# Chaitanya's Random Pages

## September 27, 2010

### My Six Favourite Formulas – #4

Filed under: mathematics — ckrao @ 2:07 pm

Here I will discuss another of my favourite formulas:

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$

It boggles the mind how the sum of reciprocals of squared natural numbers could possibly be related to $\pi$! Interestingly, the reciprocal of this number, $6/\pi^2 \approx 0.6079$, is the probability that two randomly selected positive integers are relatively prime. That is, choose two integers uniformly and independently between 1 and N, then as N tends to infinity the probability they share no common factor other than 1 is $6/\pi^2$.

Finding the infinite sum was first posed by Pietro Mengoli in 1644 and is now known as the Basel problem. It stumped the great mathematicians of the next ninety years, ranging from Leibniz to Jacob Bernoulli. Finally in 1735 Euler became famous after being the first to find the answer, aged 28. He used his immense talent in numerical analysis (using what is now known as the Euler-Maclaurin formula) to conjecture the sum to be $\frac{\pi^2}{6}$, then this motivated him to use series formulas for $\frac{\sin x}{x}$ to prove the result. We will outline three other proofs below, more proofs can be found in Robin Chapman’s collection of 14 proofs here.

### Proof 1: Elementary proof, found on MathOverflow website here.

The idea is to sandwich the sum $\displaystyle \sum_{n=1}^{N} \frac{1}{n^2}$ between two expressions which each converge to $\frac{\pi^2}{6}$ as $N \to \infty$.

$\displaystyle \sin x < x < \tan x$, where $0 < x < \pi/2$.

This can be seen by drawing a unit circle and comparing the areas of a sector with internal angle $x$ radians (area $x/2$) , a triangle with unit sides and internal angle $x$ (area $\frac{1}{2}ab\sin C = \frac{1}{2}\sin x$) and a right angle triangle with base length 1 and angle $x$ (area $\frac{1}{2}bh =\frac{1}{2}\tan x$).

Next, let $x_n = \frac{n}{2^N}\frac{\pi}{2}, n = 1,2, \ldots, 2^N - 1$. Then from the above inequality,

$\displaystyle \sum_{n=1}^{2^N-1}\frac{1}{\tan^2 x_n} < \sum_{n=1}^{2^N-1} \frac{1}{x_n^2} < \sum_{n=1}^{2^N-1} \frac{1}{\sin^2 x_n}........(*)$

Now let us evaluate the right side of this inequality, denoted by $S_N$. We sum terms symmetric about $\pi/4$ and find

$\begin{array}{lcl} \frac{1}{\sin^2 x_n} + \frac{1}{\sin^2 x_{(2^N-n)}}&=& \frac{1}{\sin^2 \frac{n}{2^N}\frac{\pi}{2}} + \frac{1}{\sin^2 \frac{2^N-n}{2^N}\frac{\pi}{2}}\\&=&\frac{1}{\sin^2 x_n} +\frac{1}{\sin^2 (\pi/2-x_n)}\\&=& \frac{\sin^2 x_n + \cos^2 x_n}{\sin^2 x_n \cos^2 x_n}\\&=&\frac{4}{\sin^2 2x_n}\\&=&\frac{4}{\sin^2 \frac{n}{2^{(N-1)}}\frac{\pi}{2}}.\end{array}$

Hence

$\begin{array}{lcl} S_N &=& \sum_{n=1}^{2^N-1} \frac{1}{\sin^2 x_n}\\&=&\sum_{n=1}^{2^{(N-1)}-1} \left(\frac{1}{\sin^2 x_n} + \frac{1}{\sin^2 x_{(2^N-n)}}\right)+\frac{1}{\sin^2 x_{2^{(N-1)}}}\\&=&\frac{1}{\sin^2 \pi/4} + \sum_{n=1}^{2^{(N-1)}-1} \frac{4}{\sin^2 \frac{n}{2^{(N-1)}}\frac{\pi}{2}}\\&=&2 + 4S_{N-1}.\end{array}$

This means we have the recurrence relation $S_N = 2 + 4S_{N-1}$ with the initial condition $S_1 = 2$ (initial terms 2, 10, 42, 170, …). Using mathematical induction or generating functions (z-transforms), one can show that this has the closed form solution $S_N = 2(4^N-1)/3$. Similarly,

$\displaystyle \sum_{n=1}^{2^N-1}\frac{1}{\tan^2 x_n} = \sum_{n=1}^{2^N-1} \frac{1-\sin^2 x_n}{\sin^2 x_n} = \frac{2(4^N-1)}{3} - (2^N-1).$

Then from (*),

$\displaystyle \frac{2(4^N-1)}{3} - (2^N-1) < \sum_{n=1}^{2^N-1} \frac{1}{\left(\frac{n}{2^N}\frac{\pi}{2}\right)^2} < \frac{2(4^N-1)}{3}.$

This implies

$\displaystyle \frac{\pi^2}{4^{N+1}}\left(\frac{2(4^N-1)}{3} - (2^N-1)\right) < \sum_{n=1}^{2^N-1} \frac{1}{n^2}< \frac{\pi^2}{4^{N+1}}\frac{2(4^N-1)}{3}.$

We see that both left and right sides tend to $\frac{\pi^2}{6}$ as $N \to \infty$, and we conclude that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$.

### Proof 2: Integration by clever change of variables, due to Calabi, Beukers and Kock.

Using the result $\displaystyle \int_0^1 x^{2r}\ dx = \frac{1}{2r+1}$,

$\begin{array}{lcl} \sum_{r=0}^{\infty} \frac{1}{(2r+1)^2} &=& \sum_{r=0}^{\infty}\left(\int_0^1 x^{2r}\ dx \right)\left(\int_0^1 y^{2r}\ dy \right)\\&=&\int_0^1 \int_0^1 \sum_{r=0}^{\infty} (x^2y^2)^r\ dx\ dy\\&=& \int_0^1 \int_0^1 \frac{1}{1-x^2y^2}\ dx\ dy.\end{array}$

Then make the remarkable substitution

$\displaystyle (x,y) = \left(\frac{\sin u}{\cos v}, \frac{\sin v}{\cos u}\right).$

We find that this sets up a bijective transformation between the unit square $\{(x,y): 0 < x < 1, 0 < y < 1\}$ and the triangle

$\displaystyle A = \{(u, v) : u > 0, v > 0, u + v < \pi/2\}.$

(Note that $\sin u/\cos v < 1$ for $0 implies $\sin u < \sin\left(\frac{\pi}{2} - v\right)$ and so $u < \pi/2 - v$ or $u + v < \pi/2$. The inverse map is $\displaystyle (u,v) = \left(\arctan x\sqrt{\frac{1-y^2}{1-x^2}}, \arctan y\sqrt{\frac{1-x^2}{1-y^2}}\right)$.)

Also we find

$\begin{array}{lcl} dx dy &=& d\left(\frac{\sin u}{\cos v}\right)\wedge d\left(\frac{\sin v}{\cos u}\right)\\&=& \left(\frac{\cos u}{\cos v}du + \frac{\sin u \sin v}{\cos^2 v}dv\right) \wedge \left(\frac{\cos v}{\cos u}dv + \frac{\sin v \sin u}{\cos^2 u}du\right)\\&=& \left( \frac{\cos u}{\cos v}\frac{\cos v}{\cos u} -\frac{\sin u \sin v}{\cos^2 v}\frac{\sin v \sin u}{\cos^2 u}\right)du dv\\&=&(1-x^2y^2)dudv.\end{array}$

This means

$\displaystyle \sum_{r=0}^{\infty} \frac{1}{(2r+1)^2} = \int\int_A du dv$, where $A$ is defined above, and has area $\pi^2/8$. Hence

$\begin{array}{lcl}\sum_{n=1}^{\infty} \frac{1}{n^2}&=& \sum_{r=1}^{\infty} \frac{1}{(2r)^2} + \sum_{r=0}^{\infty} \frac{1}{(2r+1)^2} \\&=& \frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2} + \frac{\pi^2}{8}\end{array}$

from which $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{8}/\left(1-\frac{1}{4}\right) = \frac{\pi^2}{6}$.

### Proof 3: Proof by Fourier series

This is the proof seen in many textbooks, but requires a more advanced result than the previous two, namely Parseval’s theorem. We use the fact that the complex exponentials $e_n(x) = \exp(2\pi i n x)$ where $n \in \mathbb{Z}$ form a complete orthonormal basis in the set $L^2[0,1]$ of square-integrable complex-valued functions on the unit interval [0,1]. This means such a function $f$ on the interval [0,1] can be written as

$\displaystyle f(x) = \sum_{n=-\infty}^{\infty} \alpha_n e_n(x)$,

where $\alpha_n$ are Fourier coefficients which may be found by taking the inner product of both sides of the above with $e_n$:

$\displaystyle \langle f, e_n \rangle = \sum_{m=-\infty}^{\infty} \alpha_m \langle e_m, e_n\rangle = \alpha_n.$

In linear algebra terms, we are simply projecting $f$ onto the basis vector $e_n$ to find its component in that direction. Parseval’s theorem is simply a generalisation of Pythagoras’s theorem, stating that the square of the norm of $f$ is the sum of the squared magnitudes of its components:

$\displaystyle \|f\|^2 = \langle f,f \rangle = \sum_{n= -\infty}^{\infty} \left| \langle f,e_n\rangle \right|^2.$

We now calculate this for $f(x) = x$, noting that $\langle f,g \rangle = \int_0^1 f (x)\bar{g}(x)\ dx$.

• $\displaystyle \langle f,f \rangle = \int_0^1 x^2\ dx = \frac{1}{3}$
• $\displaystyle \langle f,e_0 \rangle = \int_0^1 x\ dx = \frac{1}{2}$
• For $n \neq 0$,

$\begin{array}{lcl} \langle f,e_n \rangle &=& \int_0^1 xe^{-2\pi i n x}\ dx\\&=& \left[\frac{-xe^{-2\pi i nx}}{2\pi in} \right]_0^1 + \int_0^1 \frac{e^{-2\pi i n x}}{2\pi in}\ dx\\&=& \frac{-e^{-2\pi i n}}{2\pi i n} + \frac{1}{(2\pi i n)^2}\left(-e^{-2\pi i n} + 1\right)\\&=& \frac{-1}{2\pi i n}.\end{array}$

Combining these results in Parseval’s theorem gives

$\displaystyle \frac{1}{3} = \frac{1}{2^2} + \sum_{n \in \mathbb{Z}, n \neq 0} \frac{1}{4\pi^2 n^2} = \frac{1}{4} + \frac{1}{2\pi^2}\sum_{n=0}^{\infty} \frac{1}{n^2},$

from which $\sum_{n=1}^{\infty} \frac{1}{n^2} = 2\pi^2\left(\frac{1}{3} - \frac{1}{4}\right) = \frac{\pi^2}{6}$, as desired.

## September 19, 2010

### Number of Quarter Final Appearances of Top Tennis Players

Filed under: sport — ckrao @ 1:02 pm

These tables show the number of quarter final, semi final, final and winning appearances in grand slam singles tournaments of more recent top tennis players. I have included players who have been ranked number 1,  have won at least 2 grand slam tournaments during the Open era, or have played in many semi finals during the Open era.

A * next to a player’s name means their records prior to the Open era (1968) are also included. Statistics are current to the end of 2018, corrections welcome.

Here are a few interesting observations:

• Laver won 17 out of 18 semi finals, Borg 16 out of 17.
• Connors won 15 out of 31 semi finals.
• Chris Evert won 52 out of 54 quarter finals!
 MEN >= QF >= SF >= F W Roger Federer 53 43 30 20 Rafael Nadal 36 29 24 17 Novak Djokovic 42 33 23 14 Pete Sampras 29 23 18 14 Roy Emerson* 36 19 15 12 Rod Laver* 20 18 17 11 Björn Borg 21 17 16 11 Ivan Lendl 34 28 19 8 Ken Rosewall* 30 25 16 8 Jimmy Connors 41 31 15 8 Andre Agassi 36 26 15 8 John McEnroe 26 19 11 7 John Newcombe* 21 13 9 7 Stefan Edberg 26 19 11 6 Mats Wilander 20 14 11 6 Boris Becker 23 18 10 6 Guillermo Vilas 19 12 8 4 Jim Courier 15 11 7 4 Andy Murray 30 21 11 3 Arthur Ashe 14 10 5 3 Jan Kodeš 10 6 5 3 Stan Wawrinka 15 9 4 3 Gustavo Kuerten 8 3 3 3 Ilie Năstase 12 6 5 2 Lleyton Hewitt 15 8 4 2 Marat Safin 9 7 4 2 Patrick Rafter 7 7 4 2 Yevgeny Kafelnikov 13 6 3 2 Stan Smith 10 5 3 2 Sergi Bruguera 4 4 3 2 Johan Kriek 11 5 2 2 Andy Roddick 19 10 5 1 Juan Carlos Ferrero 9 6 3 1 Carlos Moyà 8 3 2 1 Thomas Muster 9 4 1 1 Marcelo Ríos 6 1 1 0 WOMEN >= QF >= SF >= F W Margaret Court* 43 36 29 24 Serena Williams 48 36 31 23 Steffi Graf 42 37 31 22 Chris Evert 54 52 34 18 Martina Navratilova 53 44 32 18 Billie Jean Moffitt King* 40 26 18 12 Monica Seles 31 18 13 9 Evonne Goolagong Cawley 26 22 18 7 Venus Williams 39 23 16 7 Justine Henin 19 17 12 7 Martina Hingis 24 19 12 5 Maria Sharapova 25 20 10 5 Arantxa Sánchez Vicario 35 22 12 4 Kim Clijsters 19 16 8 4 Hana Mandlíková 23 14 8 4 Lindsay Davenport 31 18 7 3 Angelique Kerber 10 7 4 3 Jennifer Capriati 23 13 3 3 Virgina Wade* 23 9 3 3 Nancy Richey Gunter* 22 13 6 2 Mary Pierce 14 6 6 2 Victoria Azarenka 16 7 4 2 Svetlana Kuznetsova 13 5 4 2 Amélie Mauresmo 17 8 3 2 Garbiñe_Muguruza 7 4 3 2 Tracy Austin 13 5 2 2 Simona Halep 11 6 4 1 Gabriela Sabatini 28 18 3 1 Caroline Wozniacki 10 7 3 1 Ana Ivanović 9 6 3 1 Dinara Safina 7 5 3 0 Pam Shriver 19 9 1 0 Jelena Janković 8 6 1 0 Karolína Plíšková 6 2 1 0

## September 14, 2010

### My Six Favourite Formulas – #3

Filed under: mathematics — ckrao @ 1:01 pm

Continuing the look at my favourite formulas, behold the following:

$\displaystyle \left[e^{{\mathbf a}.\nabla}\right]f({\mathbf x}) = f(\mathbf{ x+a})$

Here $f$ is an analytic scalar-valued function on some domain of $\mathbb{R}^n$. This is in fact the Taylor series in disguise, in which knowledge of all derivatives of a function at one point enables us to calculate its value at a different point. It may more commonly be seen in operator form as:

$\displaystyle f(\mathbf{ x+a}) = f({\mathbf x}) + ({\mathbf a}.\nabla)f({\mathbf x}) + \frac{({\mathbf a}.\nabla)^2}{2!}f({\mathbf x}) + \frac{({\mathbf a}.\nabla)^3}{3!}f({\mathbf x}) + \ldots$

or in scalar form as:

$\displaystyle f(x+a) = f(x) + af'(x) + \frac{a^2}{2!}f''(x) + \frac{a^3}{3!}f'''(x) + \ldots$

The exponent ${\mathbf a}.\nabla$$\left(= a_1 \frac{\partial}{\partial x_1} + \ldots + a_n \frac{\partial}{\partial x_n}\right)$ represents a directional derivative operator (a tangent vector), which when applied to a function gives its rate of change in the ${\mathbf a}$ direction:

$\displaystyle \left({\mathbf a}.\nabla\right) g({\mathbf x}) = \left[\frac{d}{dt} g({\mathbf x} + t{\mathbf a}) \right]_{t=0}$

Similarly $\displaystyle \left({\mathbf a}.\nabla\right)^k$ is the k’th order directional derivative.

The beautiful compact formula at the top of this post reveals another use of the exponential function apart from what we may be used to seeing. It takes us from from the infinitesimal (local) world of tangent vectors to the macroscopic (global) world of translations. Applying the exponential of the tangent vector in the direction ${\mathbf a}$ is equivalent to a translation by ${\mathbf a}$. $({\mathbf a}.{\mathbf \nabla})$ is also known as a generator of translation. In the theory of Lie algebras and groups, the exponential map generalises this concept and takes us from a Lie algebra (the space of tangent vectors at the identity element) to its corresponding Lie group. In quantum mechanics the operator $e^{-iHt/\hbar}$ generates the evolution of $\psi(x,0)$ into $\psi(x,t)$ according to the time-dependent Schrödinger equation $\displaystyle i\hbar \frac{\partial \psi}{\partial t} = H\psi$ in the same way.

1. Non-rigorous justification

One intuitive way of seeing why the Taylor series formula holds is to regard $f({\mathbf x}+{\mathbf a})$ as a large number $n$ of successive infinitesimal translations by ${\mathbf a}/n$ and then to approximate each infinitesimal translation by its linearisation

$\displaystyle f({\mathbf x} + {\mathbf a}/n) \approx Tf({\mathbf x}) := \left(1 + \frac{{\mathbf a}. {\mathbf \nabla}}{n}\right)f({\mathbf x}).$

Then

$\displaystyle \begin{array}{lcl} f({\mathbf x} + {\mathbf a})&=& f({\mathbf x} + n.{\mathbf a}/n)\\&=& \lim_{n\rightarrow \infty}T^nf({\mathbf x})\\&=&\lim_{n\rightarrow \infty} \left(1 + \frac{{\mathbf a}. {\mathbf \nabla}}{n}\right)^nf({\mathbf x})\\&=& \left[e^{{\mathbf a}.\nabla}\right]f({\mathbf x}).\end{array}$

2. Fourier transform “proof”

If we assume the Fourier transform of $f$ exists (defined by $\displaystyle F({\mathbf k}) = \frac{1}{(2\pi)^{n/2}}\int_{{\mathbb R}^n} f({\mathbf x}) e^{-i{\mathbf k}.{\mathbf x}}\ d{\mathbf x}$), another way of understanding the Taylor series formula is by using the following two properties of Fourier transform pairs:

$\displaystyle \left({\mathbf a}.\nabla\right)f({\mathbf x}) \leftrightarrow i({\mathbf a}.{\mathbf k})F({\mathbf k})$
$\displaystyle f({\mathbf x} + {\mathbf a}) \leftrightarrow e^{i {\mathbf a}.{\mathbf k}}F({\mathbf k})$

Using the power series definition of the exponential function, we can use repeated applications of the first property to write

$\displaystyle e^{{\mathbf a}.\nabla}f({\mathbf x}) \leftrightarrow e^{i {\mathbf a}.{\mathbf k}}F({\mathbf k})$

and our formula results by comparing this with the second Fourier transform property above. This equation also shows why in physics momentum (proportional to the wave number ${\mathbf k}$) is said to be the generator of translation.

3. Proof of the formula by solving a partial differential equation

Fix the vector $\mathbf{a}$ and let $g({\mathbf x},t) = \left[e^{t{\mathbf a}.\nabla}\right]f({\mathbf x})$. Partially differentiating both sides with respect to $t$ gives

$\displaystyle \begin{array}{lcl}\frac{\partial g}{\partial t} ({\mathbf x},t)&=&\frac{\partial }{\partial t}\left[e^{{\mathbf a}.\nabla}\right]f({\mathbf x})\\&=&\frac{\partial }{\partial t}\sum_{k=0}^{\infty}\frac{\left(t{\mathbf a}.{\mathbf \nabla}\right)^k}{k!}f({\mathbf x})\\&=&\sum_{k=1}^{\infty}\frac{t^{k-1}\left({\mathbf a}.{\mathbf \nabla}\right)^k}{(k-1)!}f({\mathbf x})\\&=&\left({\mathbf a}.{\mathbf \nabla}\right)\sum_{k=1}^{\infty}\frac{t^{k-1}\left({\mathbf a}.{\mathbf \nabla}\right)^{k-1}}{(k-1)!}f({\mathbf x})\\&=&\left({\mathbf a}.{\mathbf \nabla}\right)\sum_{k=0}^{\infty}\frac{t^k\left({\mathbf a}.{\mathbf \nabla}\right)^k}{k!}f({\mathbf x})\\&=&\left({\mathbf a}.{\mathbf \nabla}\right)\left[e^{{\mathbf a}.\nabla}\right]f({\mathbf x})\\&=&\left({\mathbf a}.{\mathbf \nabla}\right)g({\mathbf x},t).\end{array}$

This is a first order partial differential equation which may be solved by the method of characteristics. Introduce a parameter $u$ so that ${\mathbf x}$ and $t$ are functions of $u$. Then by the chain rule,

$\displaystyle \frac{dg}{du}={\mathbf \nabla}g.\frac{d{\mathbf x}}{du} + \frac{\partial g}{\partial t}\frac{dt}{du} \quad \left(= \sum_{i=1}^n \frac{\partial g}{\partial x_i} \frac{dx_i}{du} + \frac{\partial g}{\partial t}\frac{dt}{du}\right).$

From before, $\frac{\partial g}{\partial t}=\left({\mathbf a}.{\mathbf \nabla}\right)g = \left({\mathbf \nabla}g\right).{\mathbf a}$, so

$\displaystyle \frac{dg}{du} = {\mathbf \nabla}g.\frac{d{\mathbf x}}{du} +\left({\mathbf \nabla}g\right).{\mathbf a}\frac{dt}{du} = {\mathbf \nabla}g.\frac{d}{du}\left({\mathbf x} + t{\mathbf a} \right).$

This equation becomes 0 if $\frac{d}{du}\left({\mathbf x} + t{\mathbf a} \right) = 0$, or equivalently, $g$ is constant if ${\mathbf x} + t{\mathbf a}$ is constant. We conclude that

$\displaystyle g({\mathbf x}, t) = \Psi({\mathbf x} + t{\mathbf a})$

where $\Psi$ is a function. We know that $g({\mathbf x},t=0) = e^0 f({\mathbf x}) = f({\mathbf x})$, and so $\Psi({\mathbf x}) = \Psi({\mathbf x} + 0{\mathbf a}) = f({\mathbf x})$ for all ${\mathbf x}$.  Hence we have shown $g({\mathbf x},t) = \Psi({\mathbf x}+t{\mathbf a}) = f({\mathbf x}+t{\mathbf a})$, and it is easily verified that this satisfies $\frac{\partial g}{\partial t} ({\mathbf x},t) = \left({\mathbf a}.{\mathbf \nabla}\right)g({\mathbf x},t).$ By setting $t$ equal to 1 we conclude that $\displaystyle \left[e^{{\mathbf a}.\nabla}\right]f({\mathbf x}) = f(\mathbf{ x+a}).$

## September 5, 2010

### Sir Viv Richards stats and clips

Filed under: sport — ckrao @ 11:01 am

One of my favourite cricket players while growing up was Sir Vivian Richards of the West Indies. The first time I remember seeing him live was the 1984 Boxing Day test against Australia, in which he scored a brilliant 208, steering his team from a precarious 5-154 to 479. In an era when 70 was considered a good strike rate in one day internationals, his was 90 with an average close to 50 to boot. He was easily the most intimidating batsman of his time. While every other cricket coaching book dictates that you hit the ball on the ground, I recall his coaching manual had a section on how to go for the big one straight down the ground.

Batting aside I remember him as a fantastic close-in fielder, responsible for countless run outs through direct hits. His off-spinners were also handy especially in one-day internationals, and he even scored a century and took 5 wickets in the same game. Unfortunately I was born too late to see his early career, for which his batting statistics are particularly impressive. In 1976 he scored an astonishing 1710 runs in 11 tests and at the end of March 1981 had this imposing record in tests:

 Tests 43 Innings 67 n.o. 4 Runs 3954 HS 291 Average 62.76 100s 13 50s 16

By the end of October 1986 he had this record in one-day internationals:

 Matches 110 Innings 100 n.o. 19 Runs 4607 HS 189* Average 56.87 SR 90.35 100s 8 50s 34

Some of his highlights include:

• being voted one of the five Cricketers of the Century in 2000
• never losing a test series as captain (50 tests)
• scoring the fastest test century (in balls faced: 56)
• being ranked #1 in the Reliance Mobile ICC Player Rankings for ODIs from 23 December 1979 to 19 October 1989 (aged 37) (source: here)

He also had a distinguished career with Somerset, was one of the dominant batsmen during the World Series cricket era, and scored over 100 first class centuries.

CricInfo has a “Legends of Cricket” feature of him here.

There are a couple of nice pieces here:
Viv Richards: Antigua’s pride
Viv Richards: bowler killer

More of his stats on Cricinfo Magazine here.

Sir Viv never wore a helmet yet was fearless in playing the hook shot. One famous incident involves being struck in the face by a Rodney Hogg bouncer, and then proceeding to hook the next ball for a six!

As with a few other great batsmen it was considered counter-productive to sledge him while at the crease. During a county match between Glamorgan and Somerset, Glamorgan speedster Thomas had beaten the edge of Richards’ bat a couple of times and said to him: “It’s red, round, and weighs about five ounces, in case you were wondering.” The very next ball was hit by Sir Viv out of the ground into a river! Casually gardening the pitch, Richards said: “Greg, you know what it looks like. Now go and find it.”

Here are a few video clips of him in action. The first is from the 3rd final of the 1988/9 Benson & Hedges World Series Cup, played in Sydney.

I love his 6s at 0:41, 0:58 and 3:45, the last one inside out over cover! The remaining shots are pretty good too!

Secondly here are highlights of his highest score in one day internationals of 189*, scoring 93 in the last wicket stand alone! (The ninth wicket falls just before the 10 minute mark of the clip.)

Finally here are some of his early-career highlights, including his amazing series in England in 1976 and his 138 in the 1979 World Cup final (check out the last-ball 6 at 5:18!). He had such a good eye that he could dispatch balls outside off to the legside with ease.

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