# Lessons from a grandmaster

The next game is the only tournament game I’ve ever played against a grandmaster. My opponent was Georgian GM Mikheil Kekelidze. Not surprisingly, he outplayed me in every phase of the game. Obviously, I’ve still got a lot of work to do!

# Combinatorics, Contour Integrals, and Bessel Functions … oh my?!

Here’s an interesting exercise that can be solved using complex integration and basic combinatorics. Consider the following integral:

$$\displaystyle\int_{0}^{2\pi} \! \exp{(2\cos \theta)} \, \mathrm{d} \theta.$$

No apparent tricks from ordinary calculus appear to be helpful, so we turn instead to complex variables. The limits of integration suggest a parametrization of the unit circle. Let $$C$$ denote the unit circle, oriented counterclockwise. We’d like to convert our original integral into a contour integral of the form $$\oint_C \! f(z) \, \mathrm{d} z$$ for an appropriate function $$f(z)$$. We can parametrize the contour by letting $$z = e^{i\theta}, 0 \le \theta \le 2\pi.$$ Then $$\mathrm{d} z = ie^{i\theta} \mathrm{d} \theta$$, so that $$\mathrm{d} \theta = \frac{1}{i} e^{-i\theta} \mathrm{d} z$$, or in other words, $$\mathrm{d} \theta = \frac{1}{i} z^{-1} \mathrm{d} z.$$ We also observe that, on our contour, $$z + z^{-1} = e^{i\theta} + e^{-i\theta} = (\cos \theta + i \sin \theta) + (\cos \theta – i \sin \theta) = 2 \cos \theta$$. Hence,

$$\displaystyle\int_{0}^{2\pi} \! \exp{(2\cos \theta)} \, \mathrm{d} \theta = \displaystyle\oint_C \! \exp{(z + z^{-1})}\frac{1}{i} z^{-1} \, \mathrm{d} z.$$

Within the region bounded by the contour, the integrand has exactly one singularity. More precisely, it has an essential singularity at $$z=0$$. Thus, by Cauchy’s residue theorem, the value of the integral is equal to $$2\pi i$$ times the integrand’s residue at $$z=0$$.

Let $$f(z) = z^{-1} \exp{(z + z^{-1})}$$. To find the residue of $$f$$ at $$0$$, we consider the Laurent series for $$f$$ centered at $$0$$. (A simple approach here is to write $$\exp{(z + z^{-1})} = \exp{(z)} \exp{(z^{-1})}$$, but we shall take a more roundabout path. The reader is invited to attempt the direct route.) Using the standard series for the exponential, we obtain

$$f(z) = z^{-1} \displaystyle\sum_{n=0}^{\infty} \frac{(z + z^{-1})^n} {n!}.$$

The residue of $$f$$ at $$z = 0$$ is defined as the coefficient of the $$z^{-1}$$ term in the Laurent series for $$f$$ centered at $$0$$. From the above expansion, we see that in order to determine this coefficient, it suffices to determine the coefficient of $$z^0$$ within the infinite sum.

Fix a nonnegative integer $$n$$ and consider the binomial $$(z + z^{-1})^n$$. We wish to determine the coefficient of $$z^0$$ in the expansion of this binomial. In order to obtain a $$z^0$$ term in the first place, we see that the terms $$z$$ and $$z^{-1}$$ must multiply together in pairs. Hence, if $$n$$ is odd, the coefficient of $$z^0$$ is $$0$$. Thus, we may focus our attention on the nonnegative even integers.

Consider the binomial $$(z + z^{-1})^{2n}$$. What is the coefficient of $$z^0$$ in its expansion? If we imagine $$2n$$ copies of the expression $$(z + z^{-1})$$ being multiplied together, the only way we obtain a nonzero $$z^0$$ term is when we choose the $$z$$ and $$z^{-1}$$ terms in pairs. There will be a total of $$n$$ pairs. Of the $$2n$$ copies of $$z$$, we can choose any $$n$$ of them, and then the remaining $$n$$ choices will have to be $$z^{-1}$$ terms. Hence, the coefficient of $$z^0$$ is $${{2n}\choose{n}}$$.

Since, in our original series, each term $$(z + z^{-1})^n$$ is divided by $$n !$$, we must divide $${{2n}\choose{n}}$$ by $$(2n)!$$. By definition,  $${{2n}\choose{n}} = \frac{(2n)!}{n! n!}$$, so  $${{2n}\choose{n}}$$ divided by $$(2n)!$$ is simply $$\frac{1}{n! n!}$$.

Thus, the desired residue is seen to be

$$\displaystyle\sum_{n=0}^{\infty} \frac{1}{n! n!}.$$

It follows by the residue theorem that the value of the integral we seek is equal to

$$\displaystyle 2\pi i \frac{1}{i} \sum_{n=0}^{\infty} \frac{1}{n! n!} = 2\pi \displaystyle\sum_{n=0}^{\infty} \frac{1}{n! n!}. \, \spadesuit$$

It turns out that the original integral is not entirely unmotivated, as it can also be written as $$2\pi I_{0}(2)$$, where $$I_{n}(z)$$ denotes the modified Bessel function of the first kind. This result can be seen from the integral formulas for these functions.

# Percussion quartet

Here’s my senior composition project from Brown, written in 2008–2009. Titled “Percussion Quartet,” it’s scored for two pianists and two percussionists, much in the spirit of Bartók’s well-known piece for the same instruments. Lasting approximately 16 minutes, the piece is divided into four movements, the first two being played attacca.

Movement III Score

Movement IV Score

An analytic commentary is also available.