MaA-2025F-5.@2025-05-19T13:00D90

$ \int_0^1\sin\pi nt\cos\pi kt\mathrm dt=\frac12\int_0^1(\sin\pi(n+k)t+\sin\pi(n-k)t)\mathrm dt

$ = \frac12\int_0^1\mathrm d\left(-\frac1{n+k}\cos\pi(n+k)t\llbracket n+k\neq 0\rrbracket-\frac1{n-k}\cos\pi(n-k)t\llbracket n\neq k\rrbracket\right)

$ =\frac12\left(-\frac1{n+k}\llbracket n+k\neq0\rrbracket-\frac1{n-k}\llbracket n\neq k\rrbracket+\frac1{n+k}\llbracket n+k\neq0\rrbracket+\frac1{n-k}\llbracket n\neq k\rrbracket\right)

$ =0

$ \int_0^1\cos\pi nt\cos\pi kt\mathrm dt=\frac12\int_0^1(\cos\pi(n+k)t+\cos\pi(n-k)t)\mathrm dt

$ = \frac12\int_0^1\mathrm d\left(\frac1{n+k}\sin\pi(n+k)t\llbracket n+k\neq 0\rrbracket+\frac1{n-k}\sin\pi(n-k)t\llbracket n\neq k\rrbracket+t\llbracket n+k=0\rrbracket+t\llbracket n=k\rrbracket\right)

$ = \frac12(1-0)(\llbracket n+k=0\rrbracket+\llbracket n=k\rrbracket)

$ = \frac12(\llbracket n+k=0\rrbracket+\llbracket n=k\rrbracket)

$ \frac2T \int_0^T f(t) \cos\left(\frac{2\pi k}{T} t\right)\mathrm dt=\frac{V_0}{T}\int_0^T\cos\left(\frac{2\pi k}{T} t\right)\mathrm dt + \sum_{n\in\N} \frac2TV_n \int_0^T\cos\left(\frac{2\pi n}{T} t\right)\cos\left(\frac{2\pi k}{T} t\right)\mathrm dt + \sum_{n\in\N} \frac2Tb_n\int_0^T\sin\left(\frac{2\pi n}{T} t\right)\cos\left(\frac{2\pi k}{T} t\right)\mathrm dt

$ = 0+\sum_{n\in\N}V_n(\llbracket n+k=0\rrbracket+\llbracket n=k\rrbracket)+0

$ =V_k

$ \frac2T \int_0^T f(t) \sin\left(\frac{2\pi k}{T} t\right)\mathrm dt=\frac{V_0}{T}\int_0^T\sin\left(\frac{2\pi k}{T} t\right)\mathrm dt + \sum_{n\in\N} \frac2TV_n \int_0^T\cos\left(\frac{2\pi n}{T} t\right)\sin\left(\frac{2\pi k}{T} t\right)\mathrm dt + \sum_{n\in\N} \frac2Tb_n\int_0^T\sin\left(\frac{2\pi n}{T} t\right)\sin\left(\frac{2\pi k}{T} t\right)\mathrm dt

$ = -\frac{V_0}{2\pi k}\left(\cos 2\pi k-1\right)+0+\sum_{n\in\N}b_n(-\llbracket n+k=0\rrbracket+\llbracket n=k\rrbracket)

$ =b_k

$ \therefore V_k=\frac2T \int_0^T f(t) \cos\left(\frac{2\pi k}{T} t\right)\mathrm dt

$ \therefore b_k=\frac2T \int_0^T f(t) \sin\left(\frac{2\pi k}{T} t\right)\mathrm dt

$ \frac{V_0}{2} + \sum_{n\in\N} V_n \cos\left(\frac{2\pi n}{T} t\right) + \sum_{n\in\N} b_n \sin\left(\frac{2\pi n}{T} t\right)

$ =\sum_{n\in\Z}c_ne^{\frac{2\pi i}{T}nt}

$ \frac1T\int_0^Tf(t)e^{-i\omega kt}\mathrm dt=\frac1T\sum_{n\in\Z}c_n\int_0^Te^{i\omega(n-k)t}\mathrm dt

$ = \frac1T\sum_{n\in\Z}c_n\int_{t=0}^{t=T}\mathrm d\left(\frac{1}{i\omega}\frac1{n-k}e^{i\omega(n-k)t}\llbracket n\neq k\rrbracket+t\llbracket n=k\rrbracket\right)

$ = \frac1T\sum_{n\in\Z}c_n\left(\frac{1}{i\omega}\frac1{n-k}(e^{i\omega(n-k)T}-1)\llbracket n\neq k\rrbracket+(T-0)\llbracket n=k\rrbracket\right)

$ = \sum_{n\in\Z}c_n(0+\llbracket n=k\rrbracket)

$ =c_k

$ \therefore c_k=\frac1T\int_0^Tf(t)e^{-i\omega kt}\mathrm dt

$ \mathcal F(f)(\omega n):=\frac1T\int_0^Tf(t)e^{-i\omega nt}\mathrm dt

$ \mathcal F^{-1}(f)(t):=\sum_{n\in\Z}f(\omega n)e^{i\omega nt}

TODO: 線型空間を構成して、線型空間の記法で解き直す