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Diffstat (limited to 'source/know/concept/parsevals-theorem/index.md')
| -rw-r--r-- | source/know/concept/parsevals-theorem/index.md | 11 |
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. |