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{{Wikipedia:Reference_desk/Archives/Mathematics/2012 September 17}}

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= September 20 =
==Lanchester's Law and Game Theory==
Let us consider Lanchester's square law and three groups A,B,C which want to go to war. Making up numbers, let's say A,B,C have armies of sizes 45, 40, and 35 respectively. I am concerned with the optimal strategy for group C. From what I understand, B and C should form an alliance. They'll have a vastly superior force than A. They will beat A and both B and C will suffer casualties in proportions. After A has been beaten, then B and C can fight each other. It seems to be that if C has the smallest army, then it will always lose in the end. A,B,C can randomly all shoot at each other or B and C can form an alliance, wipe out A, and then turn against each other. Is there ever an case (with army sizes being A < B < C...notice strict inequalities) where Lanchester's Law and Multiparty Game Theory predict that C will win? A numerical counterexample is what I am looking for of course. If not then, C will always lose, right? So what is C to do? What incentive is there for C to form an alliance with B? It knows it will be killed in the end anyway so why help B (over A) survive? Is this the best strategy? If yes, then why because C's destruction seems inevitable so why should C care what it does and who it helps and what strategy it chooses?[[User:A Real Kaiser|  -  Looking for Wisdom and Insight!]] ([[User talk:A Real Kaiser|talk]]) 04:59, 20 September 2012 (UTC)

:::I assume you meant "A > B > C"  [[User:Dbfirs|''<font face="verdana"><font color="blue">D</font><font color="#00ccff">b</font><font color="#44ffcc">f</font><font color="66ff66">i</font><font color="44ee44">r</font><font color="44aa44">s</font></font>'']] 07:54, 20 September 2012 (UTC)

:I can see three possible winning strategies for C:

:1) The obvious one is to make peace with both A and B.

:2) Form an alliance with one and hope that it holds after the other is defeated.

:3) The most Machiavellian way for C to win is to secretly instigate a war between A and B, then, after one is destroyed and the other is weakened, attack. [[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 05:18, 20 September 2012 (UTC)
::See [[truel]] for something very similar. Since peace treaties as unnecessary between friends I view them as declarations of war but not yet. Option 3 certainly seems the best option for C, the problem is what happens if A or B discover it. C could also send an army of size 5 to 'help' B. It would be interesting to try and figure out the strategies to stay alive the longest if they are all fighting each other. Say army A devoted a(t) of its effort to destroying B and 1-a(t) to destroying C, so C was being destroyed by a force of A(1-a(t))+Bb(t), B by a force of C(1-c(t))+Aa(t) and A by B(1-b(t))+Cc(t). [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 12:40, 20 September 2012 (UTC)
:::Peace should break out all over in ABC world, for no one can afford to attack alone, nor be the weaker ally in an attack. ''[[User:Rich Farmbrough|Rich]]&nbsp;[[User talk:Rich Farmbrough|Farmbrough]]'', <small>21:10, 20 September 2012 (UTC).</small><br />
::::Yep as far as I can see if they have a three way war then all three will be completely destroyed at the same time if the two smaller ones could defeat the largest. What'll happen is that the smaller ones will gang up on the largest until it is no longer the largest, then the two larger will reduce each other till all three are the same size as the smallest. [[User:Dmcq|Dmcq]] ([[User talk:Dmcq|talk]]) 22:31, 20 September 2012 (UTC)

== League problem ==

For a game I'm trying to come up with a particular league system, and I'm unsure how to go about doing it.  I tried to do it manually but I couldn't come up with an algorithm that led me to a solution.  Basically I have 16 teams in a league, but for logistics' sake I need to break it up into four 4-team series.  I figure if I can break it up right, then I can each team have faced every other team in one of the series once, and only once.  So in the first leg I have four series:

{| class="wikitable"
|-
! Series 1 !! Series 2 !! Series 3 !! Series 4
|-
| Team A || Team E || Team I || Team M
|-
| Team B || Team F || Team J || Team N
|-
| Team C || Team G || Team K || Team O
|-
| Team D || Team H || Team L || Team P
|}

Each team has to face 15 other teams, and in each leg it faces 3 other teams, so then in 5 legs any given team will have faced all 15 other teams.  I can come up with two other legs: one by simply flipping it around the diagonal (so Series 1 would be teams A, E, I, and M), and the other by taking diagonals for each series (so Series 1 would be teams A, F, K, and P), but beyond that I start running into problems.  Maybe I'm backing myself into a corner, I don't know.  And I calculate that there are 2386 possible ways to sort the 16 teams into 4 series, so I'm not about to check all of those, to find five "orthogonal" legs.

Is there a better way of going about this, in a more mathematical way?  Thanks —'''[[User:Akrabbim|Akrabbim]]'''<sup>[[User talk:Akrabbim|talk]]</sup> 16:47, 20 September 2012 (UTC)

:I think your approach is reasonable.  Your 4th and 5th groups will be "what's left":

{| class="wikitable"
|-
|-
! Series 1 !! Series 2 !! Series 3 !! Series 4
|-
| Team A || Team E || Team I || Team M
|-
| Team B || Team F || Team J || Team N
|-
| Team C || Team G || Team K || Team O
|-
| Team D || Team H || Team L || Team P
|}

{| class="wikitable"
|-
|-
! Series 1 !! Series 2 !! Series 3 !! Series 4
|-
| Team A || Team B || Team C || Team D
|-
| Team E || Team F || Team G || Team H
|-
| Team I || Team J || Team K || Team L
|-
| Team M || Team N || Team O || Team P
|}

{| class="wikitable"
|-
|-
! Series 1 !! Series 2 !! Series 3 !! Series 4
|-
| Team A || Team E || Team I || Team M
|-
| Team F || Team J || Team N || Team B
|-
| Team K || Team O || Team C || Team G
|-
| Team P || Team D || Team H || Team L
|}

  .
  .
  .

:Also note that there are many possible solutions. [[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 17:12, 20 September 2012 (UTC)

::But the other diagonal will intersect.  One direction will be A,F,K,P, and the other direction will be A,N,K,H.  A and K are in both. —'''[[User:Akrabbim|Akrabbim]]'''<sup>[[User talk:Akrabbim|talk]]</sup> 17:29, 20 September 2012 (UTC)

:::Yes, you're right, I've revised my advice above.  I'm going to write a program to solve this and post it tomorrow. [[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 17:38, 20 September 2012 (UTC)

::::Thanks, I'll be interested to see what you come up with and your algorithm to solve it.  I was trying to work it out with Matlab, but it was just turning into a clumsy brute force algorithm, and I wasn't getting anywhere.  But I charted it out, and three legs that we have here are definitely not part of the solution.  For example, for A and G to meet, the only teams that haven't met A and G would be J and N, and they have both played each other already, so I right when I thought I had painted myself into a corner. —'''[[User:Akrabbim|Akrabbim]]'''<sup>[[User talk:Akrabbim|talk]]</sup> 17:56, 20 September 2012 (UTC)

(Teams have numbers rather than letters, obviously)

Series 1: 1, 2, 3, 4<br>
Series 2: 5, 6, 7, 8<br>
Series 3: 9, 10, 11, 12<br>
Series 4: 13, 14, 15, 16<br>

Series 1: 1, 5, 9, 13<br>
Series 2: 2, 6, 10, 14<br>
Series 3: 3, 7, 11, 15<br>
Series 4: 4, 8, 12, 16<br>

Series 1: 1, 6, 11, 16<br>
Series 2: 2, 5, 12, 15<br>
Series 3: 3, 8, 9, 14<br>
Series 4: 4, 7, 10, 13<br>

Series 1: 1, 7, 12, 14<br>
Series 2: 2, 8, 11, 13<br>
Series 3: 3, 5, 10, 16<br>
Series 4: 4, 6, 9, 15<br>

Series 1: 1, 8, 10, 15<br>
Series 2: 2, 7, 9, 16<br>
Series 3: 3, 6, 12, 13<br>
Series 4: 4, 5, 11, 14<br>

[[Special:Contributions/81.159.104.182|81.159.104.182]] ([[User talk:81.159.104.182|talk]]) 03:41, 21 September 2012 (UTC)

:You beat me too it, 81.  Yes, that's a valid solution.  81's version of diagonals in the third grouping seems to work, while mine does not.  (Running my program, I found one solution given his first 3 groupings, and no solution given my first 3 groupings.) [[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 06:17, 21 September 2012 (UTC)

:BTW, Akrabbim, where did the 2386 number come from ?  My program seemed to come up with 2,627,625 possible groupings.  Here's the formula:

 16!
 ---
 4!<sup>5</sup>

:[[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 06:36, 21 September 2012 (UTC)
::(formerly 81.159) I agree: 16!/(4!^5). I don't understand where that 2386 number comes from either. [[Special:Contributions/86.160.216.189|86.160.216.189]] ([[User talk:86.160.216.189|talk]]) 11:28, 21 September 2012 (UTC)

:::Hey, that's great! Thanks!  Can I ask how you figured it out?  And about the 2386:  I was thinking it would be choose 4 for the first group, then 4 for the next group from the remaining, and so on.  So (16/choose/4) * (12/choose/4) * (8/choose/4) * (4/choose/4) = 1820 * 495 * 70 * 1 = 63,063,000 (when I first punched it in accidentally did + instead of *, which is where I got 2386).  I haven't thought about probability at all in years though, so I'm not surprised I'm wrong. —'''[[User:Akrabbim|Akrabbim]]'''<sup>[[User talk:Akrabbim|talk]]</sup> 14:34, 21 September 2012 (UTC)

::::Arrangements in which the split into groups is the same but groups are listed in a different order are counted separately in your count, but only count as one arrangement in StuRat's count. So your count is 4! = 24 times as large as StuRat's count. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 14:40, 21 September 2012 (UTC)

:::::Agreed, there are the 24 ways to arrange the series, but those are all equivalent.  For example, these two are equivalent:

{| class="wikitable"
|-
! Series 1 !! Series 2 !! Series 3 !! Series 4
|-
| Team A || Team E || Team I || Team M
|-
| Team B || Team F || Team J || Team N
|-
| Team C || Team G || Team K || Team O
|-
| Team D || Team H || Team L || Team P
|}

{| class="wikitable"
|-
! Series 4 !! Series 3 !! Series 2 !! Series 1
|-
| Team A || Team E || Team I || Team M
|-
| Team B || Team F || Team J || Team N
|-
| Team C || Team G || Team K || Team O
|-
| Team D || Team H || Team L || Team P
|}

:::::As to the method I used, it wasn't very elegant, I just used brute force.  Can we mark this Q resolved ? [[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 16:31, 21 September 2012 (UTC)

:::::"''when I first punched it in accidentally did + instead of *''" ... Ooops!! [[Special:Contributions/86.160.216.189|86.160.216.189]] ([[User talk:86.160.216.189|talk]]) 17:28, 21 September 2012 (UTC)

== x/y in IV quadrant, is the given ratio negative or positive? ==

question: "Suppose that the point (x,y) is in the indicated quadrant (IV). Decide whether the given ratio is positive or negative." Is it not positive, simply by sketching? How would it be negative? thanks.[[Special:Contributions/24.139.14.254|24.139.14.254]] ([[User talk:24.139.14.254|talk]]) 20:08, 20 September 2012 (UTC)
:Just look at the signs of x and y. The ratio of two numbers with the same sign is positive; the ratio of two numbers with opposite signs is negative. [[User:Widener|Widener]] ([[User talk:Widener|talk]]) 20:28, 20 September 2012 (UTC)

= September 21 =

== Integration of 1/x in Quadrant I ==

Hi. First of all, this isn't an actual homework problem, but a conceptual question I asked a mixed crowd of people, some of whom know integration techniques and some of whom which who whom don't. Everybody seemed to know intuitively or by proof that the area underneath the curve in quadrant one is ''infinite'', because since the graph never touches either axis, the area underneath each section is infinite. This didn't make sense to me, because I intuitively compared it to a converging sum such as [[0.999...]], but apparently the hyperbolic function does ''not'' converge. Therefore, I must be severely deluded. The integration technique also relies on the fact that zero invalidates the integration, resulting in infinity; if one of the axes were shifted away from the zero-point, the area would still be infinite. This brings up the question: since shifting ''both'' axes away from zero would immediately cause the function's area to become finite, is there a tipping point of some sort? This is hillarious because the hyperbola itself is tipping pointential. Therefore, perhaps at the exact tipping point upon which the area oscillates between finite and infinite, causing the creating of a [[complex number]]. However, since this is a bi-axial method, the convergence paradigm is both null and void.

I must admit, quite embarrassingly, that I've never done [[mathematical proof]]s for more than one hour in my entire life. Therefore, I may need to rely on intuitive techniques such as ''[[visual calculus]]'' to get the point across to myself about why this function has an infinite area. I incorrectly assumed that the total area underneath the hyperbola in the first quadrant is reducible to be equivalent to the area to the bottom left of a linear function with the slope -1. However, I then remembered my [http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Mathematics/2011_September_23#The_cartesian_plane_as_a_sphere_-_theoretical_maths Cartesian plane-onto-sphere] method, which was improperly answered because my method assumes an ''infinite'' Cartesian surface, whereas stereographic projection relies on a finite Earth. It would be more like taking the membrane of an [[shape of the Universe|open universe]] and reconstructing it to make it closed. Anyway, I proved to myself that in this projection, the four points of (0,0), (x-fin,x-infinlim), (y-fin,y-infinlim) and (±∞,±∞) together comprise the [[manifold]] geodesic of a sphere in 2.5 dimensions (clarification, citation, objectivity and sanity needed), in each of the four quadrants (actually, there are a total of eight). This is where {x,y-fin/infinlim} form the continuum where the grid system transforms from finite in one direction to infinite in the opposite direction (from the vantage point of the "anti-origin"), forming two points on a sort of central meridian line, though that's irrelevant largely to solve this problem. So in the Q-I space, the first quadrant forms a circle, and that circle's area is infinite. The perimeter of this circle must also be infinite. The area under the hyperbola, thus, represents an area around the circumference of this circle, although it is widest at one point of the circle and converges at the other end of the circle. Despite the distance from this circumference getting smaller as the function recedes from the 2-space origin, the area still goes onto infinity, thus allowing the total area by integration to be infinite. Phew, though this is not as clear to most people, so there has to be a better intuitive way of proving it.

Someone enlighten me by dialing 0 on the [[Hamiltonian operator]]. Thanks. ~<font color="blue">[[User:AstroHurricane001|AH1]]</font>&nbsp;<sup>(''[[User_talk:AstroHurricane001|discuss]]!'')</sup>

:If you're having trouble seeing that the area under the curve ''y''&nbsp;=&nbsp;1/''x'' in the first quadrant is infinite, consider the following picture, similar to those commonly drawn to illustrate [[Riemann sum]]s:
:[[File:Right Riemann sum of y=x^(-1).png|500px|center]]
:All of the blue rectangles are included in the area under the red curve ''y''&nbsp;=&nbsp;1/''x'', so the area under the curve must be at least as large as the total area of all the rectangles. Every rectangle has width&nbsp;1. The first rectangle has height&nbsp;1, the second has height&nbsp;1/2, the third has height&nbsp;1/3, and so on. So the total area of the rectangles is the sum of the [[harmonic series]] 1&nbsp;+&nbsp;1/2&nbsp;+&nbsp;1/3&nbsp;+&nbsp;…, and this sum is infinite:
::<math>\begin{align}
A_{\textrm{rect}}&=1+\frac12+\left(\frac13+\frac14\right)+\left(\frac15+\frac16+\frac17+\frac18\right)+\left(\frac19+\frac1{10}+\frac1{11}+\cdots+\frac1{16}\right)+\cdots\\
&\ge1+\frac12+\left(\frac14+\frac14\right)+\left(\frac18+\frac18+\frac18+\frac18\right)+\left(\frac1{16}+\frac1{16}+\frac1{16}+\cdots+\frac1{16}\right)+\cdots\\
&=1+\frac12+\frac12+\frac12+\cdots.
\end{align}</math>
:Since the total area of the blue rectangles is infinite, the area under the curve ''y''&nbsp;=&nbsp;1/''x'' must also be infinite. —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 08:21, 21 September 2012 (UTC)

:: See also [[Gabriel's Horn]] for the nice paradox that the "horn" formed by rotation of y = 1/x (x>1) about the x axis has finite volume but infinite area, so it can be filled with paint but its surface can't be painted. [[User:AndrewWTaylor|AndrewWTaylor]] ([[User talk:AndrewWTaylor|talk]]) 09:51, 21 September 2012 (UTC)
:::Good example, but Gabriel's horn has '''x^2''' in the denominator, and 1/x^2 has a finite integral from 1 to infinity. [[User:SemanticMantis|SemanticMantis]] ([[User talk:SemanticMantis|talk]]) 15:15, 21 September 2012 (UTC)
::::No, AndrewWTaylor was correct when he said that Gabriel's horn is constructed by rotating ''y''&nbsp;=&nbsp;1/''x'' about the ''x''-axis. The volume integral has a 1/''x''<sup>2</sup> in it, but that's not the defining curve. —[[User:Bkell|Bkell]] ([[User talk:Bkell|talk]]) 02:59, 22 September 2012 (UTC)
*Also, the "graph never touches either axis" is '''not''' a valid argument for infinite area, so perhaps your reticence to accept it was more correct than you knew! Consider a piecewise defined function, f=1/x^(0.5), x in (0,1), f=1/x^2, x in [1,infinity). This function has a ''finite'' integral over (0,infinity), and the graph never touches either axis. [[User:SemanticMantis|SemanticMantis]] ([[User talk:SemanticMantis|talk]]) 15:15, 21 September 2012 (UTC)

== Limits of Integration ==

Is it true in general that <math>\lim_{n \rightarrow \infty} \int_{a(n)}^{b(n)} f(x,n)dx = \lim_{n \rightarrow \infty} \int_{\lim_{m\rightarrow \infty}a(m)}^{\lim_{m\rightarrow \infty}b(m)} f(x,n)dx</math>?<br /><br /><br />
[[User:Widener|Widener]] ([[User talk:Widener|talk]]) 02:57, 21 September 2012 (UTC)
: Consider <br> &nbsp; &nbsp; &nbsp; <math>a_n = n,\ b_n = a_n+1,\ f = 1\ \dots</math> <br/>Or <br/> &nbsp; &nbsp; &nbsp; <math>a_n=0,\ b_n=n,\ f(x,n)=\tfrac 1n</math><br/>[[User:CiaPan|CiaPan]] ([[User talk:CiaPan|talk]]) 05:30, 21 September 2012 (UTC)
::Good answer. [[Interchange_of_limiting_operations]] has scant information, and could use some improving if anyone is interested. [[User:SemanticMantis|SemanticMantis]] ([[User talk:SemanticMantis|talk]]) 15:03, 21 September 2012 (UTC)
::Oh, of course, but what if you add the condition that the limits <math>\lim_{n \rightarrow \infty} a_n</math> and <math>\lim_{n \rightarrow \infty} b_n</math> exist? [[User:Widener|Widener]] ([[User talk:Widener|talk]]) 16:34, 21 September 2012 (UTC)
:::<math>a_n=-\tfrac1n,\ b_n=\tfrac1n,\ f=n</math>. -- [[User:Meni Rosenfeld|Meni Rosenfeld]] ([[User Talk:Meni Rosenfeld|talk]]) 17:04, 22 September 2012 (UTC)
::::Hm. Okay. I think it is true that <math>\lim_{N \rightarrow \infty} \frac{1}{2} \int_0^\frac{\pi(N+1/2)}{N+1} \frac{\sin t}{\sin(t/(2(N+1/2)))(N+1/2)} = \int_0^\pi \frac{\sin t}{t}</math> though.
::::If you can interchange the limit and integral somehow you get
::::<math>\lim_{N \rightarrow \infty} \frac{1}{2} \int_0^\frac{\pi(N+1/2)}{N+1} \frac{\sin t}{\sin(t/(2(N+1/2)))(N+1/2)}</math>
::::<math>=\lim_{N \rightarrow \infty} \frac{1}{2} \int_0^\frac{\pi(N+1/2)}{N+1} \frac{\sin t}{(t/(2(N+1/2)) + O(t^3/(8(N+1/2)^3)))(N+1/2)}</math>
::::<math>=\lim_{N \rightarrow \infty} \frac{1}{2} \int_0^\frac{\pi(N+1/2)}{N+1} \frac{\sin t}{t/2 + O(t^3/(8(N+1/2)^2)}</math>
::::<math>\stackrel ?= \frac{1}{2} \int_0^{\lim_{N \rightarrow \infty} \frac{\pi(N+1/2)}{N+1}} \lim_{N \rightarrow \infty}\frac{\sin t}{t/2 + O(t^3/(8(N+1/2)^2)}</math>
::::<math>= \frac{1}{2} \int_0^\pi \frac{\sin t}{t/2} = \int_0^\pi \frac{\sin t}{t}</math>
::::[[User:Widener|Widener]] ([[User talk:Widener|talk]]) 18:12, 22 September 2012 (UTC)
:::::<math>\lim_{\epsilon\rightarrow 0}\int_0^{1-\epsilon}f(x,\epsilon)\,dx=\lim_{\epsilon\rightarrow 0}\int_0^{1}f(x,\epsilon)\,dx-\lim_{\epsilon\rightarrow 0}\int_{1-\epsilon}^1f(x,\epsilon)\,dx</math>
:::::<math>=\int_{0}^{1}\lim_{\epsilon\rightarrow 0}f(x,\epsilon)\,dx-\lim_{\epsilon\rightarrow 0}\int_{1-\epsilon}^{1} f(x,\epsilon)\,dx</math>
:::::If the integrand in the last integral is finite (bounded) in the neighbourhood (x,ε) → (1,0), then ... — [[User_talk:Quondum|''Quondum'']] 20:56, 22 September 2012 (UTC)

== Inverse birthday problem ==

The [[birthday problem]] of course describes the probability that two people in a given set of people will share birthdays. 

But what are the odds in a group of ''n'' people that there are days where there are ''no'' birthdays? 

The birthday problem's solution implies that birthdays often cluster. Would that mean that [[unbirthday]]s are also clustered? 

This isn't homework of any sort; I'm just curious. I have ~500 Facebook "friends" — what are the odds that none of them have a birthday today? --[[User:Mr.98|Mr.98]] ([[User talk:Mr.98|talk]]) 13:39, 21 September 2012 (UTC)

:Off the top of my head, I think this is easier than the birthday problem, because we don't have to be careful about things like Alice and Bob share ''some'' birthday, OR Alice and Charles share ''some'' birthday, etc... If we assume that birthdays are uniformly distributed, and each person's birthday is independent of all the others, then we can just multiply to answer your final question. Because they are uniformly distributed, the probability that person A's birthday is ''not'' today is 364/365. Likewise for person B. Because they are independent, we can multiply to get the probability that today is not A's birthday AND is not B's birthday is 364/365 X 364/365. So the [[probability]] (not [[odds]]) that no friend has a birthday today is (364/365)^500=0.2537. Note that your last question is different from your second (i.e. some B-day free day exists, vs today is a B-day free day); that one takes a little more work. [[User:SemanticMantis|SemanticMantis]] ([[User talk:SemanticMantis|talk]]) 16:15, 21 September 2012 (UTC)

::More concisely, the last question is the same as "if 500 people roll a 365-sided die, what is the probability that none of them roll a 265 (today)", while the second question is "if n people roll a 365-sided die, what is the probability that there is some number which does not occur." [[User:SemanticMantis|SemanticMantis]] ([[User talk:SemanticMantis|talk]]) 16:20, 21 September 2012 (UTC)

::It's not so easy to work out a formula, but the probability that there is ''some'' day with no birthday is near certainty until the number of people gets well up into the thousands. [[User:Looie496|Looie496]] ([[User talk:Looie496|talk]]) 16:24, 21 September 2012 (UTC)

:::Isn't the chance that a day exists where nobody in a group of 365 or more has the same birthday given by 1-(1-(364/365)<sup>n</sup>)<sup>365</sup> ?  So, when n = 500, we get 1-(1-(364/365)<sup>500</sup>)<sup>365</sup> = 1-(1-0.2537)<sup>365</sup> = 1-(0.7463)<sup>365</sup> = 1 - 4.1779×10<sup>-47</sub>, or, for all practical purposes 1.0, a virtual certainty that at least one birthday-free day exists.  Here's the calculations with different values of n plugged in:

   n     probability of a birthday-free day existing
 -----   -------------------------------------------
   500   ≈1.0
  1000    0.99999999997
  1500    0.9975
  2000    0.78
  2500    0.32
  3000    0.093
  4000    0.0062
  5000    0.00040
  6000    0.000026
  7000    0.0000017
  8000    0.00000011
  9000    0.0000000069
 10000    0.00000000044

::::[[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 17:35, 21 September 2012 (UTC)

:::::I am immediately suspicious of that formula 1-(1-(364/365)<sup>n</sup>)<sup>365</sup> as it gives a smooth transition of probabilities at the point n = 365. [[Special:Contributions/86.160.216.189|86.160.216.189]] ([[User talk:86.160.216.189|talk]]) 17:59, 21 September 2012 (UTC)

::::::It does seem odd that it gives a probability of almost 1, but not exactly 1, when used for fewer than 365 people.  It's within 1.5×x10<sup>-78</sup> of 1, though.  Is this the chance that the universe will end before the year ends ?  :-) [[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 18:08, 21 September 2012 (UTC)

:::::::The formula is only an approximation, because it treats the days as independent, which they aren't really. [[User:Looie496|Looie496]] ([[User talk:Looie496|talk]]) 18:40, 21 September 2012 (UTC)

::::::::In what sense ? I suppose birthdays aren't completely evenly distributed, in that slightly more people are born at some parts of the year than others.  However, I don't think that's the issue here. [[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 18:44, 21 September 2012 (UTC) 

:::::::::I agree with you StuRat, independence of birthdays is not the problem with your formula (that is why I phrased the problems with dice above, just to be clear about the assumptions.) Nevertheless, I also think your formula is wrong. Why should it be right? You haven't explained your reasoning at all. Also, you can easily write a program to get some good approximate answers. I bet if you do that (correctly), you'll get answers noticeably different from your table above. [[User:SemanticMantis|SemanticMantis]] ([[User talk:SemanticMantis|talk]]) 18:53, 21 September 2012 (UTC)

::::::::::Well, if the chance of a given day being birthday free is 0.2537, then the probability that it has at least one birthday is 1-0.2537, or 0.7463.  The chances of 2 such independent days having one or more birthday each then becomes 0.7463<sup>2</sup>.  The chances of 365 independent days having one or more birthday each then becomes 0.7463<sup>365</sup>.  The chance of this not being the case then becomes 1-0.7463<sup>365</sup>. [[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 19:07, 21 September 2012 (UTC)

::::::::::I think I see what Looie means, though.  Once we know that one day contains (or doesn't contain) a birthday, this can change the count of potential birthdays that can fall on the other dates.  So, yes, in this sense my formula is an approximation, with an error under 1.5×x10<sup>-78</sup>, in the case of 364 people. [[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 19:12, 21 September 2012 (UTC)

:::::::::::I haven't actually checked anything numerically, but the fact that the formula is inexact, and the error bounds have not been mathematically quantified, leaves open the possibility that it is badly wrong for other values. The 1.5×10<sup>-78</sup> example by itself not necessarily confidence-inspiring. [[Special:Contributions/86.160.216.189|86.160.216.189]] ([[User talk:86.160.216.189|talk]]) 19:31, 21 September 2012 (UTC)

<s>Consider the total number of ways to distribute the birtdays of the n persons over the 365 possible dates. You can represent each possible distribution as a string of the form

00|0|0000|0||00|0|00|....

where each "0" represents a birtday of the person and the "|" are the separations between different days. So, on Januari 1 you have two birtdays, on Januari 2 there is one birtday etc. etc. Then everyu possible string o can write down corresponds to a valid distribution and voce versa, so you just need to count the number of strings. We have n "0"'s and we have 364 "|"'s so the total number is Binomial[364 + n,n].

Then if we demand that on every day you have at least one birtday, then this means that you must leave at least one "0" at each position in the string. This means that you have n-365 "0"'s that you can freely shuffle around. You can, of course, also consider strings with n -365 "0"s, themapping between a birthday distribution is then the number of "0"'s at some date plus one.  So, the total number of such distributions is  Binomial[n-1,n-365].

The probability that you always have at least one birtday on each date is thus Binomial[n-1,n-365]/Binomial[364 + n,n], the probability that there are days where you don't have birthdays is thus given by:

1 -  Binomial[n-1,n-365]/Binomial[364 + n,n]

[[User:Count Iblis|Count Iblis]] ([[User talk:Count Iblis|talk]]) 17:08, 21 September 2012 (UTC)</s>

*I have doubts that this is correct. For a start, you seem to be assuming that your Binomial[364 + n,n] combinations are all equally likely, but it seems to me that they are not. [[Special:Contributions/86.160.216.189|86.160.216.189]] ([[User talk:86.160.216.189|talk]]) 17:46, 21 September 2012 (UTC)
*:Yes, you are right, although it is easy to correct this by imposing an order on the birthdays. I'll correct the solution. [[User:Count Iblis|Count Iblis]] ([[User talk:Count Iblis|talk]]) 18:13, 21 September 2012 (UTC)



I corrected the above problem, I found that the probability p for there being one or more days on which there are no birthdays is given by:

:<math>p = 1 - \frac{1}{r^{n}}\sum_{k = 0}^{r}(-1)^{r-k}\binom{r}{k}k^{n}</math>

where r = 365

[[User:Count Iblis|Count Iblis]] ([[User talk:Count Iblis|talk]]) 19:50, 21 September 2012 (UTC)

:What is k ?  Also, would you like to run that against the numbers in my chart above, to see how the two compare ? [[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 23:27, 21 September 2012 (UTC)
::I tried it earlier and they seem to correspond quite well, so assuming Count Iblis' formula is correct, it seems like yours is a pretty good approximation after all. (As far as evaluating the formula is concerned, ''k'' is just an internal counter which does not need to be assigned a meaning, but probably you know that and are asking about its meaning in terms of the formula derivation.) [[Special:Contributions/86.160.216.189|86.160.216.189]] ([[User talk:86.160.216.189|talk]]) 00:17, 22 September 2012 (UTC)<br />
:Looks ok. There are <math>365^{n}</math> possibilities. To get the ones with birthdays on every day, you have to subtract the ones where at least one day has no birthdays. There are <math>364^{n}</math> ways to distribute them over 364 days, and 365 ways to choose those 364, so <math>365^{n}-365*364^{n}</math>; however, now you removed the ones with two "birthless" days twice. So you add those: there's   <math>\binom{365}{2}</math> ways to choose those days, giving you <math>365^{n}-365*364^{n}+\binom{365}{2}*363^{n}</math>. Same thing again: some are counted twice, so you must subtract  <math>\binom{365}{3}*362^{n}</math>; repeat the same proces till you arrive at all birthdays on one day, divide by total number of possibilities gives the odds of all days having birthdays, subtract from 1 to get the odds you wanted. [[User:Ssscienccce|Ssscienccce]] ([[User talk:Ssscienccce|talk]]) 01:30, 22 September 2012 (UTC)
{{od}}
About the derivation, you can use inclusion/exclusion as Ssscienccce explained above. Another way is to directly evaluate evaluate the number of ways you can have one or more birthday on each date by summing over each distribution of the birth dates over the year. So, if there are <math>k_{j}</math> birthdays on the jth date, then the total number of ways to have such a distribution is:

:<math>\frac{n!}{\prod_{j=1}^{r}k_{j}!}</math>

If we sum this over all possible values of <math>k_{j}</math> we should find <math>r^n</math>. In the summation, you have to impose the constraint <math>\sum_{j=1}^{r}k_{j}=n</math>. You can do this using generating functions. You compute the function:

:<math>f(x) = \sum_{k_{1}=0}^{\infty}\sum_{k_{2}=0}^{\infty}\cdots\sum_{k_{r}=0}^{\infty}\frac{n!}{\prod_{j=1}^{r}k_{j}!}x^{\sum_{j=1}^{r}k_{j}}</math>

The coefficent of x^n is then the desired answer. Taking out the factor n!, the summation factors into identical summations, and each summation is the series expansion of exp(x), so f(x) = n! exp(rx), the coefficient of x^n is thus r^n. If we now consider only cases where you have one or more birthdays on each date, then each of the summation starts at 1 instead of zero, so, you get f(x) = n! [exp(x) - 1]^r. The coefficient of x^n can be obtained by first expanding [exp(x) - 1]^r and then extracting the coefficient of x^n from each term, you then get the above formula.

You can also use the generating function directly to find approximations for large n, you can write the coefficient of x^n as the contour integral of f(z)/z^(n+1) around the origin and then using the saddle point method obtain approximations to that integral. [[User:Count Iblis|Count Iblis]] ([[User talk:Count Iblis|talk]]) 02:00, 22 September 2012 (UTC)
: see [[Coupon collector's problem]].  [[User:Robinh|Robinh]] ([[User talk:Robinh|talk]]) 08:28, 23 September 2012 (UTC)

::Sorry if I'm making a mistake here, but I think the formula
:::<math>p = 1 - \frac{1}{r^{n}}\sum_{k = 0}^{r}(-1)^{r-k}\binom{r}{k}k^{n}</math>
::is not correct. Let n=4 (number of people) and r=3 (number of days in the year). The number of ways to get at least one untaken day is the number of ways to get 2 untaken days (=3) plus the number of ways to get 1 untaken day (=3×2<sup>4</sup>), for a total of 51 out of 3<sup>4</sup>=81 possible arrangements. But the above formula gives 
:::<math>1 - \frac{-0+3-3 \cdot 16 + 81}{81} </math>
::due to the alternating signs term (-1)<sup>r-k</sup>, whereas every term in the numerator except the last ''should'' have a minus sign.  [[User:Duoduoduo|Duoduoduo]] ([[User talk:Duoduoduo|talk]]) 13:56, 23 September 2012 (UTC)
:::So, here the formula gives 15/27 for the probability. Let's then count the number of ways to arrange the 4 persons so that you have one or more untaken days. I can put all four on the same day, there are 3 ways to do that. Then I can divide the group of 4 into one group of 3 and one of one and put these two groups on two days. There are 4 ways to make two such groups and there are 3 ways to choose a day for the group of 3 and having made that choice, 2 ways to choose a day for the remaning person. So, there are a total of 4*3*2 = 24 ways for such a distribution. Finally we can divide the four into two groups of 2, there are binomial(4,2)/2! = 3 ways to do that (we divide by a symmetry factor of 2! so that interchanging the dates of the two groups yields configurations that are not already accounted for by the different ways one can divide the persons in the two groups). Given the two groups there are 3 ways to choose a date for one group and given that choice, 2 ways to chose a date for the other. So, there are a total of 3*3*2 = 18 such possibilities. There are thusa total of 3 + 24 + 18 = 45 possibilities, the probability is thus 45/81 = 15/27. [[User:Count Iblis|Count Iblis]] ([[User talk:Count Iblis|talk]]) 17:01, 23 September 2012 (UTC)

== cells within a table ==

{| class="wikitable"
|-
|-
| A || B
|-
| C || D
|-
| E || F
|-
| G || H
|}

The above table has 2 columns and 4 rows. In the above example I have assigned 8 different values to the 8 cells of that table. Let us say that we can rearrange the values at will. By rearranging the values, how many different arrangements are possible? [[User:Bus stop|Bus stop]] ([[User talk:Bus stop|talk]]) 19:46, 21 September 2012 (UTC)
:<math>8! = 40320</math> see [[permutation]]. [[User:Widener|Widener]] ([[User talk:Widener|talk]]) 21:53, 21 September 2012 (UTC)
::Thanks, Widener. That number surprises me as I would have guessed a far lower number. By the way I see the number at [[Factorial]]. [[User:Bus stop|Bus stop]] ([[User talk:Bus stop|talk]]) 01:31, 23 September 2012 (UTC)
:::{| class="wikitable"
|-
|-
| A || B
| C || D
| E || F
| G || H
|}
:::Followup question: Does the same calculation apply in the case of the above arrangement? [[User:Bus stop|Bus stop]] ([[User talk:Bus stop|talk]]) 11:06, 23 September 2012 (UTC)
::::Yes. The formula is not based on the physical arrangement. The idea is this: There are 8 choices for where to put A. After A has occupied a box, there are 7 choices for where to put B. So we could have had A in the first box and B in any of 7 other boxes, or we could have had A in the second box and B in any of 7 other boxes,...,so for each of 8 possibilities for A there are 7 possibilities for B, hence we have 8×7 possibilities so far. Then for each of those 8×7 possibilities there are 6 remaining possible choices for where to put C, giving 8×7×6 possibilities so far. And so on until the end, where we have 8×7×6×5×4×3×2×1 possibilities. [[User:Duoduoduo|Duoduoduo]] ([[User talk:Duoduoduo|talk]]) 13:36, 23 September 2012 (UTC) possibilities possibilities
:::::Thank you. That is really interesting. I have to think about that. [[User:Bus stop|Bus stop]] ([[User talk:Bus stop|talk]]) 14:08, 23 September 2012 (UTC)

== Friendship combinations ==
Let say there are 6 people: A, B, C, D, E, F. Each of them has 2 internet friends in the group. None of them has any internet friend outside of the group. How many different ways can this happen? I already know the answer which is 70 but I'm confused on how to find the answer. Can anyone explain to me in details how to get the answer? Thanks![[Special:Contributions/65.128.190.136|65.128.190.136]] ([[User talk:65.128.190.136|talk]]) 21:23, 21 September 2012 (UTC)

:I find problems like this usually require some fiddling around. A special case is when you have two groups of three; there are <math>10</math> of those. WLOG you can look at the group of three containing A - then you vary the other two. [[User:Widener|Widener]] ([[User talk:Widener|talk]]) 22:05, 21 September 2012 (UTC)

:You could do this systematically by starting with A, then going through all possible pairs that could be both partnered with A. Then for each person in each of those pairs, you could go through all possible pairs that could be partnered with them - this process will terminate when you get back to someone you have already considered. That's the brute-force approach; there may be a quicker approach but I'm not aware of it off the top of my head. [[User:Widener|Widener]] ([[User talk:Widener|talk]]) 22:17, 21 September 2012 (UTC)

::Think of the 6 people as [[vertex (graph)|points/nodes]]; [[friendship]] is designated by connecting two points/nodes. Given the condition that each person has exactly 2 friends, you need to connect them with lines so that every point/node is paired off with 2 other points/nodes. Then there are only two possible shapes that these nodes can take that satisfy these conditions: 2 [[circle|rings/circles]] of 3 points/nodes each, or 1 ring/circle of 6 points/nodes. Now consider each of those shapes in turn:

::* 2 rings/circles of 3 points/nodes: You need to [[combination|select]] 3 out of 6 points/nodes for the first ring/circle. Then, since there are two rings/circles and order doesn't matter, you need to divide by the possible number of ways to order those 2 rings/circles: <math>\frac{\binom {6}{3}}{2!} = \frac{20}{2} = 10</math>

::* 1 ring/circle of 6 points/nodes: The general formula for a ring [[permutation]] is (n - 1)! / 2 = 5! / 2 = 60. (To see why the formula is thus, consider that after choosing any arbitrary node in the ring as your starting point, you have 5 nodes left to order, so 5!, but then you have to account for the shape being a circle and looping back to the beginning, so you have to divide by 2 since [[clockwise]] and [[counterclockwise]] orderings are equivalent upon [[reflection (mathematics)|reflection]].)

::10 + 60 = 70, and there's your answer.

::—[[User:SeekingAnswers|SeekingAnswers]] ([[User talk:SeekingAnswers|reply]]) 22:31, 21 September 2012 (UTC)

:::Nice![[Special:Contributions/65.128.190.136|65.128.190.136]] ([[User talk:65.128.190.136|talk]]) 23:06, 21 September 2012 (UTC)

::::P.S. The problem is equivalent to asking for the number of [[undirected graph|undirected]] 2-[[regular graph|regular]] [[graph (mathematics)|graph]]s given ''n'' labeled [[node (graph)|nodes]] {{OEIS|A001205}}, where in this case ''n'' = 6. —[[User:SeekingAnswers|SeekingAnswers]] ([[User talk:SeekingAnswers|reply]]) 04:28, 22 September 2012 (UTC)

== nPr? ==
How do I calculate nPr by hand? Let say in Ti 89 calculator, I type in nPr(9 , 2) = 72. So how can I do it without the calculator? Thanks![[Special:Contributions/65.128.190.136|65.128.190.136]] ([[User talk:65.128.190.136|talk]]) 23:06, 21 September 2012 (UTC)
:<math>\frac{n!}{(n-r)!}</math>, or <math>\prod_{i=n-r+1}^n i</math> or <math>\prod_{i=1}^{r} (n-i+1)</math>. [[User:Widener|Widener]] ([[User talk:Widener|talk]]) 23:23, 21 September 2012 (UTC)<br />

:Or in simpler terms: P(15,4)=15*14*13*12, just multiply, first number is where you start with, second one is the number of factors to use. P(12,3)= 12*11*10 you get the idea. [[User:Ssscienccce|Ssscienccce]] ([[User talk:Ssscienccce|talk]]) 01:41, 22 September 2012 (UTC)

::<small>Wasn't this discussed on [[NPR]] ? :-) [[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 01:44, 22 September 2012 (UTC) </small>
:::<small>{{#switch: {{lc:{{{1}}}}} | supreme | sfod=[[File:Facepalm3.svg|20px]] '''Supreme [[Template:facepalm|facepalm]] of destiny''' | #default=[[File:Facepalm3.svg|15px]] '''[[Template:Facepalm|Facepalm]]'''}}... all right, I laughed a little. --<font face="Book Antiqua">[[User:Kinu|<font color="blue"><strong>Kinu</strong></font>]]&nbsp;<sup>[[User_talk:Kinu|<font color="red">''t''</font>]]</sup>/<sub>[[Special:Contributions/Kinu|<font color="red">''c''</font>]]</sub></font> 04:36, 22 September 2012 (UTC)</small>

= September 22 =

== a? ==
[http://s1172.photobucket.com/albums/r574/englishwikipedia/?action=view&current=IMG_0083_zps2be7fe73.jpg problem 20]. I saw the answer key but still unable to understand what is going on. I understand the binary expansion part. I get lost in the statement "it follows that the coefficient of x^2012 is equal to 2^0 + 2^1 + 2^5 = 2^6. [http://s1172.photobucket.com/albums/r574/englishwikipedia/?action=view&current=IMG_0081_zps19f3eb5a.jpg answer key part 1], [http://s1172.photobucket.com/albums/r574/englishwikipedia/?action=view&current=IMG_0080_zpsde051df7.jpg answer key part 2]. Thanks![[User:Pendragon5|Pendragon5]] ([[User talk:Pendragon5|talk]]) 00:02, 22 September 2012 (UTC)<br />
:A product of sums is the sum of all the possible products; say (a+b)*(c+d)*(e+f): the result will contain every product you can make by picking one of each sum,so for example a*d*e, b*c*e, b*c*f and five others.
:In your polynomial, you know which factors supply the power of x part, so the rest must provide the coefficient: showing only the parts that make up the x<sup>2012</sup> term :  (x<sup>1024</sup>+..)*(x<sup>512</sup>+..)*(x<sup>256</sup>+..)*(x<sup>128</sup>+..)*(x<sup>64</sup>+..)*'''(..+ 32)'''*(x<sup>16</sup>+..)*(x<sup>8</sup>+..)*(x<sup>4</sup>+..)*'''(..+2)*(..+1)'''. And 1*2*32=64=2<sup>6</sup> [[User:Ssscienccce|Ssscienccce]] ([[User talk:Ssscienccce|talk]]) 02:11, 22 September 2012 (UTC)
::Oh okay I think I got it now. Thanks! I also have another question. How can I convert 2012 into binary number without the answer key?[[User:Pendragon5|Pendragon5]] ([[User talk:Pendragon5|talk]]) 19:37, 22 September 2012 (UTC)
:::[[Binary_numeral_system#Conversion_to_and_from_other_numeral_systems]], first paragraph. — [[User_talk:Quondum|''Quondum'']] 21:02, 22 September 2012 (UTC)

== Commutability of integration ==

Is <math>\int (\int f(r) dr) dv</math> the same as <math>\int (\int f(r) dv) dr</math>? [[Special:Contributions/71.207.151.227|71.207.151.227]] ([[User talk:71.207.151.227|talk]]) 21:04, 22 September 2012 (UTC)

:See [[Fubini's theorem]].  [[User:Sławomir Biały|<span style="text-shadow:grey 0.3em 0.3em 0.1em; class=texhtml">Sławomir Biały</span>]] ([[User talk:Sławomir Biały|talk]]) 21:09, 22 September 2012 (UTC)

= September 23 =

== Seeking function ==

Does anyone have any feeling whether any function ''f'' exists to satisfy the following:

:<math>\int_0^1\frac{x}{z^2}f(x)f\left(x\left(\frac{1}{z}-1\right)\right)\,dx=1</math>

:for all 1/2 < z < 1

Of course, actual solution(s) would be even better, but failing that an idea of whether there are likely to be no solutions, one solution, or many solutions would be of interest. [[Special:Contributions/81.159.105.97|81.159.105.97]] ([[User talk:81.159.105.97|talk]]) 00:59, 23 September 2012 (UTC)

== <s>Topology</s> Norm on a space over a complex field ==

Is there any "recognized" <s>metric</s> [norm] which satisfies the Pythagorean Theorem even when the <s>lengths of</s> [vectors representing] sides in a triangle are <s>not real</s> [complex] numbers? (e.g. the hypotenuse will be 0 <s>in</s> [by] that <s>metric</s> norm if the <s>lenghs of</s> [vectors representing] the other sides are 1 and i; In other words: the length of the vector (1,i) will be zero, because 1<sup>2</sup> + i<sup>2</sup> = 0<sup>2</sup>). Additionally, can such a <s>metric</s> [norm] be regarded as Euclidean (or pseudo-Euclidean) in some sense? Or rather, can it even be regarded as the ''simple Euclidean'' <s>metric</s> [norm] on a space over the ''complex field''? Is it a legitimate "recognized" <s>metric</s> [norm]? [[Special:Contributions/77.126.50.25|77.126.50.25]] ([[User talk:77.126.50.25|talk]]) 08:58, 23 September 2012 (UTC)

: Would the pseudo-Euclidean Lorentz metric ([[Minkowski space]]) qualify? This still uses real numbers, but modifies signs in the Pythagorean theorem. The question is essentially circular, inasmuch as what is considered to be orthogonal (a "right angle") is ''defined'' by the metric (usually a [[symmetric bilinear form]]), including in any number of dimensions. Pure imaginary numbers can be used to produce the same effect. — [[User_talk:Quondum|''Quondum'']] 10:11, 23 September 2012 (UTC)
::The circularity you've pointed at can be avoided by defining the space over the complex field (rather than over the real field), so that the two vectors (x,0) (0,y) be orthogonal - because their inner product is zero, yet x may equal 1 and y may equal i, in which case - the metric I'm looking for - determines the length of their sum (1,i) to be zero (because 1<sup>2</sup> + i<sup>2</sup> = 0<sup>2</sup>).
::I suspect Minkowski space won't be sufficient, because of two reasons: First, it's not defined over the complex field - as opposed to what's expected from the space having the metric I'm looking for; Second, even if one realy tried to use Minkowski space over the complex field, its metric still doesn't fully obey the Pythagorean Theorem - but rather modifies signs - as you have already pointed out.
::As long as I think about this, it seems that the metric I'm looking for is the ''simple Euclidean'' metric on a space over the complex field, isnt it? [[Special:Contributions/77.126.50.25|77.126.50.25]] ([[User talk:77.126.50.25|talk]]) 11:32, 23 September 2012 (UTC)

:::You are describing a space of two real dimmensions with a metric. If you choose to separately consider it as the complex field, this is irrelevant, since the complex multiplication is not used here except as you choose within the metric, and you are choosing to dismember it into it real and imaginary parts (i.e. two one-dimensional spaces over the real numbers) before defining the metric in terms of the pieces. You are separately defining (x,0) and (0,y) be orthogonal; you can't do this, it is determined by the metric (i.e. you have to find whether they are orthogonal under the metric you've defined). The metric you have described is exactly the Lorentz metric in two dimensions, where you have simply labelled the units on one axis with an "i".  The two axes do turn out to be orthogonal, but vectors away from the axes behave a little counterintuitively, with those at 45° being self-orthogonal. This matches the behaviour of [[split-complex number]]s rather than the normal complex numbers.  If you were to try to define a "simple Euclidean metric" over the complex field in two dimensions, each "axis" would allow values from the whole field (you'd be dealing with 4 real dimensions). This construction works perfectly well, but I doubt that it is what you're looking for. — [[User_talk:Quondum|''Quondum'']] 13:00, 23 September 2012 (UTC)
::::Thanks to your comment, I've just fixed some mistakes in the old version of my original question, in order to make it clear that what I've been trying to describe - was '''not''' "''a space of two real dimensions''" - nor "''two one-dimensional spaces over the real numbers''", but rather one space of two ''complex dimensions''. Btw, I didn't understand why I couldn't state that (x,0) and (0,y) are orthogonal; Doesn't this follow from the very definition of the orthogonality and the inner product? As to the name of the section, I've just changed it. [[Special:Contributions/77.126.50.25|77.126.50.25]] ([[User talk:77.126.50.25|talk]]) 14:14, 23 September 2012 (UTC)
: The complex metric is "unrecognised" not because it is impossible to construct. It is "unrecognised" because fails to satisfy the [[triangle inequality]], if not to mention that almost certainly evokes a lot of [[null vector]]s. And the existence of non-zero null vectors is a hint to impossibility of the Pythagorean Theorem in complex-valued "metric spaces" (which are actually not so by their core properties). <small>BTW, I still do not realize why this section is named "Topology".</small> [[User:Incnis Mrsi|Incnis Mrsi]] ([[User talk:Incnis Mrsi|talk]]) 13:28, 23 September 2012 (UTC)
::As Quondum has indicated, physicists typically may not require the triangle inequality of a "metric" (although mathematicians typically do). [[Special:Contributions/77.126.50.25|77.126.50.25]] ([[User talk:77.126.50.25|talk]]) 14:38, 23 September 2012 (UTC)
::For this I implicity assumed a definition of "metric" as ''quadratic form''. Mathematicians typically require the triangle inequality of a "metric", whereas physicists don't. And in physics, the Lorentz "metric" is most certainly recognized and is used extensively. <small>and yes, the section is misnamed, care to suggest another?</small> — [[User_talk:Quondum|''Quondum'']] 14:08, 23 September 2012 (UTC)
:::See my response to your previous comment. [[Special:Contributions/77.126.50.25|77.126.50.25]] ([[User talk:77.126.50.25|talk]]) 14:15, 23 September 2012 (UTC)

Let's take a step back, and try to determine what you are looking for.
* We could define a [[quadratic form]] on an ''n''-dimensional complex vector space.  This will assign a complex value to each vector in this space. It could look mathematically very similar to the classical sum-of-squares in geometry, only with complex values (or the dot product on complexified vectors). It is full of null vectors, but that never bothered relativists. Regarded as a real vector space, it has a [[metric signature|signature]] of (''n'',''n'').
* Alternatively, we can define a [[sesquilinear form]] (or else its associated "quadratic" form) on the same space. This will assign a real number to each vector. The end result is rather like (actually indistinguishable from) a 2''n''-dimensional (real) Euclidean space, with a signature of (2''n'',0).
I would assume from your description that the first is what you mean. There's nothing particularly recognized or not about it; it would be known as "an ''n''-dimensional complex vector space with a nondegenerate quadratic form", or, if you're referring to the "metric" itself, "a nondegenerate quadratic form on an ''n''-dimensional complex vector space". It trivially satisfies the Pythagorean identity for the addition of orthogonal vectors. — [[User_talk:Quondum|''Quondum'']] 15:15, 23 September 2012 (UTC)
:Thank you very much, both for the information you presented - and for your patience; Yes, it seems like the first option is what I'm looking for (since the "magnitude"-function I'm looking for is supposed to assign a complex number, which - of course - may not be real). However, something still bothers me: I'm looking for a "magnitude"-function - being identical to the well celebrated Euclidean formula for calculating the length of the hypotenuse: this formula is supposed to look like a square root of sum of squares, but that formula is not a "quadratic form", but rather "a square root of a quadratic form". So are you still sure the first option may point at the correct direction for satisfying my requirement? [[Special:Contributions/77.126.50.25|77.126.50.25]] ([[User talk:77.126.50.25|talk]]) 18:50, 23 September 2012 (UTC)

== Cantor set and compactness ==

The article on the Cantor set says that every compact metric space is the continuous image of the Cantor set. If I do not misunderstand it means that for every compact space X there is a continuous onto function with domain the Cantor set and range X. But won't the cardinality of X be at most the continuum as if f is an onto function from A to B then |B|<=|A|? What if X has cardinality more then the continuum?-[[User:Shahab|Shahab]] ([[User talk:Shahab|talk]]) 15:42, 23 September 2012 (UTC)