[VA] SRC#001 - Spiky Integral

[Albert Chan]

(07-10-2018 10:10 PM)Valentin Albillo Wrote:  Hi all, welcome to my SRC#001 - Spiky Integral:

Here I'll deal with a real-world math problem not of my own making but which did appear at a certain math competition. The problem introduces this "spiky" integral:

            I(N) = \(\int_{0}^{2\pi} cos(x)\, cos(2x)\, cos(3x)\, ...\, cos(Nx)\, dx \)

and asks  for which values of  N  in the interval  [1,10]  does  I(N)  have a non-zero value.

It then specifies that the result must include proof of correctness

It seems the problem is easier to solve with symmetry.

let F = cos(x) cos(2x) cos(3x) ... cos(Nx)

I(N) = \(\int_{0}^{2 \pi} F dx \) = 2 \(\int_{0}^{\pi} F dx \) = 2 \(\int_{0}^{\pi/2} F dx \) + 2 \(\int_{\pi/2}^{\pi} F dx \)

If F is symmetric around x = Pi/2, the two terms are same sized and same sign.

A simple test is when x=Pi, F=1, which imply even number of odd values between 1 to N:
(since cos(odd Pi) = -1, even numbers of such terms restored the symmetry)

--> If N mod 4 = 0 or 3, I(N) = 4 \(\int_{0}^{\pi/2} F dx \)

Since F is maximized (= 1.0) when x=0, above should always be positive.
As N increases, F spike is "thinner", thus smaller I(N), but still above 0

If odd number of odd values between 1 and N, symmetry is flipped.
The two terms still have same size, but opposite sign, cancelling each other.

--> If N mod 4 = 1 or 2, I(N) = 0

So, for interval [1, 10] and non-zero I(N), N = 3,4,7,8