Is it possible to find out what function given Fourier series describes?
I have the following problem:
Calculate $ \sum_{1}^{ \infty }\frac{(-1)^{n}}{2n-1} $
I know given series is expansion of constant $\pi$/4 but I can't see a way of reverse-engineering it into this function.
Best i could come up with are following equations:
$ \frac{1}{2n-1}=\int_{-\pi}^{\pi}f(x)cos(nx) dx $
$ 0=\int_{-\pi}^{\pi} f(x)dx $
I don't see a way to find f(x) from them though.
The Mercator series $$ \ln \left( {{1 \over {1 - z}}} \right) = \ln \left( {1 + {z \over {1 - z}}} \right) = \sum\limits_{0\, \le \,k} {{{z^{\,k + 1} } \over {k + 1}}} \quad \left| {\;\left| z \right| \le 1\; \wedge \;z \ne 1} \right. $$ converges for $|z| \le 1$, except when $z=1$, to the principal branch of $\ln(1/(1-z))$.
So it converges for $z=i$, which gives: $$ \eqalign{ & \ln \left( {{1 \over {1 - i}}} \right) = - \ln \sqrt 2 + i{\pi \over 4} = \cr & = \sum\limits_{0\, \le \,k} {{{i^{\,k + 1} } \over {k + 1}}} = \sum\limits_{0\, \le \,k} {{{i^{\,2n + 1} } \over {2n + 1}}} + \sum\limits_{0\, \le \,k} {{{i^{\,2n + 2} } \over {2n + 2}}} = \cr & = - \sum\limits_{0\, \le \,k} {{{i^{\,2n} } \over {2n + 2}}} + i\sum\limits_{0\, \le \,k} {{{i^{\,2n} } \over {2n + 1}}} = - \sum\limits_{0\, \le \,k} {{{\left( { - 1} \right)^{\,n} } \over {2n + 2}}} + i\sum\limits_{0\, \le \,k} {{{\left( { - 1} \right)^{\,n} } \over {2n + 1}}} \cr} $$
That means $$ \eqalign{ & {1 \over 2}\left( {\ln \left( {{1 \over {1 - z}}} \right) - \ln \left( {{1 \over {1 - \left( { - z} \right)}}} \right)} \right) = {1 \over 2}\ln \left( {{{1 + z} \over {1 - z}}} \right) = \cr & = {1 \over 2}\sum\limits_{0\, \le \,k} {{{1 - \left( { - 1} \right)^{\,k + 1} } \over {k + 1}}z^{\,k + 1} } = z\sum\limits_{0\, \le \,n} {{{z^{\,2n} } \over {2n + 1}}} \cr} $$
Therefore, putting $z=e^{it/2}$, the expression $$ \eqalign{ & F(t) = {1 \over {2e^{\,i\,t/2} }}\ln \left( {{{1 + e^{\,i\,t/2} } \over {1 - e^{\,i\,t/2} }}} \right) = \cr & = \sum\limits_{0\, \le \,n} {{{e^{\,i\,n\,t} } \over {2n + 1}}} = \sum\limits_{0\, \le \,n} {{{\cos \left( {n\,t} \right)} \over {2n + 1}}} + i\sum\limits_{0\, \le \,n} {{{\sin \left( {n\,t} \right)} \over {2n + 1}}} \cr} $$ will converge for all the values $0 < t < 2\pi$.
Finally, we arrive to $$ \bbox[lightyellow] { \eqalign{ & F(t)\quad \left| {\;0 < t < 2\pi } \right.\quad = \cr & = \sum\limits_{0\, \le \,n} {{{\cos \left( {n\,t} \right)} \over {2n + 1}}} + i\sum\limits_{0\, \le \,n} {{{\sin \left( {n\,t} \right)} \over {2n + 1}}} = \cr & = {1 \over {2e^{\,i\,t/2} }}\ln \left( {{{1 + e^{\,i\,t/2} } \over {1 - e^{\,i\,t/2} }}} \right) = {1 \over {2e^{\,i\,t/2} }}\ln \left( {{{e^{\, - \,i\,t/4} + e^{\,\,i\,t/4} } \over {e^{\, - \,i\,t/4} - e^{\,\,i\,t/4} }}} \right) = \cr & = {1 \over 2}e^{\, - \,i\,t/2} \ln \left( {{i \over {\tan \left( {t/4} \right)}}} \right) = \cr & = {1 \over 2}\left( {\cos \left( {t/2} \right) - i\sin \left( {t/2} \right)} \right) \left( { - \ln \left| {\tan \left( {t/4} \right)} \right| + i{\pi \over 2}} \right) = \cr & = {1 \over 2}\left( { - \cos \left( {t/2} \right)\ln \left| {\tan \left( {t/4} \right)} \right| + {\pi \over 2}\sin \left( {t/2} \right)} \right) + \cr & + {i \over 2}\left( {\sin \left( {t/2} \right)\ln \left| {\tan \left( {t/4} \right)} \right| + {\pi \over 2}\cos \left( {t/2} \right)} \right) \cr} }$$