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-rw-r--r--source/know/concept/parsevals-theorem/index.md10
1 files changed, 6 insertions, 4 deletions
diff --git a/source/know/concept/parsevals-theorem/index.md b/source/know/concept/parsevals-theorem/index.md
index a7ce0bf..b2db490 100644
--- a/source/know/concept/parsevals-theorem/index.md
+++ b/source/know/concept/parsevals-theorem/index.md
@@ -17,9 +17,11 @@ where $$A$$, $$B$$, and $$s$$ are constants from the FT's definition:
$$\begin{aligned}
\boxed{
\begin{aligned}
- \inprod{f(x)}{g(x)} &= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}(k)}{\tilde{g}(k)}
+ \inprod{f(x)}{g(x)}
+ &= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}(k)}{\tilde{g}(k)}
\\
- \inprod{\tilde{f}(k)}{\tilde{g}(k)} &= \frac{2 \pi A^2}{|s|} \Inprod{f(x)}{g(x)}
+ \inprod{\tilde{f}(k)}{\tilde{g}(k)}
+ &= \frac{2 \pi A^2}{|s|} \inprod{f(x)}{g(x)}
\end{aligned}
}
\end{aligned}$$
@@ -71,8 +73,8 @@ $$\begin{aligned}
For this reason, physicists like to define the Fourier transform
-with $$A\!=\!B\!=\!1 / \sqrt{2\pi}$$ and $$|s|\!=\!1$$, because then it nicely
-conserves the functions' normalization.
+with $$A\!=\!B\!=\!1 / \sqrt{2\pi}$$ and $$|s|\!=\!1$$,
+because then it nicely preserves the functions' normalization.