Fourier Series — Question 5

Question 5 – Solution

Strategy. A common origin must satisfy the phase conditions of every nonzero harmonic simultaneously.

Step 1: Derive the coefficient rotation. Angle addition gives the new coefficients $$\boxed{a_n'=a_n\cos nc+b_n\sin nc,\qquad b_n'=-a_n\sin nc+b_n\cos nc.}$$ Squaring and adding cancels the cross terms, so $(a_n')^2+(b_n')^2=a_n^2+b_n^2$. For a trigonometric polynomial, evenness is equivalent to every $b_n'$ being zero; oddness requires a zero mean and every $a_n'$ zero. Necessity follows from orthogonality, and sufficiency follows from sine/cosine parity.

Step 2: Test all evenness conditions. At frequency one, $b_1'=2(\cos c-\sin c)=0$ implies $c=\pi/4+k\pi$. At frequency two, $b_2'=-\sin 2c$ would then equal $-1$. Hence $$\boxed{\text{No origin makes }f(t+c)\text{ even}.}$$ The first frequency can be aligned, but that shift makes the second purely sine.

Step 3: Find every oddness origin. The first condition $a_1'=2(\cos c+\sin c)=0$ gives $c=3\pi/4+k\pi$. At each of these, $a_2'=\cos 2c=0$, and the mean is already zero. These are all solutions. At $c=3\pi/4$, $$\boxed{f(t+3\pi/4)=-2\sqrt 2\sin t+\sin 2t.}$$ Modulo $2\pi$ the two origins are $3\pi/4$ and $7\pi/4$.

Step 4: Change the relative harmonic phase. For $g=2\cos x+2\sin x+\sin 2x$, the first-frequency evenness condition is again $c=\pi/4+k\pi$. Now the second is $b_2'=\cos 2c=0$, which is satisfied. At $c=\pi/4$, $$\boxed{g(t+\pi/4)=2\sqrt 2\cos t+\cos 2t.}$$ All evenness origins are $\pi/4+k\pi$. Each harmonic admits its own phase alignment, but a symmetry of the full function requires the same $c$ for all frequencies. The two examples have identical harmonic amplitudes and different answers to the evenness question.