think your idea of using a logit transform should work. Use the delta method to work out the standard error of logit(mean/10), calculate a symmetric confidence interval on the logit scale under the assumption of a normal distribution, back-tranform the interval to give an asymmetric confidence interval on your original scale. Qwfp (talk) 11:22, 8 July 2010 (UTC)

LIFO v. FIFO

You have an incoming unending stream of tasks to do. The amount of time it takes to do each task varies. Which tactic would result in the least average delay in the completing of a task - last in first out, or first in first out? Is there a tactic that has less average delay than either of these? Not a homework question. Thanks. 92.15.27.146 (talk) 20:11, 7 July 2010 (UTC)

How do you define the average delay? Based only on completed tasks? Or also taking into account the backlog? Do you know ahead of time how long the task will take? How does your work output compare to volume of incoming tasks, and what is the nature of incoming tasks -- uniform or random, with what distribution? And what is the distribution of time required for tasks? Some assumptions should be made, at least. 198.161.238.19 (talk) 21:32, 7 July 2010 (UTC)

That said I think the answer is to pick the easiest task first at any given time. 198.161.238.19 (talk) 21:57, 7 July 2010 (UTC)

Scheduling (computing) discusses all this Dmcq (talk) 23:35, 7 July 2010 (UTC)

Maybe someone could translate that into the best practice for a human officer worker. 92.29.125.22 (talk) 08:37, 8 July 2010 (UTC)

"How do you define the average delay? Based only on completed tasks?" Yes. "Or also taking into account the backlog?" Every task will be eventually completed. "Do you know ahead of time how long the task will take?" Yes can be assumed. "How does your work output compare to volume of incoming tasks" In the long-run it should be the same. "What is the nature of incoming tasks -- uniform or random, with what distribution?" You have a big in-tray that always has things in it. "And what is the distribution of time required for tasks?" Random, gaussian, or random, exponential.

The scenario is that you are a busy office administrator with a pile of material in your in-tray to work through. The in-tray keps getting more material put on it. Thanks 92.29.125.22 (talk) 08:33, 8 July 2010 (UTC)

If you want minimum average delay of completed tasks just complete one small task with no delay and then go on holyday. Then your average delay of completed tasks is zero. Your manager should not request minimum average delay of completed tasks. Bo Jacoby (talk) 09:29, 8 July 2010 (UTC).

Indeed. If you want to minimize average delay, do the quickest tasks first. This may, however, delay hard tasks arbitrarily long (depending on the distribution of incoming tasks). Also, if you talk about an actual human doing the work, it's probably easier to keep up with strict LIFO. On the third, hand, what works well for me is interest-based. Just pick the most interesting tasks, and put the rest in a stack. Occasionally that stack will grow so high that it falls over. Whatever lands on the floor goes into recycling. If it would have been important, someone would have reminded me about it earlier ;-). --Stephan Schulz (talk) 09:38, 8 July 2010 (UTC)

Agree with Bo Jacoby and Stephan Schulz. If you know the duration of each task before you start it, and if your goal is to minimise the mean delay per task (with each task given equal weight), and if you must always start a new task as soon as you finish the previous one, then your best theoretical strategy is to always do the shortest task in your in-tray first - the one that will take the least time.

To see why, imagine you put your in-tray tasks in a possible sequence of execution, and calculate the total delay for all tasks if executed in this sequence. Then imagine taking two tasks A and B that are executed one after the other - A then B - in this initial sequence, and swapping them over to make a new sequence in which you do B then A. Comparing the new sequence to the old one, you have increased the delay of task A by the duration of task B - but you have also decreased the delay of task B by the duration of task A. The delays for all other tasks apart from A and B are the same in the new sequence as in the old. So the new sequence will have a lower total delay than the old sequence if and only if the duration of task B is less than the duration of task A. Therefore a sequence with minimum total delay is one in which tasks are executed in order of increasing duration.

If you are allowed to wait between tasks, then there are circumstances in which it is more favourable to wait for a new task to arrive, rather than starting the shortest task in your in-tray. Specifically, if the expected time until the arrival of the next tasks is E(t), and the expected duration of the next task is E(d), you have n tasks in your in-tray and the duration of the shortest tasks is a then you should wait for the next task if

$a>(n+1)E(t)+E(d)$

For example, if a 5-miute task arrives in your empty in-tray but you expect a 2-minute task to arrive in 1 minute, then you should wait for the 2-minute task to arrive, and then execute it first. By doing this, you have a total delay over both tasks of 3 minutes, whereas if you start the 5-minute task immediately, the total delay is 4 minutes. Gandalf61 (talk) 09:49, 8 July 2010 (UTC)

I the tasks are unending it indicates that there is more work than time to do them in and the average time to complete a task will tend to infinity. The first thing that should be done in such a situation with any incoming task is triage. Does it have to be done immediately? Is it a task which needs doing but maybe not just this instant? Or is it one which is a 'would like'? Normally the first two won't overload - if they do then more people are needed. For the 'would like' you may be able to do some but you should make it clear for the ones that obviously are too low priority or have started slipping down your list that they will not be done. If someone in authority complains point out the jobs and their priorites and time to complete according to you and let them figure out the schedule so they are completed. Dmcq (talk) 10:09, 8 July 2010 (UTC)

"If the tasks are unending it indicates that there is more work than time to do them in" - not if you have an average of forty hours work a week to do in 45 hours, for example. 92.29.125.22 (talk) 11:17, 8 July 2010 (UTC)

If you do 40 hours work in 45 hours then you have 5 free. The tasks will have ended for those 5 hours. This is about right, there should be leeway or the system is overloaded. In a computer you would aim to load it no more than 80%,i.e. not to work more than 32 hours on average in a 40 hour week. Dmcq (talk) 11:49, 8 July 2010 (UTC)

A problem is that if you have a very important task that requires a month of work, then its never going to get done under the "do the quickest thing first" tactic. 92.29.125.22 (talk) 11:20, 8 July 2010 (UTC)

You mean you go back to the beginning of the long task after getting interrupted? That would mean it took a very long time! However if there is more time available than work it will eventually be completed if the work isn't lost like that. Personally I count any quick interrupt as 15 minutes and changing tasks and then coming back again, yes that is extremely expensive. 11:49, 8 July 2010 (UTC)

So some tasks are more important than others ? I don't think you mentioned that before. Are there any other details you want to share ? Do some tasks maybe have deadlines ? Can work on a low priority task be interrupted by a higher priority task ? You will get better responses if you give the whole picture instead of drip-feeding key information. Gandalf61 (talk) 11:48, 8 July 2010 (UTC)

Have you never done office-type work? 92.28.250.159 (talk) 12:54, 8 July 2010 (UTC)

Your original question never mentioned office work. It could have been office work, a computer, a factory, a mail order dept, a harried mother, whatever. Even now you've not mentioned if it is from the point of view of a manager, a typist, a project planner or any of the others that might work in an office.. Anyway for office work the relevant article is Time management Dmcq (talk) 13:24, 8 July 2010 (UTC)

The OP did later say "The scenario is that you are a busy office administrator". -- Meni Rosenfeld (talk) 13:56, 8 July 2010 (UTC)

Reply to 92.28.250.159 - is that a rhetorical question ? As it happens, yes I have. Not sure how that helps you though. Your question has just gone right to the bottom of my personal in-tray, filed under "didn't ask nicely". Gandalf61 (talk) 13:29, 8 July 2010 (UTC)

For real office work, the criteria for performance are considerably more complicated than "least average delay". Finding the optimal solution requires knowing very particular details, and the solution can be very complicated. It's usually not worth it to try to construct a mathematical model for it (of course, you can gain some insight by modeling spherical cows). If what you really want to know is "how to be a productive office worker", this is way beyond the scope of a RefDesk question - there are entire books devoted to it (I can recommend Getting Things Done, though it focuses on issues other than you've asked here). -- Meni Rosenfeld (talk) 13:56, 8 July 2010 (UTC)

I'm interested in it from an operations research point of view. I've read GTD, and like most self-help books it was half a page of insight padded out to fill a book. 92.24.188.89 (talk) 18:10, 8 July 2010 (UTC)

I like the 'considerably more complicated'. I once had to go around training managers in project planning using a package where originally we'd got them out for a day to trial a number with a few examples and see which one they liked the best. Then we were told by the head company that they had now themselves decided on a package, and of course it was the one that scored lowest in our trials. I wish we'd never even bothered asking our managers in the first place. Dmcq (talk) 14:26, 8 July 2010 (UTC)

Poisson distribution

Flaws on a used computer tape have a Poisson distribution with an average of 1 flaw per 1200 feet. Find the probability of (a) exactly 4 flaws in 4800 feet (b) at most 2 flaws in 3000 feet

I have my teacher's answer but I am getting a different one. Actually, she only has one answer and both answers I got differ from her one answer.

(a) My reasoning is if 1200 feet is a Poisson(1), then I can consider 4800 feet as the sum of 4 independent Poisson(1)s which would be a Poisson(4). Then $P(X=4)={\frac {e^{-4}4^{4}}{4!}}\approx 0.1954$ Does this make sense?

(b) This is a bit different. Can I assume that this is a Poisson(2.5)? I am not sure. I did that and got an answer but I am not sure if that makes sense here. By looking at MGFs, I know that 1/2 a Poisson(1) is not a Poisson(1/2) so it doesn't make sense to think of this as the sum of 2 independent Poisson(1)s and a Poisson(1/2).

Any help would be appreciated. Thanks StatisticsMan (talk) 23:30, 7 July 2010 (UTC)

Did the original question use the precise words "Poisson distribution"? The correct way to refer to what it meant is Poisson process. A Poisson process is described by a single rate parameter $\lambda$, and the number of events in an interval of length $\tau$ follows a Poisson distribution with mean $\lambda \tau$ even for fractional $\tau$.

Think of it this way: Is the distribution in the first 600 feet the same as in the next 600 feet? Are they independent? If so, the number of errors in the first 1200 feet, which is distributed Poisson(1), is also the sum of two independent copies of some distribution. This distribution is Poisson(1/2). -- Meni Rosenfeld (talk) 05:11, 8 July 2010 (UTC)

L feet of tape has on average λ=L/1200 flaws. In case (a) λ=4800/1200=4 and in case (b) λ = 3000/1200 = 2.5. Case (a): The probability that X=4 is e^−λλ⁴/4!. (b): 'At most 2 flaws' means that X=0 or X=1 or X=2, and the probability for that to happen is e^−λ(λ⁰/0!+λ¹/1!+λ²/2!). Bo Jacoby (talk) 07:20, 8 July 2010 (UTC).

July 8

basic probability

I have two normally-distributed random variables X and Y. X has mean 100 and s.d. 10. Y has mean 105 and s.d. 20. Is there a clean way to find the Prob[Y < X]? I just see a real messy integral. Thanks. 71.141.88.179 (talk) 01:28, 8 July 2010 (UTC)

I assume they're independent, otherwise, I don't know. But, an easy way to do it is to consider the random variable Y - X. Perhaps you have seen that a linear combination of independent normals is normal. So, Y - X would be normal and the mean would be the mean of Y minus the mean of X and the variance would be Variance Y + Variance X. Then, you just need to do one integral from -infinity to 0 of that one random variable. StatisticsMan (talk) 01:43, 8 July 2010 (UTC)

Oh yes, of course. Then I just use an approximation to see where the cdf crosses zero. I just got discouraged when I thought of blasting it out directly. Thanks. 71.141.88.179 (talk) 03:23, 8 July 2010 (UTC)

No. You need to subtract the variables, which is completely different from subtracting the cdfs. You get a new random variable $Z=Y-X$ which is distributed normally with mean 5 and s.d. ${\sqrt {500}}$. $\mathrm {Pr} (Y<X)=\mathrm {Pr} (Z<0)$ so you need to use the standard techniques to find the probability that a normal variable is less than some value.

Again, all of this is under the unstated assumption that X and Y are independent. -- Meni Rosenfeld (talk) 04:58, 8 July 2010 (UTC)

Yes, independent variables. Yes I understood about subtracting the variables. Thanks. 71.141.88.179 (talk) 06:17, 8 July 2010 (UTC)

I like the notation (105±20)−(100±10) = (105−100)±√(20²+10²) = 5±22.4. This is < 0 when (5±22.4)/22.4 < 0 or ±1 < −0.224. Here ±1 has mean 0 and s.d. 1. Bo Jacoby (talk) 07:51, 8 July 2010 (UTC).

Ah, I get it now. I thought you meant something else, sorry. -- Meni Rosenfeld (talk) 08:09, 8 July 2010 (UTC)

If you misunderstood I might improve my communication? Bo Jacoby (talk) 09:37, 8 July 2010 (UTC).

I was responding to the OP, as can be understood from my level of Indentation. -- Meni Rosenfeld (talk) 13:37, 8 July 2010 (UTC)

Kernel and Range of an Operator

So an exercise regarding operators on a Banach space says to find the kernel and range of the linear operator

$K:C([0,1])\rightarrow C([0,1])$

defined by

$K(f(x))=\int _{0}^{1}\sin(\pi (x-y))f(y)dy.$

My problem is that the solution the teacher wrote up says to just split the integral (using the sine subtraction formula) to

$\sin(\pi x)\int _{0}^{1}\cos(\pi y)f(y)dy-\cos(\pi x)\int _{0}^{1}\sin(\pi y)f(y)dy.$

I have no problem with this. This makes sense and I agree. Then he says that looking at this you know that the range is the span (linear combination) of $\sin(\pi x)$ and $\cos(\pi x)$. This is also good. But then he says that the kernel contains all functions which satisfy

$\int _{0}^{1}\cos(\pi y)f(y)dy=0$ and $\int _{0}^{1}\sin(\pi y)f(y)dy=0$

simultaneously. I have two issues with this. First, how do we know that this is all that is in the kernel. What if we have a function where

$\sin(\pi x)\int _{0}^{1}\cos(\pi y)f(y)dy=\cos(\pi x)\int _{0}^{1}\sin(\pi y)f(y)dy$

and second, can we do a little better and figure out a better description of such functions. To me this seems like, "all functions with Kf=0 are functions in the kernel of K". Well duh, doesn't seem very enlightening to me. Thanks!-Looking for Wisdom and Insight! (talk) 05:25, 8 July 2010 (UTC)

The kernel of K is functions of x that are mapped to the zero function of x. K(f)(x) = a sin(πx) + b cos(πx) where a and b are constants. Notice that $a=\int _{0}^{1}\cos(\pi y)f(y)dy$ and $b=\int _{0}^{1}\sin(\pi y)f(y)dy$ don't depend on x. For K(f)(x) to be identically zero for all x, we need both a=0 and b=0 since sin(πx) and cos(πx) are linearly independent. Rckrone (talk) 07:54, 8 July 2010 (UTC)

Non-Hamiltonian k-critical graph

Does anyone know of an example of a k-critical graph (that is, a graph with chromatic number k such that all proper subgraphs have chromatic number less than k) that is not Hamiltonian? I found a reference in Bollobás (Extremal Graph Theory, Theorem 2.1) to a theorem of Dirac: "If G is k-critical then either G is Hamiltonian or its circumference is at least 2k − 2." Reading between the lines, this would seem to imply that some non-Hamiltonian k-critical graphs are known; otherwise I would expect to see that theorem followed by at least a conjecture that all k-critical graphs are Hamiltonian. —Bkell (talk) 07:21, 8 July 2010 (UTC)

Rational numbers with integer sums

Let (x,y,z) be an ordered triple of positive rational numbers such that x + 1/y, y + 1/z, and z + 1/x are all integers. Find all possible ordered pairs (x,y,z).

The only ones I've been able to find are (1,1,1), (1/2, 2/3, 3), and (1/3, 3/2, 2) (as well as their cyclic permutations), but I haven't been able to prove that there are no more. --138.110.206.101 (talk) 12:22, 8 July 2010 (UTC)

There is also (1, 1/2, 2) and its cyclic permutations, but then there are no more. Is this a homework? Here is an outline. Write x = a/u, y = b/v, z = c/w, with gcd(a,u) = gcd(b,v) = gcd(c,w) = 1. Since x + 1/y is an integer, ub divides ab + uv, and using coprimeness of u and b, this gives easily u = b. Symmetrically, v = c and w = a, i.e., x = a/b, y = b/c, z = c/a, and a, b, c are pairwise coprime. The assumption on x + 1/y etc. being integers means that a + b + c is divisible by lcm(a,b,c). Assuming WLOG c ≥ a, b, we have a + b + c < 3c (unless a = b = c = 1 = x = y = z), thus lcm(a,b,c) is c or 2c, which implies that a and b divide 2. This, together with c | a + b, gives finitely many choices which can be checked to give the triples written above.—Emil J. 13:04, 8 July 2010 (UTC)

(ec). Let a=x+1/y, b=y+1/z, and c=z+1/x. Solve for x.

x=a-1/y, y=b-1/z, z=c-1/x. 
x=a-1/(b-1/(c-1/x))
x(b-1/(c-1/x))=a(b-1/(c-1/x))-1
bx-x/(c-1/x)=ab-a/(c-1/x)-1
bx(c-1/x)-x=ab(c-1/x)-a-(c-1/x)
bcx-b-x=abc-ab/x-a-c+1/x
bcxx-bx-xx=abcx-ab-ax-cx+1
(bc-1)xx+(a+c-b-abc)x+ab-1=0

y,z also satisfy quadratic equations.

(ca-1)yy+(b+a-c-abc)y+bc-1=0
(ab-1)zz+(c+b-a-abc)z+ca-1=0

Then the problem is reduced to find integers a,b,c such that the quadratic equations have rational solutions. Bo Jacoby (talk) 13:55, 8 July 2010 (UTC).

probability

we have a scratch off card with 25 squares and 5 have pots of gold underneath. If you scratch off exactly 5 squares revealing 5 pots of gold you win. what is the probability of winning? —Preceding unsigned comment added by 129.120.185.7 (talk) 15:35, 8 July 2010 (UTC)

Well, the chance of hitting a pot of gold on the first scratch is 5/25. If you do that successfully, then the chance of hitting a pot of gold on the second scratch is 4/24 since there are 4 pots left under 24 possible squares. So the chance of hitting on the first 2 is (5/25)*(4/24). And then continue that way down the line for 3, 4 and 5. Rckrone (talk) 16:04, 8 July 2010 (UTC)

See binomial coefficient. The probability of winning is $\scriptstyle {\frac {1}{\binom {25}{5}}}$ = 0.0000188218. Bo Jacoby (talk) 16:39, 8 July 2010 (UTC).