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-rw-r--r--source/know/concept/parsevals-theorem/index.md11
1 files changed, 6 insertions, 5 deletions
diff --git a/source/know/concept/parsevals-theorem/index.md b/source/know/concept/parsevals-theorem/index.md
index e1d73a7..df90244 100644
--- a/source/know/concept/parsevals-theorem/index.md
+++ b/source/know/concept/parsevals-theorem/index.md
@@ -8,11 +8,11 @@ categories:
layout: "concept"
---
-**Parseval's theorem** is a relation between the inner product of two functions $f(x)$ and $g(x)$,
+**Parseval's theorem** is a relation between the inner product of two functions $$f(x)$$ and $$g(x)$$,
and the inner product of their [Fourier transforms](/know/concept/fourier-transform/)
-$\tilde{f}(k)$ and $\tilde{g}(k)$.
+$$\tilde{f}(k)$$ and $$\tilde{g}(k)$$.
There are two equivalent ways of stating it,
-where $A$, $B$, and $s$ are constants from the FT's definition:
+where $$A$$, $$B$$, and $$s$$ are constants from the FT's definition:
$$\begin{aligned}
\boxed{
@@ -48,7 +48,7 @@ $$\begin{aligned}
= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}}{\tilde{g}}
\end{aligned}$$
-Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
+Where $$\delta(k)$$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
Note that we can equally well do this proof in the opposite direction,
which yields an equivalent result:
@@ -68,11 +68,12 @@ $$\begin{aligned}
&= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x}
= \frac{2 \pi A^2}{|s|} \Inprod{f}{g}
\end{aligned}$$
+
</div>
</div>
For this reason, physicists like to define the Fourier transform
-with $A\!=\!B\!=\!1 / \sqrt{2\pi}$ and $|s|\!=\!1$, because then it nicely
+with $$A\!=\!B\!=\!1 / \sqrt{2\pi}$$ and $$|s|\!=\!1$$, because then it nicely
conserves the functions' normalization.