Let $f$ be a function in $\mathcal{S}$ whose Fourier transform $\mathcal{F}(f)$ is zero outside the segment $[-1/2, 1/2]$. According to the Fourier inversion formula, we have $$\forall x \in \mathbb{R}, \quad f(x) = \int_{-1/2}^{1/2} \mathcal{F}(f)(\xi) e^{2\pi\mathrm{i} x\xi} \mathrm{d}\xi$$ Prove that $$\forall (x, x_{0}) \in \mathbb{R}^{2}, \quad \sum_{n=0}^{+\infty} \frac{(x-x_{0})^{n}}{n!} \int_{-1/2}^{1/2} (2\pi\mathrm{i}\xi)^{n} \mathcal{F}(f)(\xi) e^{2\pi\mathrm{i} x_{0}\xi} \mathrm{d}\xi = f(x)$$
Let $f$ be a function in $\mathcal{S}$ whose Fourier transform $\mathcal{F}(f)$ is zero outside the segment $[-1/2, 1/2]$. According to the Fourier inversion formula, we have
$$\forall x \in \mathbb{R}, \quad f(x) = \int_{-1/2}^{1/2} \mathcal{F}(f)(\xi) e^{2\pi\mathrm{i} x\xi} \mathrm{d}\xi$$
Prove that
$$\forall (x, x_{0}) \in \mathbb{R}^{2}, \quad \sum_{n=0}^{+\infty} \frac{(x-x_{0})^{n}}{n!} \int_{-1/2}^{1/2} (2\pi\mathrm{i}\xi)^{n} \mathcal{F}(f)(\xi) e^{2\pi\mathrm{i} x_{0}\xi} \mathrm{d}\xi = f(x)$$