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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

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  • $\begingroup$ What were your results? $\endgroup$
    – Tim Raczkowski
    Commented Jan 18, 2015 at 3:28
  • $\begingroup$ My answer was $m = 1,2,5,6,9,10$. If you're asking about how I went about solving the problem, since it would be a lot to type in LaTeX, I'll just show you the case for $m=2$ ($m=1$ is a trivial and doesn't need work shown). $\int_0^\pi{(\cos{x})(\cos{2x})}\,dx = \frac{1}{2} \int_0^\pi{(\cos{3x}) + (\cos{x})} = \frac{1}{2}[\frac{1}{3} \sin{3x} + \sin{x}]_0^\pi = 0$. $\endgroup$
    – A is for Ambition
    Commented Jan 18, 2015 at 3:30
  • $\begingroup$ The only thing I can think of right now is writing $\cos x$ in complex form $\cos x={e^{ix}+e^{-ix}\over 2}$ but I don't think that will simplify things much, and it may not be something your familiar with. $\endgroup$
    – Tim Raczkowski
    Commented Jan 18, 2015 at 3:37
  • $\begingroup$ Could you show your work for one value of $m$ to clarify what you mean? I'm having a little trouble using that to work towards a solution. $\endgroup$
    – A is for Ambition
    Commented Jan 18, 2015 at 3:40
  • $\begingroup$ I hadn't really worked out an example as I wasn't sure it was something you could use. How familiar are you with complex numbers? $\endgroup$
    – Tim Raczkowski
    Commented Jan 18, 2015 at 3:42

1 Answer 1

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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 $$I_{m}=0\Longrightarrow m=1,2,5,6,9,10$$

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    $\begingroup$ My teacher who had given me this problem told me that this problem was from the Stanford Math Tournament. Stanford's solution to this problem is found here: sumo.stanford.edu//pdfs/smt2013/… 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. $\endgroup$
    – A is for Ambition
    Commented Jan 18, 2015 at 4:13
  • $\begingroup$ oh,you are right,it's $1,2,5,6,9,10$ $\endgroup$
    – math110
    Commented Jan 18, 2015 at 4:18

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