diff options
Diffstat (limited to 'content/know/concept/parsevals-theorem')
| -rw-r--r-- | content/know/concept/parsevals-theorem/index.pdc | 32 |
1 files changed, 19 insertions, 13 deletions
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: +<div class="accordion"> +<input type="checkbox" id="proof-fourier"/> +<label for="proof-fourier">Proof</label> +<div class="hidden"> +<label for="proof-fourier">Proof.</label> +We insert the inverse FT into the defintion of the inner product: $$\begin{aligned} \braket{f}{g} @@ -54,7 +54,7 @@ $$\begin{aligned} \end{aligned}$$ Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/). -Note that we can equally well do the proof in the opposite direction, +Note that we can equally well do this proof in the opposite direction, which yields an equivalent result: $$\begin{aligned} @@ -73,6 +73,12 @@ $$\begin{aligned} &= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x} = \frac{2 \pi A^2}{|s|} \braket{f}{g} \end{aligned}$$ +</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 +conserves the functions' normalization. |