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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 (edit in Nov 2018: Kohli has overtaken him). 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.

Last updated: May 07 2020

Player Innings Not out Runs HS Average Balls faced Strike rate 100s 50s 0s Time period
Kohli* 100 20 5834 183 72.93 6012 97.04 24 24 2 Oct 14-Dec 19
de Villiers 100 21 5454 162* 69.04 4935 110.52 20 27 1 Nov 09-Oct 15
Sharma* 100 11 5688 264 63.09 5634 96.55 25 21 5 Aug 14-Jan 20
Taylor* 100 24 4795 181* 63.09 5634 85.11 14 27 2 Oct 13-Feb 20
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
Root* 100 17 4802 133* 57.86 5369 89.44 15 28 2 Sep 14-Jun 19
Richards 100 19 4611 189* 56.93 5103 90.36 8 34 4 Jun 75-Nov 86
du Plessis* 100 15 4593 185 54.04 5225 87.90 12 29 3 Jul 13-Jul 19
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
Javed Miandad 100 25 3989 119* 53.19 5476 72.85 6 30 2 Sep 82-Dec 88
Williamson* 100 9 4826 148 53.03 5768 83.67 10 33 1 Jun 13-Jun 19
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
Warner* 100 6 4802 179 51.09 4930 97.40 18 18 1 Feb 12-Jan 20
Mathews* 100 25 3764 139* 50.19 4482 83.98 3 25 6 Nov 13-Jul 19
Klusener 100 46 2705 103* 50.09 2929 92.35 2 14 9 Apr 98-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.

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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