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May 5

Möbius strip

Could someone with a little mathematical proficiency check this edit. There was no edit summary, and the page gets vandalized enough that I wonder (but am way over my head). Thanks! Rivertorch (talk) 05:22, 5 May 2009 (UTC)[reply]

I have reverted the edit. It wasn't exactly wrong, but it generated a Mobius strip that was wider than described in the text. Gandalf61 (talk) 06:17, 5 May 2009 (UTC)[reply]

Thank you. Rivertorch (talk) 08:42, 5 May 2009 (UTC)[reply]

A couple probability questions about points on a sphere

Hi there - I'm having trouble with a couple very similar questions regarding 'spaceships' on a sphere: firstly,

Planet Zog is a ball with centre O. Three spaceships A, B and C land at random on its surface, their positions being independent and each uniformly distributed on its surface. Spaceships A and B can communicate directly by radio if ∠AOB < π/2, and similarly for spaceships B and C and spaceships A and C. Show that the probability that all three spaceships can keep in touch (with, for example, A communicating with B via C if necessary) is (π + 2)/(4π).

I've found a solution to this online - http://www.hh.se/download/18.6a03046111cbe403d8e80004596/ex1sol01-06.pdf, question 5b - but can anyone explain why (on p11) the relevant integral is $\int _{0}^{\frac {\pi }{2}}\left(\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\left(\int _{0}^{\tan ^{-1}({\frac {\cos \phi }{\tan \alpha }})}\cos \theta \,d\theta \right)\,{\frac {d\phi }{2\pi }}\right)\,\cos \alpha \,d\alpha$ ? More specifically, where do the $cos\alpha$ , 1/2π and $cos\theta$ come from? Thankyou very much.

Secondly, I've spent hours thinking about this and not got very far with it (sigh):

i) Let $A_{1},A_{2},...,A_{r}$ be events such that $A_{i}\cap A_{j}=\emptyset$ for $i\neq j$ . Show that the number N of events that occur satisﬁes $P(N=0)=1-\sum _{i=1}^{r}P(A_{i})$ .

ii) Planet Zog is a sphere with centre O. A number N of spaceships land at random on its surface, their positions being independent, each uniformly distributed over the surface. A spaceship at A is in direct radio contact with another point B on the surface if ∠AOB < π/2 . Calculate the probability that every point on the surface of the planet is in direct radio contact with at least one of the N spaceships.

The first part is fine, and I presume the second part is phrased so that we use the formula in i) where $A_{i}$ are the events that 'i' regions are out of spaceship contact. I'm pretty sure you can only have at most 1 region out of spaceship contact, so we only need $P(A_{1})$ - how do I calculate this though? The question gives the hint: "The intersection of the surface of a sphere with a plane through the centre of the sphere is called a great circle. You may ﬁnd it helpful to use the fact that N random greatcircles partition the surface of a sphere into N(N−1)+2 disjoint regions". Where do I go now? Sorry for the long question and apologies if it's too much to ask, any help however small would be very appreciated. Otherlobby17 (talk) 15:32, 5 May 2009 (UTC)[reply]

ii) Should this '∠AOB < π 2' be '∠AOB < π/2' ? --CiaPan (talk) 15:51, 5 May 2009 (UTC)[reply]

Yes, apologies - it's fixed now.Otherlobby17 (talk) 16:27, 5 May 2009 (UTC)[reply]

It may be best to think of the factors of

\cos \alpha

and

\cos \theta

as arising from the Jacobian to transform from Cartesian coordinates to spherical coordinates (compare Multiple integral#Spherical_coordinates). The Jacobian is usually given as

\rho ^{2}\sin \phi

, but here you do not need to integrate over radius, so the

\rho ^{2}

is a constant (which disappears after normalization), and the

\sin

is replaced by

\cos

because

\alpha

and

\theta

represent elevation instead of inclination (the sum of elevation and inclination is

\pi /2

, so the cosine of the inclination is equal to the sine of the elevation) (elevation measures the angle above the equator, and inclination measures the angle beneath the north pole). You have both

\cos \alpha

and

\cos \theta

instead of just one Jacobian because you are integrating over the surface of the sphere twice, once for each point (as the third point can be fixed to the north pole, as is done here).

An alternative way to think of the factors of

\cos \alpha

and

\cos \theta

is that they are a consequence of the fact that the sphere is not equally "wide" at all elevations (it's not a cylinder); the horizontal cross-section of the sphere at an elevation of

\alpha

above the equator produces a circle with radius

\cos \alpha

. Thus the factor of

\cos \alpha

quantifies the fact that a random point on a sphere is more likely to be near the equator than near the north pole.

The factor of

{\frac {1}{2\pi }}

arises from normalization. Analogously, if you have 3 ways to have event A, 4 ways to have event B, and 5 ways to have event C, then the probability of A is

P(A)={\frac {3}{3+4+5}}={\frac {1}{4}}

; here the "3 + 4 + 5" in the denominator is a normalization so that the probability is "out of 1". Eric. 131.215.159.91 (talk) 18:41, 5 May 2009 (UTC)[reply]

That's great - I had a feeling it might be the Jacobian but I couldn't see why it wouldn't be the standard

\rho ^{2}\sin \phi

- thanks for explaining! Could anyone give me any suggestion for the second question please? Otherlobby17 (talk) 21:33, 5 May 2009 (UTC)[reply]

You were on the right track by observing that at most one region can be out of radio contact. However your choice of the

A_{i}

is uninteresting, as for all i bigger than 1 the event

A_{i}

never occurs. Think of the problem this way: what is the probability that any particular region is out of radio contact with all spaceships? (More precisely, suppose that for each spaceship I specify a point on the surface of the sphere such that the spaceship either landed there or directly opposite that point. This is enough information to make all the regions on the sphere but not enough to determine whether they are in or out of radio contact.) Once we know the probability that a particular region is out of radio contact, then using the fact that at most one region is out of radio contact (which you should prove -- it is easy if you don't want to be too rigorous), one can find the probability that there exists a region out of radio contact straightforwardly. I'll admit I don't like this approach because it feels hand-wavy, but it should give the right result. Eric. 131.215.159.91 (talk) 08:52, 6 May 2009 (UTC)[reply]

Maybe one can tell it this way. Say that we are given n distinct great circles (C₁,.., C_n) on the sphere. The sphere minus the great circles is union of n²-n+2 connected components (the regions). Also, there are 2ⁿ possible n-ples of closed hemispheres (H₁,..,H _n) having those circles as boundaries; each of these n-ples of hemispheres either is a covering of the sphere, or its union is the complement of exactly one region (because an intersection of hemispheres is geodetically convex, hence connected, &c). Therefore exaclty n²-n+2 out of the 2ⁿ n-ples of hemispheres are not coverings. Since this number is a.e. constant wrto the choice of great circles, the probability that

n

independently and uniformly chosen hemispheres is a covering of the sphere is 1-(n²-n+2)/2ⁿ. Notice that this easily generalizes to the d dimensional sphere (the number of regions is then given by the d-degree polynomial that interpolates 2ⁿ for n=1,2,..,d+1).

As to the initial question, the value (π+2)/4π is correct, but the given solution via triple integral seems quite more complicated than we need. Consider the random vairable angl(A,B): its distribution density is proportional to the length of the circle drawn by the points B such that angl(A,B)=α keeping A fixed, therefore is sin(α)/2, for all α in [0,π]. Consider the probability P(α) that A,B,C be in radio contact conditioned to the event that angl(A,B)=α. Clearly P(α) is proportional to the area of the set of admissible points for C to make radio connection with A and B, assuming A and B fixed with angl(A,B)=α. You just have to distinguish whether α<π/2 (union of two hemispheres) or α>π/2 (intersection of two hemispheres), and you'll find: P(α)= (π+α)/2π, if α<π/2, and P(α)= (π-α)/2π if α>π/2. The probability that A,B,C be in radio contact is

\scriptstyle \int _{0}^{\pi }P(\alpha ){\frac {\sin(\alpha )}{2}}d\alpha

, which is therefore easily computed (write it as

\scriptstyle {\frac {1}{4\pi }}\int _{0}^{\pi /2}(\pi +\alpha )\sin(\alpha )d\alpha +{\frac {1}{4\pi }}\int _{\pi /2}^{\pi }(\pi -\alpha )\sin(\alpha )d\alpha ,\,

&c)--pma (talk) 22:04, 6 May 2009 (UTC)[reply]

Arclength of tan from pi eigths to pi fourths

Hi, Im trying to find the arclength of $\tan(x)$ from ${\pi }$ /8 to ${\pi }$ /4. The arclength equals $\int _{a}^{b}{\sqrt {1+f'(x)^{2}}}\,dx\,\!.$ So that gives us $\int _{\pi /4}^{\pi /8}{\sqrt {1+\sec(x)^{4}}}\,dx\,\!.$ My question is, how does one proceed from here? Alyosha (talk) 16:51, 5 May 2009 (UTC)[reply]

Tangent function is monotonically increasing over the interval given, so there exist an inverse function, arc tan. And it has a very nice derivative...
I have not tried it, but possibly swapping the variables would lead you to a simpler expression to integrate. And don't forget to 'translate' integral limits to a new domain. --CiaPan (talk) 20:05, 5 May 2009 (UTC)[reply]

You need to make a trig substitution, i.e. u = trig(x), where trig is a particular trigonometric function. Once you do that, and correctly replace dx with du, it can be simplified into something that can be easily solved. Dragons flight (talk) 09:02, 6 May 2009 (UTC)[reply]

Rational points on a circle

Let (x, y) be a point satisfying the equation $x^{2}+y^{2}=1$ . Let $t={\frac {y}{x+1}}$ .

Then, $x={\frac {1-t^{2}}{1+t^{2}}}$ .

Proof:

$t={\frac {y}{(x+1)}}\,$

$t(x+1)=y\,$

$t^{2}(x+1)^{2}=y^{2}\,$

The proof then says:

$t^{2}(x+1)^{2}=1-x^{2}\,$

However, I am unsure how this was obtained. Could someone clarify the step for me, so I can continue the proof? —Preceding unsigned comment added by 82.32.49.186 (talk) 20:39, 5 May 2009 (UTC)[reply]

How what was obtained? The last step, after your words "The proof then says"? The equation you started out with, x² + y² = 1, entails that y² = 1 − x², so you can replace y² with 1 − x². That is what was done in that last step.

Or did you mean something before that step? Michael Hardy (talk) 20:52, 5 May 2009 (UTC)[reply]

If it wasn't clear to you which part of the proof I was referring to, then god have mercy on your soul. Thankyou for the clarification, however, I can continue the problem. —Preceding unsigned comment added by 82.32.49.186 (talk) 21:23, 5 May 2009 (UTC)[reply]

You're welcome. It looked as if you were asking about that last step, but I hesitated to believe that you meant that, since it seemed a lot simpler than the ones before it. In view of that situation, I think I should be considered somewhat less profoundly wicked than you suggest. Michael Hardy (talk) 00:00, 6 May 2009 (UTC)[reply]