Chaitanya's Random Pages

July 28, 2011

Recap of the 2011 Tour de France

Filed under: sport — ckrao @ 10:15 pm

I thoroughly enjoyed following this year’s Tour de France and it was capped off by seeing an Australian win the event for the first time. Here are some memorable moments for me.

  • The many falls of the first half of the race, the one featuring the car taking out Johnny Hoogerland and Juan Antonio Flecha was the most dramatic. Hoogerland was catapulted into a barbed wire yet managed to finish that stage and claim the polka dot jersey!
  • Mark Cavendish winning 5 stages – in the last four years he has won 4, 6, 5 and 5 TdF stages for 20 in total!
  • Norway having only 2 riders (Edvald Boasson Hagen and Thor Hushovd) yet they each won two stage wins.
  • Thomas Voeckler holding on to yellow day after day from Stage 12 to 18 after claiming it in Stage 9. Amazing.
  • Stage 18 was amazing with a successful breakaway by Andy Schleck on the second of three HC climbs (the Col d’Izoard) with some 60km remaining! His lead climbed to around 4 and a half minutes before being reeled in partly by the chase led by Cadel Evans. Against all odds Voeckler (whose previous best result in the Tour de France in six tours was 66th) managed to cling yet again to the yellow jersey. I regarded Frank Schleck as the GC favourite after his less energy-sapping ride.
  • Stage 19 had another early attack – this time by Alberto Contador less than 20km into the 109km race! He didn’t manage to hold on for the win, but his ride took a lot out of the legs of the main contenders in trying to chase him down. This tour ended an amazing run of six Grand Tour victories in a row by Alberto Contador, in Grand Tours he contested.A Frenchman (Pierre Roland) won atop Alpe d’Huez for the first time since 1986 and ended up claiming the white jersey for best young rider. Enthralling to watch.
  • The final time trial with Evans putting in a brilliant performance despite having what must have been weary legs to take the overall lead by a comfortable margin in the end (2nd in the stage, 2 and a half minutes faster over 42km than the Schleck brothers). He was the oldest to win the Tour since 1923 – yes even older than Lance Armstrong when he won for the seventh time. It is viewed by many including myself as one of the top sporting achievements by an Australian.

Here are some articles I enjoyed reading:

Podium Cafe – Poll: You Decide. The 2011 TdF Queen stage is…?

Podium Cafe – Stage 18 preview – the roof of the Alps – Agnel, Izoard and Galibier

Wikipedia – 2011 Tour de France, Stage 1 to Stage 11

Wikipedia – 2011 Tour de France, Stage 12 to Stage 21 – A view from a Sports Science point of view, including stats on the strength of the cyclists in watts/kg.

Inside the Tour with John Wilcockson: Voeckler the survivor has restored romanticism to the Tour

Tour de France: stage 18 – as it happened | Barry Glendenning | Sport |

Podium Cafe – The Ultimate Endurance Sport – 3170km in just 8 stages in the 1914 Giro – just 8 out of 81 competitors finished! (hardest stages of grand tour history: English translation here via Google Translate)

July 24, 2011

Tannery’s theorem

Filed under: mathematics — ckrao @ 12:19 pm

Here is a result that I feel deserves to be better known.

Tannery’s Theorem (series form): Suppose that

  1. \displaystyle s(n) = \sum_{k=1}^{\infty} f_k(n) is a convergent sum for all n \geq 1.
  2. \displaystyle \lim_{n \rightarrow \infty} f_k(n) exists and is equal to f_k for all k \geq 1.
  3. \displaystyle |f_k(n)| \leq M_k for all k \geq 1, n \geq 1.
  4. \sum_{k=1}^{\infty} M_k converges.

Then \displaystyle \sum_{k=1}^{\infty} f_k converges and is equal to \lim_{n\rightarrow \infty} s(n). In other words, both sides of the following are well-defined and equal:

\displaystyle \lim_{n\rightarrow \infty} \sum_{k=1}^{\infty} f_k(n) = \sum_{k=1}^{\infty} \lim_{n\rightarrow \infty} f_k(n)

Note the similarity with the Weierstrass M-test which states that if f_k(x) \leq M_k for all k and if \sum_{k=1}^{\infty} M_k converges, then \sum_{k=1}^{\infty} f_k(x) converges uniformly to s(x) = \sum_{k=1}^{\infty} f_k(x). If each f_k is continuous, so is s. Tannery’s theorem follows from this by the continuity of f_k and s at \infty [1]. It can be proved by standard \epsilon- type arguments.

Tannery’s theorem can also be considered a consequence of Lebesgue’s dominant convergence theorem applied to the sequence space \ell^1 with counting measure (hence integrals become sums). It gives a condition (namely, dominance by a convergent series) under which the order of limits and infinite sums may be interchanged.

One common application of this result in showing that the following two expressions for exp(x) are equal:

\displaystyle \lim_{n \rightarrow \infty} \left(1 + \frac{x}{n}\right)^n = \sum_{k = 0}^{\infty} \frac{x^k}{k!}.

Setting f_k(n) = \binom{n}{k}\frac{x^k}{n^k} it can be shown that f_k = \lim_{n \rightarrow \infty} f_k(n) = \frac{x^k}{k!} and Tannery’s theorem can be applied with the upper bound M_k = \frac{|x|^k}{k!}.

Tannery’s theorem can also exist in product form:

Tannery’s Theorem (product form): Suppose that

  1. \displaystyle p(n) = \prod_{k=1}^{\infty} (1 + f_k(n)) is a convergent product for all n \geq 1.
  2. \displaystyle \lim_{n \rightarrow \infty} f_k(n) exists and is equal to f_k for all k \geq 1.
  3. \displaystyle |f_k(n)| \leq M_k for all k \geq 1, n \geq 1.
  4. \sum_{k=1}^{\infty} M_k converges.

Then \displaystyle \prod_{k=1}^{\infty} f_k converges and is equal to \lim_{n\rightarrow \infty} p(n). In other words, both sides of the following are well-defined and equal:

\displaystyle \lim_{n\rightarrow \infty} \prod_{k=1}^{\infty} (1 + f_k(n)) = \prod_{k=1}^{\infty} \lim_{n\rightarrow \infty} (1 + f_k(n))

This follows from the fact that \prod(1 + a_n) converges absolutely if and only if \sum a_n converges absolutely (taking logarithms helps to see this).

One application of this result is in proving Euler’s infinite product formula for the sine function:

\displaystyle \sin \pi z = \pi z \prod_{k=1}^{\infty} \left(1-\frac{z^2}{k^2}\right)

To prove this (ref: [2]), firstly, it can be shown that for any complex z,

\displaystyle \sin z = \lim_{n\rightarrow\infty} F_n(z),

where n = 2m+1 is odd and

\displaystyle F_n(z) := \frac{1}{2i} \left \{\left(1 + \frac{iz}{n}\right)^2 - \left(1 - \frac{iz}{n}\right)^n\right \} = z\prod_{k=1}^m \left(1 - \frac{z^2}{n^2 \tan^2 (k \pi/n)}\right).

[In brief, set z to n \tan \theta and show F_n(n\tan \theta) = \sec^n \theta \sin(n\theta).  Hence the 2m+1 zeros of F_n are at n\tan \left(\frac{k\pi}{2m+1}\right). Finally construct the polynomial with these numbers as roots.]

Applying the product form of Tannery’s theorem with f_k(n) = - \frac{z^2}{n^2\tan^2 (k\pi/n)} and M_k = \frac{|z|^2}{k^2\pi^2} gives

\displaystyle \sin z = z \lim_{m \rightarrow \infty} \prod_{k=1}^{\infty} (1 + f_k(m)) = z \prod_{k=1}^{\infty} \lim_{m \rightarrow \infty} (1 + f_k(m)) = z\prod_{k=1}^{\infty}\left(1 - \frac{z^2}{k^2\pi^2}\right).

Finally, replace z with \pi z for the desired result. In my next mathematical post I plan to show some consequences of this beautiful formula for sine.


[1] J. Hofbauer, A Simple Proof of 1 + 1/2^2 + 1/3^2 + ⋯ = π^2/6 and Related Identities, The American Mathematical Monthly, Vol. 109, No. 2 (Feb., 2002), pp. 196-200

[2] P. Loya, Amazing and Aesthetic Aspects of Analysis: On the incredible infinite, available at

July 17, 2011

Men’s tennis rivalries

Filed under: sport — ckrao @ 12:35 pm

Here is a graphic showing the head-to-head records (all matches in blue, grand slam matches in green) of some top men’s singles tennis players of recent past and present. A line joins two players if they have played at least 8 times. The thicker the line joining two players, the more often they have played against each other. The image is current to the end of 2012 2013.


The pairs with the most matches between them are Lendl-McEnroe, followed by Edberg-Becker and Lendl-Connors. The trio with the most matches between them is Connors-Lendl-McEnroe which is hardly a surprise given their careers overlapped heavily (despite Connors starting much earlier) and they lead the list of most ATP singles titles won in the Open era (109, 94, 77 respectively).

A reason for generating the graphic was to see how the current quartet of Federer-Nadal-Djokovic-Murray compares with quartets of the past. These four have been ranked 1 to 4 in some order for 141 of the past 155 weeks (to the end of 2011). Curiously all of Murray’s grand slam matches against Djokovic and Federer have been in finals, while none of his 7 grand slam matches against Nadal were in a final.

It looks like the previous quartet to stay at the top for so long was McEnroe-Lendl-Connors-Wilander who were in the top 4 in some order from 1983 to 1986. Interestingly Connors and Wilander only met 5 times since they would have been in opposite sides of the draw for much of that time (ranked 3 and 4), much like Murray and Djokovic prior to this year.

In terms of matches played the biggest quartets here appear to be Lendl-McEnroe-Connors-Edberg and Lendl-McEnroe-Wilander-Edberg (where at least 13 matches were played between any pair). It will be fascinating to see how the present-day quartet ends up.

Also of note is that all of Wilander’s 3 career wins against Becker came in grand slam matches. Becker also had losing records against his other main foes in grand slam matches except for Lendl.

Some other head to head records among the players shown above are as follows:

Agassi-Wilander: 5-2
Connors-Becker: 0-6
Connors-Wilander: 0-5
Agassi-McEnroe: 2-2
Sampras-Wilander: 2-1
Sampras-McEnroe: 3-0
Sampras-Connors: 2-0
Connors-Agassi: 0-2
Agassi-Nadal: 0-2
Sampras-Federer: 0-1
Borg-Wilander: 1-0

Edit: here is how the image looks as of the end of 2016.




Consecutive Weeks Player Groups: July 1984 to present

Head to Head GS Matches


July 7, 2011

The number 3456

Filed under: mathematics — ckrao @ 10:07 pm

Here is a cute numerical fact that I did not know before. Not only is the number 3456 easy to remember in having its digits in arithmetic sequence, but its smallest prime factor is particularly small:

\displaystyle 3456 = 2^7 \times 3^3

That’s right, its only prime factors are 2 and 3. 🙂 (See OEIS A033845 for other numbers of this form.) One can easily read off that it has 32 factors. It is by no means the smallest integer to have this many factors (that honour belongs to 840), but it is the smallest to have 30 non-prime factors (see OEIS A055079). The sum of the cubes of its digits is 27 + 64 + 125 + 216 = 432, which curiously is a factor of 3456 (OEIS A034088).

Halving this number gives 1728 = 12^3 which is one less than the Hardy-Ramanujan number 1729, the smallest number that can be written as the sum of two positive cubes in two different ways. Having known this previously, I guess I should have seen earlier in my life that 3456 is double 1728 and therefore has small prime factors!

Finally, using the above factorisation and knowing that 128 = 101 + 27 allows us to write 3456 in this attractive form (see OEIS A105814):

\displaystyle 3456 = 27\times 27 + 2727

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