October | 2015 | 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.

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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