# Chaitanya's Random Pages

## January 31, 2016

### Most wickets after n test matches from debut

Filed under: cricket,sport — ckrao @ 12:18 am

Here is a list of the leading test cricket wicket takers after playing n matches. A few current-day players feature and I hope to update this list over time (last updated January 31 2016). It was created largely manually using ESPNCricinfo’s Statsguru starting from [1] and [2], and using lists of the fastest to multiples of 50 wickets here.  Corrections are more than welcome (updated: March 25 2017).

#### References

[1] Most wkts in consec Tests from debut – Google Groups

[2] Top 10 bowlers with most wickets in 10 Tests or less (crictracker.com)

## January 30, 2016

### Patterns early in the digits of pi

Filed under: mathematics — ckrao @ 8:59 pm

Recently when taking a look at the early decimal digits of $\pi$ I made the following observations:

3.141592653589793238462643383279502884197169399375105820974944592307816406286…

• The first run of seven distinct digits (8327950, shown underlined) appears in the 26th to 32nd decimal place. Curiously the third such run (5923078, also underlined) in decimal places 61 to 67 contains the same seven digits. (There is also a run of seven distinct digits in places 51 to 57 with 5820974.)
• Decimal digits 60 to 69 (shown in bold) are distinct (i.e. all digits are represented once in this streak). The same is true for digits 61 to 70 as both digits 60 and 70 are ‘4’.

Assume the digits of $\pi$ are generated independently from a uniform distribution. Firstly, how often would we expect to see a run of 7 distinct digits? Places $k$ to $k+6$ are distinct with probability

$\displaystyle \frac{9}{10} \times \frac{8}{10} \times \frac{7}{10} \times \frac{6}{10} \times \frac{5}{10} \times \frac{4}{10} = \frac{9!}{3.10^6} = \frac{189}{3125}.$

Hence we expect runs of 7 distinct digits to appear $3125/189 \approx 16.5$ places apart. This includes the possibility of runs such as 12345678 which contain two runs of 7 distinct digits that are only 1 place apart.

How often would we expect to see the same 7 digits appearing in a run as we did in places 26-32 and 61-67? Furthermore let’s assume the two runs have no overlap, so we discount possibilities such as 12345678 which have a six-digit overlap. We expect a given sequence (e.g. 1234567, in that order) to appear $1/10^7$ of the time. There are $7! = 5040$ permutations of such a sequence, but of these $1$ has overlap 6 (2345671), $2!$ has overlap 5 (3456712 or 3456721), $3!$ has overlap 4, …, $6!$ has overlap 1 with the original sequence. This leaves us with $5040 - (1 + 2 + 6 + 24 + 120 + 720) = 4167$ possible choices of the next run to have the same 7 digits but non-overlapping (or to appear in precisely the same order – overlap 7). Hence we expect the same 7 digits to recur (no overlap with the original run) after approximately $10^7/4167 \approx 2400$ places apart so what we saw in the first 100 places was remarkable.

Now let us turn to runs of all ten distinct digits. Repeating the argument above, such runs occur every $10^{10}/10! \approx 2756$ places. According to [1] the next time we see ten distinct digits is in decimal places 5470 to 5479.

To answer the question of when we would expect to see the first occurrence of ten distinct digits, we adopt an argument from renewal-reward theory based on Sec 7.9.2 of [2] (there also exist approaches based on setting up recurrence relations, or martingale theory (modelling a fair casino), see [3]-[4]). Firstly we let $T$ be the first time we get 10 consecutive distinct values – we wish to find $E[T]$ where $E$ denotes the expected value operator. Note that this will be more than the 2756 answer we obtained above since we make no assumption of starting with a run of ten distinct digits – there is no way $T$ could be 1 for example, but we could have two runs of ten distinct digits that are 1 apart.

From a sequence of digits we first define a renewal process in which after we get 10 consecutive distinct values (at time $T$) we start over and wait for the next run of 10 consecutive distinct values without using any of the values  up to time $T$. Such a process will then have an expected length of cycle of $E[T]$.

Next, suppose we earn a reward of $1 every time the last 10 digits of the sequence are distinct (so we would have obtained$1 at each of decimal places 69 and 70 in the $\pi$ example). By an important result in renewal-reward theory, the long run average reward is equal to the expected reward in a cycle divided by the expected length of a cycle.

In a cycle we will obtain

• $1 at the end •$1 at time 1 in the cycle with probability $1/10$ (if that digit is the same as ten digits before it)
• $1 at time 2 in the cycle with probability $2/100$ (if the last two digits match those ten places before it) •$1 at time 9 in the cycle with probability $9!/10^9$ (if the last nine digits match those ten places before it)

Hence the expected reward in a cycle is given by

$\displaystyle 1 + \sum_{i=1}^9 \frac{i!}{10^i} = \sum_{i=0}^9 \frac{i!}{10^i}.$

We have already seen that the long run average reward is $10!/10^{10}$ at each decimal place. Hence the expected length of a cycle $E[T]$ (i.e. the expected number of digits before we expect the first run of ten consecutive digits) is given by

$\frac{10^{10}}{10!}\sum_{i=0}^9 \frac{i!}{10^i} \approx 3118.$

Hence it is pretty cool that we see it so early in the decimal digits of $\pi$. 🙂

#### References

[2] S. Ross, Introduction to Probability Models, Academic Press, 2014.

[4] A Collection of Dice Problemsmadandmoonly.com

## December 30, 2015

### Novak Djokovic against younger players

Filed under: sport — ckrao @ 9:11 pm

#### Reference

[1] Răzvan Gelca and Titu Andreescu, Putnam and Beyond, Springer, 2007.

## 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.

## September 28, 2015

### First space probes to visit bodies of the Solar System

Filed under: science — ckrao @ 11:11 am

The table below shows the first space probes to visit various bodies of the Solar System (and the year) by mission type (flyby, orbit, impact or soft landing). Notably 2015 had three events: MESSENGER ended its four year orbit of Mercury with the first impact on the planet, Dawn became the first spacecraft to orbit a dwarf planet (Ceres, in the asteroid belt), and New Horizons flew by Pluto.

 flyby orbit impact soft landing Sun Helios 2 1976 (within 43m km) Luna 1 1959 Mercury Mariner 10 1974 MESSENGER 2011 MESSENGER 2015 Venus Venera 1 1961 Venera 9 1975 Venera 3 1966 Venera 9 1975 Mars Mariner 4 1965 Mariner 9 1971 Mars 2 1971 Mars 3 1971 Jupiter Pioneer 10 1973 Galileo 1995 Galileo 1995 Saturn Pioneer 11 1979 Cassini 2004 Uranus Voyager 2 1986 Neptune Voyager 2 1989 Pluto New Horizons 2015 Ceres Dawn 2015 Moon Luna 1 1959 Luna 10 1966 Luna 2 1959 Luna 9 1966 Titan Huygens 2005 asteroid Galileo asteroid 951 Gaspra – 1991 NEAR Shoemaker asteroid 433 Eros – 2000 NEAR Shoemaker asteroid 433 Eros – 2000 comet ICE comet Giacobini-Zinner – 1985 Rosetta comet Churyumov-Gerasimenko – 2014 Deep Impact – Impactor comet Tempel -2005 Philae comet Churyumov-Gersimenko – 2014

## September 20, 2015

### The simplest Heronian triangles

Filed under: mathematics — ckrao @ 12:05 pm

Heronian triangles are those whose side lengths and area have integer value. Most of the basic ones are formed either by right-angled triangles of integer sides, or by two such triangles joined together. Following the proof in [1] it is not difficult to show that such triangles have side lengths proportional to $(x,y,z) = (n(m^2 + h^2), m(n^2 + h^2), (m+n)(mn-h^2))$ where $m,n$ and $h$ are integers with $mn > h^2$.  Firstly, if a triangle has integer side lengths and area, its altitudes must be rational, being twice the area divided by a side length. Also by the cosine rule, the cosine of its angles must be rational, so $z_1$ and $z_2$ in the diagram below are rational too (here assume $z$ is the longest side, so that the altitude is inside the triangle).

This gives us the equations

$\displaystyle h^2 = x^2 - z_1^2 = y^2 - z_2^2, z_1 + z_2 = z,\quad \quad (1)$

where $h, z_1, z_2 \in \mathbb{Q}$. Letting $x + z_1 = m$ and $y + z_2 = n$ it follows from the above equations that $x - z_1 = h^2/m, y-z^2 = h^2/n$ from which

$\displaystyle (x,y,z) = \left(\left(\frac{1}{2}(m + \frac{h^2}{m}\right), \frac{1}{2}\left(n + \frac{h^2}{n}\right), \frac{1}{2}\left( m - \frac{h^2}{m} + (n - \frac{h^2}{n}\right)\right). \quad\quad (2)$

Scaling the sides up by a factor of $2mn$, the sides are proportional to

$(x',y',z') = (n(m^2 + h^2), m(n^2 + h^2), (m+n)(mn-h^2)).\quad\quad(3)$

Next, letting $d$ be the common denominator of the rational numbers $h, z_1$ and $z_2$, we multiply the rational solution $(x', y', z')$ in (3) each by $d^3$ to obtain an integral solution. The altitude upon side length $z$ is proportional to $2hmn$ and the area is $hmn(m+n)(mn-h^2)$. Hence if we start with positive $m,n,h$ with no common factor and with $mn > h^2$, then (3) gives the side lengths of a Heronian triangle that can then be made primitive by dividing by a common factor.

Below the 20 primitive Heronian triangles with area less than 100 are illustrated to scale, where the first row has been doubled in size for easier viewing (a larger list is here). Note that all but one of them is either an integer right-angled triangle or decomposable into two such triangles as indicated by the blue numbers and sides. Refer to [2] for more on triangles which are not decomposable into two integer right-angled triangles. Here are the primitive Pythagorean triples that feature in the triangles:

• 3-4-5
• 5-12-13
• 8-15-17
• 20-21-29
• 7-24-25
• 28-45-53

#### References

[1] Carmichael, R. D., 1914, “Diophantine Analysis”, pp.11-13; in R. D. Carmichael, 1959, The Theory of Numbers and Diophantine Analysis, Dover Publications, Inc.

[2] Yiu, Paul (2008), Heron triangles which cannot be decomposed into two integer right triangles (PDF), 41st Meeting of Florida Section of Mathematical Association of America.

« Previous PageNext Page »

Blog at WordPress.com.