Rearranging fourier expansion to find solution to a series

f(x) = x+ \pi

Fourier expansion is

Unparseable latex formula:

f(x) = \pi - 2 \displaystyle\sum_{n=1}^\infty} \dfrac{(-1)^n sin(nx)}{n}

and I'm finding

Unparseable latex formula:

\displaystyle\sum_{n=0}^\infty} \dfrac{(-1)^k}{2k+1}

So at

x =\dfrac{\pi}{2}

we have

Unparseable latex formula:

\pi + \dfrac{\pi}{2} = \pi - 2 \displaystyle\sum_{n=1}^\infty} \dfrac{(-1)^n sin(\dfrac{n\pi}{2})}{n}

Unparseable latex formula:

\dfrac{\pi}{2} = - 2 \displaystyle\sum_{n=1}^\infty} \dfrac{(-1)^n sin(\dfrac{n\pi}{2})}{n}

But

sin(\dfrac{n\pi}{2})

when

n=2k+1

gives

sin(\dfrac{n\pi}{2}) = -1 \ , \ n \geq 1 , \ n \in \mathbb{Z} , \ k \in \mathbb{N}

Hence

Unparseable latex formula:

\dfrac{\pi}{2} = + 2 \displaystyle\sum_{n=0}^\infty} \dfrac{(-1)^{2k+1}}{2k+1}

But the issue here is that

(-1)^{2k+1} = -1

so I'm not getting the correct series. What am I doing wrong? My brain's dead

(edited 13 years ago)

When n = 2k, then sin(n*pi/2) = sin(k*pi) = 0.

When n = 2k + 1, then sin(n*pi/2) = sin((2k+1)*pi/2) = (-1)^k.

Original post by Glutamic Acid

When n = 2k, then sin(n*pi/2) = sin(k*pi) = 0.

When n = 2k + 1, then sin(n*pi/2) = sin((2k+1)*pi/2) = (-1)^k.

cheers, taking a second look at the poor sine curve I drew, I now realise I need sleep

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