## Monthly Archives: January 2016

### Steve Jobs quote about TV

You go to the TV to switch off your brain, you go to the computer to turn on your brain.

—- Steve Jobs.

### Solution to puzzle of white tailed cats

The following solution is reproduced from Prof Ian Stewart’s book “Cabinet of Mathematical Curiosities”.

Suppose that there are c cats, of which w have white tails. There are $c(c-1)$ ordered pairs of distinct cats, and $w(w-1)$ ordered pairs of white tailed cats. (You can choose the first cat of the pair in c ways, but the second in only $c-1$ ways since you have used up one cat. Ditto for white tailed cats. By ordered, what is meant is choosing first cat A and then cat B is considered to be different from B first and then A. If you don’t like that, then both formulae have to be halved — with the same result.)

This means that the probability of both cats having white tails is $\frac{w(w-1)}{c(c-1)}$ and this must be 1/2. Therefore, $c(c-1)=w(w-1)$ with c and w being whole numbers. The smallest solution is $c=4$, and $w=3$. The next smallest turns out to  be $c=21$ and $w=15$. Since Ms Smith has fewer than 20 cats, she must have 4 cats, of which three have white tails.

More fun later,

Nalin Pithwa

### Expanding Universe

Reference: Ian Stewart’s Cabinet of Mathematical Curiosities

The starship Indefensible starts from the center of a spherical universe of radius 1,000 light years, and travels radially at a speed of one light year per year — the speed of light. How long will it take to reach the edge of the universe? Clearly, 1000 years. Except that I forgot to tell you that this universe is expanding. Every year, the universe expands its radius instantly by precisely 1000 light years. Now, how long will it take to reach the edge? (Assume that the first such expansion happens exactly one year after the Indefensible starts the voyage, and successive expansions occur at intervals of exactly one year.)

It might seem that the Indefensible never gets to the edge, because that is receding faster than the ship can move. but at the instant that the universe expands, the ship is carried along with the space in which it sits, so its distance from the centre expands proportionately. To make these conditions clearly, let’s look at what happens for the first few years.

In the first year, the ship travels 1 light year, and there are 999 light years left to traverse. Then, the universe instantly expands to a radius of 2000 light years, and the ship moves with it. so it is then 2 light years from the center, and has 1998 left to travel.

In the next year, it travels a further light year, to a distance of 3 light years, leaving 1997. But then the universe expands to a radius of 3000 light years, multiplying its radius by 1.5 so the ship ends 4.5 light years from the center, and the remaining distance increases by 2995.5 light years.

Does the ship ever get to the edge? If so, how long does it take?

Hint:

It will be useful to know that the nth harmonic number $H_{n}=1-1/2+1/3-1/4+ \ldots + 1/n$

is approximately equal to $\log {n} +\gamma$

where $\gamma$ is Euler’s constant, which is roughly 0.5772156649.

More fun later,

Nalin Pithwa

### To take natural talent forward — T. R. Ramadas, Distinguished Prof, Chennai Mathematical Institute

(The following article appeared by the author T. R. Ramadas in today’s prominent newspaper, The Hindu, in print. The author is a Distinguished Professor with the Chennai Mathematical Institute, Chennai.)

Manjul Bhargava, born of Indian parents and raised in Canada, the U.S. and India, was known to the mathematical community before he was awarded the Fields Medal in 2014. An accomplished table player, he was invited to address the Sadas of The Music Academy as the Sangita Kalanidhi title was conferred to Sanjay Subrahmanyan. Bright young Indians will thrill to the words of Professor Bhargava that speak of the deep connections between Indian classical music and mathematics. They well do well to also read the text of Subrahmanyan’s wonderful acceptance speech, which is available at the website of the academy.

There are a number of lessons to be learnt from the lives and careers of these two extraordinary people — Bhargava, 42 years of age and Subrahmanyan, 47.

Let me spell them out.

First, arrange to have plenty of talent. Then,

1. Make sure to be born in a middle class family with a tradition of learning.
2. Find your metier at a young age. In magical fantasy as well as in real life, this calls for the intervention of a mentor — a parent, a close relative or a teacher besotted with the subject.
3. Be lucky enough that charlatans or self-deluded messiah do not guide you down blind-paths. (This will be before you have acquired the good sense to distinguish chaff from wheat.)
5. Acquire the habits of hard work and application; learn the craft and then the art.

There is little one can do regarding points 1 and 2 — unless the state takes its responsibilities seriously and reverses decades of neglect of the school system.

Both for music and mathematics (and for that matter science in general) there are world class institutions in India — try your hardest to gravitate to one of these.

And, as Subrahmanyan says the Internet is the great leveller as far as access is concerned, whether it is music or math. There is an abundance of material available in the form of lectures and performances as well as textbooks and lecture notes, though ill-curated.

Look for something that engages you, seek advice when possible. Always keep  your nonsense-detector on. Learn to focus. Know that progress will be slow, but (hopefully) with sudden epiphanies. Once you find the resources you seek, shut off the computer and work with your teacher (if you are lucky enough to find one) at the desk with paper and pencil, in the laboratory of a clean well-lit room with the tanpura.

(If you like this advice from Prof. T. R. Ramadas, please send a thank you note to him at The Chennai Mathematical Institute. I too thank him. This will help my students immensely.)

Regards,

Nalin Pithwa

### Spelling Mistakes

‘Thare are five mistukes im this centence.’

True or False?

Regards,

Nalin Pithwa

### New Merology

Reference: Prof Ian Stewart’s Cabinet of Mathematical Curiosities

New Merology

Let him that hath understanding count the number of the beast; for it is the number of a man, and his number is Six hundred threescore and six. Revelation of St. John 13:18

Or may be not. The Oxyrhynchus Papyri — ancient documents found at Oxyrhynchus in Upper Egypt — include a fragment of the Book of Revelation from the third or fourth century which contains the earliest version of some sections. The number that this papyrus assigns to the Beast is 616, not 666. So much for barcodes being symbols of evil. No  matter, for this puzzle is not about the Beast. It is about an idea that its inventor Lee Sallows, called “New metrology”. Let it be clear that his proposal was not serious, except as a mathematical problem.

The traditional methods for assigning numbers to names, known as gematria, sets $A=1$, $B=2$ upto $Z=26$. Then you add up all the numbers corresponding to the letters of the name. But there are lots of different systems of this kind, and lots of alphabets. Sallows suggested a more rational method based on words that denote numbers. For instance, with the numbering just described, the word ONE becomes $15+14+5=34$. However, the number corresponding to ONE surely  ought to be 1. Worse, no English number denotes its numerological total, a property we will call “perfect”.

Sallows wondered what happens if assign a whole number to each letter, so that as many as, possible of the number words ONE, TWO, and so on are perfect. To make the problem interesting, different letters must be given different values. So you get a whole pile of equations like $O+N+E=1$ $T+W+O=2$ $T+H+R+E+E=3$

In algebraic unknowns O, N, E, T, W, H, R …And, you must solve them in integers, all distinct.

The equation $O+N+E=1$ tells us that some of the numbers have to be negative. Suppose, for example, that $E=1$ and $N=2$. Then, the equation for ONE tells us that $O=-2$, and similar equations with other number-words imply that $I=4$, $T=7$ and $W=-3$. To make THREE perfect, we must assign values to H and R. If $H=3$, then R has to be -9. FOUR involves two more new letters, F and U. If $F=5$, then $U=10$. Now, $F+I+V+E=5$ leads to $I=-5$. Since SIX contains two new letters, we try SEVEN first, which tells us that $S=8$. Then, we can fill in X from SIX, getting $X=-6$. The equation for EIGHT leads to $G=-7$. Now, all the number names from ONE to TEN are perfect.

The only extra letter in ELEVEN and TWELVE is L. Remarkably, L=11 makes them both perfect. But $T+H+I+R+T+E+E+N=7+3+4+(-9)+7+1+1+2$, which is 16, so we get stuck at this point.

In fact, we always get stuck at this point. If THIRTEEN is perfect, then $THREE+TEN=THIRTEEN$

and we can remove common letters from both sides. This leads to $E=I$, violating the rule that different letters get different values.

However, we can go the other way and try to make ZERO perfect as well as ONE to TWELVE. Using the choices above, $Z+E+R+O=0$ leads to $Z=10$, but that’s the same as U.

Can you find a different assignment of positive or negative whole-number values to letters, so that all words from ZERO to TWELVE are perfect?

More fun later,

Nalin Pithwa

### Solution to another S L Loney problem for IITJEE Advanced Mathematics

Exercise XXVII. Problem #32. The sides of a triangle are in AP and the greatest and least angles are $\theta$ and $\phi$. Prove that $4(1-\cos{\theta})(1-\cos{\phi})=\cos{\theta}+\cos{\phi}$

Proof:

Let a, b, c be in AP. Hence, $a, which in turn implies c is greatest and a is the least.

Hence, $\theta=\angle{C}$ and $\angle{A}$.

Want: $4(1-\cos{A})(1-\cos{C})=\cos{A}+\cos{C}$

Given: $b-a=c-b$ $(b-a)^{2}=(b-c)^{2}$. Hence, $b^{2}-2ab+a^{2}=b^{2}+c^{2}-2bc$ $b^{2}+a^{2}-c^{2}=b^{2}-2bc+2ab$ $2b=a+c$ $4b^{2}=a^{2}+c^{2}+2ac$ $a^{2}+c^{2}-b^{2}=3b^{2}-2ac$ $a^{2}+b^{2}-c^{2}=b^{2}-2bc+2ab$ $b^{2}+c^{2}-a^{2}=b^{2}-2ab+2bc$

Now, we have $1-\cos{A}=1-\frac{b^{2}+c^{2}-a^{2}}{2bc}$ and this is equal to the following: $\frac{2bc-(b^{2}+c^{2}-a^{2})}{2bc}=\frac{2bc-(b^{2}-2ab+2bc)}{2bc}$ $= \frac{2bc-b^{2}+2ab-2bc}{2bc}=\frac{2ab-b^{2}}{2bc}$.

Also, similarly, we have the following: $1-\cos{C}=1-\frac{a^{2}+b^{2}-c^{2}}{2ab}=1-\frac{b^{2}-2bc+2ab}{2ab}=\frac{2bc-b^{2}}{2ab}$ $\cos{A}+\cos{C}=\frac{b^{2}+c^{2}-a^{2}}{}+\frac{a^{2}+b^{2}-c^{2}}{2ab}$ $= \frac{b^{2}-2ab+2bc}{2bc}+\frac{b^{2}-2bc+2ab}{2ab}$ $=\frac{a(b^{2}-2ab+2bc)+c(b^{2}-2bc+2ab)}{2abc}$ $=\frac{ab^{2}-2a^{2}b+2abc+b^{2}c-2bc^{2}+2abc}{2abc}$ $=\frac{ab^{2}+b^{2}c-2a^{2}b-2c^{2}b+4abc}{2abc}$ $= \frac{ab+bc-2a^{2}-2c^{2}+4ac}{2ac}$ $=\frac{b(a+c)-2(a-c)^{2}}{2ac}$ $=\frac{2b^{2}-2(a-c)^{2}}{2ac}$ $=\frac{b^{2}-(a-c)^{2}}{2ac}$ $\frac{(b+a-c)}{(b-a+c)}{ac}$

but it is given that $b-a=c-b$, hence, $a+c=2b$, $a+c-b=b$. So the above expression changes to $= \frac{(b+a-c)(c-b+c)}{ac}$ $= \frac{(2c-b)(a+b-c)}{ac}$ $= \frac{(2c-b)(a+b-\overline{2b-a})}{ac}$ $= \frac{(2c-b)(a-b+c)}{ac}$ $= \frac{(2c-b)(2a-b)}{ac}$ $= \frac{4ac-2bc-2ab+b^{2}}{ac}$ $= 4(1-\cos{A})(1-\cos{C})$.

QED.