Discussion
The Fourier series expansion of \(x^{3}\) in the interval \(-1 \le x \lt 1\) with periodic continuation has
Answer: A
The series has only sine terms.
\(x^{3}\) is an odd function, and the interval \(-1 \le x \lt 1\) is symmetric about zero, so its periodic continuation is also odd. For an odd function the constant term \(a_{0}\) and every cosine coefficient \(a_{n} = \int_{-1}^{1} x^{3}\cos(n\pi x)\,dx\) vanish, because each integrand is odd. Only the sine coefficients survive.
Only cosine terms would need an even function, both kinds would need a function with no symmetry, and a non-zero constant would need a non-zero average over the period, which an odd function on a symmetric interval does not have.
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