WO.3 · Fourier symmetry and wave superposition

For a sufficiently regular real 2π-periodic function, write f(x)=a0/2+Σ[a_n cos(nx)+b_n sin(nx)]. The coefficients are a_n=(1/π)∫from−πtoπ f(x)cos(nx)dx and b_n=(1/π)∫from−πtoπ f(x)sin(nx)dx, with a0 using n=0. Cosines and sines are orthogonal on a full period. The mean is a0/2, not a0; retain the constant term even when all nonzero-frequency coefficients of one family vanish.

If f is even about the chosen origin, f(x)sin(nx) is odd and its symmetric integral is zero, so all b_n vanish. If f is odd, its mean and all a_n vanish. Symmetry depends on the origin: shifting the same physical signal can mix sine and cosine coefficients while leaving its harmonic frequencies unchanged. A nonnegative triangular waveform symmetric about x=0 can therefore have cosine harmonics and a nonzero mean without any sine harmonics.

For the 2π-periodic extension of f(x)=|x| on [−π,π], the mean is π/2. Integrating x cos(nx) by parts on [0,π] gives a_n=2[(-1)^n−1]/(πn²), so even-n cosine coefficients are zero and odd-n coefficients are −4/(πn²). All sine coefficients are zero. The 1/n² decay reflects a continuous function with a slope discontinuity. A jump discontinuity often gives slower coefficient decay and partial-sum overshoot; do not infer the same convergence behaviour for every waveform.

A Fourier sum adds amplitudes with their phases. For equal coherent monochromatic amplitudes A meeting with phase difference φ, resultant squared amplitude is 2A²(1+cosφ), so intensity is 2I0(1+cosφ). It is 4I0 in phase and zero at φ=π. Incoherent averaging removes the cross term, yielding 2I0. Harmonics at different frequencies can construct a shape over time; their instantaneous sum is not the sum of their individual intensities. Identify whether the question concerns a waveform, time-average power or coherent interference.