Uncategorized | Ewan Delanoy's mathematical blog

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Better to have more input variables and less output variables

October 15, 2012

When trying to describe an invariant map $\phi : U \times V \to W$ , where $U,V$ and $W$ are finite-dimensional spaces, $\phi$ is “complicated” because its values are vector-valued. It is sometimes useful to look at $\phi^{\sharp} : U \times V \times W^{*} \to {\mathbb K}$ , defined by $\phi^{\sharp}(u,v,w^*)=w^*(\phi(u,v))$ . This map has more variables but takes scalar values.

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Vector whose coordinates are pairwise independent

October 8, 2011

Consider the following twelve vectors in $\lbrace 0,1 \rbrace^4$ :

$\begin{array}{l} V_1=(0,0,0,0),V_2=(0,0,0,1),V_3=(0,0,0,0), \\ V_4=(0,1,0,1),V_5=(0,1,1,0),V_6=(0,1,1,1), \\ V_7=(1,0,0,1),V_8=(1,0,1,0),V_9=(1,0,1,1), \\ V_{10}=(1,1,0,0),V_{11}=(1,1,0,0),V_{12}=(1,1,1,1). \\ \end{array}$

Then the multiset $V=\lbrace V_k\rbrace_{1 \leq k \leq 12}$ (note that $V_{11}=V_{10}$ ) has the following property : if $\overrightarrow{X}$ is a random vector whose distribution is unform in $V$ , then the coordinates of $\overrightarrow{X}$ are pairwise independent. Moreover, $V$ is minimal with respect to this property. Can $V$ be “explained” rather than just enumerated?

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Changing the measure

February 27, 2011

Roth’s function $g(n)$ is defined as the largest size of a subset of $\lbrace 1,2, \ldots ,n \rbrace$ without arithmetic progressions. We may “force” the arithmetic progressions into the measure, and instead of taking the usual measure $\mu(A)=|A|$ we may consider $\nu(A)=|A|-\begin{displaystyle}\sum_{T\in{\cal T}}c_T1_{T \subseteq A}\end{displaystyle}$ where $\cal T$ denotes the set of all AP’s in $\lbrace 1,2, \ldots ,n \rbrace$ and $1_{T \subseteq A}$ equals $1$ if $T \subseteq A$ , $0$ otherwise, and the $c_T$ are constants. The advantage of doing this is that we now have an optimization problem on all subsets of $\lbrace 1,2, \ldots ,n \rbrace$ instead of the subsets without APs.
For example, we may take all the $c_T$ ‘s equal to $1$ , but the extremal subsets (those which maximize $\nu$ ) do not appear to exhibit a very interesting structure. Perhaps there are better choices for the $c_T$ constants.

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Decomposing sets without arithmetic progressions

February 23, 2011

Let $A$ be a subset of $\lbrace 1,2, \ldots ,8 \rbrace$ containing no arithmetic progressions. Then one of the following two must hold :

(1) $|A \cap \lbrace 3,5 \rbrace | \leq 1$ and $|A \cap \lbrace 1,2,4,6,7,8 \rbrace | \leq 1$ .

(2) $|A \cap \lbrace 4,6 \rbrace | \leq 1$ and $|A \cap \lbrace 1,2,3,5,7,8 \rbrace | \leq 1$ .

Let $A$ be a subset of $\lbrace 1,2, \ldots ,10 \rbrace$ containing no arithmetic progressions. Then one of $|A \cap \lbrace 1,2, \ldots ,7 \rbrace|$ and $|A \cap \lbrace 4,5, \ldots ,10 \rbrace|$ is at most $3$.

Let $A$ be a subset of $\lbrace 1,2, \ldots ,13 \rbrace$ containing no arithmetic progressions. Then one of the following two must hold :

(1) $|A \cap \lbrace 1,2,3,4,6 \rbrace | \leq 3$ and $|A \cap \lbrace 5,7,8,9,10,11,12,13 \rbrace | \leq 4$ .

(2) $|A \cap \lbrace 8,10,11,12,13 \rbrace | \leq 3$ and $|A \cap \lbrace 1,2,3,4,5,6,7,9 \rbrace | \leq 4$ .

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Sets without arithmetic progressions in [1..7]

January 7, 2011

Let $A$ be a subset of $\lbrace 1,2, \ldots ,7 \rbrace$ containing no arithmetic progressions. Then one of the following seven must hold :

(1) $A \cap \lbrace 1,4,7 \rbrace=\emptyset$ .

(2) $A \cap \lbrace 3,4,5 \rbrace=\emptyset$ .

(3) $|A \cap T | \leq 1$ for any $T\in \lbrace \lbrace 1,2,3 \rbrace, \lbrace 1,3,5 \rbrace\rbrace$ .

(4) $|A \cap T | \leq 1$ for any $T\in \lbrace \lbrace 3,5,7 \rbrace, \lbrace 5,6,7 \rbrace\rbrace$ .

(5) $|A \cap T | \leq 1$ for any $T\in \lbrace \lbrace 1,3,5 \rbrace, \lbrace 1,4,7 \rbrace , \lbrace 3,4,5 \rbrace, \lbrace 4,5,6 \rbrace \rbrace$ .

(6) $|A \cap T | \leq 1$ for any $T\in \lbrace \lbrace 1,4,7 \rbrace, \lbrace 3,5,7 \rbrace , \lbrace 2,3,4 \rbrace, \lbrace 3,4,5 \rbrace \rbrace$ .

(7) $|A \cap T | \leq 1$ for any $T\in \lbrace \lbrace 2,3,4 \rbrace, \lbrace 3,4,5 \rbrace , \lbrace 4,5,6 \rbrace, \lbrace 2,4,6 \rbrace \rbrace$ .

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Center of gravity of a trapezium

November 14, 2009

It is worth noting that the center of gravity of a trapezium $ABCD$ does not coincide with the center of gravities of the set $\lbrace A,B,C,D \rbrace$ . Indeed, if we consider the special case where the coordinates of $A,B,C,D$ are $(-\alpha,0),(\alpha,0),(\beta,h)$ and $(-\beta,h)$ , respectively, the coordinates of the first center of gravity (let’s call it $G_1$ ) are $(0,\frac{(\alpha+2\beta)h}{3(\alpha+\beta)})$ while the coordinates of the second (let’s call it $G_2$ ) are $(0,\frac{h}{2})$ . If $\alpha>\beta$ , then $G_1$ is below $G_2$ . Also, $G_1$ satisifies $(2\alpha+beta)(\overrightarrow{GA}+\overrightarrow{GB})+(\alpha+2\beta)(\overrightarrow{GC}+\overrightarrow{GD})=\overrightarrow{0}$ while $G_2$ satisifies $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}=\overrightarrow{0}$ .