Fourier Series — Question 8

Question 8 – Solution

Strategy. Correlation retains harmonic energies but discards their individual phases.

Step 1: Evaluate the correlation. Different frequencies are orthogonal. At frequency $n$, a pair $a_n\cos nx+b_n\sin nx$ contributes $(a_n^2+b_n^2)\cos(n\tau)/2$; the two mixed sine terms cancel. The constant contributes the square of the mean. Thus $$\boxed{R(\tau)=1+\frac 52\cos\tau+\frac 12\cos 2\tau.}$$ In particular, $R(0)=4$ is the mean square of $f$, not the square of its mean. The correlation is even regardless of the parity of $f$.

Step 2: Optimize over all shifts. Put $t=\cos\tau\in[-1,1]$. Then $R=t^2+(5/2)t+1/2$, whose derivative $2t+5/2$ is positive throughout this interval. Hence $$\boxed{\max R=4\quad(\tau=0\bmod 2\pi),\qquad \min R=-1\quad(\tau=\pi\bmod 2\pi).}$$ A correlation can be negative here; the original signal is not required to be nonnegative.

Step 3: Exhibit phase ambiguity beyond translation. Take $g(x)=1+\sqrt 5\cos x+\cos 2x$. Its mean is one and its harmonic squared amplitudes are $5$ and $1$, so it gives exactly the same $R$. If $g(x)=f(x+c)$, equality at frequency two requires $\cos 2c=1$ and $\sin 2c=0$, so $c=k\pi$. But then the first sine coefficient of $f(x+c)$ is $2(-1)^k\ne 0$, unlike $g$. Thus $g$ is not a translate. Correlation does not determine relative harmonic phases.

Step 4: Recover the exact translation invariance. Translation preserves $\int f^2$. Expanding the squared difference gives $$\boxed{\frac 1{2\pi}\int_0^{2\pi}[f(x+\tau)-f(x)]^2\,dx =2\bigl(R(0)-R(\tau)\bigr).}$$ This is zero precisely at $\tau=0$ modulo $2\pi$. Zero integral of a continuous nonnegative square means equality at every point. Therefore the invariant shifts are exactly $2k\pi$, and the fundamental period is $2\pi$.