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For $V(t)=V_0\max (\sin \omega t,0)$, with period $2\pi /\omega$, half-wave rectification givesThe Fourier coefficients are $a_0=2V_0/\pi$, $b_1=V_0/2$, $a_{2m}=-2V_0/[\pi (4m^2-1)]$, with all remaining coefficients zero. To derive the cosine coefficients, integrate $\sin s\cos ns$ over $0<s<\pi$ and use the product-to-sum identity. The sine coefficients follow from orthogonality. A half-period delay reverses only the fundamental sine term and leaves the constant and even-frequency cosine terms unchanged, by translation of a Fourier series.

The given identity immediately yields $V_2=V_1-V_0\sin \omega t$. Thus only the fundamental sine coefficient changes: $V_0/2$ becomes $-V_0/2$. No new integral is needed. The first five nonzero terms areThis also follows from translation of a Fourier series: a half-period delay multiplies the $n$th harmonic by $(-1{)}^n$, leaving the constant and even harmonics unchanged and reversing the only odd harmonic present.

Use the periodic extension of $f$ when the translated argument falls outside the original interval. With $k_n=2\pi n/L$, the sine subtraction identity givesThis translation of a Fourier series generally introduces cosine terms: a translated odd function need not be odd about the original origin. Equality is understood wherever the original Fourier series represents its periodic extension, with midpoint values at jumps.