I could only solve this problem via brute force, trying every value from $m = 1$ to $10$...
What is the more efficient and proper method of approach?
(Note: my method involved repeated usage of the product-to-sum trig identity to achieve a simplified result of the integrand; but is there a way to generalize this simplified result for all $m \in [1,10]$?)
Thanks
Let $$I_{m}=\int_{0}^{\pi}\prod_{k=1}^{m}\left(\dfrac{e^{ikx}+e^{-ikx}}{2}\right)dx=2^{-m}\sum_{\xi=\pm 1}\int_{0}^{\pi}e^{i(\xi_{1}+2\xi_{2}+\cdots+m\xi_{m})x}dx$$ since $$\int_{0}^{\pi}e^{itx}dx=\begin{cases}\pi&t=0\\ 0&\textbf{otherwise} \end{cases}$$ then $I_{m}\neq 0$, if and only if $0$ can written as $$\xi_{1}+2\xi_{2}+\cdots+m\xi_{m},\xi_{i}\in\{-1,1\},i=1,2,\cdots,m$$ It is to easy to find when $$m\equiv 3,4\pmod 4$$ because $$0=\xi_{1}+2\xi_{2}+\cdots+m\xi_{m}=1+2+\cdots+m=\dfrac{m(m+1)}{2}\pmod 2$$ so $$\Longrightarrow m(m+1)\equiv 0\pmod 4\Longrightarrow m\equiv 3,4\pmod 4$$then satisfying this condition except are $3,4,7,8$, so if $$I_{m}=0\Longrightarrow m=1,2,5,6,9,10$$
Stanford's solution to this problem is found here: https://sumo.stanford.edu//pdfs/smt2013/calculus-tiebreaker-solutions.pdf
As you can see, their answer is the opposite of yours (the numbers $1$ to $10$ not including $3,4,7,8$). I worked through your solution, but couldn't figure out what was incorrect about it.
– A is for Ambition Jan 18 '15 at 04:13
$\int_0^\pi{(\cos{x})(\cos{2x})},dx = \frac{1}{2} \int_0^\pi{(\cos{3x}) + (\cos{x})} = \frac{1}{2}[\frac{1}{3} \sin{3x} + \sin{x}]_0^\pi = 0$.
– A is for Ambition Jan 18 '15 at 03:30