Hallo,
ich bin noch neu im Wikipedia-Geschäft, freue mich aber immer über Anregungen. --Joko31 22:57, 22. Jul. 2007 (CEST)
e i π = − 1 {\displaystyle {\begin{matrix}e^{\mathrm {i} \,\pi }=\end{matrix}}-1\;}
f , g ∈ R → R {\displaystyle f,g\in R\to R}
( e i x ) ′ = ( sin x ) ′ + ( i cos x ) ′ = cos x − i sin x {\displaystyle (e^{\mathrm {i} x})'=(\sin x)'+(i\cos x)'=\cos x-i\sin x}
1 π ⋅ ∫ − π π cos n x ⋅ cos m x d x = { 0 falls n ≠ m 1 falls n = m > 0 2 falls n = m = 0 {\displaystyle {1 \over \pi }\cdot \int _{-\pi }^{\pi }\cos nx\cdot \cos mx\ dx={\begin{cases}0&{\text{falls}}\qquad n\neq m\\1&{\text{falls}}\qquad n=m>0\\2&{\text{falls}}\qquad n=m=0\end{cases}}}
1 π ⋅ ∫ − π π sin n x ⋅ cos m x d x = 0 {\displaystyle {1 \over \pi }\cdot \int _{-\pi }^{\pi }\sin nx\cdot \cos mx\ dx=0}
a 0 ( x ) = 1 a 1 ( x ) = cos x a 2 ( x ) = sin x a 3 ( x ) = cos 2 x a 2 ( x ) = sin 2 x a 5 ( x ) = cos 3 x a 4 ( x ) = sin 3 x ⋮ ⋮ {\displaystyle {\begin{alignedat}{2}a_{0}(x)&=1\\a_{1}(x)&=\cos x&\quad &a_{2}(x)=\sin x\\a_{3}(x)&=\cos 2x&\quad &a_{2}(x)=\sin 2x\\a_{5}(x)&=\cos 3x&\quad &a_{4}(x)=\sin 3x\\&\vdots &\qquad &\qquad \vdots \end{alignedat}}}
f ( t ) = ∑ n = − ∞ ∞ e i n ω t T ∫ 0 T f ( t ) e − i n ω t d t . {\displaystyle f(t)=\sum _{n=-\infty }^{\infty }{e^{\mathrm {i} n\omega t} \over T}\int _{0}^{T}f(t)e^{-\mathrm {i} n\omega t}dt.}
