- #1
redtree
- 322
- 13
- TL;DR Summary
- Is the Fourier transform of a function whose argument is a vector equivalent to performing a Fourier transform on the function of each vector component separately?
Given ##f(\vec{x})##, where the Fourier transform ##\mathcal{F}(f(\vec{x}))= \hat{f}(\vec{k})##.
Given ##\vec{x}=[x_1,x_2,x_3]## and ##\vec{k}=[k_1,k_2,k_3]##, is the following true?
\begin{equation}
\begin{split}
\mathcal{F}(f(x_1))&= \hat{f}(k_1)
\\
\mathcal{F}(f(x_2))&= \hat{f}(k_2)
\\
\mathcal{F}(f(x_3))&= \hat{f}(k_3)
\end{split}
\end{equation}
such that
\begin{equation}
\begin{split}
f(\vec{k})&= [\hat{f}(k_1), \hat{f}(k_1), \hat{f}(k_1)]
\end{split}
\end{equation}
Given ##\vec{x}=[x_1,x_2,x_3]## and ##\vec{k}=[k_1,k_2,k_3]##, is the following true?
\begin{equation}
\begin{split}
\mathcal{F}(f(x_1))&= \hat{f}(k_1)
\\
\mathcal{F}(f(x_2))&= \hat{f}(k_2)
\\
\mathcal{F}(f(x_3))&= \hat{f}(k_3)
\end{split}
\end{equation}
such that
\begin{equation}
\begin{split}
f(\vec{k})&= [\hat{f}(k_1), \hat{f}(k_1), \hat{f}(k_1)]
\end{split}
\end{equation}