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November 16

What is the pre-fix to indicate you have 1 & 1/2 of something?

I know in genetics, that they use terms like haploid for 1/2, and diploid for 2, and monoploid for 1; however, what theoretically might I call something that is 1.5-ploid? 완젬스 (talk) 03:19, 16 November 2011 (UTC)

I don't know much about genetics but ploidy appears to have different definitons and only allow natural numbers, and it seems those terms are only for use in genetics. More generally (but rarely used), number prefix mentions sesqui- at 1½. PrimeHunter (talk) 03:31, 16 November 2011 (UTC)

Yes! Sesqui- thank you very much. It's actually a question I had regarding what is halfway in between octofinals and double-octofinals [at this article] and it make me deathly curious what therefore I might call a "Round of 24" so by my reasoning, I can call it sesqui-octofinals, lol. This word will impress my Korean friends who are not good at English--I will make them memorize my new word on a flashcard as part of a flash-card series I'm helping my younger brother learn this "difficult" language. Apparently to him, it's a word as common as building or car is for us! 완젬스 (talk) 04:01, 16 November 2011 (UTC)

The premise here is wrong -- haploid means one copy of each chromosome, diploid means two. The term "monoploid" does not fit on this scale. Diploid is the most common condition, so an organism with 1 1/2 times the usual number of chromosomes is triploid. To my knowledge there is no such thing as an organism that is 1.5-ploid. Looie496 (talk) 17:10, 17 November 2011 (UTC)

Point symmetry - the improper rotation symmetry element

When describing the improper rotation symmetry element (in 3 dimensional Euclidean space) it is always specified that the reflection plane be perpendicular to the rotation axis. What is the reason for this restriction? What about a transformation consisting of a rotation of θ=360°/n (n = a positive integer)combined with a reflection through a plane inclined to the rotation axis by an arbitrary angle α? I am well aware of the fact that for α=0 (i.e. when the rotation axis is in the reflection plane) the combination of the above operations amounts to a simple reflection but what about 0<α<90°? Can this be a symmetry element?Shlomo Levinger (talk) 11:00, 16 November 2011 (UTC)

Any rotation (improper or otherwise) that isn't the identity or a simple reflection sends a unique line to itself. For proper rotations, this is the axis of rotation and the points are fixed by the rotation. For improper rotations, the points are flipped about the origin. It is this line that allows an improper rotation to be decomposed into a proper rotation about the line followed by a reflection that flips the line over (i.e., in a plane perpendicular to the line). For your second question, if the rotation and reflection are not correlated, then they still describe a symmetry element. In fact, any symmetry element can be described in an infinity of ways in this case. But every symmetry element has a unique normal form, in which it is given as a rotation followed (if necessary) by a reflection in the plane of that rotation. Sławomir Biały (talk) 12:28, 16 November 2011 (UTC)

Thank you Sławomir for clarifying the uniqueness of a reflection in the plane of rotation. Still, one point bothers me here. The presence of a rotation-reflection symmetry element (with the mirror plane perpendicular to the rotaion axis), Sn, of an even order (n is even) does not imply the presence of a simple plane of symmetry for a 3-dimensional object. Can this be claimed in the case of an object possessing an oblique rotation-reflection symmetry element (i.e. where the mirror plane is inclined to the rotation axis)? My "intuition" is that such a symmetry element, if at all present, will necessarily imply either the presence of a simple plane of symmetry or that of another rotation-reflection symmetry element having the mirror plane perpendicular to the rotaion axis. This would mean that such an oblique element is not an "independent" one. However, I can not yet prove this. Your comment will be appreciated.Shlomo Levinger (talk) 11:05, 17 November 2011 (UTC)

If you find the mid points between original points and where they go to that will define a subspace, here a plane unless it is just a point. That's the reflection plane. Dmcq (talk) 12:20, 17 November 2011 (UTC)

You can prove that all such improper rotations can be described in terms of a rotation in and reflection in the same plane using geometric algebra. Something like this:

Suppose your rotation is in a plane with bivector B through angle θ. Then the rotation is given by the rotor

$R=e^{\mathbf {B} \theta /2}$

The important thing is this is an even element of the algebra so has a scalar and bivector part. It acts on a vector to rotate it like so

$\mathbf {v} '=R\mathbf {v} R^{-1}$

where all products are the geometric product. The reflection is generated by the vector in the direction of the reflection so perpendicular to the plane of reflection, n say, as follows

$\mathbf {v} ''=-\mathbf {nv'n}$

The rotation and reflection together are then

$\mathbf {v} ''=-\mathbf {n} R\mathbf {v} R^{-1}\mathbf {n}$

The quantity nR, the product of a vector and a rotor, is an odd element of the algebra and it describes the improper rotation. What's more the product of an orthogonal vector and bivector produce a particular result. If n is perpendicular to the plane of rotation then it is orthogonal to B and so to the bivector part of the rotor. It's product with that produces a trivector/pseudoscalar. The other part of the product, with the scalar part of the rotor, produces a vector parallel to n.

More generally if your planes of rotation and reflection are not parallel the products are still composed the same way (though order matters in such a case). As n is a vector and R is a rotor nR is still an odd element of the algebra, with a vector an trivector part.

The resulting improper rotation can be read off this product. The vector part is parallel to the reflection vector, i.e. the normal vector to the reflection/rotation plane. And the angle of rotation in that plane is given by the ratio of the trivector and vector parts. If there's no vector part then you have the central inversion which has any plane of rotation/reflection. If there's no trivector part then it's a pure reflection with rotation angle zero.--JohnBlackburne^words_deeds 13:17, 17 November 2011 (UTC)

shortest connections graph

working in n dimensions, and with y nodes, i'm trying to find a sequence that will tell you how many graphs are possible if its built according to these rules.

1)each node only connects to the closest node/s 2)it is possible to go from each node to any other node following the connectors i.e. the nodes are all connected to each other 3) a graph is considored distinct if nodes connect to different nodes, but not if it's of a different shape, or has a longer line somewhere.

in 1 dimension, only 1 graph is possible. in 2, then for each extra node it goes 1,1,2,5,12... (i think) and in three 1,1,2,6,17... (again, i think). in 4, instead of 17, its 18. is this possible? has it been done before?109.148.193.179 (talk) 21:04, 16 November 2011 (UTC)

Rule 1 needs specification. If each node can ONLY connect to the closest nodes, then assume I have three nodes, A, B, and C. A and B are 1 unit of distance from each other. C is 2 units from B and 3 units from C. So, A and B connect. That is a given. C will want to connect to B, but B cannot connect to C. So, how is this handled? Is B forced to connect to C since B is closest to C? -- kainaw™ 21:11, 16 November 2011 (U

they do connect

OK, so you are thinking of some kind of recursive scheme, perhaps. Another question: You write "working in n dimensions, and with n nodes,...". Is n=n here? In other words, is the number of nodes equal to the dimension? JoergenB (talk) 19:36, 17 November 2011 (UTC)

no. I've cfhanged it now.

Derivative of the Pythagorean theorem with respect to time

What does ${\operatorname {d} b \over \operatorname {d} t}$ equal when $b={\sqrt {c^{2}-a^{2}}}$, assuming that $c$ and $a$ are functions of time and ${\operatorname {d} c \over \operatorname {d} t}=1$ and ${\operatorname {d} a \over \operatorname {d} t}=3\times {\operatorname {d} b \over \operatorname {d} t}$? Do the variables $c$ and $a$ need to be known? I've tried the chain rule, but I cannot determine how to differentiate $c^{2}-a^{2}$. --Melab±1 ☎ 22:43, 16 November 2011 (UTC)

Our article, total derivative, explains how to compute this.

(ec) By specifying that a is also a function of b, you have described a nonlinear partial differential equation. Nimur (talk) 22:56, 16 November 2011 (UTC)

That doesn't sound good. --Melab±1 ☎ 22:58, 16 November 2011 (UTC)

$a$ is a function of time. --Melab±1 ☎ 23:03, 16 November 2011 (UTC)

"but I cannot determine how to differentiate $c^{2}-a^{2}$" :
I don't understand; ${\operatorname {d} (c^{2}-a^{2}) \over \operatorname {d} t}$ is just ${\operatorname {d} (c^{2}) \over \operatorname {d} t}-{\operatorname {d} (a^{2}) \over \operatorname {d} t}$. The chain rule should work, just keep drilling down and substituting. Caveat: I'm an engineer, not a mathematician. There's a non-zero chance I'm wrong. --Floquenbeam (talk) 23:28, 16 November 2011 (UTC)

It equals one over the square root of ten. Integrate your given identities with respect to t, to find c in terms of t, and a in terms of b. Substitute the answers into the equation, rearange, and differenciate. Plasmic Physics (talk) 00:34, 17 November 2011 (UTC)

But, ${\operatorname {d} (a) \over \operatorname {d} t}$ uses ${\operatorname {d} (b) \over \operatorname {d} t}$ in its definition, which in turn has ${\operatorname {d} (a) \over \operatorname {d} t}$ in its definition. How does it work out like that? Also, $a=4$ and $b=3$ when I am looking for ${\operatorname {d} (a) \over \operatorname {d} t}$. --Melab±1 ☎ 01:38, 17 November 2011 (UTC)

If the derivative of c with respect to t is 1, then its integral is c = (t + C) , where C is the constant of integration. If the derivative of a with respect to t is thrice the derivative of b with respect to t, then its integral is a = 3b. Now substitute these two values into the original equation and rearange and solve for b, then find the derivative. Plasmic Physics (talk) 02:16, 17 November 2011 (UTC)

We have not gotten into integrals yet. --Melab±1 ☎ 02:44, 17 November 2011 (UTC)

I figured it out. ${\operatorname {d} b \over \operatorname {d} t}={1 \over 3}$ --Melab±1 ☎ 03:51, 17 November 2011 (UTC)

Wait, what? How? Plasmic Physics (talk) 04:05, 17 November 2011 (UTC)

${\operatorname {d} b \over \operatorname {d} t}={1 \over {\sqrt {10}}}$ Plasmic Physics (talk) 04:19, 17 November 2011 (UTC)

You're ignoring a constant of integration when you get that answer: a = 3b + D. Solving with this will get a non-constant derivative. D=1, t = 1 will get 1/3, which might how that answer was arrived at.--121.74.125.249 (talk) 05:32, 17 November 2011 (UTC)

There, the constant of integration is implicit. ${\operatorname {d} b \over \operatorname {d} t}$ integrated equals ${b}$, not ${b+D}$. Plasmic Physics (talk) 07:39, 17 November 2011 (UTC)

That is one solution, but not all. You're assuming that ${\operatorname {d} a \over \operatorname {d} t}=3\times {\operatorname {d} b \over \operatorname {d} t}$ implies a = 3b, but that's not necessarily true. If you let c = t and a = 3b+1, you can solve for b as a function of time and get a different solution which satisfies everything given.--121.74.125.249 (talk) 07:56, 17 November 2011 (UTC)

(edit conflict)Let ${y=ax+b}$, then ${\operatorname {d} y \over \operatorname {d} x}={a}$. The antiderivative of ${\operatorname {d} y \over \operatorname {d} x}$ does not equal ${y+c}$, because ${(ax+b)+c}$ does not equal to ${ax+b}$ unless ${c=0}$. Therefore, the antiderivative of ${\operatorname {d} y \over \operatorname {d} x}$ equals ${y}$. Plasmic Physics (talk) 07:58, 17 November 2011 (UTC)

Differentiation is a many-to-one operator, and so you cannot speak of "the antiderivative", at least not as a unique function. Anti-differentiation takes a function to a family of functions, not a single one. Try repeating the above argument starting with the equation $y+c=ax+b$.--121.74.125.249 (talk) 08:18, 17 November 2011 (UTC)

Let ${y+c=ax+b}$, then ${\operatorname {d} (y+c) \over \operatorname {d} x}={a}$. The antiderivative of ${\operatorname {d} (y+c) \over \operatorname {d} x}$ does not equal ${y+c+d}$, because ${(ax+b)+d}$ does not equal to ${ax+b}$ unless ${d=0}$. Therefore, the antiderivative of ${\operatorname {d} (y+c) \over \operatorname {d} x}$ equals ${y+c}$. Plasmic Physics (talk) 08:36, 17 November 2011 (UTC)

By definition, the antiderivative of a derivative cannot equal anything other than the original function. When the original function is known, the constant of integration is made to equal zero, for the answer to be true. Otherwise an false equality is produced, whereupon ${y=y+c}$, where ${c}$ is non-zero. Plasmic Physics (talk) 08:49, 17 November 2011 (UTC)

And yet simultaneously, if $y+c=ax+b$, then ${\operatorname {d} y \over \operatorname {d} x}=a$. So since they have the same derivative, $y+c=y$?

Again, you mistakenly believe the antiderivative of a function to be a unique function. The antiderivative of a function is a family of functions. Once again, a = 3b+1 can give you consistent solutions to every requirement in the original problem.--121.74.125.249 (talk) 09:07, 17 November 2011 (UTC)

Does it not seem strange to you that your logic indicates that ${y=y+c}$? It's like saying ${2=1}$, in non-mathematical terms: I'm the only person in the room, and you are also the only person in the same room. You can believe me when I say that that the antiderivative of a derivative of a function, is equal to the function and only the function, I didn't get a B+ in university-level Calculus for nothing. Plasmic Physics (talk) 09:20, 17 November 2011 (UTC)

Try this, you are a forensic detective, you find a foot print and you want to describe the person who made the foot print. All you can say is that the foot print was made by a person with a specific size foot, but the person can be any height.

Next, you find a foot print and you're told that a person who is 1.84 m tall made the foot print. Now, you can determine that the foot print was made by a person with a specific size foot, and a specific height. You can no longer say that the person can be any height, because you already know that the person is 1.84 m tall. Only a person who is 1.84 m tall could have made that particular foot print. Plasmic Physics (talk) 09:43, 17 November 2011 (UTC)

Likewise, you are given ${b}$. Now you can determine that the derivative is of a function with a rate of change and a specific constant. You can no longer say that the function can have any constant, because you already know that the constant is ${b}$. Plasmic Physics (talk) 09:48, 17 November 2011 (UTC)

You're still wrong. Watch this function: f(t) = 2×t + 4 has a derivative k(t) = f' (t) = 2. You say that having k(t) alone you know its antiderivative is f(t) = 2×t + 4. 'Because you already know the constant is 4.'
However, the function g(t) = 2×t + 7 has exactly the same derivative k(t) while having different constant term. Now what is the antiderivative of k(t) — is it f(t) or g(t)...? --CiaPan (talk) 07:22, 18 November 2011 (UTC)

(ec)The equations are P_i = 0 where the polynomials P_i are

P₁ = a²+b²−c²

P₂ = dc−dt

P₃ = da−3db

differentiate:

P₄ = dP₁/2 = ada+bdb−cdc

eliminate da:

P₅ = P₄−aP₃ = 3adb+bdb−cdc

eliminate a:

P₆ = (3adb−(bdb−cdc))P₅ = 9a²db²−(bdb−cdc)² = 9a²db²−b²db²−c²dc²+2bcdbdc

P₇ = 9db²P₁ = 9db²a²+9db²b²−9db²c²

P₈ = P₇−P₆ = 9db²b²−9db²c²+b²db²+c²dc²−2bcdbdc = (10b²−9c²)db²−2bcdbdc+c²dc²

The equation P₈ = 0 is a nonlinear differential equation

$(10b^{2}-9c^{2})\left({\frac {db}{dc}}\right)^{2}-2bc{\frac {db}{dc}}+c^{2}=0$

or, setting c=t:

${\frac {db}{dt}}={\frac {bt}{10b^{2}-9t^{2}}}\pm {\sqrt {\left({\frac {bt}{10b^{2}-9t^{2}}}\right)^{2}-{\frac {t^{2}}{10b^{2}-9t^{2}}}}}$

Bo Jacoby (talk) 10:30, 17 November 2011 (UTC).

You failed to eliminate ${b}$ from the equation. Plasmic Physics (talk) 10:32, 17 November 2011 (UTC)

I don't think you can eliminate $b$ without reintroducing $a$ . Not every differential equation can be solved by elementary integration. Bo Jacoby (talk) 10:56, 17 November 2011 (UTC).

Since $a=\int {\frac {da}{dt}}\,dt\!$

$a=\int 3{\frac {db}{dt}}\,dt\!=3b$

Therefore $b={\sqrt {c^{2}-(3b)^{2}}}={\frac {c}{\sqrt {10}}}$

Therefore ${\frac {db}{dt}}={\frac {1}{\sqrt {10}}}{\frac {dc}{dt}}={\frac {1}{\sqrt {10}}}$

Simple, and true. Plasmic Physics (talk) 11:28, 17 November 2011 (UTC)

Nice! c²=a²+b²=(3b)²+b²=10b². But a=3b+K is also a possibility for K≠0. Then c²=a²+b²=(3b+K)²+b²=10b²+K²+6Kb. Bo Jacoby (talk) 13:40, 17 November 2011 (UTC).

You are making the same mistake as 121.74.125.249 above, when integrating the derivative of a known function, the constant of integration must always equal to zero for the integral to equal the known function, otherwise an stuation is created where $y+k=y$ which is false for all $y$. Let $y=1$, and $k=1$, then $2=1$. See the point? Only when the original function is unknown can we introduce $k$. Plasmic Physics (talk) 20:35, 17 November 2011 (UTC)

This is still wrong. Suppose I have two functions, $f(x)=x$ and $g(x)=3x+1$. I don't tell you what the functions are, but I do tell you that ${\operatorname {d} g \over \operatorname {d} x}=3\times {\operatorname {d} f \over \operatorname {d} x}$. I have not lied to you. If you conclude that $g(x)=3f(x)$, you're wrong.--130.195.2.100 (talk) 00:57, 18 November 2011 (UTC)

Well, the problem was that the sides of a triangle increased in such a way that ${\operatorname {d} c \over \operatorname {d} t}=1$ and ${\operatorname {d} a \over \operatorname {d} t}=3\times {\operatorname {d} a \over \operatorname {d} t}$. I used related rates. I'll post my solution in a bit. --Melab±1 ☎ 16:10, 17 November 2011 (UTC)

$a^{2}+b^{2}=c^{2}$, so differentiating implicitly, $2a{\frac {da}{dt}}+2b{\frac {db}{dt}}=2c{\frac {dc}{dt}}$.

At the desired value of t, ${\frac {dc}{dt}}=1$ and ${\frac {da}{dt}}=3{\frac {db}{dt}}$ so

${\frac {db}{dt}}(6a+2b)=2c$

Also at the desired value of t, $a=4,b=3$ so $c=5$ because $a^{2}+b^{2}=c^{2}$.

${\frac {db}{dt}}(24+6)=10$

${\frac {db}{dt}}={\frac {1}{3}}$

No integrating required. Readro (talk) 16:26, 17 November 2011 (UTC)

My solution works like this:

$c={\sqrt {a^{2}+b^{2}}}=(a^{2}+b^{2})^{1 \over 2}$

${\operatorname {d} c \over \operatorname {d} t}={1 \over 2}\times (a^{2}+b^{2})^{-{1 \over 2}}\times ([2\times a]\times {\operatorname {d} a \over \operatorname {d} t}+[2\times b]\times {\operatorname {d} b \over \operatorname {d} t})$

${\operatorname {d} c \over \operatorname {d} t}={1 \over 2}\times (a^{2}+b^{2})^{-{1 \over 2}}\times ([2\times a]\times (3\times {\operatorname {d} b \over \operatorname {d} t})+[2\times b]\times {\operatorname {d} b \over \operatorname {d} t})$

$1={1 \over 2}\times (4^{2}+3^{2})^{-{1 \over 2}}\times ([2\times 4]\times (3\times {\operatorname {d} b \over \operatorname {d} t})+[2\times 3]\times {\operatorname {d} b \over \operatorname {d} t})$

$1={1 \over 10}\times (30\times {\operatorname {d} b \over \operatorname {d} t})$

$10=30\times {\operatorname {d} b \over \operatorname {d} t}$

${\operatorname {d} b \over \operatorname {d} t}={1 \over 3}$

--Melab±1 ☎ 21:25, 17 November 2011 (UTC)

Where are you getting $a=4,b=3$? That wasn't in the original description of the problem.--130.195.2.100 (talk) 00:57, 18 November 2011 (UTC)

I know they weren't, but I didn't want to leave all of the work to someone else. --Melab±1 ☎ 02:56, 18 November 2011 (UTC)

That seems correct, but so does mine. How can there be two correct answers? Plasmic Physics (talk) 20:10, 17 November 2011 (UTC)

Yes, that also seems correct. Still, how can there be two correct answers? Plasmic Physics (talk) 22:15, 17 November 2011 (UTC)

Because your answer assumes the constant $C$ in $a=3b+C$ is zero. Apparently there was additional information in the statement of the problem that allows one to conclude that $C=-5$.--130.195.2.100 (talk) 00:57, 18 November 2011 (UTC)

Actually, you are wrong. $a$ does not equal $3b$, instead the derivative of $a$ equals 3 times the derivative of $b$. --Melab±1 ☎ 02:42, 18 November 2011 (UTC)

If ${\frac {da}{dt}}=3{\frac {db}{dt}}$, then $\int {\frac {da}{dt}}\,dt\!=\int 3{\frac {db}{dt}}\,dt\!$.

$\int {\frac {da}{dt}}\,dt\!=a$

$\int 3{\frac {db}{dt}}\,dt\!=3\int {\frac {db}{dt}}\,dt\!=3b$.

Therefore $a=3b$, showing that I am indead correct. In short, the $a=3b$, because $\int {\frac {da}{dt}}\,dt\!=\int 3{\frac {db}{dt}}\,dt\!$. Plasmic Physics (talk) 06:52, 18 November 2011 (UTC)

C must be zero for $b=b+C$ to be true. Note that C ≠ c, one is a constant of integration, the other is a function of $t$. Plasmic Physics (talk) 01:14, 18 November 2011 (UTC)

I'll use prime for a derivative, We have:
1. c' = 1
2. a' = 3×b'
3. a² + b² = c².

From 1. we have
4. c = t + K (for some constant K)

From 2.
5. a = 3×b + L (for some constant L)

Now from 3 and 5:
6. 10×b² + 6L×b + L² = (t + K)²

Solve the quadratic equation 6. with respect to b
7. b = [ −3L ± sqrt(10(t + K)² − L²) ] / 10

and finally differentiate b(t).

For L=0, i.e. a=3b, equation 7. simplifies to
b = |t + K| / √10,
so assuming positive t and K
b = (t + K) / √10 and finally b = 1 / √10 — but L=0 is not that obvious to me, so the general answer would look much more ugly...
CiaPan (talk) 07:10, 18 November 2011 (UTC)

The only case when $L$ would equal a non-zero number, would be if $\int {\frac {da}{dt}}\,d{\frac {db}{dt}}\!$. Plasmic Physics (talk) 07:44, 18 November 2011 (UTC)

Plasmic Physics, you are mistaken. Consider the case a=4, b=3, c=5. Then L=a-3b=-5. This triangle can in dt seconds expand to c=5+dt, a=4+da=4+3db, b=3+db, satisfying 0=c²-a²-b²=25+10dt-16-24db-9-6db=10dt-30db, so in this special case db/dt=1/3. Bo Jacoby (talk) 08:16, 18 November 2011 (UTC).

If that is the case then you are suggesting that $b=b-5$. Go ahead and solve for $b$. Plasmic Physics (talk) 08:33, 18 November 2011 (UTC)

No, I am merely suggesting that if a=3b-5 then da/dt=3db/dt. Bo Jacoby (talk) 09:48, 18aNovember 2011 (UTC).

General solution is

$a={\frac {3t}{\sqrt {10}}}+3k$

$b={\frac {t}{\sqrt {10}}}+k$

$c=t+{\sqrt {10}}k$

where k is an arbitrary constant representing the value of b when t = 0. Gandalf61 (talk) 12:11, 18 November 2011 (UTC)

That is a solution, but not the general solution, because it always statisfies the condition that a=3b, which is not generally true for right-angled triangles. Bo Jacoby (talk) 14:16, 18 November 2011 (UTC).

I didn't start by assuming a=3b - this just fell out of the solution. I assumed that c = t + c0 for some constant c0, and then that a and b were polynomial functions of t. It is fairly easy to see that a and b must in fact be linear functions of t, so a = 3rt + a0 and b = rt + b0. Equating coefficients of t^2 in c^2 = a^2 + b^2 then gives 10r^2 = 1 and hence the value of r. Equating the coefficients of t gives an expression for c0 in terms of a0 and b0. Using c0^2 = a0^2 + b0^2 (which we know is true by setting t to 0) to eliminate c0 gives a quadratic in a0/b0. It just so happens that this quadratic has two equal roots, which are both 3, so a0 = 3b0, and so a = 3b for all values of t.

If you think there is a more general solution, can you show us what it is ? Gandalf61 (talk) 14:56, 18 November 2011 (UTC)

You have shown that the solution for a and b is not polynomial functions unless a=3b. The differential equation above can perhaps not be solved by elementary functions. Bo Jacoby (talk) 17:16, 18 November 2011 (UTC).

Even if we drop the assumption that a and b are polynomial functions of t, we can still show that a = 3b. We know that c = t + c0 because dc/dt=1. And because da/dt = 3 db/dt for all t, we know that a(t) = 3b(t) + k for some constant k; this is true whatever type of functions a(t) and b(t) may be. We have

$c(t)^{2}=a(t)^{2}+b(t)^{2}$

for all t, so we must have

$c(-c_{0})^{2}=a(-c_{0})^{2}+b(-c_{0})^{2}$

but c(-c0) = 0, so a(-c0) = b(-c0) = 0. But a(-c0) = 3b(-c0) + k, therefore k = 0, therefore a(t) = 3b(t) for all t. Gandalf61 (talk) 20:50, 18 November 2011 (UTC)

If you assume these are intended to hold for all t. Often questions like this are only for certain t, for example non-negative t. In particular, since the OP was told to investigate when b = 3 and a = 4, that must be the case.--121.74.125.249 (talk) 21:37, 18 November 2011 (UTC)

As has been shown, if we assume a and b are polynomial functions of t for t > 0 then a = 3b and there is no solution with b=3 and a=4 at time 0. This is because the solution set has only one degree of freedom - you can specify a value for a or b at time 0 (or, indeed, for c) but not for both a and b at the same time. If we go beyond polynomial functions then we are into the very speculative territory of non-elementary functions that have to satisfy the functional equation given by Bo below. I very much doubt that a student who does not know how to calculate a total derivative (which is where we came in) would be expected to venture into such uncharted territory. It is much more likely that they have misunderstood or mis-stated the problem. Gandalf61 (talk) 10:21, 19 November 2011 (UTC)

Related rates problems are a standard feature of a calculus sequence; indeed, that's how the OP went on to solve it. There's no need to determine the functions to solve such a problem.--121.74.125.249 (talk) 11:02, 19 November 2011 (UTC)

The OP appears to believe that db/dt = 1/3, which is definitely incorrect. If db/dt is constant then both a and b are linear functions of t, and it is immediately clear by differentiating c^2 = a^2 + b^2 that 10(db/dt)^2 = 1 and so db/dt = 1/sqrt(10). Deriving a, b and c as explicit functions of t in this case shows that the solution set only has one degree of freedom, and there is no solution in which b=3 and a=4 at the same time. If db/dt is not constant then we are into Bo's territory of non-elementary functions that satisfy a quadratic functional equation. Gandalf61 (talk) 09:13, 20 November 2011 (UTC)

The unknown sides of the triangle a,b,c satisfy a²+b²=c², and at time t=0 they have the known values a₀,b₀,c₀, satisfying a₀²+b₀²=c₀². Set c=c₀+t satisfying dc=dt, and set a=a₀+3x and b=b₀+x satisfying da=3db. We have (a₀+3x)²+(b₀+x)²=(c₀+t)². Or 10x²+(6a₀+2b₀)x=2c₀t+t². Solve this second degree equation and substitute the solution x into the expressions for a and b. Bo Jacoby (talk) 00:17, 19 November 2011 (UTC). For 0<t<<c₀ the square terms are neglegible and (6a₀+2b₀)x=2c₀t, such that db/dt=dx/dt=2c₀/(6a₀+2b₀). For c₀<<t the linear terms are neglegible and 10x²=t², such that db/dt=dx/dt=1/sqrt(10). Bo Jacoby (talk) 08:19, 19 November 2011 (UTC).

November 17

Probability Question

Hi I'm preparing for a stats test and I'm stumped on a problem I found in my text (paraphrased below):

2. There are n children in a classroom and each child has exactly one toy (so there are n toys in total). They break for recess, leaving their toys in the classroom. When they come back to the classroom, each child picks a toy by random. a) What the expected number of children who pick the toy they had before leaving for recess? b) What's the probability no child picks the toy they had initially? — Preceding unsigned comment added by Rain titan (talk • contribs) 10:14, 17 November 2011 (UTC)

See Derangement. Bo Jacoby (talk) 10:39, 17 November 2011 (UTC).

The wording of the question is not very clear: what does it mean for the children to pick a toy at random? Do they each independently name the one they want (so that a toy could be chosen by more than one child), or do they choose randomly in sequence from the toys that are left after those before them in the queue have chosen? (The reality is of course likely to be more disorganised and tearful than either of these.) AndrewWTaylor (talk) 12:22, 17 November 2011 (UTC)

Agreed, bring cookies to help distract the children.Naraht (talk) 16:53, 17 November 2011 (UTC)

True. However, this wording or something rather similar is standard in text books introducing derangements. JoergenB (talk) 19:30, 17 November 2011 (UTC)

Sorry for the late reply. Each child chooses a top left after those before them in the queue have chosen (so each child will have exactly one toy). Thanks! — Preceding unsigned comment added by Rain titan (talk • contribs) 04:38, 18 November 2011 (UTC)

See Derangement. --COVIZAPIBETEFOKY (talk) 14:14, 18 November 2011 (UTC)

Inscribed circle question

Take a semi-circle whose flat side has a length of 4 units. The largest circle which can be inscribed in the semicircle touches the middle of the arc and the middle of the flat side and has radius 1 (diameter 2). What is the radius of the inscribed circle in each of the two shark fin shaped pieces between the Semi-circle and the inscribed diameter 1 circle?Naraht (talk) 16:51, 17 November 2011 (UTC)

See Descartes' theorem, especially the section on special cases.--RDBury (talk) 21:11, 17 November 2011 (UTC)

Ignore that, it doesn't apply in this case.--RDBury (talk) 21:21, 17 November 2011 (UTC)

This is an example of the Problem of Apollonius. There may be a theorem that covers this case nicely but if there is I can't find it at the moment. In any case, the problem can be solved with Inversive geometry. Apply the inversion with respect to the semicircle, it and the diameter remain fixed and the smaller circle becomes a line parallel to the diameter and tangent to the semicircle. You must find a circle tangent to the lines, so radius 1, and tangent to the semicircle, so the distance from the center to the center of the semicircle is 3. Inverting again to get the solution to the original problem, the desired circle has a diameter whose endpoints are at distance 1 and 2 from the center of the semicircle and when extended it passes through that point. So the radius of the desired circle is 1/2.--RDBury (talk) 21:57, 17 November 2011 (UTC)

Put the drawing into Cartesian coordinates, so the diameter is in the X axis and its middle point is (0,0). Let x, y, r be the coordinates of the centre and the radius of a circle S wee seek. S is tangent to two given circles and the X axis, so x, y, r must satisfy
$\left\{{\begin{array}{ccc}x^{2}+y^{2}&=&(2-r)^{2}\\x^{2}+(y-1)^{2}&=&(1+r)^{2}\\y&=&r\end{array}}\right.$
This results (if I did my calculations right) in solutions: $(x,y,z)=(\pm {\sqrt {2}},1/2,1/2)$. Replacing the third equation with $|y|=r$, which is more general expression for 'circle being tangent to the X axis' results in thhe third solution $(0,-1,1)$, which is, of course, outside the semicircle. CiaPan (talk) 06:59, 21 November 2011 (UTC)

November 18

Probability and boolean events

Suppose you want to buy a toy for your daughter and you want to make sure she will like it. To make sure she'll like it, you decide to ask her friends (for this problem you can imagine she has infinity friends). Each friend, independently, can tell you whether she will like it or not correctly with probability 2/3 (each friend gives a boolean answer though).

How can you find out your daughter will like the toy you are about to buy with probability >= 1 - 2^(1/n) where n is the number of friends you will ask? — Preceding unsigned comment added by Rain titan (talk • contribs) 04:39, 18 November 2011 (UTC)

First, I think you meant 1 - (1/2)^n. 1 - 2^(1/n) is always negative. Assuming you did, this is called probability amplification. Short answer: choose k bigger than $n/\log _{2}(9/8)$, and ask 2k friends. If more than half say yes, assume she likes it. Otherwise, assume she doesn't.

Justification: if you made the wrong decision, then at most half her friends gave the right answer. The odds of this happening is $\sum _{i=0}^{k}{2k \choose i}(2/3)^{i}(1/3)^{2k-i}\leq \sum _{i=0}^{k}{2k \choose i}(2/3)^{k}(1/3)^{k}\leq (2/9)^{k}\sum _{i=0}^{k}{2k \choose i}\leq (2/9)^{k}2^{2k}=(2/9)^{k}4^{k}=(8/9)^{k}$.

So you need $(8/9)^{k}<2^{-n}$ So $k\log _{2}(8/9)<-n$, and thus $k>n/\log _{2}(9/8)$.--121.74.125.249 (talk) 05:42, 18 November 2011 (UTC)

Just noticed you specified that you can only ask n friends. In that case, it's not possible. The best you can possibly do is majority vote, and your certainty doesn't increase that fast. For example, with three friends, you only have probability (20/27) of making the right choice.--121.74.125.249 (talk) 05:48, 18 November 2011 (UTC)

This is a frequentist solution. For a Bayesian solution You need to know the prior probability. If A = "Your daughter likes it" and D = "m out of n friends asked says she likes it", then

$\mathrm {Pr} (A|D)={\frac {\mathrm {Pr} (A)\mathrm {Pr} (D|A)}{\mathrm {Pr} (D)}}={\frac {\mathrm {Pr} (A){\binom {n}{m}}(2/3)^{m}(1/3)^{n-m}}{\mathrm {Pr} (A){\binom {n}{m}}(2/3)^{m}(1/3)^{n-m}+(1-\mathrm {Pr} (A)){\binom {n}{m}}(1/3)^{m}(2/3)^{n-m}}}={\frac {1}{1+{\frac {1-\mathrm {Pr} (A)}{\mathrm {Pr} (A)}}2^{n-2m}}}$

If we assume for simplicity that $\mathrm {Pr} (A)=1/2$ and that m is high, then this is roughly $1-2^{n-2m}$. So only in the highly unlikely case that everyone says yes you'll have $1-2^{-n}$ confidence.

This could have been solved more simply by considering odds ratios. Every friend who says yes multiplies the odds ratio by 2, every one who says no divides it by 2. -- Meni Rosenfeld (talk) 08:31, 18 November 2011 (UTC)

Euler's Identity Question

Just a quick question, something I don't understand about Euler's identity.

Why does

$e^{i\pi }=-1\!$

when

$e^{i\pi }=e^{(\pi \times i)}=e^{3.1415i}=23.1406926{i}\!$?

174.93.63.116 (talk) 15:53, 18 November 2011 (UTC)

It comes from the identity $e^{i\theta }=\cos(\theta )+i\sin(\theta )$. Plug in $\theta =\pi$ and you get the desired result. Readro (talk) 16:06, 18 November 2011 (UTC)

(edit conflict) How do you get to 23.1406926i ? By playing fast and loose with re-arrangements of infinite series we can informally "prove" Euler's identity as follows:

$e^{i\pi }=1+(i\pi )+{\frac {(i\pi )^{2}}{2!}}+{\frac {(i\pi )^{3}}{3!}}+{\frac {(i\pi )^{4}}{4!}}+{\frac {(i\pi )^{5}}{5!}}\dots$

$=(1-{\frac {\pi ^{2}}{2!}}+{\frac {\pi ^{4}}{4!}}\dots )+i(\pi -{\frac {\pi ^{3}}{3!}}+{\frac {\pi ^{5}}{5!}}\dots )$

$=\cos(\pi )+i\sin(\pi )=-1$

Gandalf61 (talk) 16:07, 18 November 2011 (UTC)

$e^{i\pi }$ ≈ ${23.1406926}^{i}$, not $23.1406926{i}$. At this point in the calculation there is no reason to believe that the right-hand side is not equal to -1. If you want to go this route you have to keep working on the right-hand side until it has the form $a+bi$. 98.248.42.252 (talk) 16:24, 18 November 2011 (UTC)

The error is that $e^{3.1415i}=23.1406926^{i}$, not $23.1406926{i}$. Looie496 (talk) 16:28, 18 November 2011 (UTC)

Probability of certain factors existing

Hi, is there any way of estimating the probability that a randomly chosen n-digit number will have at least one factor whose length is between p and q digits? Assume n is sufficiently large. 81.159.104.115 (talk) 21:00, 18 November 2011 (UTC)

The expected number of factors between $10^{p}$ and $10^{q}$ doesn't depend on $n$ (as long as $n$ is sufficiently large) and is

$\sum _{k=10^{p}}^{10^{q}}1/k$ ≈ $\ln(10^{q})-\ln(10^{p})$ = $(q-p)\ln(10)$.

The expected number of prime factors between $10^{p}$ and $10^{q}$ is

$\sum _{k=10^{p}}^{10^{q}}1/(k\ln k)$ ≈ $\ln(\ln(10^{q}))-\ln(\ln(10^{p}))$ = $\ln(q/p)$.

You asked for a probability, not the expected value. Assuming that $n$ is sufficiently large, the number of prime factors between $10^{p}$ and $10^{q}$ should follow a Poisson distribution, so the probability of having at least one prime factor between $10^{p}$ and $10^{q}$ is $1-e^{-\ln(q/p)}$ = $1-p/q$. For the probability with composite factors allowed, I don't know. The number of factors doesn't follow a Poisson distribution because the events are not very independent. I don't know what distribution it follows so I'll let someone else take it from here. 98.248.42.252 (talk) 07:03, 19 November 2011 (UTC)

Power series for arcsine

What's a relatively straightforward way to find a power series for arcsin(x)? I'm stumped! — Trevor K. — 21:49, 18 November 2011 (UTC) — Preceding unsigned comment added by Yakeyglee (talk • contribs)

This may be a bit dodgy, but how about:

$\arcsin(x)=\int _{0}^{x}{\frac {dy}{\sqrt {1-y^{2}}}}$

I'm not sure what happens when $x=\pm 1$, as arsin(x) is defined but this integral isn't? Given that |x|<1 we can use the Maclaurin series for the square root (from Taylor_series:

$(1+x)^{-1/2}=\sum _{n=0}^{\infty }{-{\frac {1}{2}} \choose n}x^{n}{\text{ for }}|x|\leq 1$

$\arcsin(x)=\int _{0}^{x}dy\sum _{n=0}^{\infty }{-{\frac {1}{2}} \choose n}(-y^{2})^{n}$

$\arcsin(x)=\sum _{n=0}^{\infty }{-{\frac {1}{2}} \choose n}(-1)^{2n}{\frac {1}{2n+1}}x^{2n+1}$

(I think) Not very straightforward though... 77.86.108.27 (talk) 22:33, 18 November 2011 (UTC)

Impressive! I like it! It's reminiscent of the polynomial expansion of the Lorentz Factor from special relativity! I am satisfied! :P — Trevor K. — 22:45, 18 November 2011 (UTC) — Preceding unsigned comment added by Yakeyglee (talk • contribs)

The integral exists for $x=\pm 1$, see Improper integral. -- Meni Rosenfeld (talk) 16:55, 19 November 2011 (UTC)

The Lagrange inversion theorem is relevant. -- Meni Rosenfeld (talk) 16:55, 19 November 2011 (UTC)

November 19

Differentiability

Define

$f(x):=\left\{{\begin{array}{ll}\cos \left({\frac {1}{x}}\right)&x\neq 0\\0&x=0\end{array}}\right.$

and

$F(x):=\int _{0}^{x}f(y)\,\operatorname {d} \!y.$

Is F differentiable at 0? My gut says no, but unsure of how to establish it. Sorry for the sloppy TeX. 1.41.7.35 (talk) 13:32, 19 November 2011 (UTC)

(I've fixed the TeX for you — Fly by Night (talk) 19:45, 19 November 2011 (UTC))

F is in fact differentiable, and $F'(0)=0$. Currently I know this only by numerical observation, but to show this formally you'll probably want to use the definition of derivative and the substitution $u=1/x$ to make the integral more manageable. -- Meni Rosenfeld (talk) 16:48, 19 November 2011 (UTC)

I tried integrating using Maple and got

$F(x)=x\cos({\frac {1}{x}})+\int _{0}^{1/x}{\frac {\sin(t)}{t}}dt+c$

The second integral is the Sine integral (see Trigonometric integral) Si(1/x) and

$c=-\int _{0}^{\infty }{\frac {\sin(t)}{t}}dt.$

Since F(0)=0 we have:

$F^{\prime }(0)=\lim _{h\to 0}{\frac {F(h)-F(0)}{h}}$

$F^{\prime }(0)=\lim _{h\to 0}{\frac {F(h)}{h}}$

$F^{\prime }(0)=\lim _{h\to 0}\cos({\frac {1}{h}})-{\frac {1}{h}}\int _{1/h}^{\infty }{\frac {\sin(t)}{t}}dt$

Not sure this helps at all... 77.86.108.27 (talk) 19:06, 19 November 2011 (UTC)

Note that if we break f into "lobes", F is an alternating series. Also, the norm of the sum of an alternating series is at most the norm of the first term. So for ${\frac {2}{\pi (2n+3)}}\leq x<{\frac {2}{\pi (2n+1)}}$, we know $|F(x)|\leq \left|\int _{\frac {2}{\pi (2n+3)}}^{\frac {2}{\pi (2n+1)}}\cos(1/t)\operatorname {d} t\right|<{\frac {2}{\pi (2n+1)}}-{\frac {2}{\pi (2n+3)}}\sim {\frac {1}{\pi n^{2}}}\sim {\frac {\pi x^{2}}{4}}$. Using this, it's easy to see that the difference quotient goes to 0.--121.74.125.249 (talk) 21:27, 19 November 2011 (UTC)

I think I misunderstood the function definition when I commented earlier (I didn't appreciate that f(0) was 'defined' to be zero. Still if you plot

$\int _{0}^{x}\cos(1/t)dt$

it is discontinuous at the origin: approaching from 0+ the function is near the origin, whereas from 0- it's near $-\pi$.

${\text{Let }}G(x)=\int _{0}^{x}\cos(1/t)dt$

$G(x)-G(-x)=\int _{-x}^{x}\cos(1/t)dt$

$G(x)-G(-x)=2x\cos(1/x)+\int _{-1/x}^{1/x}{\frac {\sin(t)}{t}}dt$

So $G(0)-G(-0)=\int _{-\infty }^{\infty }{\frac {\sin(t)}{t}}dt=Pi$ — Preceding unsigned comment added by 77.86.108.27 (talk) 10:15, 20 November 2011 (UTC)

Something is wrong with your plotting. Cosine is even, so F(x) is odd.--121.74.125.249 (talk) 10:34, 20 November 2011 (UTC)

Obviously $\lim _{x\to 0^{-}}F(x)=0$. The mistake (which Mathematica did for me, and probably Maple did for you) is that when you try to express the function with a sine integral, the constant is different for positive or negative x, and the generated expression assumes positive x. Anyway, since the function was defined only for nonnegative x it is understood that we are after the right-derivative, though the derivative is still defined and 0 if you consider both positive and negative x.

Ah okay, so F(x) is (implicitly) defined only for non-negative x, because for negative x we'd want:

$F(x)=\int _{x}^{0}f(y)dy$ 77.86.108.27 (talk) 13:43, 20 November 2011 (UTC)

Actually, I was in error - I meant that f was only defined for $x\geq 0$, but looking again it was actually defined for all x. So F is defined for all x too, because by convention $\int _{a}^{b}f(t)\ dt=-\int _{b}^{a}f(t)\ dt$ for $b<a$.

The point is valid, though - in this light it is still clear that $\lim _{x\to 0}F(x)=0$. -- Meni Rosenfeld (talk) 18:47, 20 November 2011 (UTC)

Also, the value of f at 0 is completely irrelevant to the question. -- Meni Rosenfeld (talk) 11:04, 20 November 2011 (UTC)

Self-similarity

The logarithmic spiral appears often in nature, as the article can attest to. This spiral has the property of self-similarity, which, I'm guessing, plays a role in why it emerges often in nature. Is there a way of "seeing" why nature would tend to be self-similar? 74.15.139.235 (talk) 22:02, 19 November 2011 (UTC)

That is a very good question and this answer is not final. You should look at it the other way round. There is no need for a reason for something to be self-similar, but there must be a reason for it not being self-similar. In that case it must have some characteristic length. (size, diameter, period ...). If the mechanism for creating or developing the item does not define a characteristic length, then the item will be self-similar. Bo Jacoby (talk) 23:33, 19 November 2011 (UTC).

Thank you, your answer has given me a lot to think about! 74.15.139.235 (talk) 00:10, 20 November 2011 (UTC)

To expand on that thought a bit, imagine your circulatory system. It has veins, arteries and capillaries of various sizes. If your DNA contained separate code for the design of each segment, you would need far more DNA. The only way it can work, then, is to have one basic design, and alter it the fit the situation. The same is true of many other biological systems. StuRat (talk) 00:21, 20 November 2011 (UTC)

Woah. 74.15.139.235 (talk) 07:01, 22 November 2011 (UTC)

Benoit Mandelbrot wrote extensively on self-similarity in nature. His book The Fractal Geometry of Nature is very interesting, and accessible to non-mathematicians. Logarithmic spirals are comparatively simple, though: they tend to arise in systems where the rate of growth is proportional to the current size. Looie496 (talk) 17:48, 20 November 2011 (UTC)

Thanks, I'll be sure to take a look at it. 74.15.139.235 (talk) 07:01, 22 November 2011 (UTC)

November 20

Extended Schur's theorem

Hello all,

I'm trying to deduce from Ramsey's theorem that whenever $\mathbb {N}$ is finitely coloured there exist x, y, z with $\{x,\,y,\,z,\,x+y,\,y+z,\,x+y+z\}$ monochromatic.

I've seen a proof of Schur's theorem (effectively finding x, y, x+y with $\{x,\,y,\,x+y\}$ mono) using Ramsey's theorem before, and this seems like a similar but adapted version. Could you prove the above (which seems to be a strengthened Schur) using Ramsey's theorem too? I've tried but haven't gotten anywhere, could anyone please give me a hint? Thank you! Frimgandango (talk) 13:55, 20 November 2011 (UTC)

It's certain true; it's called Folkman's theorem. I don't know how the proof goes, though.--121.74.125.249 (talk) 20:57, 20 November 2011 (UTC)

Yes, I've seen Folkman's theorem before. However, I've been told to "deduce it from Ramsey's theorem", presumably rather than using Folkman's theorem which I also covered in my lectures later on; it becomes trivial very quickly using Folkman's theorem of course. Frimgandango (talk) 21:04, 20 November 2011 (UTC)

Addition Problems

What is one add one? I.e. 1+1? This is not a homework question. Thankyou for your time. 94.195.251.61 (talk) 16:30, 20 November 2011 (UTC)

There are 10 kinds of people in the world: those who understand binary and those who don't. Dbfirs 17:08, 20 November 2011 (UTC)

Usually 2, of course. It depends on your definitions of 1 and +. You might be interested in the classic book Principia Mathematica, or the articles on modular arithmetic, group theory, or the binary numeral system. Bobmath (talk) 18:02, 20 November 2011 (UTC)

I've said it before, and I'll say it again - it's -1 in $\mathbb {Z} _{3}$, 0 in $\mathbb {Z} _{2}$, 1 in boolean algebra, 2 in $\mathbb {Z}$, 10 in binary and 11 in Gray code. -- Meni Rosenfeld (talk) 18:42, 20 November 2011 (UTC)

1+1 is a sum. Hope that helps, Qwfp (talk) 19:20, 20 November 2011 (UTC)

This question has been asked on the reference desk before. Please search the archives before asking a new question. Widener (talk) 02:08, 21 November 2011 (UTC)

Schaum's Topology

There is a new edition of Schaum's General Topology by Seymour Lipschutz, but when I look at it on Amazon, it seems to give me only the old version. Has anyone had any experience with the new version, and is it better? I only do this stuff in my spare time as a hobby (and I don't get much time for it) so I'm looking for something that avoids excessive theory. I also have the Schaum's set theory guide, so I don't need more stuff on cardinality. I'm interested in stuff like connectedness and compactness etc., if that helps. Thanks, IBE (talk) 20:05, 20 November 2011 (UTC)

November 21

Probability Zero and One

My friend asked me a rather interesting question and I am curious to think what others here think. Can an event with probability zero ever happen? Can an event with probability ever not happen at all? It is related to the whole discussion of moving away from religion/spirituality/fate/destiny and accepting the role of randomness in our lives (not trying to drag religion here on this desk but just providing context). I have thought about this question but not settled on an answer yet (if there is an objective one). I am thinking of probability zero being zero except on a set of measure zero so an zero-probability event is possible. But if it is possible then the probability isn't zero. HELP! Thank you. - Looking for Wisdom and Insight! (talk) 06:27, 21 November 2011 (UTC)

See Almost never -- Widener (talk) 07:24, 21 November 2011 (UTC)

You say if it is possible the the probability isn't zero. Who sez?

I mean, don't tell me literally who sez. Probably Laplace sez. Probably Fermat sez. But they were working in very simple models — finite atomic Boolean algebras where all the atoms have the same probability. First die comes up 3, second die comes up 2; that's an atom, and it has the same probability as, First die comes up 6, second die comes up 1.

In those cases, it's true that all possible events have nonzero probability. But why should it be true in general? Can you elucidate the intuition that is breaking for you? Or are you just being held back by what was taught in elementary math classes, where the models are like the ones Laplace and Fermat used? --Trovatore (talk) 08:08, 21 November 2011 (UTC)

Under the frequentist definition of probability, a probability is defined as $P(x)=\lim _{n_{t}\rightarrow \infty }{\frac {n_{x}}{n_{t}}}$ where $n_{t}$ is the total number of trials and $n_{x}$ is the number of trials where the event $x$ occurred. It is possible for this limit to be zero even if $n_{x}$ is nonzero for some $n_{t}$. Widener (talk) 08:52, 21 November 2011 (UTC)

Well, honestly, that's not much of a definition (which is partly why I don't think very highly of frequentism). It's only with probability 1 that that limit equals the desired value (or indeed even exists), which makes the whole thing sort of circular. Also it only works if the trials are independent, and in that case, for any finite $n_{t}$, if the probability per trial is zero, then the probability that $n_{x}$ is anything but zero, is itself zero. Sorry for the convoluted sentence. --Trovatore (talk) 09:52, 21 November 2011 (UTC)

If you pick a number between 1 and 2 at random it will start say 1.3, 1/10 of them start that way, then perhaps 1.36, 1/100 start that way. The digits are all random so getting one that starts 1.367842 has a 1/1000000 chance. Going on and on like that to infinity leads to a probability that is less than any positive number you care to mention. A departed quantity in Berkeley's terms. And every single number between 1 and 2 would have that same chance of being picked and yet there is a probability of 1 that one of them will be picked. Measure theory is how people cope with the problem. Dmcq (talk) 11:31, 21 November 2011 (UTC)

So can an event with zero probability ever happen? - Looking for Wisdom and Insight! (talk) 07:27, 22 November 2011 (UTC)

The answer is yes. Widener (talk) 07:31, 22 November 2011 (UTC)

Another answer is no. The event that a random variable assumes a specific real value will never happen, because it takes forever to identify a real number precisely, as it has an infinite number of digits. A random real number is a useful theoretical concept, but it is not a practical concept. Bo Jacoby (talk) 13:41, 22 November 2011 (UTC).

Another answer is indeed "no". However, that answer is incorrect. --Trovatore (talk) 17:48, 22 November 2011 (UTC)

Not if you take a second for the first digit, a half a second for the next digit, a quarter for the next etc. Anyway I can wait. Dmcq (talk) 14:42, 22 November 2011 (UTC)

p.s. you might like Conformal Cyclic Cosmology where Roger Penrose suggests after an infinite time another universe may come into being. So even infinite time may not be all of time. Dmcq (talk) 14:46, 22 November 2011 (UTC)

Does an event, that happens after an infinite time, ever happen? Bo Jacoby (talk) 17:02, 22 November 2011 (UTC).

According to that it may already have happened an infinite number of times. Dmcq (talk) 17:20, 22 November 2011 (UTC)

parabola or cylinder

in my intermediate i learnt that y² = 4ax is a parabola,now i cant digest that the same thing is a parabolic cylinder. please,make me understand with necessary eqations if needed.59.165.108.89 (talk) 12:15, 21 November 2011 (UTC)

I'm not going to force you to understand, I prefer the softly softly approach ;-) If you draw a parabola on the ground and then start building upwards by placing bricks along it you get part of a parabolic cylinder. It is simply that equation in 3D where z can be anything so making a sheet with a cross section that is a parabola. A cylinder is used here as a general term for a 2Dshape drawn out into the third dimension. Dmcq (talk) 13:06, 21 November 2011 (UTC)

(ec) Technically the graph of that thing is a parabola (or parabolic cylinder). The equation itself is just an equation. When you want to figure out what the graph is, first you need to decide what variables are in play, and in particular how many dimensions you want the graph to live in. In this case you can decide that you're only considering 2 variables x and y, in which case you graph it in 2 dimensions and you get a parabola. But you can also imagine that you have 3 variables x, y, and z, where z just happens to be absent from your equation. In that case you graph it in 3 dimensions (one for each variable), and you get the parabolic cylinder. (By the way, these are not the only choices- you can imagine any number of additional variables and get graphs that live in higher dimensions.) Staecker (talk) 13:09, 21 November 2011 (UTC)

so its a hyper parabolic cylinder? or something like that? — Preceding unsigned comment added by 109.156.115.144 (talk) 13:15, 21 November 2011 (UTC) it means in 3d a parabola acquires a cylindrical form like a circle becomes a sphere. then every thing can be studied about a cylinder just including one more variable likewise we do in a circle for studying a sphere.nd thanx 4 reply,i think i got what u said.59.165.108.89 (talk) 12:16, 22 November 2011 (UTC)

November 22

Football (soccer) league problem

A football (soccer) league comprises 10 teams. The league is conducted on a single round-robin basis, which means that each team has to play against every opponent once.

The league uses the current evaluating system (3 pts for a win, 1 pt for a draw, 0 pts for a defeat), and after the completion of all matches Team A finishes first.

If the league were to use the former evaluating system (2 pts for a win, 1 pt for a draw, 0 pts for a defeat) and all results were to be repeated, Team A would finish last.

Identify the results.

It was reported in the news that this problem was given at some kind of International Mathematical Olympiad for high school students held in Nepal a few days ago. I'm very curious to know the answer, but I'm not sure I can figure it myself, so I'm asking you for assistance. --Theurgist (talk) 02:22, 22 November 2011 (UTC)

Can we assume that there is no tie with A in either first or last place ? Otherwise we could just make all games a draw and thus A would be both first and last under both systems. StuRat (talk) 02:35, 22 November 2011 (UTC)

I suppose this can be assumed. The above description is what a contestant retells in brief in this TV reportage (in Bulgarian). I guess the exact wording of the problem might have been published somewhere by the organisers of the Olympiad, but I haven't so far been able to discover it. But I personally find it unlikely that the problem should not have ruled out scenarios like all matches ending as draws and all teams being both first and last under both systems. --Theurgist (talk) 02:53, 22 November 2011 (UTC)

Write X->Y if X defeats Y and let the outcomes be

A->B->C->A

A->D->E->A

A->F->G->A

A->H->I->A

J->A

with every other game being a draw. Then I get for 3-point wins scores of A-12, J-11, others-10 so A is first; and for 2-point wins A-8, J-10, others-9, so A is last. Not sure if it is unique in any sense so I don't think it completely solves the problem, but it does show there are outcomes that satisfy the conditions.--RDBury (talk) 06:53, 22 November 2011 (UTC)

Thanks a lot, this is indeed a solution to the problem. And maybe this is the solution to the problem, as it very nicely makes use of the only essential difference between the two systems - namely that, for two teams playing against each other, the old system distributes a total of 2 points per game in all cases, while the new system distributes 3 points if either team wins and only 2 if the game is a tie. So the final standings under each of the two systems are those:

3-point wins Pos Team W D L Pts 1 Team A 4 0 5 12 2 Team J 1 8 0 11 3-10 all others 1 7 1 10

2-point wins Pos Team W D L Pts 1 Team J 1 8 0 10 2-9 all others 1 7 1 9 10 Team A 4 0 5 8

They say in the video material that all contestants had to solve this problem within 15 minutes, and that one of the boys appearing in the material was the only one who managed to do so.

As demonstrated here, the "three points for a win" system does have its cons - a team that loses more than 50% of its games can still triumph with the title eventually, which is mathematically impossible under the old system. :)

--Theurgist (talk) 02:44, 23 November 2011 (UTC)

Isometry group of the k dimensional torus

Put the cubical metric on the k-torus (i.e. the obvious cover is a k-cube, not some k-cuboid.) It seems intuitively clear that the isometry group of the k-torus should arise from translations in the k directions of the k-cube cover (inducing `rotations' about the various loops in the torus), and the symmetry group of the k-cube. (In fact, the group must be a semi-direct product of these two subgroups, surely?). The trouble is, I can't seem to find this written down anywhere. Is my intuition correct? And where can I find a reference on this? Thanks for your help, Icthyos (talk) 16:40, 22 November 2011 (UTC)

November 23

is this quadratic?

Hi I'm just wondering if you can help me find how to get the answers for this formula.

0=-4000+2000/(1+r)+4000/(1+r)^2

The following is not homework questions specifically but old notes which already say that the answers are:

• -1/4(square root of 17)-3/4
• 1/4(square root of 17)-3/4

My limited knowledge tells me the following equation is quadratic, but I'm at a loss how to do this at all. SQRT(number) means square root the number:

• 0=ax^2+bx+c
• 0=4000/(1+r)^2+2000/(1+r)-4000

Quadratic formula

• 0=-b{+/-}SQRT((b^2-4ac)/2a)
• 0=-2000{+/-}SQRT((2000^2-(4*4000*(-4000)/2*4000)
• 0=-2000{+/-}SQRT((4000000+32000000)/8000))
• 0=(x-1932.92)(x-2067.08)

Aaaand that's how far I got. I'm sorry I'm pretty bad at this, but how do I make 1/(1+r)=x to solve for r? --Thebackofmymind (talk) 00:32, 23 November 2011 (UTC)

You have misstated the formula. If $ax^{2}+bx+c=0$ then $x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$. Widener (talk) 01:02, 23 November 2011 (UTC)

If you could tell me what I need to fix that would be very helpful thanks. --Thebackofmymind (talk) 02:16, 23 November 2011 (UTC)