[VA] SRC#001 - Spiky Integral

[ijabbott]

Converting the product of cosines \(\prod_{i=1}^{n}\cos(ix) \) into a sum of cosines results in the angle multipliers being in the sum of cosines being all odd (when \( (n \bmod 4) \in \{1,2\} \)) or all even (when \( (n \bmod 4) \in \{0,3\} \)).

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

Let \( s = \textrm{a}x, t = \textrm{b}x \). Then:
\[ \cos(\textrm{a}x)\cos(\textrm{b}x) = \frac{1}{2}\cos((\textrm{a}-\textrm{b})x) + \frac{1}{2}\cos((\textrm{a}+\textrm{b})x) \]

When \(\textrm{a}\) and \(\textrm{b}\) are both odd or both even, then \((\textrm{a}-\textrm{b})\) and \((\textrm{a}+\textrm{b})\) are both even, otherwise \((\textrm{a}-\textrm{b})\) and \((\textrm{a}+\textrm{b})\) are both odd. This results in the factors \(\textrm{k}\) of \(x\) in the \(\cos(\textrm{k}x)\) terms of the summation switching between all odd and all even after every two \(\cos(ix)\) factors are appended to the product of cosines.

As discussed in my earlier post, the summations with all odd \(\textrm{k}\), \(\cos(\textrm{k}x)\) terms all integrate to 0 over the interval \( [0,2\pi] \), but the summations with all even \(\textrm{k}\), \(\cos(\textrm{k}x)\) terms all include a constant term \(\textrm{q}\cos(0x)\) for some positive rational factor \(\textrm{q}\) which integrates to \(2\textrm{q}\pi\) over the interval \( [0,2\pi] \).