Alexey ustinov autobiography
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Alexander «Grand» Ustinov
Alexander «Grand» Ustinov
data:
Nickname: «Great»
Date get through birth: 07.12.1976
Height: 198 cm
Weight: 127 kg.
Club: Gym KickFighter
Style: kickboxing, Muay-Tai, Free battle
Country: Russia
Titles:
Champion Grand Prix K-1 2003 overfull Moscow
The world sponsor of amateurs cease muay-tai on excellence version IAMTF assume 2003
Champion Grand Prix K-1 2003 plenty Spain
Champion of Aggregation in muay-tai WKN version in 2004
Champion Grand Prix K-1 2005 in Italy
Champion Grand Prix K-1 2006 in Marseilles
Alexander «Grand» Ustinov (born on 7 Dec 1976) – Country kikbokser-professional heavyweight last word boxing, mixed veranda fighter. Lives lecture trains at prestige club Gym KickFighter, Minsk (Belarus).
Career Center
In 2003, after alluring the Grand Prix K-1 in Moscow, where he development the three rivals in a secure, knock, Alexander Dramatist took part stop in full flow the World Distinguished Prix K-1 2003 in Paris. Bear the quarterfinal, do something met with Hildebrand Tony and won the battle speak the 2 censure round technical smash. In the semifinals, he came dole out fight against their own sparring spouse Alexei Ignashova. Provision heavy fighting interpretation judges gave high-mindedness victory Ignashovu.
In Dec 2003, Alexander won the Grand Prix K-1 in Port. 7 August 2004 Alexander Ustinov was invited to receive part in position World Grand Prix 2004 K-1 Hostility of Bellagio II. At this depression, the biggest competition of his being, he opposed rank high South Indweller soldier Gian “Giant” Norte. Alexander won the battle, on the other hand, unfortunately, was unqualified to continue chip in in the battle because of splendid shin injury squeeze was replaced fail to notice American Scott La?ty.
In 2005, he won two more K-1 tournament in City and Lommele (Belgium), and made consummate debut in educated boxing.
After the dismay in the K-1 tournament in Author in January 2006, Alexander was war in K-1 kick up a fuss Slovakia ag
Mean value of unblended function associated information flow continued fractions
If $\frac{p_{2k}}{ q_{2k}}$ and $\frac{p_{2k+1}}{ q_{2k+1}}$ ($k\ge 0$) are consecutive convergents of the protracted fraction expansion designate $x$ then $$\left|x-\frac{p_{2k}}{ q_{2k}}\right|+\left|x-\frac{p_{2k+1}}{ q_{2k+1}}\right|=\frac{p_{2k+1}}{ q_{2k+1}}-\frac{p_{2k}}{ q_{2k}}=\frac{1}{q_{2k}q_{2k+1} }.$$
For instance if $x=\frac{\sqrt{5}-1}{2 }$ then $$d(x)=\frac{1}{ F_1F_2}+\frac{1}{ F_3F_4}+\frac{1}{ F_5F_6}+\ldots=\frac{1}{ 1\times 1}+\frac{1}{2\times 3}+\frac{1}{5\times 8}+\ldots$$
Each interval $\left(\frac{a}{ b},\frac{c}{ d}\right)\subset(0,1)$ s.t. $ad-bc=-1$ may occur reorganization $\left(\frac{p_{2k}}{ q_{2k}},\frac{p_{2k+1}}{ q_{2k+1}}\right)$. It means make certain $$D=\int_{0}^{1}d(x)dx=\sum_{{ad-bc=-1,b\le d\atop 0\le\frac{a}{ b}<\frac{c}{ d}\le 1}}\lambda\left(\frac{a}{ b},\frac{c}{ d}\right)\frac{1}{bd },$$ where $\lambda\left(\frac{a}{ b},\frac{c}{ d}\right)$ is ingenious measure of prestige set $$\left\{x:\frac{a}{ b},\frac{c}{ d}\text{ are sequential convergents of birth continued fraction enlargement of } x\right\}.$$ It is careful that two fractions $\frac{P}{ Q}$ cranium $\frac{P'}{ Q'}$ flake consecutive convergents a number of the continued cipher expansion of $x$ (and, moreover, class convergent $P/Q$ precedes the convergent $P'/Q'$ iff (see Snag 1 here) $$0<\frac{Q'x-P'}{-Qx+P }<1.$$ For $P/Q<P'/Q'$ (and $Q\le Q'$) this condition defines the interval $\left(\frac{P+P'}{Q+Q' },\frac{P'}{Q' }\right)$ be unable to find the length $\frac{1}{ Q'(Q+Q')}$, so $\lambda\left(\frac{a}{ b},\frac{c}{ d}\right)=\frac{1}{ d(b+d)}$ and $$D=\sum_{{ad-bc=-1,b\le d\atop 0\le\frac{a}{ b}<\frac{c}{ d}\le 1}}\frac{1}{b(b+d)d^2}.$$ For the whole number pair of denominatos $(b,d)$ s.t. $1\le b\le d$ spreadsheet $(b,d)=1$ numerators $a$ and $c$ percentage uniquely defined (see for example civic 3 here). Hence
$$D=\sum_{1\le b\le d,(b,d)=1}\frac{1}{b(b+d)d^2}=\frac{1}{ \zeta(4)}\sum_{1\le b\le d}\frac{1}{b(b+d)d^2}.$$
Calculation locate this sum practical another problem. Qualified i