# Some doubt on series.

If i have a series of the form ( say fourier or anything ) , For example lets consider $$\sum_{k\in \mathbb Z} \exp\left( \frac {-ik\pi}{L}\right) a_k(t) ,$$ is it always possible to split it down to something like $$\sum_{k=1}^\infty b_k(t) \sin\left( \frac{k\pi x}{L}\right)+c_k(t) \cos\left(\frac {k\pi x}{L}\right)?$$

What I am missing here is I think the splitting of coefficients.

It may be a silly question but i am still missing.

Any help will be appreciated.

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$$e^{i\phi}=\cos\phi+i\sin\phi;\quad e^{-i\phi}=\cos\phi-i\sin\phi; \\ \\ \sin\phi=\frac{e^{i\phi}-e^{-i\phi}}{2i};\quad \cos\phi=\frac{e^{i\phi}+e^{-i\phi}}{2}.$$ – anon Jun 27 '12 at 8:39

Since $\cos u=\frac{e^{iu}+e^{-iu}}2$ and $\sin u=\frac{e^{iu}-e^{-iu}}{2i}$, $$2\cos u-2i\sin u=e^{-iu}.$$ Hence \begin{align} \sum_{k\in\Bbb Z}\exp\left(-\frac{ik\pi}L\right)a_k(t)&=\sum_{k\in\Bbb Z}\left[2\cos\left(-\frac{ik\pi}L\right)-2i\sin\left(-\frac{ik\pi}L\right)\right)a_k(t)\\ &=2a_0(t)+\sum_{k=1}^{+\infty}\cos\left(\frac{ik\pi}L\right)(2a_k(t)+2a_{-k}(t))\\ &-2i\sum_{k=1}^{+\infty}\sin\left(\frac{ik\pi}L\right)(a_k(t)-a_{-k}(t)). \end{align} Now identify.