Fourier coefficient

Remarks

In terms of the Fourier monomials $e_n$, $\hat f(n) = \langle f, e_n \rangle$ for the inner product $\langle f, g \rangle = \frac{1}{2\pi} \int_{-\pi}^{\pi} f \overline{g}$. By periodicity the integral may be taken over any interval of length $2\pi$. The bound $\abs{\hat f(n)} \le \frac{1}{2\pi} \int_{-\pi}^{\pi} \abs{f}$ holds for every $n$. Example: for $f(x) = \cos x$, the formula $\cos x = \frac{1}{2}(e^{ix} + e^{-ix})$ and the orthogonality of the $e_n$ give $\hat f(1) = \hat f(-1) = \frac{1}{2}$ and $\hat f(n) = 0$ for $\abs{n} \ne 1$. Non-example: the $2\pi$-periodic function equal to $1/x$ on $(0, 2\pi]$ is not integrable on a period, so its Fourier coefficients are not defined. The real coefficients $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx \, dx$ and $b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx \, dx$ satisfy $a_n = \hat f(n) + \hat f(-n)$ and $b_n = i(\hat f(n) - \hat f(-n))$. Other notations are $c_n$ and $\mathcal{F}f(n)$; for period $T$ the coefficient is $\frac{1}{T} \int_0^T f(x) e^{-2\pi i n x / T} \, dx$.