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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.

Comments (7)

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.

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

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

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

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