Chaitanya's Random Pages

October 31, 2015

Highest ODI batting averages over 100 consecutive innings

Filed under: cricket,sport — ckrao @ 11:13 am

The 3 centuries by A B de Villiers in the most recent ODI series against India has given him a sizeable lead over the rest in the following list of highest batting averages in 100 consecutive one day international cricket innings. On 92 out of those 100 innings he came in at number 4 or lower (and never opened in that time span), and still he managed to score over 54 runs per innings. All of his ODI hundreds to date have been scored at a strike rate of at least 100. Some very impressive numbers have been posted for other batsmen in the list too.

 

Player Innings Not out Runs HS Average Balls faced Strike rate 100s 50s 0s Time period
de Villiers* 100 21 5454 162* 69.04 4935 110.52 20 27 1 Nov 09-Oct 15
Bevan 100 40 3726 108* 62.10 4844 76.92 3 26 2 Sep 94-Aug 99
Dhoni* 100 34 4027 139* 61.02 4587 87.79 5 28 2 Feb 09-Jan 14
Amla* 100 7 5388 159 57.94 5991 89.93 20 27 2 Nov 08-Mar 15
Richards 100 19 4611 189* 56.93 5103 90.36 8 34 4 Jun 75-Nov 86
Kohli* 100 16 4668 183 55.57 5016 93.06 17 21 7 Mar 11-Mar 15
Tendulkar 100 11 4789 186* 53.81 5273 90.82 18 16 5 Apr 98-Jun 02
Hussey 100 33 3574 109* 53.34 4114 86.87 2 27 2 Feb 05-Nov 09
Jones 100 19 4317 145 53.30 5667 76.18 7 32 1 Jan 85-Feb 91
Sangakkara 100 10 4736 169 52.62 5417 87.43 14 28 5 Aug 11-Mar 15
Chanderpaul 100 21 4144 149* 52.46 5627 73.64 8 30 2 Sep 04-Jun 10
Lara 100 11 4575 169 51.40 5738 79.73 11 30 3 Mar 92-Dec 97
Kallis 100 21 4044 139 51.19 5474 73.88 7 27 3 Mar 00-Feb 04
Ponting 100 11 4397 164 49.40 5068 86.76 11 29 4 Nov 03-Dec 07
Greenidge 100 11 4382 133* 49.24 6583 66.57 10 27 3 Dec 75-Dec 85

Statistics are from ESPN Cricinfo. I would be interested to know if there are others who averaged 50+ over 100 consecutive ODI innings.

October 28, 2015

The product of distances to a point from vertices of a regular polygon

Filed under: mathematics — ckrao @ 11:12 am

Here is a cool trigonometric identity I recently encountered:

\displaystyle \prod_{k=1}^n \sin \frac{(2k-1)\pi}{4n} = \prod_{k=1}^n \cos \frac{(2k-1)\pi}{4n} = \frac{\sqrt{2}}{2^n}.

For example, for n = 9:

\displaystyle \sin 5^{\circ} \sin 15^{\circ} \sin 25^{\circ} \ldots \sin 85^{\circ} = \frac{\sqrt{2}}{2^9}.

After thinking about it for some time I realised that the terms on the left side can each be seen as half the lengths of chords of a unit circle with 4n evenly spaced points that can then be rearranged to be distances from a point on the unit circle to half of the points of a regular (2n)-gon, as shown in the figure below.

equalchords

With the insight of this figure we then write

\begin{aligned} \prod_{k=1}^n \sin \frac{(2k-1)\pi}{4n} &= \left(\prod_{k=1}^{2n} \left| \sin \frac{(2k-1)\pi}{4n} \right|\right)^{1/2} \quad \text{ (all terms are positive)}\\ &= \prod_{k=1}^{2n} \left| \frac{\exp(i\frac{(2k-1)\pi}{4n}) - \exp(-i\frac{(2k-1)\pi}{4n})}{2i} \right|^{1/2} \\ &= \frac{1}{2^n} \prod_{k=1}^{2n} \left| \exp\left(-i \frac{(2k+1)\pi}{4n}\right) \left(\exp\left(i\frac{4k\pi}{4n}\right) - \exp\left(i\frac{2\pi}{4n}\right)\right) \right|^{1/2}\\ &= \frac{1}{2^n} \prod_{k=1}^{2n} \left| \exp\left(-i \frac{(2k+1)\pi}{4n}\right) \right|^{1/2} \left| \left(\exp\left(i\frac{4k\pi}{4n}\right) - \exp\left(i\frac{2\pi}{4n}\right)\right) \right|^{1/2}\\ &= \frac{1}{2^n} \prod_{k=1}^{2n} \left| \left(\exp\left(i\frac{2k\pi}{2n}\right) - \exp\left(i\frac{\pi}{2n}\right)\right) \right|^{1/2}\\ &= \frac{1}{2^n} \left|\prod_{k=1}^{2n} \left(z-\exp\left(i\frac{2k\pi}{2n}\right)\right) \right|^{1/2} \quad \text{where }z = \exp\left(i\frac{\pi}{2n}\right)\\ &= \frac{1}{2^n} |(z^{2n}-1)|^{1/2}\\ &= \frac{1}{2^n} |-1-1|^{1/2}\\ &= \frac{\sqrt{2}}{2^n}. \end{aligned}

(The cosine formula can be derived in a similar manner.)

In general, the product of the distances of any point z in the complex plane to the n roots of unity \omega_n  is

\displaystyle \prod_{k=0}^{n-1} |z-\omega_n| = |z^n - 1|.

The above case was where z^n = -1. Two more cases are illustrated below, this time for n = 10. In the left example the product of distances is

\displaystyle \prod_{k=0}^9 |(1+i)-\exp(2\pi i k/10)| = |(1+i)^{10}-1| = 5\sqrt{41}

while for the right example it is

\displaystyle \prod_{k=0}^9 |1/2-\exp(2\pi i k/10)| = |(1/2)^{10}-1| = 1023/1024.

equalchords2

Note that earlier in the year I posted on the distances to a line from vertices of a regular polygon.

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