Chaitanya's Random Pages

May 31, 2013

ATP World Tour players of the year

Filed under: sport — ckrao @ 12:32 pm

Below is a list of ATP World Tour tennis players of the year dating back to 1989. In most cases their ranking had suffered the previous year due to injury. Information here is from Wikipedia and atpworldtour.com. There are several big names featuring here. I was led to this list after thinking about how well Tommy Haas has been doing in recent times after all the injuries he has been through.

 

Year Player Ranking at end of year Ranking at end of previous year Notes
1989 Goran Prpić 28 226  
1990 Thomas Muster 7 21 struck by a car in 1989
1991 Jimmy Connors 49 936 reached SF of US Open at age 39
1992 Henri Leconte 61 159 reached SF in French Open
1993 Mikael Pernfors 32 237  
1994 Guy Forget 40 158  
1995 Derrick Rostagno 110 no ranking  
1996 Stephane Simian 92 333  
1997 Sergi Bruguera 8 82 reached final of French Open
1998 Younes El Aynaoui 45 237  
1999 Chris Woodruff 51 1342  
2000 Sergi Bruguera (2) 85 378  
2001 Guillermo Cañas 15 231  
2002 Richard Krajicek 112 no ranking  
2003 Mark Philippoussis 9 80 reached Wimbledon final
2004 Tommy Haas 17 no ranking  
2005 James Blake 23 97 had broken neck, shingles
2006 Mardy Fish 47 225  
2007 Igor Andreev 33 91  
2008 Rainer Schüttler 33 99  
2009 Marco Chiudinelli 56 779  
2010 Robin Haase 65 451  
2011 Juan Martín del Potro 11 258  
2012 Tommy Haas (2) 21 205  

May 25, 2013

Sine of half angles

Filed under: mathematics — ckrao @ 8:34 am

I found the following cool-looking formulas in [1].

\displaystyle \sin \frac{45^{\circ}}{2^n} = \frac{\sqrt{2 - \sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}}{2}
(n + 1 nested roots)

\displaystyle \sin \frac{15^{\circ}}{2^n} = \frac{\sqrt{2 - \sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}}{2}
(n + 2 nested roots)

\displaystyle \sin \frac{18^{\circ}}{2^n} = \frac{\sqrt{8 - 2\sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}}{4}
(n + 2 nested roots)

To prove them, we use the following form double angle formula for cosine.

\displaystyle 4 \sin^2 x = 2(1 - \cos 2x) = 2 - \sqrt{4 - 4 \sin^2 2x }\quad \quad (1)

Now we evaluate this for different values of x. Firstly \sin^2 45^{\circ} = 1/2 implying from (1) that

4 \sin^2 22.5^{\circ} = 2 - \sqrt{2}.\quad \quad \quad \quad (2)

Secondly,

\begin{aligned}4 \sin^2 15^{\circ} &= 2(1 - \cos 30^{\circ})\\ &= 2 - \sqrt{3},\end{aligned}

so that from (1),

\displaystyle 4 \sin^2 7.5^{\circ} = 2 - \sqrt{4 - (2 - \sqrt{3}} = 2 - \sqrt{2 + \sqrt{3}}. \quad \quad (3)

To find \sin 18^{\circ} one could apply Ptolemy’s theorem to four points of a regular pentagon or alternatively set y = 36^{\circ} and write

\begin{aligned} \cos 3y &= -\cos (180^{\circ} - 3y)\\&= - \cos 2y.\end{aligned}

Using the double and triple angle formulae for cosine this becomes

\begin{aligned} 4 \cos^3 y - 3\cos y = 1 - 2\cos^2 y \\ \hbox{i.e.}\ 4\cos^3 y + 2\cos^2 y - 3\cos y - 1 &= 0\\ \hbox{i.e.} \ (\cos y + 1) (4 \cos^2 y - 2 \cos y - 1) &= 0.\end{aligned}

From this the only valid (positive) solution is \cos y = (2 + 2\sqrt{5})/8, so from (1)

16 \sin^2 18^{\circ} = 8(1 - \cos 36^{\circ}) = 6 - 2\sqrt{5}.

Hence applying (1) again,

\displaystyle 16 \sin^2 9^{\circ} = 2(4 - \sqrt{16 - 16 \sin^2 18^{\circ} }) = 8 - 2\sqrt{10 + 2\sqrt{5}}.\quad \quad (4)

Now we set up a hypothesis for mathematical induction. Assume that

\displaystyle 4 \sin^2 \frac{45^{\circ}}{2^k} = 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}_{\hbox{k nested roots}}
\displaystyle 4 \sin^2 \frac{15^{\circ}}{2^k} = 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}_{\hbox{k+1 nested roots}}
\displaystyle 16 \sin^2 \frac{18^{\circ}}{2^k} = 8 - 2\underbrace{\sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}_{\hbox{k+1 nested roots}}

Note that these are equivalent to our initial three formulas. In (2)-(4) we have seen that these results are true for k = 1. Then using (1) and the inductive hypothesis,

\begin{aligned} 4 \sin^2 \frac{45^{\circ}}{2^{k+1}} &= 2 - \sqrt{4 - \sin^2 \frac{45^{\circ}}{2^{k}}}\\&= 2 - \sqrt{4 - \left(2 - \underbrace{ \sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}_{\hbox{k nested roots}}\right)} \\ &= 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}_{\hbox{k+1 nested roots}}\end{aligned}

Similarly,

\begin{aligned} 4 \sin^2 \frac{15^{\circ}}{2^{k+1}} &= 2 - \sqrt{4 - \sin^2 \frac{15^{\circ}}{2^{k}}}\\&= 2 - \sqrt{4 - \left(2 - \underbrace{ \sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}_{\hbox{k+1 nested roots}}\right)} \\ &= 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}_{\hbox{k+2 nested roots}}\end{aligned}

and

\begin{aligned} 16 \sin^2 \frac{18^{\circ}}{2^{k+1}} &= 8 - 2\sqrt{16 - 16 \sin^2 \frac{18^{\circ}}{2^{k}}}\\&= 8 - 2\sqrt{16 - \left(8 - 2\underbrace{ \sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}_{\hbox{k+1 nested roots}}\right)} \\ &= 8 - 2\underbrace{\sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}_{\hbox{k+2 nested roots}}.\end{aligned}

In each case we have shown that if the result is true for n = k, it is true for n = k+1. By the principle of mathematical induction, the result is true for n = 1,2, \ldots and we are done.

Reference

[1] E. Maor, Trigonometric Delights, Princeton University Press, 1998.

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