Expand knowledge base

1 files changed, 138 insertions, 12 deletions

diff --git a/content/know/concept/fourier-transform/index.pdc b/content/know/concept/fourier-transform/index.pdc
index 3be47ff..1d1e27d 100644
--- a/content/know/concept/fourier-transform/index.pdc
+++ b/content/know/concept/fourier-transform/index.pdc
@@ -25,8 +25,8 @@ The **forward** FT is defined as follows, where $A$, $B$, and $s$ are unspecifie
 $$\begin{aligned}
     \boxed{
         \tilde{f}(k)
-        = \hat{\mathcal{F}}\{f(x)\}
-        = A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x}
+        \equiv \hat{\mathcal{F}}\{f(x)\}
+        \equiv A \int_{-\infty}^\infty f(x) \exp\!(i s k x) \dd{x}
     }
 \end{aligned}$$
 
@@ -35,8 +35,8 @@ The **inverse Fourier transform** (iFT) undoes the forward FT operation:
 $$\begin{aligned}
     \boxed{
         f(x)
-        = \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\}
-        = B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k x) \dd{k}
+        \equiv \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\}
+        \equiv B \int_{-\infty}^\infty \tilde{f}(k) \exp\!(- i s k x) \dd{k}
     }
 \end{aligned}$$
 
@@ -46,9 +46,9 @@ again. Let us verify this, by rearranging the integrals to get the
 
 $$\begin{aligned}
     \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{f(x)\}\}
-    &= A B \int_{-\infty}^\infty \exp(-i s k x) \int_{-\infty}^\infty f(x') \exp(i s k x') \dd{x'} \dd{k}
+    &= A B \int_{-\infty}^\infty \exp\!(-i s k x) \int_{-\infty}^\infty f(x') \exp\!(i s k x') \dd{x'} \dd{k}
     \\
-    &= 2 \pi A B \int_{-\infty}^\infty f(x') \Big(\frac{1}{2\pi} \int_{-\infty}^\infty \exp(i s k (x' - x)) \dd{k} \Big) \dd{x'}
+    &= 2 \pi A B \int_{-\infty}^\infty f(x') \Big(\frac{1}{2\pi} \int_{-\infty}^\infty \exp\!(i s k (x' - x)) \dd{k} \Big) \dd{x'}
     \\
     &= 2 \pi A B \int_{-\infty}^\infty f(x') \: \delta(s(x' - x)) \dd{x'}
     = \frac{2 \pi A B}{|s|} f(x)
@@ -74,15 +74,15 @@ on whether the analysis is for forward ($s > 0$) or backward-propagating
 
 ## Derivatives
 
-The FT of a derivative has a very interesting property.
+The FT of a derivative has a very useful property.
 Below, after integrating by parts, we remove the boundary term by
 assuming that $f(x)$ is localized, i.e. $f(x) \to 0$ for $x \to \pm \infty$:
 
 $$\begin{aligned}
     \hat{\mathcal{F}}\{f'(x)\}
-    &= A \int_{-\infty}^\infty f'(x) \exp(i s k x) \dd{x}
+    &= A \int_{-\infty}^\infty f'(x) \exp\!(i s k x) \dd{x}
     \\
-    &= A \big[ f(x) \exp(i s k x) \big]_{-\infty}^\infty - i s k A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x}
+    &= A \big[ f(x) \exp\!(i s k x) \big]_{-\infty}^\infty - i s k A \int_{-\infty}^\infty f(x) \exp\!(i s k x) \dd{x}
     \\
     &= (- i s k) \tilde{f}(k)
 \end{aligned}$$
@@ -110,13 +110,139 @@ $$\begin{aligned}
 Derivatives in the frequency domain have an analogous property:
 
 $$\begin{aligned}
+    \dv[n]{\tilde{f}}{k}
+    &= A \dv[n]{k} \int_{-\infty}^\infty f(x) \exp\!(i s k x) \dd{x}
+    \\
+    &= A \int_{-\infty}^\infty (i s x)^n f(x) \exp\!(i s k x) \dd{x}
+    = \hat{\mathcal{F}}\{ (i s x)^n f(x) \}
+\end{aligned}$$
+
+
+## Multiple dimensions
+
+The Fourier transform is straightforward to generalize to $N$ dimensions.
+Given a scalar field $f(\vb{x})$ with $\vb{x} = (x_1, ..., x_N)$,
+its FT $\tilde{f}(\vb{k})$ is defined as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \tilde{f}(\vb{k})
+        \equiv \hat{\mathcal{F}}\{f(\vb{x})\}
+        \equiv A \int_{-\infty}^\infty f(\vb{x}) \exp\!(i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{x}}
+    }
+\end{aligned}$$
+
+Where the wavevector $\vb{k} = (k_1, ..., k_N)$.
+Likewise, the inverse FT is given by:
+
+$$\begin{aligned}
     \boxed{
-        \dv[n]{\tilde{f}}{k}
-        = A \int_{-\infty}^\infty (i s x)^n f(x) \exp(i s k x) \dd{x}
-        = \hat{\mathcal{F}}\{ (i s x)^n f(x) \}
+        f(\vb{x})
+        \equiv \hat{\mathcal{F}}^{-1}\{\tilde{f}(\vb{k})\}
+        \equiv B \int_{-\infty}^\infty \tilde{f}(\vb{k}) \exp\!(- i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{k}}
     }
 \end{aligned}$$
 
+In practice, in $N$D, there is not as much disagreement about
+the constants $A$, $B$ and $s$ as in 1D:
+typically $A = 1$ and $B = 1 / (2 \pi)^N$, with $s = \pm 1$.
+Any choice will do, as long as:
+
+$$\begin{aligned}
+    \boxed{
+        A B
+        = \bigg( \frac{|s|}{2 \pi} \bigg)^{\!N}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-constants-ND"/>
+<label for="proof-constants-ND">Proof</label>
+<div class="hidden">
+<label for="proof-constants-ND">Proof.</label>
+The inverse FT of the forward FT of $f(\vb{x})$ must be equal to $f(\vb{x})$ again, so:
+
+$$\begin{aligned}
+    \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\}
+    &= A B \int \exp\!(- i s \vb{k} \cdot \vb{x})
+    \int f(\vb{x}') \exp\!(i s \vb{k} \cdot \vb{x}') \dd[N]{\vb{x}'} \dd[N]{\vb{k}}
+    \\
+    &= (2 \pi)^N A B \int f(\vb{x}')
+    \Big( \frac{1}{(2 \pi)^N} \int \exp\!(i s \vb{k} \cdot (\vb{x}' - \vb{x})) \dd[N]{\vb{k}} \Big) \dd[N]{\vb{x}'}
+    \\
+    &= (2 \pi)^N A B \int f(\vb{x}')
+    \Big( \prod_{n = 1}^N \frac{1}{2 \pi} \int \exp\!(i s k_n (x_n' - x_n)) \dd{k_n} \Big) \dd[N]{\vb{x}'}
+\end{aligned}$$
+
+Here, we recognize the definition of the Dirac delta function again,
+leading to:
+
+$$\begin{aligned}
+    \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\}
+    &= (2 \pi)^N A B \int f(\vb{x}')
+    \Big( \prod_{n = 1}^N \delta(s(x_n' - x_n)) \Big) \dd[N]{\vb{x}'}
+    \\
+    &= \frac{(2 \pi)^N A B}{|s|^N} \int f(\vb{x}') \: \delta(\vb{x}' - \vb{x}) \dd[N]{\vb{x}'}
+    = \frac{(2 \pi)^N A B}{|s|^N} f(\vb{x})
+\end{aligned}$$
+</div>
+</div>
+
+Differentiation is more complicated for $N > 1$,
+but the FT is still useful,
+notably for the Laplacian $\nabla^2 \equiv \dv*[2]{x_1} + ... + \dv*[2]{x_N}$.
+Let $|\vb{k}|$ be the norm of $\vb{k}$,
+then for a localized $f$:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\mathcal{F}}\{\nabla^2 f(\vb{x})\}
+        = - s^2 |\vb{k}|^2 \tilde{f}(\vb{k})
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-laplacian"/>
+<label for="proof-laplacian">Proof</label>
+<div class="hidden">
+<label for="proof-laplacian">Proof.</label>
+We insert $\nabla^2 f$ into the FT,
+decompose the exponential and the Laplacian,
+and then integrate by parts (limits $\pm \infty$ omitted):
+
+$$\begin{aligned}
+    \hat{\mathcal{F}}\{\nabla^2 f\}
+    &= A \int \big( \nabla^2 f \big) \exp\!(i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{x}}
+    \\
+    &= A \int \Big( \sum_{n = 1}^N \pdv[2]{f}{x_n} \Big) \Big( \prod_{m = 1}^N \exp\!(i s k_m x_m) \Big) \dd[N]{\vb{x}}
+    \\
+    &= A \sum_{n = 1}^N \bigg[ \pdv{f}{x_n} \exp\!(i s \vb{k} \cdot \vb{x}) \bigg]
+    - A \sum_{n = 1}^N i s k_n \int \pdv{f}{x_n} \exp\!(i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{x}}
+\end{aligned}$$
+
+Just like in 1D, we get rid of the boundary term
+by assuming that all derivatives $\dv*{f}{x_n}$ are nicely localized.
+To proceed, we then integrate by parts again:
+
+$$\begin{aligned}
+    \hat{\mathcal{F}}\{\nabla^2 f\}
+    &= - A \sum_{n = 1}^N i s k_n \int \pdv{f}{x_n} \Big( \prod_{m = 1}^N \exp\!(i s k_m x_m) \Big) \dd[N]{\vb{x}}
+    \\
+    &= - A \sum_{n = 1}^N i s k_n \bigg[ f \exp\!(i s \vb{k} \cdot \vb{x}) \bigg]
+    + A \sum_{n = 1}^N (i s k_n)^2 \int f \exp\!(i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{x}}
+\end{aligned}$$
+
+Once again, we remove the boundary term
+by assuming that $f$ is localized, yielding:
+
+$$\begin{aligned}
+    \hat{\mathcal{F}}\{\nabla^2 f\}
+    &= - A s^2 \sum_{n = 1}^N k_n^2 \int f \exp\!(i s \vb{k} \cdot \vb{x}) \dd[N]{\vb{x}}
+    = - s^2 \sum_{n = 1}^N k_n^2 \tilde{f}
+\end{aligned}$$
+</div>
+</div>
+
 
 
 ## References