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Derivative's Fourier Coefficients 📂Fourier Analysis

Derivative's Fourier Coefficients

Formula

Given that a function ff defined on the interval [L, L)[-L,\ L) is continuous and piecewise smooth, then the Fourier coefficients of ff^{\prime} are as follows.

an=nπLbn a^{\prime}_{n}=\dfrac{n\pi}{L}b_{n}

bn=nπLan b^{\prime}_{n}=-\dfrac{n\pi}{L}a_{n}

cn=inπLcn c^{\prime}_{n}=\dfrac{in\pi}{L}c_{n} Here, an, bna_{n},\ b_{n} are the Fourier coefficients of ff, and cnc_{n} are the complex Fourier coefficients of ff.

Proof

cn=12LLLf(t)einπtLdt=12L[f(t)einπtL]LL+inπL12LLLf(t)einπLtdt=12Lf(t)[einπeinπ]+inπLcn=12Lf(t)[(1)n(1)n]+inπLcn=12Lf(t)(1)n[(1)2n1]+inπLcn=inπLcn \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.

an=1LLLf(t)cosnπtLdt=1L[f(t)cosnπtL]LL+nπL1LLLf(t)sinnπtLdt=1Lf(t)(cosnπcosnπ)+nπLbn=nπLbn \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.

bn=1LLLf(t)sinnπtLdt=1L[f(t)sinnπtL]LLnπL1LLLf(t)cosnπtLdt=1Lf(t)(sinnπ+sinnπ)nπLan=nπLan\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.