Derivative's Fourier Coefficients
Formula
Given that a function $f$ defined on the interval $[-L,\ L)$ is continuous and piecewise smooth, then the Fourier coefficients of $f^{\prime}$ are as follows.
$$ a^{\prime}_{n}=\dfrac{n\pi}{L}b_{n} $$
$$ b^{\prime}_{n}=-\dfrac{n\pi}{L}a_{n} $$
$$ c^{\prime}_{n}=\dfrac{in\pi}{L}c_{n} $$ Here, $a_{n},\ b_{n}$ are the Fourier coefficients of $f$, and $c_{n}$ are the complex Fourier coefficients of $f$.
Proof
$$ \begin{align*} c^{\prime}_{n} &=\dfrac{1}{2L}\int _{-L}^{L} f^{\prime}(t)e^{-i\frac{n\pi t}{L}}dt \\ &= \dfrac{1}{2L}\left[ f(t)e^{-i\frac{n\pi t}{L}} \right]_{-L}^{L} +\dfrac{in \pi}{L}\dfrac{1}{2L}\int_{-L}^{L} f(t)e^{-i\frac{n \pi}{L}t} dt \\ &= \dfrac{1}{2L}f(t)\left[ e^{-in\pi} -e^{in\pi}\right] +\dfrac{in \pi}{L}c_{n} \\ &= \dfrac{1}{2L}f(t)\left[ (-1)^{-n} -(-1)^{n}\right] +\dfrac{in \pi}{L}c_{n} \\ &= \dfrac{1}{2L}f(t)(-1)^{n}\left[ (-1)^{-2n} -1 \right] +\dfrac{in \pi}{L}c_{n} \\ &= \dfrac{in \pi}{L}c_{n} \end{align*} $$
The second equality is due to the integration by parts.
$$ \begin{align*} a^{\prime}_{n} &= \dfrac{1}{L}\int _{-L}^{L} f^{\prime}(t)\cos \frac{n\pi t}{L} dt \\ &= \dfrac{1}{L}\left[ f(t)\cos \dfrac{n\pi t}{L} \right]_{-L}^{L} +\dfrac{n\pi}{L}\dfrac{1}{L}\int _{-L}^{L} f(t)\sin \dfrac{n\pi t}{L} dt \\ &= \dfrac{1}{L}f(t)\left( \cos n\pi -\cos n\pi \right) +\dfrac{n\pi}{L}b_{n} \\ &= \dfrac{n\pi}{L}b_{n} \end{align*} $$
The second equality is due to the integration by parts.
$$\begin{align*} b^{\prime}_{n} &= \dfrac{1}{L}\int _{-L}^{L} f^{\prime}(t)\sin \dfrac{n\pi t}{L} dt \\ &= \dfrac{1}{L}\left[ f(t)\sin \dfrac{n\pi t}{L} \right]_{-L}^{L} -\dfrac{n\pi}{L}\dfrac{1}{L}\int _{-L}^{L} f(t)\cos \dfrac{n\pi t}{L} dt \\ &= \dfrac{1}{L}f(t)\left( \sin n\pi + \sin n\pi \right) -\dfrac{n\pi}{L}a_{n} \\ &= -\dfrac{n\pi}{L}a_{n} \end{align*} $$
The second equality is due to the integration by parts.
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