> [!NOTE] > Let $\phi:\mathbb{R}\to \mathbb{R}$ be $2\pi$-periodic (that is, $\phi(2\pi+x)=\phi (x),\forall x\in \mathbb{R}$) such that its [[Fourier Transform|Fourier expansion]] exists. If $\phi$ is even then its Fourier coefficients satisfy $\hat{\phi}(k)=\hat{\phi}(-k)$ its Fourier expansion satisfies $S_{n}[\phi](x)=\hat{\phi}(0)+ 2 \sum_{k=1}^{\infty} \hat{\phi}(k) \cos(kx)$otherwise, if $\phi$ is odd then $\hat{\phi}(-k)=-\hat{\phi}(k)$ and $S_{n}[\phi](x)= 2i\sum_{k=1} \hat{\phi}(k)\sin(kx).$ ###### Proof Suppose $\phi$ is even. Substituting $y=-x$ gives $\begin{align} \hat{\phi}(k) = \int_{-\pi}^{\pi} \phi(x)e^{-ikx} \, dx &= \int_{\pi}^{-\pi} \phi(-y) e^{-i(-k)y}\, (-dy) \\ &= \int_{-\pi}^{\pi} \phi(y) e^{-i(-k)y} \, dy \\ &= \hat{\phi}(-k). \end{align}$Therefore $\begin{align} S_{n}[\phi](x) &= \sum_{k=-n}^{n} \hat{\phi}(k) e^{ikx} = \hat{\phi}(0) + \sum_{k=1}^\infty \hat{\phi}(k)e^{ikx} + \hat{\phi}(-k)e^{-ikx} \\ &= \hat{\phi}(0) + \sum_{k=1}^{\infty} \hat{\phi}(k) (e^{ikx} +e^{-ikx}) = \hat{\phi}(0) +\sum_{k=1}^\infty \hat{\phi}(k) 2\cos(kx) \\ &= \hat{\phi}(0)+ 2 \sum_{k=1}^{\infty} \hat{\phi}(k) \cos(kx). \end{align}$