# A quick dollar_latex test of the back-slash LaTeX command

Will it compile? — $\LaTeX$

# Exploring mathjax latex etc. (I hope for etc too, the sooner the better)

I am looking at web page:

Wait, is this the way to insert links here? What about my own wlod.net? let me try:

Well?–Nothing! WordPress is pathetic, ridiculous!

Let me try again, by hand:

Hm, well?–this time I got working links!!! And the first one is correct, while the second one resulted not in the wlod.net home page but in my blastog. Still not too bad :-).

Fine, back to mathjax $\LaTeX$ (this in-line TeX-code perhaps doesn’t work here?)–let me check the slash-bracket tags:

$\forall_{a\ b\ >\ 0}\ \ a^{\log(b)}\ =\ b^{\log(a)}$

How is it?–Works like a charm!

Hm, on that page they propose a different way to install mathjax on wordpress.org blog. I’ll stay with what I have already. They show how to install it under different conditions, also on an html page, when one has an access to the header. Thus on ipages I can use both the method which I have already tried in wiggles, and their. Theirs has actually two script-tagged commands, not one, i.e. the pairs of script tags occur twice.

# Great moment! :-) MathJax $$\LaTeX$$ etc.

Finally somehow I’ve found how to install it for html files on ipage, and also for the whole wordpress.org blog like this one! (but no such luck for wordpress.com blogs, too bad). Let me try mathjax latex:

$$3^2 – 2^3 = 1$$

OK, let me have a look at it. ………. YES!!!!! It works!!!!!

Let’s compare:

• wordpress latex:   $\qquad\frac 53-\frac 85 = \frac 1{15}$
• mathjax   $$\LaTeX:\qquad\frac 53-\frac 85 = \frac 1{15}$$

I need to learn more of MathJax. Double\$ tags display the text inside on a separate line, or rather not in-line. It’s strange that wordpress latex doesn’t understand the command which displays sign LaTeX. Let me try differently: … — well? No, it doesn’t.

# Euclidean ball and sphere

Let

•   $B^n\ :=\ \,\{x\in R^n : |x| \le 1\}$
•   $S^{n-1} := \{x\in R^n : |x| = 1\}$
•   $\Delta^n := \{(x\ x) : x \in B^n\}$

I’d like to compute explicitly the linearly determined function  $F : B^n\times B^n \setminus \Delta^n \rightarrow S^{n-1}$ such that  $F(a\ b) = b$  whenever  $b\in S^{n-1}$. I want to show its well known continuity.

Let  $a\ b\ \in\ B^n$,  and  $a \ne b$.  Let

$c_t\ :=\ (1-t)\cdot a + t\cdot b$

I’d like to find  $t := t_a \le 0$  such that  $|c_t|=1$,  i.e.  $c_t^2=1$.  This last equation is equivalent to the following ones:

•   $(1-t)^2\cdot a^2 + 2\cdot(1-t)\cdot t\cdot a\cdot b + t^2\cdot b^2\ \ =\ \ 1$
•   $(a - b)^2\cdot t^2 + 2\cdot a\cdot (b-a)\cdot t + a^2-1\ \ =\ \ 0$
•   $t^2 + 2\cdot\frac{a\cdot (b-a)}{(a-b)^2}\cdot t + \frac{a^2-1}{(a-b)^2}\ \ =\ \ 0$

Define   $\delta\ :=\ \left(\frac{a\cdot (b-a)}{(a-b)^2}\right)^2 + \frac{1 - a^2}{(a-b)^2}$.  Both summands are non-negative (since  $|a| \le 1$).  Thus  $\delta$  is a non-negative real number. The two solutions  $t_a\ t_b$  of the above quadratic equation are given by

•   $t_a := -\frac{a\cdot (b-a)}{(a-b)^2} - \delta^{\frac 12}$
•   $t_b := -\frac{a\cdot (b-a)}{(a-b)^2} + \delta^{\frac 12}$

Thus  $t_a \le 0$  (of course), and  $t_a = 0$  ⇔  $|a| = 1$   —   indeed, when   $|a| = 1 \ge |b|$   then

$a\cdot(b-a) = a\cdot b - a^2 = a\cdot b - 1 \le |a|\cdot|b| - 1 \le 0$

and it’s clear that  $t_a = 0$. The inverse implication is simple.

# Acceleration of convergence of slow series

It’s fun (to me too) to accelarate classical slow series like

$\log(2)\ =\ \frac 11 - \frac 12 + \frac 13 - \ldots$

or

$\frac{\pi}4\ =\ \frac 11 - \frac 13 + \frac 15 - \ldots$

The latter is called Gregory-Leibniz series. What about the former? Is it Gregory-Leibniz too? Anyway, I played with them in the past and have reproduced, after Euler, versions which converge almost as fast as linear combinations of Taylor series. Possibly my web pages are still somewhere buried deep quietly.

Let  $a_n\ (n=0\ 1\ \ldots)$  be an arithmetic progression of positive real numbers. Let  $d := a_1-a_2$  be the difference of this progression. Finally, let  $k$  be a positive integer. Then

$\sum_{n=0}^\infty\ \frac 1{\prod_{j=0}^k a_{n+j}}\ \ =\ \ \frac1{k\cdot d}\cdot\frac1{\prod_{j=0}^{k-1} a_{n+j}}$

This allows to accelerate similar series by approximating them by the ones based on arithmetic series as above. A given series gets split into one of the above and the difference, which is a new series. But the new series converges faster. Euler was able to iterate this simple acceleration operation infinitely many times in a way which resulted in fast converging series.

This time I’d like to look at this method a bit more systematically than in the past. Instead of selecting the above easily summable auxiliary series artistically I’d like to do it perhaps optimally (or to make a compromise but a conscientious compromise).

Let me give an idea of what I am talking about on the initial examples. Thus first let’s transform the above G-L series::

$\log(2)\ \ =\ \ \sum_{n=0}^\infty\ \frac 1{(2\cdot n+1)\cdot(2\cdot n+2)}$

$\frac{\pi}4\ \ =\ \ \sum_{n=0}^\infty\ \frac 1{(4\cdot n+1)\cdot(4\cdot n+3)}\ \ =\ \ 2\cdot\sum_{n=1}^\infty\ \frac 1{(4\cdot n-1)\cdot(4\cdot n+1)}$

Now

$\frac 1{(2\cdot n+1)\cdot(2\cdot n+2)}\ \ =\ \ \frac 1 {(2\cdot n+\frac 12)\cdot(2\cdot n+\frac 52)}\,\ -\,\ \frac 34\cdot\frac 1{(2\cdot n+\frac 12)\cdot(2\cdot n+1)\cdot(2\cdot n+2)\cdot(2\cdot n+\frac 52)}$

hence

$\log(2)\ \ =\ \ 1\ -\ 3\cdot\sum_{n=0}^\infty \frac 1{(2\cdot n+1)\cdot(2\cdot n+2)\cdot(4\cdot n+1)\cdot(4\cdot n+5)}$

# Pinning–the very start, cnt. 3 (Certain 2-colorings)

Given integers  $A\ B\ C\ D$,  we can easily and always uniquely recover integers  $0 \le a < b < c < d < n$  such that

$A=\sigma(T_a)\qquad B=\sigma(T_b)\qquad C=\sigma(T_c)\qquad D=\sigma(T_d)$

It is enough to define:

$a:=\tau-A,\qquad b:=\tau-B,\qquad c:=\tau-C,\qquad d:=\tau-D$

where  $\tau$  is defined below, and the existence is subjected to the conditions:

•     $\tau := \frac{A + B + C + D}3$   is an integer
•     $3 \le D < C < B < A \le 3\cdot(n-2)$
•     $B+C+D \ge 2\cdot A$

# Pinning–the very start, cnt 2 (Certain 2-colorings)

It’s Sunday, h21:19. Starbucks. I’ll have to leave in less than 40m. The time is passing by awfully fast.

I am looking for the 2-colorings of triangles, which avoid unicolor tetrahedra. I’ll try the colorings  $\gamma := F \circ \sigma$  which are split into a composition of

$\sigma : \binom X3 \rightarrow \{3\ \ldots\ 3\cdot(n-1)\}$

and

$F : \{3\ \ldots\ 3\cdot(n-1)\} \rightarrow \{0\ 1\}$

where

$X := \{0\ \ldots\ n\!-\!1\}$

and

$\sigma(a\ b\ c) := a+b+c$

Then for every tetrahedron  $T := \{a\ b\ c\ d\}$,  and $\tau := a+b+c+d$,  we obtain the following $\sigma$-values of its walls

$T_x := T\setminus\{x\}$

namely:

$\sigma(T_x) = \tau - x$

Assume  $a < b < c < d$.  Then

•     $\sigma(T_a) + \sigma(T_b) + \sigma(T_x) + \sigma(T_x) = 3\cdot \tau$   (is divisible by 3)
•     $3 \le \sigma(T_d) < \sigma(T_c) < \sigma(T_b) < \sigma(T_a) \le 3\cdot(n-2)$
•     $n > \sigma(T_a) - \sigma(T_d) \ge 3$

# Pinning–the very start, cnt 1

Let

$X := \{0\ 1\ 2\ 3\ 4\ 5\}$

Let’s define a coloring  $\gamma : \binom X3 \rightarrow Z_2$  as follows:

$\gamma(\{a\ b\ c\})\ \ :=\ \ a+b+c \mod 2$

Then it’s easy to show that there is no unicolor tetrahedron.

(This is a very weak result but still a nice exercise :-)).

# Pinning–the very start (very modest)

The following trivial observation (OK, it’s a version of Axiom of Choice :-)) is still useful:

every family admits a pinning of cardinality not higher than that of the family itself.

It follows easily from my earlier intro how small pinnings lead to cliques. But here let me write down my first steps. Consider 2-coloring:

$\gamma : \binom X3 \rightarrow \{0 1\}$

What can we say about the pinnings when there is no unicolor tetrahedron?

• When $|X|=4$ then it is necessary and sufficient that there is at least one triangle of each color; and it’s easily possible, hence there may be no unicolor tetrahedron (there is one altogether anyway :-)).
• Now let  $|X| = 5$.  Any two triangles have at least one vertex in common (because  $|X| = 5$)).&nbsp Thus if no more than 2 triangles are painted in one of the colors then they admit a 1-element pinning, hence the tetrahedron of the remainng 4 vertices is unicolored (has all its triangular faces colored by the other color). Thus the assumption about the absence of the unicolored tetrahedrons implies that there are at least 3 different triangles painted in each of the colors. And indeed, such coloring, say in  $X := \{0\ 1\ 2\ 3\ 4\}$,  free of unicolored tetrahedrons, is possible:

blue  $\triangle\!-\!s:\ \ \ \{0\ 1\ 2\}\ \ \{0\ 1\ 3\}\ \ \{2\ 3\ 4\}$

red  $\triangle\!-\!s:\ \ \ \{1\ 2\ 4\}\ \ \{1\ 3\ 4\}\ \ \{0\ 2\ 3\}$

And the remaining 4 triangles (for a total of 10) can be painted arbitrarily.

This is already a result of interest. When  $X$  has any finite number of elements then any 5-element subset houses at least 3 triangles of each color which is a significant restriction.

• Now let  $|X| = 6$,  say  $X := \{0\ 1\ 2\ 3\ 4\ 5\}$.  Then  $\{0\ 1\ 2\ 3\ 4\}$,  contains  $k \ge 3$  different triangles colored blue. If  $k = 3$  then there is a vertex, say vertex  $0$,  which pins at least 2 blue triangles, thus leaving at the most one blue triangle (inside  $\{0\ 1\ 2\ 3\ 4\}$).  Thus there are at least 2 more blue triangles within  $\{1\ 2\ 3\ 4\ 5\}$,  for a total of at least 5 different blue triangles in the whole  $X$.

Next, let  $\{0\ 1\ 2\ 3\ 4\}$,  contain  $k = 4$  different triangles colored blue. Then there are 4*3 = 12 incident pairs blue_triangle-vertex, and  $\frac{12}5 > 2$,  which means that there is a vertex, say vertex  $0$,  which pins at least 3 blue triangles, thus leaving at the most one blue triangle (inside  $\{0\ 1\ 2\ 3\ 4\}$).  Thus there are at least 2 more blue triangles within  $\{1\ 2\ 3\ 4\ 5\}$,  for a total of at least 6 different blue triangles in the whole  $X$.

In the remaining case of  $k \ge 5$  there are at least 5 different blue triangles in  $X$  right away.

Thus there are at least 5 different blue triangles always (under the assumption of no unicolored tetrahedron).

Everything that was said about blue holds also for red. Thus if  $|X| = 6$,  and there is no unicolor tetrahedron, then there are at least 5 different triangles of each color.

# Bigos (mish-mash)

He had his surgery at a different hospital, on an out-patient basis–no wonder I (or rather police, after I told them about the possibility of that other hospital) didn’t get any information from them.

All that my worrying and good will amounted to was unnecessarily bothering a poor fellow, weak after his surgery. I am not making too much sense.

I should and want to think intensively about applying pinnings to Ramsey Theory. And I should continue to write about it despite having only and barely some initial observations and computations. Perhaps I should still focus on the triangles-tetraedra case (the 2-color case).

I’ve written to MK about my willingness to write for Delta (in Polish) on unicolor triangles and on the minusops-cyclands-Inteland theme. We are old friends from student years, in the same group, and student military service (almost an oxymoron :-)). He passed my case to his young Delta coworkers.

Here at Starbucks I talked to many people about their activities. One of them happened to be molecular biologist (neurons, etc.) B.Y. from UM. After the initial improvised meeting we made an appointment for today, and had here a working session. B.Y. has presented some of his research topics and open questions. It gave me also an occasion–during the second part of the meeting–to get acquainted with his small daughter. She was quite bored by our conversation though :-). I hope for an interesting development involving applications of mathematics and parallel processing (or computers in general) to biology. A lot of it was done already over many years but it is still as interesting and exciting as always.

There were today, and there always are, many more silly things to mention but I already have written way too much.