\(f\left( x \right) = \frac{L}{2} + \frac{{2L}}{\pi }\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {\left( {2n - 1} \right)\pi x} \right)}}{{2n - 1}}} \Leftrightarrow \sum\limits_{n = 1}^\infty {\frac{{\sin \left( {\left( {2n - 1} \right)\pi x} \right)}}{{2n - 1}}} = \frac{\pi }{{2L}}\left( {f\left( x \right) - \frac{L}{2}} \right)\)
Verschuiven van de index:
\(\sum\limits_{n = 0}^\infty {\frac{{\sin \left( {\left( {2n + 1} \right)\pi x} \right)}}{{2n + 1}}} = \frac{\pi }{{2L}}\left( {f\left( x \right) - \frac{L}{2}} \right)\)
Kies L = 2:
\(\sum\limits_{n = 0}^\infty {\frac{{\sin \left( {\left( {2n + 1} \right)\pi x} \right)}}{{2n + 1}}} = \frac{\pi }{4}\left( {f\left( x \right) - 1} \right)\)
Dan is f:
\(f\left( x \right) = \left\{ {\begin{array}{*{20}l} 0 & \; {\mbox{elders}} \\ 2 & \; {0 < x < 2} \\\end{array}} \right.\)
Voor x = 1/2 is f(x) = 2 en geldt:
\(\sum\limits_{n = 0}^\infty {\frac{{\sin \left( {\left( {2n + 1} \right)\frac{\pi }{2}} \right)}}{{2n + 1}}} = \frac{\pi }{4} \Rightarrow \sum\limits_{n = 0}^\infty {\frac{{\left( { - 1} \right)^n }}{{2n + 1}}} = \frac{\pi }{4}\)
