Comments for The polymath blog http://polymathprojects.org Massively collaborative mathematical projects Sat, 28 Apr 2012 14:00:10 +0000 hourly 1 http://wordpress.com/ Comment on About by Owning Your Words: Personal Clouds Build Professional Reputations | Matias Vangsnes http://polymathprojects.org/about/#comment-5472 Sat, 28 Apr 2012 14:00:10 +0000 #comment-5472 [...] a large group of mathematicians, and use that group mind to solve hard problems. Thus was born the Polymath project, chronicled by Michael Nielsen on his blog and in his important new book, Reinventing [...]

]]> Comment on About by Un inicio « la ciencia hacker http://polymathprojects.org/about/#comment-5378 Mon, 09 Apr 2012 17:27:02 +0000 #comment-5378 [...] and collectively make easy work of his hard mathematical problem. He dubbed the experiment the Polymath Project. [...]

]]> Comment on Minipolymath3 project: 2011 IMO by twio http://polymathprojects.org/2011/07/19/minipolymath3-project-2011-imo/#comment-5359 Mon, 02 Apr 2012 13:20:37 +0000 http://polymathprojects.org/?p=249#comment-5359 brandon wrote:
Let set A have this property
Let T be a triangle whose vertexes are in A, and has the remaining points in A in its interior.
Assume T exists for any A (A has no 3 co-linear points, so this should be easy to prove)

A counterexample is easy to find. Let A be the points of a square. There is no T with your desired property.

Also, later, in your proof for Lemma 1, you use C to represent both a circle and a point.

]]> Comment on Minipolymath3 project: 2011 IMO by brandon http://polymathprojects.org/2011/07/19/minipolymath3-project-2011-imo/#comment-5356 Mon, 02 Apr 2012 05:21:14 +0000 http://polymathprojects.org/?p=249#comment-5356 Lemma 1:
Let set A have this property
Let T be a triangle whose vertexes are in A, and has the remaining points in A in its interior.
Assume T exists for any A (A has no 3 co-linear points, so this should be easy to prove)
Let P be a point outside of T
A union P has this property

Proof of Lemma 1:
Let circle C circumscribe T
Let L be the line that is windmilling through A
Let I be the intersection of C and L
As L turns clockwise, its angle increases, and I moves clockwise
Both the angle and the location of I are continuas.
In order for a cycle to occur in A, L must be at all angles, and intersect C at all points
(Otherwise it would need to jump or move counter-clockwise to reach the original state)
L must intersect P (See Lemma 2)
Let B be the pivot point of L prior to it hitting P.
Let C be the point that L would hit following B, if not for P
The distance between L and C decreases as L pivots around P, until L intersects with C.
L cannot intersect with another point first, or else it would have done so if P were not there as well
The cycle resumes as it would if not for the presence of P, until it has hit all points in A, and repeats.
QED
(I’m not entirely convinced the last two lines of this are necessarily true.)

Lemma 2:
Assume line L intersects circle C at point I.
Assume the angle of L is continuas, increasing, and will hit all angles.
Assume I will be at all points on C, and is continuas.
Assume point P is outside of C
L intersects P

Proof of Lemma 2:
Let C` go through P, and have the same center as C.
Let I` be the intersection of L and C`
Let Z be the distance between I` and P
If I were to move along C at a constant angle, we can see that Z will decrease until it reaches 0;
If the angle of L were to increase, with a constant I, we can see that Z will decrease until it reaches 0;
As both the angle L, and the posistion of I are continuas, when they both increase, Z will decrease until it reaches 0

Main Proof
Let T(A) represent a triangle whose vertices are in set A, and which surrounds all other points in A.
Select a subset A of S, of size 3, such that T(A) does not surround a member of S.
We can see that A has the described property
Add a point P, that is a member of S, to A, such that T(A) does not surround a member of S which is not in A.
A still has the described property.
Repeat until A=S
QED

]]> Comment on How to use LaTeX in comments by sporepigfish http://polymathprojects.org/how-to-use-latex-in-comments/#comment-5343 Fri, 30 Mar 2012 15:11:40 +0000 http://polymathprojects.wordpress.com/?page_id=10#comment-5343 parallel and perp does not work

]]> Comment on Minipolymath3 project: 2011 IMO by mkhida http://polymathprojects.org/2011/07/19/minipolymath3-project-2011-imo/#comment-5337 Wed, 28 Mar 2012 14:09:14 +0000 http://polymathprojects.org/?p=249#comment-5337 sorry for my English,I have just a remark on this problem,I just tried to see this problem in dimension three (in the space), so I remarked that all points on a sphere verify the condition :each three point are not collinear,it means that there is a plane passing through each set of three points.Now if we consider starting point P and a plane witch contain it .then we can do the same procedure and try to meet all remaining points.the problem is to define the movement of our plane such that:
in the second step is to consider the stereographic projection and see the behaviour of the movement of our line(it will be a line after projection and try to conclude).

]]> Comment on Draft version of polymath4 paper by Warren D Smith http://polymathprojects.org/2010/06/29/draft-version-of-polymath4-paper/#comment-5325 Fri, 23 Mar 2012 23:58:22 +0000 http://polymathprojects.org/?p=167#comment-5325 Hello, I was reading http://arxiv.org/pdf/1009.3956v3.pdf and have some comments.

One of your key identities EQ2.1
2^w(n)= sum(d^2|n)of mobius(d) * tau(n/d^2)
can be generalized. Let m be any integer with m>=0.
2^(m*w(n)) =
m-fold-sum( d1^2|n, d2^2|n, …, dm^2|n )of
mobius(d1)*mobius(d2)*…*mobius(dm) * tau(n/d1^2) * tau(n/d2^2) *…* tau(n/dm^2).
PROOF SKETCH:
It is easy to see this identity is valid in the following easy cases:
(i) m=0 when it is just 1=1, we agree 0-fold-sum is 1.
(ii) m=1 when it is your old identity.
(iii) n=squarefree when only summand comes from d1=d2=…=dm=1.
(iv) n=p^k=prime power when all dj=1 or p:
If k=0 we get n=1 and 2^(m*w(1))=2^0=1=1.
If k=1 we get 2^m=2^m.
If k>=2 we get 2^m=([k+1]-[k-1])^m=2^m.
Now use coprime-multiplicativity to see the easy cases imply validity for all cases.
QED.

This would enable you to count primes in [a,b] not mod 2, but in fact mod 2^m,
for any desired m>=0. If m>log2(b-a+1) this would count the primes, full stop.

Can this be made efficient? I have not tried to figure that out.
Warren D Smith, warren.wds AT gmail.com

]]> Comment on How to use LaTeX in comments by Anonymous http://polymathprojects.org/how-to-use-latex-in-comments/#comment-5239 Thu, 08 Mar 2012 01:57:30 +0000 http://polymathprojects.wordpress.com/?page_id=10#comment-5239 functional of the probability of pulling a jewel of value v, is

I = \int_0^\infty dv \, p(v) \, ln \, p(v)

The constraints are that the probabilities must sum to one,

1 = \int_0^\infty dv \, p(v)

and that the expected value is 10,

10 = \int_0^\infty dv \, v \, p(v)

Using Lagrange multipliers, we solve for the probability distribution extremizing

\int_0^\infty dv \{ p \, ln \, p + \lambda p + \mu v p \}

The variation of this is

\int_0^\infty dv \, \delta p \{ ln \, p + 1 + \lambda + \mu v \}

which vanishes for

p(v) = p_0 e^{-\mu v}

We can solve for the constants by satisfying the constraints, which demand p_0 = \mu and \mu = 1/10. And so we have the precise probability distribution consistent with our knowledge:

p(v) = \frac{1}{10} e^{- \frac{v}{10}}

What’s the probability of pulling a jeweled coin valued at least $100?

P(v \geq 100) = \int_{100}^\infty dv p(v) = \int_{100}^\infty dv \frac{1}{10} e^{- \frac{v}{10}} = e^{-10} \simeq 0.000045

]]> Comment on How to use LaTeX in comments by Anonymous http://polymathprojects.org/how-to-use-latex-in-comments/#comment-5238 Thu, 08 Mar 2012 01:53:35 +0000 http://polymathprojects.wordpress.com/?page_id=10#comment-5238 Heh, fail. value v, is

I = \int_0^\infty dv p(v) ln p(v)

The constraints are that the probabilities must sum to one,

1 = \int_0^\infty dv p(v)

and that the expected value is 10,

10 = \int_0^\infty dv v p(v)

Using Lagrange multipliers, we solve for the probability distribution extremizing

\int_0^\infty dv \{ p ln p + \lambda p + \mu v p \}

The variation of this is

\int_0^\infty dv \delta p \{ ln p + 1 + \lambda + \mu v \}

which vanishes for

p(v) = p_0 e^{-mu v}

We can solve for the constants by satisfying the constraints, which demand p_0 = \mu and \mu = 1/10. And so we have the probability distribution consistent with our knowledge,

p(v) = \frac{1}{10} e^{- \frac{v}{10}}

which we can use to answer questions. What’s the probability of pulling a jeweled coin valued at least $100?

P(v \geq 100) = \int_100^\infty dv p(v) = \int_100^\infty dv \frac{1}{10} e^{- \frac{v}{10}} = e^{-10} \simeq 0.000045

]]> Comment on (Research thread II) Deterministic way to find primes by Phillip Brix http://polymathprojects.org/2009/08/09/research-thread-ii-deterministic-way-to-find-primes/#comment-5228 Tue, 06 Mar 2012 18:51:40 +0000 http://polymathprojects.org/?p=97#comment-5228 one (dumb) approach to solving problem 1 and 2, assuming you can detect whether or not a number is prime fairly quickly, would be the following.
start off with a binary number c bits long, containing all 1′s. test for prime. change the lowest bit to 0. test for prime. change the second lowest bit to 0. test for prime. change the lowest bit back to 1, test for prime. continue changing 1 bit at a time and testing for prime. in the worst case, it will take 2^(c/2) changes to find a prime. for problem 1, this means that you can find a prime of size log_2(k) in k time. for problem 2, this means you can find the largest prime of size log_2(k) in k steps.

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