Evaluate the DFT of the vector $(f(x_0),\dots,f(x_{18}),f(x_{19}))$

Question

Let $f(x)=1-x^2$, with $x \in [0,1)$.

1. Evaluate $\hat{f}(x)=\int_0^1 f(y)e^{-2 \pi ixy} \,dy$ (the Fourier transform of $f$).

2. Let $x_j=\frac{k}{10}$, with $k=0,\dots,18,19$.

   a. Evaluate the DFT of the vector $(f(x_0),\dots,f(x_{18}),f(x_{19}))$

   b. How is that related to the Fourier transform of $f$?

$\textbf{Evaluate $\hat{f}(x)=\int_0^1 f(y)e^{-2 \pi ixy} \,dy$}$

$$\hat{f}(x) =\int_0^1 (1-y^2)e^{-2 \pi ixy} \,dy =\int_0^1 e^{-2 \pi ixy} \,dy - \int_0^1 y^2e^{-2 \pi ixy} \,dy $$

\begin{array}{l|l} y^2 & e^{-2 \pi ixy} \\ \hline 2y & \frac{1}{-2 \pi ix}e^{-2 \pi ixy} \\ \hline 2 & \frac{1}{(-2 \pi ix)^2}e^{-2 \pi ixy} \\ \hline 0 & \frac{1}{(-2 \pi ix)^3}e^{-2 \pi ixy} \\ \end{array}

\begin{equation*} \begin{split} \hat{f}(x) & =e^{-2 \pi ixy}(\frac{-1}{2 \pi ix}+\frac{y^2}{2 \pi ix}+\frac{2y}{(-2 \pi ix)^2}-\frac{2}{(-2 \pi ix)^3})]_0^1 \\ & =e^{-2 \pi ixy}(-\frac{1}{2 \pi ix}+\frac{1}{2 \pi ix}+\frac{2}{(-2 \pi ix)^2}-\frac{2}{(-2 \pi ix)^3}+\frac{1}{2 \pi ix}-\frac{2}{(-2 \pi ix)^3}) \\ & =e^{-2 \pi ixy}(\frac{2}{(-2 \pi ix)^2}-\frac{2}{(-2 \pi ix)^3})+\frac{1}{2 \pi ix}-\frac{2}{(-2 \pi ix)^3} \\ \end{split} \end{equation*}

How would I do Part 2?

Answer 1

$\textbf{Evaluate the DFT of the vector $(f(x_0),\dots,f(x_{18}),f(x_{19}))$}$

$\textbf{Definition:}$ The DFT of a vector is $X_k \equiv \sum_{n=0}^{N-1}x_ne^\frac{-2\pi kni}{N}$ where $k \in \mathbb{Z}$

The DFT for vector $(f(x_0),\dots,f(x_{18}),f(x_{19}))$ is $$\sum_{n=0}^{19}x_ne^\frac{-2\pi kni}{5}=1+\frac{399}{400}e^\frac{-2\pi ki}{20}+\frac{396}{400}e^\frac{-4\pi ki}{20}+\frac{391}{400}e^\frac{-6\pi ki}{20}+\dots+\frac{39}{400}e^\frac{-38\pi ki}{20}$$

$\textbf{How is that related to the Fourier transform of $f$?}$

The summation of the DFT is a periodic summation, where the Fourier transform has one peak.