1 files changed, 11 insertions, 9 deletions
diff --git a/latex/know/concept/parsevals-theorem/source.md b/latex/know/concept/parsevals-theorem/source.md
index 41af734..9f406da 100644
--- a/latex/know/concept/parsevals-theorem/source.md
+++ b/latex/know/concept/parsevals-theorem/source.md
@@ -3,23 +3,24 @@
 
 # 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/).
+**Parseval's theorem** relates the inner product of two functions $f(x)$ and $g(x)$ to 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:
 
 $$\begin{aligned}
     \boxed{
-        \braket{f(x)}{g(x)} = \frac{2 \pi B^2}{|s|} \braket{\tilde{f}(k)}{\tilde{g}(k)}
+        \braket{f(x)}{g(x)} = \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}(k)}{\tilde{g}(k)}
     }
     \\
     \boxed{
-        \braket{\tilde{f}(k)}{\tilde{g}(k)} = \frac{2 \pi A^2}{|s|} \braket{f(x)}{g(x)}
+        \braket*{\tilde{f}(k)}{\tilde{g}(k)} = \frac{2 \pi A^2}{|s|} \braket{f(x)}{g(x)}
     }
 \end{aligned}$$
 
-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
+For this reason, physicists like to define their Fourier transform
+with $A = B = 1 / \sqrt{2\pi}$ and $|s| = 1$, because then the FT nicely
 conserves the total probability (quantum mechanics) or the total energy
 (optics).
 
@@ -40,14 +41,15 @@ $$\begin{aligned}
     &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \: \tilde{g}(k) \: \delta(s (k_1 - k)) \dd{k_1} \dd{k}
     \\
     &= \frac{2 \pi B^2}{|s|} \int_{-\infty}^\infty \tilde{f}^*(k) \: \tilde{g}(k) \dd{k}
-    = \frac{2 \pi B^2}{|s|} \braket{\tilde{f}}{\tilde{g}}
+    = \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}}{\tilde{g}}
 \end{aligned}$$
 
 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:
+Note that we can just as well do it in the opposite direction,
+which yields an equivalent result:
 
 $$\begin{aligned}
-    \braket{\tilde{f}}{\tilde{g}}
+    \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