Fourier Coefficients of Odd Functions
📂Fourier Analysis Fourier Coefficients of Odd Functions Theorem If a function f f f 2 L 2L 2 L antisymmetric , then the Fourier coefficients of f f f
a 0 = 0 a n = { 2 L ∫ 0 L f ( t ) cos n π t L d t ( n = 1 , 3 , ⋯ ) 0 ( n = 0 , 2 , ⋯ ) b n = { 2 L ∫ 0 L f ( t ) sin n π t L d t ( n = 1 , 3 , ⋯ ) 0 ( n = 0 , 2 , ⋯ ) c n = { 1 L ∫ 0 L f ( t ) e − i n π t L d t ( n = ± 1 , ± 3 , ⋯ ) 0 ( n = ± 2 , ± 4 , ⋯ )
\begin{align*}
a_{0} &= 0
\\ a_{n} &= \begin{cases} \dfrac{2}{L} {\displaystyle \int_{0}^{L}} f(t) \cos \frac{n \pi t}{L} dt & (n=1, 3, \cdots )
\\ 0 & (n=0, 2, \cdots )\end{cases}
\\ b_{n} &= \begin{cases} \dfrac{2}{L} {\displaystyle \int_{0}^{L}} f(t) \sin \frac{n \pi t}{L} dt & (n=1, 3, \cdots )
\\ 0 & (n=0, 2, \cdots )\end{cases}
\\ c_{n} &= \begin{cases} \dfrac{1}{L} {\displaystyle \int_{0}^{L} } f(t)e^{-i\frac{n \pi t}{L} } dt & (n=\pm 1, \pm 3, \cdots)
\\ 0 &( n=\pm 2, \pm 4, \cdots )\ \end{cases}
\end{align*}
a 0 a n b n c n = 0 = ⎩ ⎨ ⎧ L 2 ∫ 0 L f ( t ) cos L nπ t d t 0 ( n = 1 , 3 , ⋯ ) ( n = 0 , 2 , ⋯ ) = ⎩ ⎨ ⎧ L 2 ∫ 0 L f ( t ) sin L nπ t d t 0 ( n = 1 , 3 , ⋯ ) ( n = 0 , 2 , ⋯ ) = ⎩ ⎨ ⎧ L 1 ∫ 0 L f ( t ) e − i L nπ t d t 0 ( n = ± 1 , ± 3 , ⋯ ) ( n = ± 2 , ± 4 , ⋯ )
c n c_{n} c n complex Fourier coefficient .
Proof a 0 = 1 L ∫ − L L f ( t ) d t = 1 L ∫ 0 L f ( t ) d t + 1 L ∫ − L 0 f ( t ) d t = 1 L ∫ 0 L f ( t ) d t − 1 L ∫ − L 0 f ( t + L ) d t ( ∵ f ( t ) = − f ( t + L ) ) = 1 L ∫ 0 L f ( t ) d t − 1 L ∫ 0 L f ( t ) d t ( change of variable t + L = t ) = 0
\begin{align*}
a_{0} &= \dfrac{1}{L} {\displaystyle \int_{-L}^{L} } f(t) dt
\\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L} } f(t) dt + \dfrac{1}{L} {\displaystyle \int_{-L}^{0} } f(t) dt
\\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L} } f(t) dt - \dfrac{1}{L} {\displaystyle \int_{-L}^{0} } f(t+L) dt \Big( \because f(t)=-f(t+L) \Big)
\\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L} } f(t) dt - \dfrac{1}{L} {\displaystyle \int_{0}^{L} } f(t) dt \quad (\text{change of variable }t+L = t)
\\ &= 0
\end{align*}
a 0 = L 1 ∫ − L L f ( t ) d t = L 1 ∫ 0 L f ( t ) d t + L 1 ∫ − L 0 f ( t ) d t = L 1 ∫ 0 L f ( t ) d t − L 1 ∫ − L 0 f ( t + L ) d t ( ∵ f ( t ) = − f ( t + L ) ) = L 1 ∫ 0 L f ( t ) d t − L 1 ∫ 0 L f ( t ) d t ( change of variable t + L = t ) = 0
a n = 1 L ∫ − L L f ( t ) cos n π t L d t = 1 L ∫ 0 L f ( t ) cos n π t L d t + 1 L ∫ − L 0 f ( t ) cos n π t L d t = 1 L ∫ 0 L f ( t ) cos n π t L d t − 1 L ∫ − L 0 f ( t + L ) cos n π t L d t ( ∵ f ( t ) = − f ( t + L ) ) = 1 L ∫ 0 L f ( t ) cos n π t L d t − 1 L ∫ 0 L f ( t ) cos n π ( t − L ) L d t ( change of variable t + L = t ) = 1 L ∫ 0 L f ( t ) cos n π t L d t − 1 L ∫ 0 L f ( t ) ( cos n π t L cos ( n π ) + sin n π t L sin n π ) d t = 1 L ∫ 0 L f ( t ) cos n π t L d t − ( − 1 ) n 1 L ∫ 0 L f ( t ) cos n π t L d t = { 2 L ∫ 0 L f ( t ) cos n π t L d t ( n = 1 , 3 , ⋯ ) 0 ( n = 0 , 2 , ⋯ )
\begin{align*}
a_{n} &= \dfrac{1}{L}\int_{-L}^{L} f(t) \cos \frac{n \pi t}{L} dt
\\ &= {\displaystyle \dfrac{1}{L}\int_{0}^{L} }f(t) \cos \frac{n \pi t}{L} dt + \dfrac{1}{L}{\displaystyle \int_{-L}^{0}} f(t) \cos \frac{n \pi t}{L} dt
\\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \cos \frac{n \pi t}{L} dt - \dfrac{1}{L}{\displaystyle \int_{-L}^{0}} f(t+L) \cos \frac{n \pi t}{L} dt \quad \Big( \because f(t)=-f(t+L) \Big)
\\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \cos \frac{n \pi t}{L} dt - \dfrac{1}{L}{\displaystyle \int_{0}^{L}} f(t) \cos \frac{n \pi (t-L)}{L} dt \quad (\text{change of variable }t+L = t)
\\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \cos \frac{n \pi t}{L} dt - \dfrac{1}{L}{\displaystyle \int_{0}^{L}} f(t) \left( \cos\frac{n \pi t}{L}\cos (n \pi)+\sin\frac{n \pi t}{L}\sin n\pi \right) dt
\\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \cos \frac{n \pi t}{L} dt - (-1)^n\dfrac{1}{L}{\displaystyle \int_{0}^{L}} f(t) \cos\frac{n \pi t}{L} dt
\\ &= \begin{cases} \dfrac{2}{L} {\displaystyle \int_{0}^{L}} f(t) \cos \frac{n \pi t}{L} dt & (n=1, 3, \cdots )
\\ 0 & (n=0, 2, \cdots )\end{cases}
\end{align*}
a n = L 1 ∫ − L L f ( t ) cos L nπ t d t = L 1 ∫ 0 L f ( t ) cos L nπ t d t + L 1 ∫ − L 0 f ( t ) cos L nπ t d t = L 1 ∫ 0 L f ( t ) cos L nπ t d t − L 1 ∫ − L 0 f ( t + L ) cos L nπ t d t ( ∵ f ( t ) = − f ( t + L ) ) = L 1 ∫ 0 L f ( t ) cos L nπ t d t − L 1 ∫ 0 L f ( t ) cos L nπ ( t − L ) d t ( change of variable t + L = t ) = L 1 ∫ 0 L f ( t ) cos L nπ t d t − L 1 ∫ 0 L f ( t ) ( cos L nπ t cos ( nπ ) + sin L nπ t sin nπ ) d t = L 1 ∫ 0 L f ( t ) cos L nπ t d t − ( − 1 ) n L 1 ∫ 0 L f ( t ) cos L nπ t d t = ⎩ ⎨ ⎧ L 2 ∫ 0 L f ( t ) cos L nπ t d t 0 ( n = 1 , 3 , ⋯ ) ( n = 0 , 2 , ⋯ )
b n = 1 L ∫ − L L f ( t ) sin n π t L d t = 1 L ∫ 0 L f ( t ) sin n π t L d t + 1 L ∫ − L 0 f ( t ) sin n π t L d t = 1 L ∫ 0 L f ( t ) sin n π t L d t − 1 L ∫ − L 0 f ( t + L ) sin n π t L d t ( ∵ f ( t ) = − f ( t + L ) ) = 1 L ∫ 0 L f ( t ) sin n π t L d t − 1 L ∫ 0 L f ( t ) sin n π ( t − L ) L d t ( change of variable t + L = t ) = 1 L ∫ 0 L f ( t ) sin n π t L d t − 1 L ∫ 0 L f ( t ) ( sin n π t L cos ( n π ) + cos n π t L sin n π ) d t = 1 L ∫ 0 L f ( t ) sin n π t L d t − ( − 1 ) n 1 L ∫ 0 L f ( t ) sin n π t L d t = { 2 L ∫ 0 L f ( t ) sin n π t L d t ( n = 1 , 3 , ⋯ ) 0 ( n = 0 , 2 , ⋯ )
\begin{align*}
b_{n} &= \dfrac{1}{L}\int_{-L}^{L} f(t) \sin \frac{n \pi t}{L} dt
\\ &= {\displaystyle \dfrac{1}{L}\int_{0}^{L} }f(t) \sin \frac{n \pi t}{L} dt + \dfrac{1}{L}{\displaystyle \int_{-L}^{0}} f(t) \sin \frac{n \pi t}{L} dt
\\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \sin \frac{n \pi t}{L} dt - \dfrac{1}{L}{\displaystyle \int_{-L}^{0}} f(t+L) \sin \frac{n \pi t}{L} dt \quad \Big( \because f(t)=-f(t+L) \Big)
\\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \sin \frac{n \pi t}{L} dt - \dfrac{1}{L}{\displaystyle \int_{0}^{L}} f(t) \sin \frac{n \pi (t-L)}{L} dt \quad (\text{change of variable }t+L = t)
\\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \sin \frac{n \pi t}{L} dt - \dfrac{1}{L}{\displaystyle \int_{0}^{L}} f(t) \left( \sin\frac{n \pi t}{L}\cos (n \pi)+\cos\frac{n \pi t}{L}\sin n\pi \right) dt
\\ &= \dfrac{1}{L} {\displaystyle \int_{0}^{L}} f(t) \sin \frac{n \pi t}{L} dt - (-1)^n\dfrac{1}{L}{\displaystyle \int_{0}^{L}} f(t) \sin\frac{n \pi t}{L} dt
\\ &= \begin{cases} \dfrac{2}{L} {\displaystyle \int_{0}^{L}} f(t) \sin \frac{n \pi t}{L} dt & (n=1, 3, \cdots )
\\ 0 & (n=0, 2, \cdots )\end{cases}
\end{align*}
b n = L 1 ∫ − L L f ( t ) sin L nπ t d t = L 1 ∫ 0 L f ( t ) sin L nπ t d t + L 1 ∫ − L 0 f ( t ) sin L nπ t d t = L 1 ∫ 0 L f ( t ) sin L nπ t d t − L 1 ∫ − L 0 f ( t + L ) sin L nπ t d t ( ∵ f ( t ) = − f ( t + L ) ) = L 1 ∫ 0 L f ( t ) sin L nπ t d t − L 1 ∫ 0 L f ( t ) sin L nπ ( t − L ) d t ( change of variable t + L = t ) = L 1 ∫ 0 L f ( t ) sin L nπ t d t − L 1 ∫ 0 L f ( t ) ( sin L nπ t cos ( nπ ) + cos L nπ t sin nπ ) d t = L 1 ∫ 0 L f ( t ) sin L nπ t d t − ( − 1 ) n L 1 ∫ 0 L f ( t ) sin L nπ t d t = ⎩ ⎨ ⎧ L 2 ∫ 0 L f ( t ) sin L nπ t d t 0 ( n = 1 , 3 , ⋯ ) ( n = 0 , 2 , ⋯ )
c n = 1 2 ( a n − i b n ) = 1 2 2 L ∫ 0 L f ( t ) ( cos n π t L − i sin n π t L ) d t = 1 L ∫ 0 L f ( t ) e − i n π t L d t
\begin{align*}
c_{n} &= \dfrac{1}{2}(a_{n}-ib_{n})
\\ &= \dfrac{1}{2}\dfrac{2}{L}{\displaystyle \int_{0}^{L} }f(t) \left( \cos\frac{n \pi t}{L} -i\sin \frac{n \pi t}{L} \right)dt
\\ &= \dfrac{1}{L}{\displaystyle \int_{0}^{L} }f(t) e^{-i\frac{n \pi t }{L}} dt
\end{align*}
c n = 2 1 ( a n − i b n ) = 2 1 L 2 ∫ 0 L f ( t ) ( cos L nπ t − i sin L nπ t ) d t = L 1 ∫ 0 L f ( t ) e − i L nπ t d t
Due to the conditions of a n a_{n} a n b n b_{n} b n
c n = { 1 L ∫ 0 L f ( t ) e − i n π t L d t ( n = ± 1 , ± 3 , ⋯ ) 0 ( n = ± 2 , ± 4 , ⋯ )
c_{n} = \begin{cases} \dfrac{1}{L} {\displaystyle \int_{0}^{L} } f(t)e^{-i\frac{n \pi t}{L} } dt & (n=\pm 1, \pm 3, \cdots) \\ 0 &( n=\pm 2, \pm 4, \cdots )\ \end{cases}
c n = ⎩ ⎨ ⎧ L 1 ∫ 0 L f ( t ) e − i L nπ t d t 0 ( n = ± 1 , ± 3 , ⋯ ) ( n = ± 2 , ± 4 , ⋯ )
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