Last year I worked on a interesting problem of the possibility of partitioning a convex set in a plane in to internally disjoint convex pieces of equal area and perimeter. The problem has since been proved when is a prime power, by Aronov and Hubard and Roman Karasev for all dimensions. Here in this post I discuss a crucial lemma i worked while attempting to prove the n is a “power of two” case. In the simplest case, the lemma goes like this: given a continuous function which has the -antipodal property: , there is a pair of separators of a torus along its “waist” that map to zero under . More specifically, there is a subset of the zero set of which is such that the projection map restricted to is not null homotopic. For larger powers of , the notation gets a little unwieldy but the proof scales up well. Lets tackle notation first:

Label the coordinates of in a binary system, with words using the letters , and . The words that can be formed using these letters can be treated as vertices in a binary tree, with the word as the root node. The root node has a unique child vertex as the word . Other vertices consist of words in letters ( of length ) each having child nodes and . For example, the binary tree for would have vertices as below:

If we impose the dictionary ordering on such words, , then each word of length uniquely identifies a coordinate in . For example, a point in can be written as .

Let be the set of words representing indices to coordinates of points in that have letters in 0,1 and 2, with 0 appearing only as the starting letter and are of length not more than . For each word of length we define to be the projection map: . Given , we define a map as follows:

and for every other word , .

For instance, for and , we have,

Thus, changes the sign of the coordinate with index and swaps coordinates with indexes with , leaving all the other coordinates fixed. If represents a subword of consisting of first letters from , then using the definition of , we note in particular, that . Finally let to be the map that gives the length of a word in .

Definition 1 (-antipodal map) For , we say that a map is -antipodal, if,

.

Theorem 2 Let be a continuous map with the -antipodal property for each word . Suppose there exists such that . Then the zero set of contains a connected subset which can be approximated arbitrarily closely in the Hausdorff distance by a smooth loop , for which the projection map: , is not null-homotopic.

The second lemma is a Borsuk Ulam type theorem which has a cleaner and simpler proof:

Lemma 2 Let be a real valued function defined on the connected closed subset on the torus such that projection defined by is not null-homotopic. Then there exists two points and in such that .

Lets tackle this in a more natural setting shall we? which is a pdf document rather than html. Here it is.

are called -binomial coefficients. In keeping with the existing convention we will use as the indeterminate instead of and write for

It was established in an earlier post that:

Using this identity we may reduce the -binomial coefficient to:

We also made use of the identity:

If we put in the above equation, we get

We now discuss a few consequences of the above identities:

1. For any natural number , and , is divisible by . For the special case where is a prime, is divisible by ( for which we will use the shorter notation as in wikipedia). Consequently we have,

   

   For odd primes , modulo . Thus,

   

   Also, observe that last term in the above product : , reduces to modulo . So we may further simplify the above identity to,

   

   We emphasize again that this is true only for odd primes .

2. The exponents in the identity (1)

   

   are all either or , according as the the remainder when divides is greater or smaller than remainder when divides . In particular, since , it follows that,

   

   Stated differently, this means that

   

   Using the above result, a primality condition can be given, similar to the ones involving standard binomial coefficients can be given in terms of -binomials as well. A number is prime if and only if:

   

]]> http://mymbl.wordpress.com/2011/09/17/q-binomials/feed/ 0 mymbl http://mymbl.wordpress.com/2011/05/01/ponder-problem-now-on-last-legs/ http://mymbl.wordpress.com/2011/05/01/ponder-problem-now-on-last-legs/#comments Sun, 01 May 2011 05:37:41 +0000 mymbl http://mymbl.wordpress.com/?p=390 ]]> This post culminates our efforts in solving the problem:

If are positive integers with and satisfying the following is divisible by for all positive integers , then is a power of .

First recap a few facts from this post: We began by showing that:

is divisible by for all and therefore by the lcm,

We however wanted a stronger divisibility condition that should be divisible by,

For then, the sequence,

consists of integers only, and if we let , this will in turn show that the sequence also consists of integers only. But as, converges to as , this will prove that for large enough integer , vanishes, showing that . For the skeptics amongst us, here is a quick proof of the fact that as . We recall from our analysis courses that the product of

converges absolutely iff converges absolutely, provided for all large i. Now the result that as follows by verifying these facts and the fact that .

So to wind up the proof, we need to show that . Since we know that we will show that , by proving that each prime power dividing also divides

The last post allowed us to write and in terms of cyclotomic polynomials:

and

In an earlier post we also showed that for distinct integers , is a prime number or . If not , it is the prime only if for some . Also if then using the formula,

and going modulo , we get

Let be a prime such that and for positive integers . The highest power of in the product,

is,

where is the highest power on dividing and is the largest integer () such that . The number is bounded by the integer satisfying .

On the other hand, the highest power of the prime dividing is . To show that it is enough to show that for each such , the highest prime power of in

also divides .

From our first post on the problem, if then . If is the smallest integer for which , there exists a unique in , such that . In fact, if is largest integer such that , then by the minimality of . Now consider the numbers,

where is an integer such that . It is clear that the numbers are integers. Taking any consecutive integer values of , we get distinct numbers no two of which are same modulo , for otherwise, would not be the largest prime power dividing . Thus, there is such that . There are disjoint groups of consecutive terms of size , in the product each giving us an additional factor of in the product. Likewise, if we consider numbers where is any integer for which , repeating the arguments in the previous paragraph, for every consecutive values of , there exists such that is a multiple of . Since there are groups of consecutive terms of size in the product , giving us additional prime powers in . We can continue this process until we get all the additional prime powers

in , where is such that . This prime power is clearly divisible by the prime power of in in (1).

]]> http://mymbl.wordpress.com/2011/05/01/ponder-problem-now-on-last-legs/feed/ 0 mymbl http://mymbl.wordpress.com/2011/04/26/another-nugget-concerning-cyclotomic-polynomials/ http://mymbl.wordpress.com/2011/04/26/another-nugget-concerning-cyclotomic-polynomials/#comments Tue, 26 Apr 2011 18:06:09 +0000 mymbl http://mymbl.wordpress.com/?p=377 ]]> In our build up to a proof of this problem., we established that the product

is a multiple of each of the numbers for . But we wished to show the stronger result that is divisible by . One way to proceed to estimate how the lcm,

differs from the product and see if also divides . We take the cyclotomic route, using the identity,

to rewrite as,

We also write using cyclotomic polynomials as,

To see this, we write,

where is the Mobius function. Then the power on in the product,

is

The later sum is of the form

with and , reminding us, that we can use the Mobius inversion formula to obtain,

Thus,

Our next task is to estimate which we will do in the next post.

Theorem: For distinct positive integers the GCD of the values of cyclotomic polynomials and , is only if for some integer . On the other hand, if for some , .

Proof: () We show that for some if . We will start by proving that the gcd . Then show that is a prime power. Let and assume to the contrary that . The cyclotomic polynomials and divide and resp. But, and are relatively prime (easy exercise), and so their divisors and are also relatively prime. This shows that if for , the values of cyclotomic polynomials and have a factor in common then .

Now suppose and that, a prime , divides the values of cyclotomic polynomials and . But divides

and so also divides it. Going modulo , we see that . This proves that is a multiple of . Next we show that is a power of .

Let be the largest power of dividing and be the largest power of that divides . Now using Thm 1.1 of \cite{YVES}, we have,

Now, and . Since and , divides both and . From the arguments in the first paragraph of this proof, this is possible only if or . Hence .

() Let (). From the proof in the “only if” direction is the only prime that can possibly divide . To show that

it will be enough to show that

First suppose . Since , we must have, and so we may write , for some . Also, as , it is enough to show that to show that . Consider,

Thus . If with , and is an even number then . If is odd then,

. The last value modulo is an odd number and so is not a multiple of . Hence in any case

This shows that and that .

{9} \bibitem{YVES} Y. Gallot, Cyclotomic polynomials and Prime Numbers, \url{http://perso.orange.fr/yves.gallot/papers/cyclotomic.pdf}

]]> http://mymbl.wordpress.com/2011/04/24/a-result-concerning-cyclotomic-polymials/feed/ 1 mymbl http://mymbl.wordpress.com/2011/04/22/an-old-ponder-problem/ http://mymbl.wordpress.com/2011/04/22/an-old-ponder-problem/#comments Fri, 22 Apr 2011 17:15:15 +0000 mymbl http://mymbl.wordpress.com/?p=305 ]]> It is well-known that if natural numbers and satisfy , for some , then is a multiple of for all ,. April 2005 ponder asks whether the converse is true, that is if for integers , the number is a multiple of for all , then a power of .

While trying to prove this I stumbled on this interesting polynomial identity:

where

looks similar to the binomial coefficient . Importantly, is an integer for integer values of .

To use this identity for the problem at hand, we first note that setting causes LHS to vanish and so we may replace each of the terms in RHS by without changing the value of the expansion. But wait, is a multiple of for by hypothesis. We may pull out the denominator and the numerator in the RHS to rewrite this identity as:

It is now apparent that each of the terms in the summand in RHS is a multiple of and therefore product

In fact, one easily sees that,

for each in . If we up our optimism a bit, we are led to suspect that

Among the factors of that are relatively prime to , one candidate is where . But is ?. We can squeeze out this easy fact from the hypothesis: Every prime that does not divide also does not divide . For if then for some and so and therefore cannot divide . Stated differently, this says that every prime that divides also divides . Given that both are greater than , this means that .

It is still not clear whether , but if we take this to be true how can we prove the ponder problem?. To see this, observe that

approaches zero as , so if we show that each of the terms in the sequence is an integer, it would mean that the numerator of vanishes for some large , proving that for some .

Given that and are relatively prime, we just need to show that is an integer in order to show that is an integer. Which is what we will proceed to do in the next few posts.

]]> http://mymbl.wordpress.com/2011/04/22/an-old-ponder-problem/feed/ 5 mymbl http://mymbl.wordpress.com/2009/12/30/a-problem/ http://mymbl.wordpress.com/2009/12/30/a-problem/#comments Wed, 30 Dec 2009 19:27:05 +0000 mymbl http://mymbl.wordpress.com/?p=269 ]]>

Let A and B be connected subsets of [0,1]² where π(A) = [0,1] = π(B) where π is the projection map π(x,y) = x. We construct a new set C out of A and B as follows.
Let C = {(x,y,z) | (x,y) ∈ A and (x,z) ∈ B}.

Is C connected always? If not then are there counterexamples?
Read below to see an answer.

Here is a counter example to the statement where the set C generated by A and B as in figure below, is not connected:

Now comes the interesting question. Given such sets A and B, can we always find a connected set whose projections are A and B respectively.? The answer is provided here at m.o.

Given a matrix of trace zero can we construct two matrices and of determinant , such that . Here is a proposed way of finding matrices and such that .

It turns out that this problem has a variety of applications and it goes by the name Sylvester equation (after fixing one matrix say A). For example, it is used in Computer Vision to estimate the pose(position, orientation ) of an object based on a series of images. See video here

We first try a simple case: Suppose is matrix with non-zero entries only in the diagonal and at locations for (like in the Jordan Normal form) with or . We want to find a matrix such that where is a matrix with trace zero. We equate the entries on both sides to get,

expanding we get

Now if denotes the operator that flattens the rectangular(column-wise) matrix to a column vector then we need to solve the system.

where,

We can solve the system provided has maximum rank. We can reconstruct from but this does not ensure that the matrix is of determinant . We can ensure that has determinant by addition of a suitable scalar times identity matrix ( which must also be a solution trivially)

Wikipedia mentions that there is a better way to represent this matrix using tensor product ( also called Kronecker product of matrices http://en.wikipedia.org/wiki/Kronecker_product):

where is the identity matrix.

0.1. Related results

On a related note, here is another question. Can you produce matrices and such that the commutator is the sum of two other commutators and ? . Using the methods above we can find and indirectly, but is there a simple representation of and in terms of ? I do not know. However here are some insights:

Suppose denotes the matrix with only at the position , (), in the matrix and the rest filled by zeros. We use the fact that the set of matrices which are commutators is a linear space with basis:

and

I will leave this to the diligent reader to see why this is true. This says that any commutator is a linear combination of these basis vectors. Armed with this result we can break the original problem down to the following two problems:

• Given matrices matrices with commutator , the commutator ( ) and a scalar , find and such that

  

  .

• Similarly given a scalar , and matrices , for each , find matrices such that

  

I could not give a simple representation of in terms of . For for some special cases, it seems possible. Suppose you know that for the index , is the only column of containing non-zero entries. Now choose constants , , such that .

We next note the identity , for and , using which we can write:

where denotes the entry at the location in the matrix . Now let . Next using the fact that is , we have and , so the above expression for can be rewritten as:

or,

where . Now if , then

Thus,

Here are other examples: For , , ,

Also, for the case where and , we have:

]]> http://mymbl.wordpress.com/2009/12/26/matrix-commutators/feed/ 0 mymbl syl http://mymbl.wordpress.com/2009/12/23/two-non-idiotic-guesses/ http://mymbl.wordpress.com/2009/12/23/two-non-idiotic-guesses/#comments Wed, 23 Dec 2009 19:27:55 +0000 mymbl http://mymbl.wordpress.com/?p=188 ]]> Three problems to ponder for the next 100 years:

• First one may seem contrived but thinking about this would hightlight some major properties of primes. So here it is : The number 82 is the largest positive number that cannot be written as sum of two square free numbers each having odd number of prime factors.
  Motivation for this guess comes from the result of the following python program that was run on sage to compute the number of such pairs of numbers summing upto a given number.

  def sqfreesum(n):
   sum = 0;
   for i in range(1,n ):
    sum +=moebius(i)^2* moebius(n-i)^2
      *( ( 1 + moebius(i))*( 1 + moebius(n-i) )/4 );
   return sum;

  
  The graph which shows the number of ways of writing a number as a sum of two such square free numbers, reveals a trend:it increasingly stays away from zero. In fact after taking off from 82, where it is zero, it never seems to drop down.

  Note that we can always write a number as a sum of two square free numbers. The numbers of ways of writing them is:
  . Have to warn you that I have not seen an actual proof but I have seen it mentioned here:
  A098235

• The second one is in the same spirit as the previous one. Any even number > 10 can be written as the sum of two square free numbers with one having odd number of prime factors and the other having even number of prime factors.
  Motivation for this guess comes from the the result of the following python program that was run on sage to compute the number of such pairs of numbers summing upto a given number.

  def sqfreesum(n):
   sum = 0;
   for i in range(1,n ):
    sum +=moebius(i)^2* moebius(n-i)^2
      *( ( 1 + moebius(i))*( 1 – moebius(n-i) )/4 + ( 1 – moebius(i))*( 1 + moebius(n-i) )/4);
   return sum;

  
  As with the previous conjecture the graph which shows the number of ways of writing a number as sum of such square free numbers. has only a few hiccups taking off, and at n=10 it achieves escape velocity. I should also mention here that any even number according to Chen Jigrun can be written as the sum of prime and semi-prime(product of two primes) or a sum of two primes. So when diluted, this result that says if Goldbach conjecture were false, we atleast know that every even positive number can be written as sum of two square free positive numbers one with odd number of prime factors( infact 1) and the other with even number of factors ( in fact 2).

• Treat the following with caution please. If denotes the Euler totient function, then the following number is always greater than 2n.
  .
  Proving this will prove the Goldbach conjecture. This is because, it is only when both i and 2n-i are prime that: is 1. There are better ways to reformulate the Golbach conjecture. One other way is the following:
  Given an even number $2n$, there exists a number such that
  

is convergent for each and . The standard bounds for matrix norm given in the text for , are and , for spaces. We give yet another bound using the power mean inequality:

1. Power Mean Inequality :

Consider positive weights , which have total mass , then for nonnegative real numbers , and for all one has

2. Bound for matrix norm

Consider,

where for some fixed and , .

Now we use power mean inequality with { and } , { and } . So,

Now if we take,

Then we have . So we have infinite number of bounds for the norm as we vary among positive reals (of course, not all values of might be useful )

Example : Consider the matrix

For this matrix and is also infinity, however is finite with . This is because and . Thus .

You can further see that is a better bound than always, since if we take and , we have

where we have used the fact that .