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diff --git a/source/know/concept/parsevals-theorem/index.md b/source/know/concept/parsevals-theorem/index.md new file mode 100644 index 0000000..43d3717 --- /dev/null +++ b/source/know/concept/parsevals-theorem/index.md @@ -0,0 +1,82 @@ +--- +title: "Parseval's theorem" +date: 2021-02-22 +categories: +- Mathematics +- Physics +layout: "concept" +--- + +**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 $\tilde{g}(k)$. +There are two equivalent ways of stating it, +where $A$, $B$, and $s$ are constants from the FT's definition: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \Inprod{f(x)}{g(x)} &= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}(k)}{\tilde{g}(k)} + \\ + \inprod{\tilde{f}(k)}{\tilde{g}(k)} &= \frac{2 \pi A^2}{|s|} \Inprod{f(x)}{g(x)} + \end{aligned} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-fourier"/> +<label for="proof-fourier">Proof</label> +<div class="hidden" markdown="1"> +<label for="proof-fourier">Proof.</label> +We insert the inverse FT into the defintion of the inner product: + +$$\begin{aligned} + \Inprod{f}{g} + &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\}\big)^* \: \hat{\mathcal{F}}^{-1}\{\tilde{g}(k)\} \dd{x} + \\ + &= B^2 \int + \Big( \int \tilde{f}^*(k_1) \exp(i s k_1 x) \dd{k_1} \Big) + \Big( \int \tilde{g}(k) \exp(- i s k x) \dd{k} \Big) + \dd{x} + \\ + &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \tilde{g}(k) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s x (k_1 - k)) \dd{x} \Big) \dd{k_1} \dd{k} + \\ + &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \: \tilde{g}(k) \: \delta(s (k_1 - k)) \dd{k_1} \dd{k} + \\ + &= \frac{2 \pi B^2}{|s|} \int_{-\infty}^\infty \tilde{f}^*(k) \: \tilde{g}(k) \dd{k} + = \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}}{\tilde{g}} +\end{aligned}$$ + +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: + +$$\begin{aligned} + \inprod{\tilde{f}}{\tilde{g}} + &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}\{f(x)\}\big)^* \: \hat{\mathcal{F}}\{g(x)\} \dd{k} + \\ + &= A^2 \int + \Big( \int f^*(x_1) \exp(- i s k x_1) \dd{x_1} \Big) + \Big( \int g(x) \exp(i s k x) \dd{x} \Big) + \dd{k} + \\ + &= 2 \pi A^2 \iint f^*(x_1) g(x) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s k (x_1 - x)) \dd{k} \Big) \dd{x_1} \dd{x} + \\ + &= 2 \pi A^2 \iint f^*(x_1) \: g(x) \: \delta(s (x_1 - x)) \dd{x_1} \dd{x} + \\ + &= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x} + = \frac{2 \pi A^2}{|s|} \Inprod{f}{g} +\end{aligned}$$ +</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 +conserves the functions' normalization. + + + +## References +1. O. Bang, + *Applied mathematics for physicists: lecture notes*, 2019, + unpublished. |