1

Problem A5 in the 1985 Putnam Competition: Let $I_m=\int_0^{2\pi}\cos(x)\cos(2x)\cdots\cos(mx)dx$. For which integers $m$, $1\leq m\leq10$, do we have $I_m\neq0$?

The solution rewrites $\cos(x)=\frac{e^{ikx}+e^{-ikx}}{2}$. It then says that $$I_m=\int_0^{2\pi}\prod_{k=1}^m{\biggl(\frac{e^{ikx}+e^{-ikx}}{2}\biggr)}=2^{-m}\sum_{\epsilon_{k}=\pm1}\int_0^{2\pi}{e^{i(\epsilon_1+2\epsilon_2+\cdots+m\epsilon_m)x}}$$ Where the sum ranges over the $2^m$ $m$-tuples $(\epsilon_1,\ldots,\epsilon_m)$ with $\epsilon_k=\pm1$ for every $k$.

My question is, how do you make sense of the last step? Why is this true: $$\prod_{k=1}^m{(e^{ikx}+e^{-ikx})}=\sum_{\epsilon_{k}=\pm1}{e^{i(\epsilon_1+2\epsilon_2+\cdots+m\epsilon_m)x}}$$

1 Answers1

0

We obtain \begin{align*} \color{blue}{\prod_{k=1}^m}&\color{blue}{\left(e^{ikx}+e^{-ikx}\right)}\tag{1}\\ &=\prod_{k=1}^m\left(\sum_{\varepsilon_{k}=\pm 1}e^{i\varepsilon_{k}kx}\right)\\ &=\left(\sum_{\varepsilon_{1}=\pm 1}e^{i\varepsilon_1 x}\right)\left(\sum_{\varepsilon_{2}=\pm 1}e^{i\varepsilon_22 x}\right) \cdots \left(\sum_{\varepsilon_{m}=\pm 1}e^{i\varepsilon_mm x}\right)\\ &=\sum_{\varepsilon_{1}=\pm 1}\sum_{\varepsilon_{2}=\pm 1}\cdots \sum_{\varepsilon_{m}=\pm 1}e^{i\varepsilon_1 x}e^{i\varepsilon_22 x}\cdots e^{i\varepsilon_mm x}\\ &=\sum_{{1\leq k \leq m}\atop{\varepsilon_{k}=\pm 1}}e^{i\varepsilon_1 x}e^{i\varepsilon_22 x}\cdots e^{i\varepsilon_mm x}\\ &\,\,\color{blue}{=\sum_{{1\leq k \leq m}\atop{\varepsilon_{k}=\pm 1}}e^{i\left(\varepsilon_1 +2\varepsilon_2+\cdots +m \varepsilon_{m}\right)x}}\tag{2} \end{align*} and the claim follows.

The product (1) consists of $m$ factors $e^{ikx}+e^{-ikx}$ where $1\leq k\leq m$. From each factor we select either $e^{ikx}$ or $e^{-ikx}$ giving a total of $2^m$ summands. These $2^m$ summands are explicitly stated in (2).

Markus Scheuer
  • 112,413