## Leibniz formula for π using a Fourier sine series expansion $\displaystyle\sum _{k=1}^{\infty }\dfrac{\left(-1\right)^{k-1}} {2k-1}=\dfrac {\pi }{4}$

using the Fourier sine expansion for the function $f(x)=\dfrac {1} {2}\left( \pi -x\right)$; (range: $\left[0,\pi \right]$).

To obtain the Fourier sine series expansion for $f(x)$ the coefficients $a_n$ must vanish (see below). So let $g(x)$ be the odd function extending $f(x)$ to the interval $[-\pi,0[$ defined by $g(x)=\left\{\begin{array}{l}\phantom{-}f(x)=\phantom{-}\dfrac{\pi-x}{2},\qquad\quad 0\leq x\leq\pi\\\\-f(-x)=-\dfrac{\pi+x}{2},\qquad -\pi\leq x <0\end{array}\right.$

whose graph in $[\pi,\pi]$ is shown in the following figure  $g(x)=f(x)=\dfrac{\pi -x}{2}, x\in[0,\pi[ ; \; g(x)=-f(-x)=-\dfrac{\pi +x}{2},x\in[-\pi,0]$

We know that the trigonometric Fourier series expansion for $g(x)$ in the interval $\left[ -\pi ,\pi \right]$ is given by $\dfrac{a_{0}}{2}+\displaystyle\sum_{n=1}^{\infty}\left(a_{n}\cos (nx)+b_{n}\sin (nx)\right),$

where the coefficients are the integrals $a_{n}=\dfrac{1}{\pi }\displaystyle\int_{-\pi }^{\pi }g(x)\cos (nx)\,dx,\qquad n=0,1,2,\ldots,$ $b_{n}=\dfrac{1}{\pi }\displaystyle\int_{-\pi }^{\pi }g(x)\sin (nx)\,dx,\qquad n=1,2,3,\ldots,$

The integrand $g(x)\cos (nx)$ is an odd function, while $g(x)\sin (nx)$ is even. So $a_{n}=0$ and $b_{n}=\dfrac{2}{\pi }\displaystyle\int_{0}^{\pi }g(x)\sin (nx)\,dx=\dfrac{2}{\pi }\displaystyle\int_{0}^{\pi }\dfrac{\pi -x}{2}\sin (nx)\,dx.$

Evaluating this last integral we obtain $b_{n}=\frac{1}{n}$, which proves that $f(x)=\dfrac{\pi -x}{2}=\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n}\sin (nx),\qquad 0\leq x\leq\pi,$

(the Fourier sine series for the function $f(x)$ in the interval $\left[0,\pi\right]$). We see that $f(\pi/2)=\pi/4$. We just need to confirm that for $x=\pi /2$ this last series reduces to your series. Indeed since $\sin\big(\dfrac{n\pi }{2}\big)=\left\{\begin{array}{l}\phantom{-}1,\qquad n=1,5,\ldots ,4k+1,\ldots\\\phantom{-}0,\qquad n=2,4,\ldots ,2k+2,\ldots\quad (k=0,1,2,\ldots)\\-1,\qquad n=3,5,\ldots,4k+3,\ldots,\end{array}\right.$

we have $\dfrac{\pi }{4}=\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n}\sin \big(\dfrac{n\pi }{2}\big)=1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\ldots = \displaystyle\sum_{k=1}^{\infty }\dfrac {\left( -1\right)^{k-1}} {2k-1}.$

Remark. For a Fourier cosine series we would need to extend $f(x)$ to an even function instead, because then $b_n$ would vanish. ## Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Análise de Fourier, Cálculo, Matemática, Math, Mathematics Stack Exchange, Séries com as etiquetas , , . ligação permanente.

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