= Solution
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 gives
$$
\boxed{g(x)=\sum_{n=1}^\infty b_n
\left[\sin(k_nx)\cos(k_n\ell)-\cos(k_nx)\sin(k_n\ell)\right]}.
$$
This <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.
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