# Chaitanya's Random Pages

## November 23, 2014

### Fastest to n test centuries for n sufficiently large

Filed under: cricket,sport — ckrao @ 12:04 am

My blog entry about fewest innings to n ODI centuries has encountered much traffic in recent times due to the consistent century-scoring feats of Amla, Kohli and de Villiers. As requested there, here is a similar table for test cricket.

As expected Bradman dominates the table leading the pack for n = 7 to 29 – he reached n=13, 27 and 28 roughly twice as fast as anyone else! It’s interesting to see how other players feature. Headley is in the list for every century until his last while Arthur Morris raced to 10 centuries in 36 innings before stalling. Sutcliffe, Harvey and Walcott jostle for positions after that and then Hayden and Gavaskar arrive. Tendulkar is somewhere in the top 3 from his 24th century onwards, while Ponting, Kallis and Sangakkara are present in the tail end of the list (updated: 5 October 2018).

Here are some notable streaks observed while preparing the table:

• Sir Donald Bradman scored 13 centuries in 25 innings in two separate streaks! He scored 26 of his 29 centuries between his 4th and 28th and later between his 48th and 72nd innings inclusive.
• Mohammad Yousuf entered the table during a streak of 10 centuries in 20 innings (103rd to 122nd innings).
• Sir Clyde Walcott scored 10 centuries in 23 innings (34-56).
• Sir Garfield Sobers had no centuries in his first 28 innings and then 11 in his next 31 (29-59).
• Denis Compton scored 11 centuries in 35 innings (23-57).
• Mahela Jayawardene scored 12 centuries in 37 innings (129-165).
• Sunil Gavaskar scored 13 centuries in 41 innings (61-101) and already had 19 centuries after his first 45 tests.
• Matthew Hayden scored 17 centuries in 59 innings (37-95) in his most productive patch.
• Inzamam-ul-Haq scored only 6 centuries in his first 90 innings and then 19 in his next 87 (91-177).
• Ricky Ponting had two separate amazing stretches of 11 centuries in 30 innings (87-116) and later 10 in 24 (155-178).
• Sachin Tendulkar scored 11 centuries in 34 innings (81-114) and 18 in 68 (81-148). His last 12 centuries were in the space of 39 innings (251-289).
• Kumar Sangakkara arrived late in the list while scoring 27 centuries in 110 innings (101-210).
• Jacques Kallis scored 14 centuries in 43 innings (221-263) after 20 centuries in 88 innings earlier in his career (102-189).

Candidate players were found using the best all-time ratings for batsmen on cricketratings.com. Data for the above was sourced from the original version of Statsguru, by displaying the career cumulative averages of each player. Corrections/additions are welcome!

## November 22, 2014

### A cute sum of Ramanujan

Filed under: mathematics — ckrao @ 3:09 am

Here is a beautiful sum I found in [1], apparently due to Ramanujan.

$\displaystyle \frac{1}{1^3\times 2^3} + \frac{1}{2^3 \times 3^3} + \frac{1}{3^3 \times 4^3} + \cdots = \sum_{k=1}^{\infty} \frac{1}{k^3(k+1)^3} = 10-\pi^2\quad\quad(1)$

Note that the result also demonstrates that $\pi^2$ is slightly less than 10, an alternative to the approaches in [2].

Many of his results require advanced number theory to prove, but this one is not too tricky, as long as we know the following similarly attractive result that I had previously mentioned in this blog post.

$\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}\quad\quad(2)$

To prove (1), the idea is to write $\frac{1}{k^3(k+1)^3}$ as a sum of partial fractions and then sum a telescoping series. You might wish to try this yourself before reading further.

We write

$\displaystyle \frac{1}{k^3(k+1)^3} = \frac{a(k)}{k^3} + \frac{b(k)}{(k+1)^3} = \frac{a(k)(k+1)^3 + b(k)k^3}{k^3(k+1)^3},\quad\quad(3)$

where $a(k)$ and $b(k)$ are quadratic polynomials. One way of finding $a(k)$ and $b(k)$ would be to compare coefficients of 1, $k$, $k^2$, $k^3$ and solve a system of equations. Another approach, the Extended Euclidean Algorithm, does so via finding the greatest common divisor of $k^3$ and $(k+1)^3$.

The following manipulations verify that the greatest common divisor of $k^3$ and $(k+1)^3$ is 1, where the next line computes quotients and remainders based on the previous line.

\displaystyle \begin{aligned} (k+1)^3 &= k^3 + (3k^2 + 3k + 1) & \quad (4)\\ k^3 &= \frac{k}{3}(3k^2 + 3k + 1) - \left(k^2 + \frac{k}{3}\right) & \quad (5)\\ 3k^2 + 3k+ 1 &= 3\left(k^2 + \frac{k}{3}\right) + (2k + 1) & \quad (6)\\ k^2 + \frac{k}{3} &= \frac{k}{2}(2k+1) - \frac{k}{6} & \quad (7)\\ 2k+1 &= 12 \frac{k}{6} + 1 & \quad (8) \end{aligned}

Reversing the steps of (4)-(8) gives us 1 as quadratic polynomial combinations of $k^3$ and $(k+1)^3$, thus providing us with $a(k)$ and $b(k)$.

\displaystyle \begin{aligned} 1 &= (2k+1) - 12\left[\frac{k}{2}(2k+1) - \left(k^2 + \frac{k}{3}\right)\right]\\ &= (2k+1)(1-6k) + 12\left(k^2 + \frac{k}{3}\right)\\ &= \left[3k^2 + 3k + 1 - 3\left(k^2 + \frac{k}{3}\right) \right](1-6k) + 12\left(k^2 + \frac{k}{3}\right)\\ &= (3k^2 + 3k+1)(1-6k) + \left[12 - 3(1-6k)\right]\left(k^2 + \frac{k}{3}\right)\\ &= (3k^2 + 3k+1)(1-6k) + \left[12 - 3(1-6k)\right]\left[\frac{k}{3}(3k^2+3k+1) - k^3\right]\\ &= (3k^2 + 3k + 1)\left[1-6k + 4k - k(1-6k)]\right] + \left[12 - 3(1-6k)\right]\left(-k^3\right)\\ &= (k+1)^3\left(6k^2 - 3k+ 1\right) - k^3\left(6k^2 - 3k+ 1 + 12 - 3 + 18k\right)\\ &= (k+1)^3(6k^2 - 3k+ 1) - k^3(6k^2 + 15k+10).\quad\quad(9) \end{aligned}

Using (9) we then have

\displaystyle \begin{aligned} \sum_{k=1}^{\infty} \frac{1}{k^3(k+1)^3} &= \lim_{N \rightarrow \infty} \sum_{k=1}^{N} \frac{1}{k^3(k+1)^3}\\ &= \lim_{N \rightarrow \infty} \sum_{k=1}^{N} \frac{ (k+1)^3(6k^2 - 3k+ 1) - k^3(6k^2 + 15k+10)}{k^3(k+1)^3}\\ &= \lim_{N \rightarrow \infty} \left(\sum_{k=1}^{N} \frac{ 6k^2 - 3k+ 1}{k^3} - \frac{6k^2 + 15k+10}{(k+1)^3}\right)\\ &= \lim_{N \rightarrow \infty} \sum_{k=1}^{N} \left(\frac{ 6k^2 - 3k+ 1}{k^3} - \frac{6(k+1)^2 + 3(k+1) + 1}{(k+1)^3}\right)\\ &= \lim_{N \rightarrow \infty} \sum_{k=1}^{N} \left(\frac{6}{k} - \frac{3}{k^2} + \frac{1}{k^3}\right) - \sum_{k=2}^{N+1} \left(\frac{6}{k} + \frac{3}{k^2} + \frac{1}{k^3}\right)\\ &= \left(\frac{6}{1} - \frac{3}{1} + \frac{1}{1}\right) - \lim_{N \rightarrow \infty} \sum_{k=2}^{N} \frac{6}{k^2} - \lim_{N \rightarrow \infty}\left(\frac{6}{N+1} + \frac{3}{(N+1)^2} + \frac{1}{(N+1)^3} \right)\\ &= 4 - (\pi^2 - 6) - 0 \quad\text{(using (2))}\\ &= 10-\pi^2, \end{aligned}

thus verifying (1).

The interested reader might like to find other identities in a similar manner. 🙂

#### References

[1] Clifford A. Pickover, A Passion for Mathematics: Numbers, Puzzles, Madness, Religion, and the Quest for Reality, John Wiley & Sons, 2005.

[2] Noam D. Elkies, Why is $\pi^2$ so close to 10?

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