f(x) = f(-x). Daher entfallen alle Sinus-Anteile \( b_{k} \) .
\( a_{k} = \frac{2}{T} \int\limits_{-T/2}^{+T/2} |x| * cos( \frac{2π}{T}*k*x) dx \)
wegen T = 2
\( a_{k} = \int\limits_{-1}^{+1} |x| * cos(π*k*x) dx \)
wegen cos(x) = cos(-x) und |x| = |-x|
\( a_{k} = 2*\int\limits_{0}^{+1} x * cos(π*k*x) dx = \frac{π k x sin(π k x) + cos(π k x)}{π^2 k^2} [0,1] \)
\( a_{k} = 2*( \frac{cos(π k )-1}{π^2 k^2} ) \)
k ungerade : \( cos(π*k) = -1 -> a_{k} = \frac{-4}{π^2 k^2} \)
k gerade : \( cos(π*k) = +1 -> a_{k} = 0 \)