複素Fourier係数
$ \forall T>0と可積分函数$ f: \rbrack-\frac12 T,\frac12T\lbrack\to\Complexにて、↓を複素Fourier係数と呼ぶ $ \mathcal{F}_T(f)_\bullet:\Z\ni n\mapsto\frac1T\int_{-\frac12 T}^{\frac12T}f(t)e^{-in\frac{2\pi}Tt}\mathrm dt\in\Complex
$ c_\bulletで表すことが多い
ここではFourier変換$ \mathcal Fの表記と合わせて、$ \mathcal F_Tと表現してみたtakker.icon $ fを周期$ Tの周期函数$ f(t+T)=f(t)に拡張し、かつ$ n\in\Zの時のみ積分区間を次のように変更できる $ \mathcal F_T(f)_n=\frac1T\int_0^Tf(t)e^{-in\frac{2\pi}Tt}\mathrm dt
$ \because\mathcal F_T(f)_n=\frac1T\int_{-\frac12 T}^{\frac12T}f(t)e^{-in\frac{2\pi}Tt}\mathrm dt
$ = \frac1T\int_0^{\frac12T}f(t)e^{-in\frac{2\pi}Tt}\mathrm dt+\frac1T\int_{-\frac12 T}^0f(t)e^{-in\frac{2\pi}Tt}\mathrm dt
$ = \frac1T\int_0^{\frac12T}f(t)e^{-in\frac{2\pi}Tt}\mathrm dt+e^{2\pi in}\frac1T\int_{\frac12 T}^Tf(t)e^{-in\frac{2\pi}Tt}\mathrm dt
$ \because f(t+T)=f(t)
$ = \frac1T\int_0^Tf(t)e^{-in\frac{2\pi}Tt}\mathrm dt+(e^{2\pi in}-1)\frac1T\int_{\frac12 T}^Tf(t)e^{-in\frac{2\pi}Tt}\mathrm dt
$ = \frac1T\int_0^Tf(t)e^{-in\frac{2\pi}Tt}\mathrm dt
$ \because e^{2\pi in}=1
性質
$ \mathcal F_T(f(at))_n=\frac1a\mathcal F_{aT}(f)_n
$ \mathcal F_T(f(t+\tau))_n=\mathcal F_{T+\frac12\tau}(f)_ne^{in\frac{2\pi}T\tau}
$ \mathcal F_T(f^*)_n={\mathcal F_T(f)_{-n}}^*
$ {c_n}^*=c_{-n}に対応する
$ \mathcal F_T(f(-t))_n=-\mathcal F_{-T}(f)_n=\mathcal F_T(f)_{-n}
$ \mathcal F_T(f')_n=\frac{(-1)^n}T\left(f\left(\frac12T-0\right)-f\left(-\frac12T+0\right)\right)+in\frac{2\pi}T\mathcal F_T(f)_n
$ fを周期$ Tの周期函数に拡張したときは、$ f\left(-\frac12 T\right)=f\left(T-\frac12 T\right)=f\left(\frac12 T\right)より
$ \mathcal F_T(f')_n=\frac{(-1)^n}T\left(f\left(\frac12T-0\right)-f\left(\frac12T+0\right)\right)+in\frac{2\pi}T\mathcal F_T(f)_n
ともできる
特に$ \frac12Tで連続なら
$ \mathcal F_T(f')_n=in\frac{2\pi}T\mathcal F_T(f)_n
$ \mathcal F_T(f'')_n=\frac{(-1)^n}T\left(f'\left(\frac12T-0\right)-f'\left(\frac12T+0\right)\right)+in\frac{2\pi}T\mathcal F_T(f')_n
$ =\frac{(-1)^n}T\left(f'\left(\frac12T-0\right)-f'\left(\frac12T+0\right)\right)+\frac{(-1)^n}Tin\frac{2\pi}T\left(f\left(\frac12T-0\right)-f\left(\frac12T+0\right)\right)-\left(\frac{2\pi}Tn\right)^2\mathcal F_T(f)_n
ここから、
$ \mathcal F_{c,T}(f'')_n=2\frac{(-1)^n}T\left(\Re f'\left(\frac12T-0\right)-\Re f'\left(\frac12T+0\right)\right)+2\frac{(-1)^{n+1}}Tn\frac{2\pi}T\left(\Im f\left(\frac12T-0\right)-\Im f\left(\frac12T+0\right)\right)-\left(\frac{2\pi}Tn\right)^2\mathcal F_{c,T}(f)_n
$ \mathcal F_{s,T}(f'')_n=2\frac{(-1)^{n+1}}T\left(\Im f'\left(\frac12T-0\right)-\Im f'\left(\frac12T+0\right)\right)+2\frac{(-1)^{n+1}}Tn\frac{2\pi}T\left(\Re f\left(\frac12T-0\right)-\Re f\left(\frac12T+0\right)\right)-\left(\frac{2\pi}Tn\right)^2\mathcal F_{s,T}(f)_n
$ fが実函数のときは、境界条件が$ fに絡むとき$ \mathcal F_{s,T}を、$ f'に絡むとき$ \mathcal F_{c,T}を使えばいいとわかる
$ \mathcal F_{c,T}(f'')_n=2\frac{(-1)^n}T\left(f'\left(\frac12T-0\right)-f'\left(\frac12T+0\right)\right)-\left(\frac{2\pi}Tn\right)^2\mathcal F_{c,T}(f)_n
$ \mathcal F_{s,T}(f'')_n=2\frac{(-1)^{n+1}}Tn\frac{2\pi}T\left(f\left(\frac12T-0\right)-f\left(\frac12T+0\right)\right)-\left(\frac{2\pi}Tn\right)^2\mathcal F_{s,T}(f)_n
角振動数$ \omega_T:=\frac{2\pi}Tを導入するとこうも書ける $ \mathcal{F}_T(f)_\bullet:\Z\ni n\mapsto\frac{\omega_T}{2\pi}\int_{-\frac{\pi}{\omega_T}}^{\frac{\pi}{\omega_T}}f(t)e^{-in\omega_Tt}\mathrm dt\in\Complex
$ \mathcal F_T^{-1}(c):\rbrack-\frac12T,\frac12T\lbrack\ni t\mapsto\sum_{n\in\Z}c_ne^{in\omega_Tt}\in\Complex
$ \mathcal F_T(f)_n=\frac1{\sqrt{2\pi}}\mathcal F(f)(n)*\operatorname{sinc}\left(\frac12Tn\right)
$ =\frac1{\sqrt{2\pi}}\int_R\mathcal F(f)(\omega)\operatorname{sinc}\left(\frac12T(n-\omega)\right)\mathrm d\omega
$ \mathcal F_T(f)_n=\frac12\mathcal F_T(f(t)+f(-t))_n=\frac12(\mathcal F_T(f)_n+\mathcal F_T(f)_{-n})=\frac1T\int_{-\frac12 T}^{\frac12T}f(t)\cos\left(\frac{2\pi}Tnt\right)\mathrm dt
$ \mathcal F_T(f)_n=\frac12(\mathcal F_T(f)_n-\mathcal F_T(f)_{-n})=-\frac iT\int_{-\frac12 T}^{\frac12T}f(t)\sin\left(\frac{2\pi}Tnt\right)\mathrm dt
$ \mathcal F_T(f)_n=\frac12(\mathcal F_{c,T}(f)_n-i\mathcal F_{s,T}(f)_n)とすると、
$ fが偶函数のとき$ \mathcal F_T(f)_n=\frac12\mathcal F_{c,T}(f)_n $ fが奇函数のとき$ \mathcal F_T(f)_n=-\frac i2\mathcal F_{s,T}(f)_n $ \mathcal F_{c,T}(f)_n=2\Re\mathcal F_T(f)_n=\frac2T\int_{-\frac12 T}^{\frac12T}f(t)\cos\left(\frac{2\pi}Tnt\right)\mathrm dtをFourier余弦係数という $ \mathcal F_{s,T}(f)_n=-2\Im\mathcal F_T(f)_n=\frac2T\int_{-\frac12 T}^{\frac12T}f(t)\sin\left(\frac{2\pi}Tnt\right)\mathrm dtをFourier正弦係数という $ f(t)=\sum_{n\in\Z}\mathcal F_T(f)_ne^{in\frac{2\pi}Tt}
$ = \mathcal F_T(f)_0+\sum_{n\in\N}\left(\mathcal F_T(f)_ne^{in\frac{2\pi}Tt}+\mathcal F_T(f)_{-n}e^{-in\frac{2\pi}Tt}\right)
$ = \mathcal F_T(f)_0+\sum_{n\in\N}\left(\mathcal F_T(f)_ne^{in\frac{2\pi}Tt}+{\mathcal F_T(f^*)_n}^*{e^{in\frac{2\pi}Tt}}^*\right)
$ = \mathcal F_T(f)_0+\sum_{n\in\N}\left((\mathcal F_T(f)_n+\mathcal F_T(f^*)_n^*)\cos\left(\frac{2\pi}Tnt\right)+i(\mathcal F_T(f)_n-\mathcal F_T(f^*)_n^*)\sin\left(\frac{2\pi}Tnt\right)\right)
$ = \mathcal F_T(f)_0+\sum_{n\in\N}\left((\mathcal F_T(\Re f)_n+\mathcal F_T(\Re f)_n^*+i(\mathcal F_T(\Im f)_n+\mathcal F_T(\Im f)_n^*))\cos\left(\frac{2\pi}Tnt\right)+i(\mathcal F_T(\Re f)_n-\mathcal F_T(\Re f)_n^*+i(\mathcal F_T(\Im f)_n-\mathcal F_T(\Im f)_n^*))\sin\left(\frac{2\pi}Tnt\right)\right)
$ = \frac12\mathcal F_{c,T}(f)_0+\sum_{n\in\N}\left((\mathcal F_{c,T}(\Re f)_n+i\mathcal F_{c,T}(\Im f)_n)\cos\left(\frac{2\pi}Tnt\right)+i(-i\mathcal F_{s,T}(\Re f)_n+\mathcal F_{s,T}(\Im f)_n)\sin\left(\frac{2\pi}Tnt\right)\right)
$ = \frac12\mathcal F_{c,T}(f)_0+\sum_{n\in\N}\left(\mathcal F_{c,T}(\Re f)_n\cos\left(\frac{2\pi}Tnt\right)+\mathcal F_{s,T}(\Re f)_n\sin\left(\frac{2\pi}Tnt\right)\right)+i\sum_{n\in\N}\left(\mathcal F_{c,T}(\Im f)_n\cos\left(\frac{2\pi}Tnt\right)+\mathcal F_{s,T}(\Im f)_n\sin\left(\frac{2\pi}Tnt\right)\right)
$ \therefore\Re f(t)=\frac12\mathcal F_{c,T}(f)_0+\sum_{n\in\N}\left(\mathcal F_{c,T}(\Re f)_n\cos\left(\frac{2\pi}Tnt\right)+\mathcal F_{s,T}(\Re f)_n\sin\left(\frac{2\pi}Tnt\right)\right)
$ \therefore\Im f(t)=-\frac12\mathcal F_{s,T}(f)_0+\sum_{n\in\N}\left(\mathcal F_{c,T}(\Im f)_n\cos\left(\frac{2\pi}Tnt\right)+\mathcal F_{s,T}(\Im f)_n\sin\left(\frac{2\pi}Tnt\right)\right)
$ \mathcal F_{s,T}(f)_0=0なのでなくてもいい
$ \mathcal{F}_T(f)_\bullet:\Z\ni n\mapsto\frac1{\sqrt{2\pi}}\int_{-\frac12 T}^{\frac12T}f(t)e^{-in\frac{2\pi}Tt}\mathrm dt\in\Complex
$ \mathcal F_T^{-1}(c):\rbrack-\frac12T,\frac12T\lbrack\ni t\mapsto\frac1{\sqrt{2\pi}}\sum_{n\in\Z}c_ne^{in\frac{2\pi}{T}t}\frac{2\pi}T\in\Complex
$ \mathcal F_T(f)_\bullet:\Z\ni n\mapsto\frac1{\sqrt{2\pi}}\int_{0}^{T}f(t)e^{-in\frac{2\pi}{T}t}\mathrm dt\in[0,T]
本では周期を$ 2\piに固定しているが、ここでは$ Tに拡張した定義を示している
$ \mathcal{F}_T^o(f)_\bullet:\Z\ni n\mapsto\frac1T\int_{-\frac12 T}^{\frac12T}f(t)e^{-i\frac{2n+1}{2}\frac{\pi}Tt}\mathrm dt\in\Complexとするとどうなる？
$ \mathcal{F}_T^o(f)_n=\mathcal{F}_T(f)_{\frac{2n+1}{4}}
References