From cc295b5da8e3db4417523a507caf106d5839d989 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Wed, 2 Jun 2021 13:28:53 +0200 Subject: Introduce collapsible proofs to some articles --- content/know/concept/parsevals-theorem/index.pdc | 32 ++++++++++++++---------- 1 file changed, 19 insertions(+), 13 deletions(-) (limited to 'content/know/concept/parsevals-theorem/index.pdc') diff --git a/content/know/concept/parsevals-theorem/index.pdc b/content/know/concept/parsevals-theorem/index.pdc index 824afa6..9f440f2 100644 --- a/content/know/concept/parsevals-theorem/index.pdc +++ b/content/know/concept/parsevals-theorem/index.pdc @@ -17,24 +17,24 @@ markup: pandoc and the inner product of their [Fourier transforms](/know/concept/fourier-transform/) $\tilde{f}(k)$ and $\tilde{g}(k)$. There are two equivalent ways of stating it, -where $A$, $B$, and $s$ are constants from the Fourier transform's definition: +where $A$, $B$, and $s$ are constants from the FT's definition: $$\begin{aligned} \boxed{ - \braket{f(x)}{g(x)} = \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}(k)}{\tilde{g}(k)} - } - \\ - \boxed{ - \braket*{\tilde{f}(k)}{\tilde{g}(k)} = \frac{2 \pi A^2}{|s|} \braket{f(x)}{g(x)} + \begin{aligned} + \braket{f(x)}{g(x)} &= \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}(k)}{\tilde{g}(k)} + \\ + \braket*{\tilde{f}(k)}{\tilde{g}(k)} &= \frac{2 \pi A^2}{|s|} \braket{f(x)}{g(x)} + \end{aligned} } \end{aligned}$$ -For this reason, physicists like to define the Fourier transform -with $A\!=\!B\!=\!1 / \sqrt{2\pi}$ and $|s|\!=\!1$, because then it nicely -conserves the functions' normalization. - -To prove the theorem, we insert the inverse FT into the inner product -definition: +
+ + + +
+ +For this reason, physicists like to define the Fourier transform +with $A\!=\!B\!=\!1 / \sqrt{2\pi}$ and $|s|\!=\!1$, because then it nicely +conserves the functions' normalization. -- cgit v1.2.3