Fourier Series — Question 10

Question 10 – Solution

Strategy. One delay rotates each harmonic by a different multiple of the same angle, creating a coupled optimization.

Step 1: Compute distance through correlation. Both signals have squared norm $2\pi$. Angle addition and orthogonality give $$\int_0^{2\pi} f(x-\delta)g(x)\,dx =\pi(\sin\delta+\cos 2\delta).$$ Therefore $$\boxed{\frac{D(\delta)}\pi=4-2(\sin\delta+\cos 2\delta).}$$ The sine sign follows from $\cos(x-\delta)=\cos x\cos\delta+\sin x\sin\delta$.

Step 2: Optimize over the whole circle. Put $t=\sin\delta\in[-1,1]$. Since $\cos 2\delta=1-2t^2$, $$\boxed{\frac{D(\delta)}\pi=\frac 74+4\left(t-\frac 14\right)^2.}$$ The feasible value $t=1/4$ is the unique minimizing sine value. Consequently $$\boxed{D_{\min}=\frac{7\pi}{4},\qquad \delta=\arcsin(1/4)\ \text{or}\ \pi-\arcsin(1/4)\pmod{2\pi}.}$$ Completing the square proves these are all global minimizers.

Step 3: Compare two tempting alignments. At $\delta=0$, the second harmonics agree and $D=2\pi$. At $\delta=\pi/2$, the first harmonic matches $\sin x$, but the second becomes $-\cos 2x$, so $D=4\pi$. Exactly matching one harmonic can worsen the full error. The optimum balances their contributions and improves on both choices.

Step 4: Interpret the two minima. The minimum is strictly positive, so no delay makes the signals identical. The objective depends only on $\sin\delta$; the two distinct delays with sine $1/4$ give the same compromise. They do not produce the same shifted function: their first cosine coefficients are opposite and nonzero. Thus nonuniqueness of the best delay does not imply exact agreement or identical aligned waveforms. The graph shows both minima in $[0,2\pi)$.