summaryrefslogtreecommitdiff
path: root/source/know/concept/parsevals-theorem/index.md
blob: 43d3717a9722c7b4a0ef9ce312cff160d69ee8a7 (plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
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.