You are currently browsing the monthly archive for February 2008.

When the mathematician proposed to his girlfriend, why did she turn him down?

Because he offered her a ring but she wanted a field!

(Sarah Brown)

Less than a couple of months ago, we heard of the (untimely) death of Bobby Fischer, undoubtedly the greatest chess player who ever lived on earth. For a lot of people, in his later years he was a raving arrogant “lunatic.” But few people knew/know about his human side, which was brought out in a moving eulogy on Fischer, by Dick Cavett, titled Was It Only a Game? written for The New York Times. The accompanying video in that article shows how “normal” Fischer was, just like you and me.

Here is a wonderful video of Fischer as a 15-yr old kid appearing in a game show I’ve Got a Secret.

The following is a casual interview in which Fischer smiles and laughs as never seen before.

And here is a short documentary on Fischer’s world championship match with Boris Spassky in 1972 and how he beat the gargantuan Soviet chess machine.

This is the first part of a TV documentary titled Fermat’s Last Theorem
made by the famous British science author, Simon Singh, along with John Lynch in 1996. More information on the documentary can be found here.

To quote Simon Singh,

In 1996, working with John Lynch, I made a documentary about Fermat’s Last Theorem for the BBC series Horizon. It was 50 minutes of mathematicians talking about mathematics, which is not the obvious recipe for a TV blockbuster, but the result was a programme that captured the public imagination and which received critical acclaim.

The programme won the BAFTA for best documentary, a Priz Italia, other international prizes and an Emmy nomination – this proves that mathematics can be as emotional and as gripping as any other subject on the planet…

You should definitely watch the opening scene which became very famous later on!
Part 1:

There are five parts in all. The rest of the four may be found by visiting YouTube.

Ok, this one looks like it’s from XKCD but I am not entirely sure. If you do happen to know the source, I will greatly appreciate if you would tell me about it, so that I can properly cite it.

[ Thanks to Dr Armstrong for providing the correct link. ]

This is the first part of the video of a lecture given by Prof Timothy Gowers at the millennium meeting of the Clay Mathematics Institute at College de France, Paris on May 24, 2000. Prof Gowers mentions on his webpage,

As its title suggests, it is an attempt to justify pure mathematics. The lecture was aimed at the general public and assumes very little mathematical background.

There are eight parts in all. The rest of the seven may be found on YouTube.

A transcript of the lecture (without the illustrations) can be found here.

In an earlier issue of Mathematical Reflections, Iurie Boreico (from Harvard) proposed the following problem.

Problem: A polynomial $p \in \mathbb{R}[X]$ is called a “mirror” if $|p(x)| = |p(-x)|$ . Let $f \in \mathbb{R}[X]$ and consider polynomials $p, q \in \mathbb{R}[X]$ such that $p(x) - p'(x) = f(x)$, and $q(x) + q'(x) = f(x)$. Prove that $p + q$ is a mirror iff $f$ is a mirror.

Two solutions – the pdf file size is around 1 Mb – to the above problem were proposed, and the one by the author is very close to the one I had worked out myself (partially) earlier but never really got around finishing it. So, I will post my solution here but at a slightly later time. In the meantime, you might be interested in finding a solution yourself.

Those who love (elementary) problem-solving eventually come across the Sophie Germain identity. It has lots of applications in elementary number theory, algebra and so on. The identity states

$x^4 + 4y^4 = (x^2 - 2xy + 2y^2)(x^2 + 2xy + 2y^2)$.

Indeed, note that

$x^4 + 4y^4$

$= (x^2)^2 + (2y^2)^2 + 2\cdot x^2 \cdot 2y^2 - 2\cdot x^2 \cdot 2y^2$

$= (x^2 + 2y^2)^2 - (2xy)^2$

$= (x^2 - 2xy + 2y^2)(x^2 + 2xy + 2y^2)$

More can read about Marie-Sophie Germain, a brilliant mathematician, here and here.

Let us use the above identity in solving a couple of problems. Here is the first one.

Problem 1: Evaluate $\displaystyle \sum_{k=1}^{n}\frac{4k}{4k^4 + 1}$.

Solution: A first glance tells us that the sum should somehow “telescope.” But the denominator looks somewhat nasty! And, it is here that the above identity comes to our rescue. Using the Sophie Germain identity, we first note that

$1 + 4k^4 = (1 -2k + 2k^2)(1 + 2k + 2k^2).$

We thus have

$\displaystyle \sum_{k=1}^{n} \frac{4k}{4k^4 + 1}$

$\displaystyle = \sum_{k=1}^{n} \left(\frac{(1 + 2k + 2k^2)-(1 -2k + 2k^2)}{(1 -2k + 2k^2)(1 + 2k + 2k^2)}\right)$

$\displaystyle = \sum_{k=1}^{n} \left(\frac1{1 -2k + 2k^2} - \frac1{1 + 2k + 2k^2}\right)$

$\displaystyle = \sum_{k=1}^{n} \left(\frac1{1 -2k + 2k^2} - \frac1{1 - 2(k+1) + 2(k+1)^2} \right)$

$\displaystyle = 1 - \frac1{1 + 2n + 2n^2}$.

Here is another one.

Problem 2: Show that $n^4 + 4^n$ is a prime iff $n=1$, where $n \in \mathbb{N}$.

Solution: Note that if $n$ is even, then the expression is clearly composite. If $n$ is odd, say, $2k+1$ for some $k \in \mathbb{N}$, then we have

$n^4 + 4^n$

$= n^4 + 4^{2k+1}$

$= n^4 + 4\cdot (2^k)^4$

$= (n^2 - n\cdot 2^{k+1} + 2^{2k+1})(n^2 + n\cdot 2^{k+1} + 2^{2k+1})$

And, if $n > 1$, then both the factors above are greater than $1$, and hence the expression is composite. Moreover, if $n = 1$, we have $1^4 + 4^n = 5$, which is prime. And, we are done.

Here are three “little” problems that my friend John (from UK) gave me yesterday.

$(1)$ How many 5-digit numbers can be constructed using only $1, 2$ and $3$ such that each of those three digits is used at least once?

$(2)$ Find all pairs of positive integers $(m,n)$ such that $1 + 5\cdot 2^m = n^2$.

$(3)$ Find all positive integers $a, b$ such that $a^b = b^a$.

The following fun problem was posed in one of the issues of the American Mathematical Monthly (if I am not wrong). I don’t remember the exact issue or the author, but here is the problem anyway.

Prove that $\sqrt[n]{2}$ is irrational for all $n > 2$ and $n \in \mathbb{N}$.

Slick solution: We could either use Euclid’s arguments or invoke the rational root theorem to prove the above statement. However, there is a slicker proof!

Assume, for the sake of contradiction, that $\sqrt[n]{2} = p/q$, where $p, q \in \mathbb{N}$ and $p \ne 0$. Then, we have $2 = (p/q)^n$ which implies $q^n + q^n = p^n$. But this contradicts Fermat’s Last Theorem! And, we are done.

There is immense joy and thrill in discovering that one of the world’s greatest mathematicians (and probably many more like him) got interested in mathematics, just as I did as a kid, after reading one of Yakov I. Perelman‘s popular science books. Physics for Entertainment is the book and Grisha Perelman is the mathematician I am referring to!

Yakov I. Perelman’s books on physics, mathematics and astronomy were written in a style that brought out many of the aspects of the aforesaid subjects in the most enjoyable way. He breathed life into every page of his books and made mathematics and physics accessible to any kid in a way that brought sheer joy to the soul! At the same time, his writings provided a glimpse of the amazing way that physics helps us understand and study the nature around us. Physics Can Be Fun, Mathematics Can be Fun and Astronomy for Entertainment are the titles of some of his other books that were immensely popular among young students who read them. The credit for my long-lasting interest/passion in physics and mathematics solely goes to him.

I think all twelve-year olds should be given copies of his books as birthday gifts instead of toys! Surprisingly, there is an online copy of Physics For Entertainment here. I am not sure if any copyright laws will be violated if you download the online version but it seems that the site is a “genuine” one.

• 346,162 hits