[VA] SRC#001 - Spiky Integral

[ijabbott]

You don't really need to do any integration to solve this problem. You just need to use the trig identities to convert the product of cosines into a sum of cosines. When there is a non-zero constant term in the sum, the integral from \( 0 \) to \( 2\pi \) will be non-zero. The integrals of all the \( \cos(kx) \) terms from \( 0 \) to \( 2\pi \) will be zero.

The general form for converting the product of cosines to a sum of cosines is:

\[ \cos(s)\cos(t) = \frac{1}{2}\cos(s-t) + \frac{1}{2}\cos(s+t) \]

So:

\[ \begin{eqnarray}
\cos(1x) &=& \cos(1x)
\end{eqnarray} \]

\[ \begin{eqnarray}
\cos(1x)\cos(2x) &=& \frac{1}{2}\cos(1x) + \frac{1}{2}\cos(3x)
\end{eqnarray} \]

\[ \begin{eqnarray}
\cos(1x)\cos(2x)\cos(3x) &=& \frac{1}{2}\cos(1x)\cos(3x) + \frac{1}{2}\cos(3x)\cos(3x) \\
&=& \frac{1}{2}\left(\frac{1}{2}\cos(2x) + \frac{1}{2}\cos(4x)\right) + \frac{1}{2}\left(\frac{1}{2}\cos(0x) + \frac{1}{2}\cos(6x)\right) \\
&=& \frac{1}{4}\cos(2x) + \frac{1}{4}\cos(4x) + \frac{1}{4}\cos(0x) + \frac{1}{4}\cos(6x) \\
&=& \frac{1}{4}\cos(0x) + \frac{1}{4}\cos(2x) + \frac{1}{4}\cos(4x) + \frac{1}{4}\cos(6x)
\end{eqnarray} \]

\[ \begin{eqnarray}
\cos(1x)\cos(2x)\cos(3x)\cos(4x) &=& \frac{1}{4}\cos(0x)\cos(4x) + \frac{1}{4}\cos(2x)\cos(4x) + \frac{1}{4}\cos(4x)\cos(4x) + \frac{1}{4}\cos(6x)\cos(4x) \\
&=& \frac{1}{4}\left(\frac{1}{2}\cos(4x) + \frac{1}{2}\cos(4x)\right) + \frac{1}{4}\left(\frac{1}{2}\cos(2x) + \frac{1}{2}\cos(6x)\right) + \frac{1}{4}\left(\frac{1}{2}\cos(0x) + \frac{1}{2}\cos(8x)\right) + \frac{1}{4}\left(\frac{1}{2}\cos(2x) + \frac{1}{2}\cos(10x)\right) \\
&=& \frac{1}{8}\cos(4x) + \frac{1}{8}\cos(4x) + \frac{1}{8}\cos(2x) + \frac{1}{8}\cos(6x) + \frac{1}{8}\cos(0x) + \frac{1}{8}\cos(8x) + \frac{1}{8}\cos(2x) + \frac{1}{8}\cos(10x) \\
&=& \frac{1}{8}\cos(0x) + \frac{1}{4}\cos(2x) + \frac{1}{4}\cos(4x) + \frac{1}{8}\cos(6x) + \frac{1}{8}\cos(8x) + \frac{1}{8}\cos(10x)
\end{eqnarray} \]

\[ \begin{eqnarray}
\cos(1x)\cos(2x)\cos(3x)\cos(4x)\cos(5x) &=& \frac{1}{8}\cos(0x)\cos(5x) + \frac{1}{4}\cos(2x)\cos(5x) + \frac{1}{4}\cos(4x)\cos(5x) + \frac{1}{8}\cos(6x)\cos(5x) + \frac{1}{8}\cos(8x)\cos(5x) + \frac{1}{8}\cos(10x)\cos(5x) \\
&=& \frac{1}{8}\left(\frac{1}{2}\cos(5x) + \frac{1}{2}\cos(5x)\right) + \frac{1}{4}\left(\frac{1}{2}\cos(3x) + \frac{1}{2}\cos(7x)\right) + \frac{1}{4}\left(\frac{1}{2}\cos(1x) + \frac{1}{2}\cos(9x)\right) + \frac{1}{8}\left(\frac{1}{2}\cos(1x) + \frac{1}{2}\cos(11x)\right) + \frac{1}{8}\left(\frac{1}{2}\cos(3x) + \frac{1}{2}\cos(13x)\right) + \frac{1}{8}\left(\frac{1}{2}\cos(5x) + \frac{1}{2}\cos(15x)\right) \\
&=& \frac{1}{16}\cos(5x) + \frac{1}{16}\cos(5x) + \frac{1}{8}\cos(3x) + \frac{1}{8}\cos(7x) + \frac{1}{8}\cos(1x) + \frac{1}{8}\cos(9x) + \frac{1}{16}\cos(1x) + \frac{1}{16}\cos(11x) + \frac{1}{16}\cos(3x) + \frac{1}{16}\cos(13x) + \frac{1}{16}\cos(5x) + \frac{1}{16}\cos(15x) \\
&=& \frac{3}{16}\cos(1x) + \frac{3}{16}\cos(3x) + \frac{3}{16}\cos(5x) + \frac{1}{8}\cos(7x) + \frac{1}{8}\cos(9x) + \frac{1}{16}\cos(11x) + \frac{1}{16}\cos(13x) + \frac{1}{16}\cos(15x)
\end{eqnarray} \]

It's a bit tedious to continue, but you can see a constant term \( \frac{1}{4}\cos(0x) = \frac{1}{4} \) in the expansion of \( \prod_{i=1}^{3} \cos(ix) \), a constant term of \( \frac{1}{8}\cos(0x) = \frac{1}{8} \) in the expansion of \( \prod_{i=1}^{4} \cos(ix) \), and the constant term goes away in the expansion of \( \prod_{i=1}^{5} \cos(ix) \). The constant terms are the coefficients of \( \cos(0x) \) in those expansions whose terms consist of cosines of even multiples of \( x \).

Your challenge, should you choose to accept it, is to come up with a general formula for each term (or a summation formula for all the terms).

Of course, once you've found the coefficient for the constant \(\cos(0x)\) term, calculating the integral from \(0\) to \(2\pi\) is a piece of cake! (Just multiply the coefficient by \(2\pi\).)