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April 4

Calculating Probabilities while playing "Lets Make a Deal.

While playing Lets Make a Deal, you are shown 3 doors. Behind one door is a car, but there are goats behind the other two doors. You pick door #1. Then the host shows you that there is a goat behind door #3. He then asks you if you want to keep door #1, or change to door #2. Should you stay with your first pick, or pick the other door. Which has a higher probability of winning the car? — Preceding unsigned comment added by Psychoshag (talk • contribs) 03:29, 4 April 2012 (UTC)[reply]

We have an article on the Monty Hall problem, which will likely tell you more than you want to know. --Trovatore (talk) 03:30, 4 April 2012 (UTC)[reply]

I want to thank you for this link. It indeed has a lot of information, to which I found very useful. Thanks again! — Preceding unsigned comment added by Psychoshag (talk • contribs) 06:20, 5 April 2012 (UTC)[reply]

there are 4 kinds of topology in Set{1,2},and 29 in Set{1,2,3}, is exist a Formula in a Set {1,2,3,4,5,6...n}?

there are 4 kinds of topology in Set{1,2},and 29 in Set{1,2,3}, is exist a Formula in a Set {1,2,3,4,5,6...n}? — Preceding unsigned comment added by Cjsh716 (talk • contribs) 06:26, 4 April 2012 (UTC)[reply]

The number of topologies on a set with n labelled elements is sequence A000798 at OEIS. But I can't see anything there that suggests there is a known formula. Gandalf61 (talk) 09:26, 4 April 2012 (UTC)[reply]

thank you! — Preceding unsigned comment added by Cjsh716 (talk • contribs) 04:27, 5 April 2012 (UTC)[reply]

Evaluation of Residue using Analytic Continuation

Starting from the Euler integral representation of the Gamma Function, I have derived the expression $\Gamma (z)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(z+n)n!}}+\int _{1}^{\infty }e^{-t}t^{z-1}dt$ and have to use this to find the residue of the Euler integral at z=-m, m an integer. From the way the question is worded, I don't think this should be a difficult task but I haven't evaluated residues in this manner before and need some help in finding the correct approach. Thanks. meromorphic [talk to me] 09:52, 4 April 2012 (UTC)[reply]

The integral is holomorphic at

z=-m

, so only the term in the summation contributes to the residue. Sławomir Biały (talk) 12:21, 4 April 2012 (UTC)[reply]

(At the risk of asking a potentially obvious question...) So the residue is just

\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(n-m)n!}}

? meromorphic [talk to me] 14:24, 4 April 2012 (UTC)[reply]

No, it's just (-1)^m/m! Sławomir Biały (talk) 14:39, 4 April 2012 (UTC)[reply]

Ah, I'm with you now. Many thanks. meromorphic [talk to me] 14:44, 4 April 2012 (UTC)[reply]

Bounding a tricky series

I'm trying to show the following series is O(log(log(x))) for all x > e.

$f(x)=\sum _{n=1}^{\infty }{\frac {1}{n}}\sin {\frac {x}{4^{n}}}$

So far, I have tried, separately: (1) Euler-Maclauren summation, (2) expressing the summand using Vieta's formula [1], (3) writing the sine factor as an exponential and looking for telescoping terms in the series, and (4) trying to relate the sum to an entire function whose order necessarily contains a loglog term (see [2]) . I get the feeling I'm closest with the first two, but I must be missing something.

After that, I'd like to show the existence of a strictly increasing sequence $(x_{n})_{n=1}^{\infty }$ (converging to zero) such that $f(x_{n})\geq c\,\log(\log(x_{n}))$ , where c is a positive constant independent of n. I wasn't sure how to approach this, but it reminded me of the construction of an entire function with prescribed zeros.

But now I am stumped. Any pointers would be greatly appreciated. Korokke (talk) 10:53, 4 April 2012 (UTC)[reply]

I'm not sure if it helps, but writing

y=\log \log x

transforms the summation into

\sum _{n=1}^{\infty }{\frac {1}{n}}\sin {\frac {e^{e^{y}}}{4^{n}}}

and the problem is now to show that this is O(y) for

-\infty <y<\infty

. Sławomir Biały (talk) 12:27, 4 April 2012 (UTC)[reply]

Hint: Decompose the sum to two parts, one bounded by a constant and one bounded by the harmonic series. -- Meni Rosenfeld (talk) 17:37, 4 April 2012 (UTC)[reply]

Evaluation of an Integral using Special Functions

I have to evaluate the integral $\int _{-1}^{1}\!{\frac {1}{(1+x)^{2/3}(1-x)^{1/3}}}\,dx\,$ using the substitution x=2t-1, which transforms the integral to $\int _{0}^{1}\!{\frac {1}{t^{2/3}(1-t)^{1/3}}}\,dt\,$ and any standard properties of special functions that I like. Now, given the limits, I expect that the special function I want to use is the hypergeometric function with a=0, b=1/3 and c=1 but I cannot see the direct link. I'm tempted to say that I want to use the relation $F(a,b;c;z)={\frac {\Gamma (c)}{{\Gamma (c-b)}{\Gamma (b)}}}\int _{0}^{1}\!t^{b-1}(1-t)^{c-b-1}(1-zt)^{-a}dt\,$ but I am not least confused by the lack of the variable z in the integral I want to compute. Any ideas? Thanks. meromorphic [talk to me] 11:15, 4 April 2012 (UTC)[reply]

I've no idea if what you are doing is the best approach, but surely you can just use that relation with z=1? 130.88.73.65 (talk) 13:27, 4 April 2012 (UTC)[reply]

Thanks. I realised that using the relation

F(a,b;c;1)={\frac {{\Gamma (c)}{\Gamma (c-a-b)}}{{\Gamma (c-a)}{\Gamma (c-b)}}}

, which in this case leads to

F(0,{\frac {1}{3}};1;1)=1

, and then making the appropriate rearrangements and using the reflection formula for the gamma function leads to the answer. meromorphic [talk to me] 14:21, 4 April 2012 (UTC)[reply]

Billiard ball accident

Instead of a car being rear-ended, with its complicating factors like friction and crumple zones, suppose there is a stationary billiard ball A with mass m_a in a frictionless vacuum, and suppose it receives a direct hit at time t=0 from billiard ball B with mass m_b and pre-collision velocity v>0. Suppose both balls are perfectly hard, totally incompressible.

(1) Am I right that regardless of the relative masses and regardless of B's prior velocity, A will move forward and henceforth have a positive velocity?

(2) For any t>0, am I right that the post-collision velocity of A is constant? If so, then must there be a discontinuity in its velocity at time t=0 -- that is, a moment of infinite acceleration?

(3) If the mass of B is much greater than the mass of A, then it seems to me that the post-collision velocity of B remains positive; but if the mass of B is sufficiently small then the post-collision velocity of B must be negative. Is this right? If so, then what relationship between the relative masses and the original velocity of B characterizes the in-between situation in which B ends up with zero velocity? Duoduoduo (talk) 15:37, 4 April 2012 (UTC)[reply]

Yes, all three observations are correct (in an ideal no friction no compression situation). In case of a full-blown collision, i.e. if B initially moves exactly in A's direction, B will end up with zero velocity if and only if A and B have equal masses, regardless of the initial velocity of B. This follows from conservation of momentum (m1*v1 + m2*v2 = constant) and conservation of kinetic energy (0.5*m1*v1^2 + 0.5*m2*v2^2 = constant). All kinetic energy and momentum of B will be transferred to A, so A will move forward with whatever velocity B had before. -- Lindert (talk) 16:10, 4 April 2012 (UTC)[reply]

For part 2, it's quite obvious that we can't have infinite acceleration, even in theory, as that would then require infinite force or zero mass, accoding to F = ma. So, this means that the billiard balls actually deform a bit as they collide, and accelerate rapidly over some short time period, as they rebound from this deformation. StuRat (talk) 16:24, 4 April 2012 (UTC)[reply]

We can definitely have infinite acceleration in theory, specifically a Dirac delta. -- Meni Rosenfeld (talk) 17:32, 4 April 2012 (UTC)[reply]

It is useful to consider the frame of reference of the center of mass in collisions. In this frame, each ball will have its velocity negated after the collision. If the masses are equal, B's speed is v/2, so the change in his velocity is of magnitude v, meaning he will come to rest in the original frame. -- Meni Rosenfeld (talk) 17:32, 4 April 2012 (UTC)[reply]

Creating a minimum spanning tree with arbitrary points allowed - name for this problem?

Years ago, I took a Discrete Mathematics course, in which the professor discussed what was he said was an unsolved problem. I don't know if he was right, or if the problem has been solved since then, and I want to find any information on the problem... but I can't remember the terminology to even do a search.

Basically, consider a 2d plane (or 3d, or nd, but for now 2d) on which are a set of points. The problem is to create a minimum spanning tree that includes those points-- but! it can include arbitrary additional points if it will help reduce the tree. For example, consider the case with three points-- if you only include those points and have links between only those, than the minimum spanning tree would be a triangle minus a leg. But it may be able to decrease the tree size by having the three points you must include connect to a fourth arbitrary point in the center of the three instead.

This problem is useful when trying to create, for example, a transportation network that minimizes road length while still connecting everything. Or for circuit board design. Anyone know what this problem is called? Fieari (talk) 19:42, 4 April 2012 (UTC)[reply]

See Steiner tree problem. -- Meni Rosenfeld (talk) 19:58, 4 April 2012 (UTC)[reply]

That's the one! Thank you so much! Fieari (talk) 20:06, 4 April 2012 (UTC)[reply]

You're welcome. -- Meni Rosenfeld (talk) 06:51, 5 April 2012 (UTC)[reply]