From c2327bcc3571ead88ba2b0ce40656211a888f640 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sun, 21 Feb 2021 16:46:21 +0100
Subject: Add "Convolution theorem" and "Parseval's theorem"

---
 latex/know/concept/parsevals-theorem/source.md | 64 ++++++++++++++++++++++++++
 1 file changed, 64 insertions(+)
 create mode 100644 latex/know/concept/parsevals-theorem/source.md

(limited to 'latex/know/concept/parsevals-theorem/source.md')

diff --git a/latex/know/concept/parsevals-theorem/source.md b/latex/know/concept/parsevals-theorem/source.md
new file mode 100644
index 0000000..41af734
--- /dev/null
+++ b/latex/know/concept/parsevals-theorem/source.md
@@ -0,0 +1,64 @@
+% Parseval's theorem
+
+
+# Parseval's theorem
+
+**Parseval's theorem** relates the inner product of two functions to the
+inner product of their [Fourier transforms](/know/concept/fourier-transform/).
+There are two equivalent ways of stating it,
+where $A$, $B$, and $s$ are constants from the Fourier transform's definition:
+
+$$\begin{aligned}
+    \boxed{
+        \braket{f(x)}{g(x)} = \frac{2 \pi B^2}{|s|} \braket{\tilde{f}(k)}{\tilde{g}(k)}
+    }
+    \\
+    \boxed{
+        \braket{\tilde{f}(k)}{\tilde{g}(k)} = \frac{2 \pi A^2}{|s|} \braket{f(x)}{g(x)}
+    }
+\end{aligned}$$
+
+For this reason many physicists like to define their Fourier transform
+with $|s| = 1$ and $A = B = 1 / \sqrt{2\pi}$, because then the FT nicely
+conserves the total probability (quantum mechanics) or the total energy
+(optics).
+
+To prove this, we insert the inverse FT into the inner product
+definition:
+
+$$\begin{aligned}
+    \braket{f}{g}
+    &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\}\big)^* \: \hat{\mathcal{F}}^{-1}\{\tilde{g}(k)\} \dd{x}
+    \\
+    &= B^2 \int
+    \Big( \int \tilde{f}^*(k_1) \exp(i s k_1 x) \dd{k_1} \Big)
+    \Big( \int \tilde{g}(k) \exp(- i s k x) \dd{k} \Big)
+    \dd{x}
+    \\
+    &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \tilde{g}(k) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s x (k_1 - k)) \dd{x} \Big) \dd{k_1} \dd{k}
+    \\
+    &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \: \tilde{g}(k) \: \delta(s (k_1 - k)) \dd{k_1} \dd{k}
+    \\
+    &= \frac{2 \pi B^2}{|s|} \int_{-\infty}^\infty \tilde{f}^*(k) \: \tilde{g}(k) \dd{k}
+    = \frac{2 \pi B^2}{|s|} \braket{\tilde{f}}{\tilde{g}}
+\end{aligned}$$
+
+Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
+We can just as well do it in the opposite direction, with equivalent results:
+
+$$\begin{aligned}
+    \braket{\tilde{f}}{\tilde{g}}
+    &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}\{f(x)\}\big)^* \: \hat{\mathcal{F}}\{g(x)\} \dd{k}
+    \\
+    &= A^2 \int
+    \Big( \int f^*(x_1) \exp(- i s k x_1) \dd{x_1} \Big)
+    \Big( \int g(x) \exp(i s k x) \dd{x} \Big)
+    \dd{k}
+    \\
+    &= 2 \pi A^2 \iint f^*(x_1) g(x) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s k (x_1 - x)) \dd{k} \Big) \dd{x_1} \dd{x}
+    \\
+    &= 2 \pi A^2 \iint f^*(x_1) \: g(x) \: \delta(s (x_1 - x)) \dd{x_1} \dd{x}
+    \\
+    &= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x}
+    = \frac{2 \pi A^2}{|s|} \braket{f}{g}
+\end{aligned}$$
-- 
cgit v1.2.3