Editing Parseval's theorem (section)

== Statement of Parseval's theorem ==

Suppose that <math>A(x)</math> and <math>B(x)</math> are two complex-valued functions on <math>\mathbb{R}</math> of [[periodic function|period]] <math>2 \pi</math> that are [[square integrable]] (with respect to the [[Lebesgue measure]]) over intervals of period length, with [[Fourier series]]

:<math>A(x)=\sum_{n=-\infty}^\infty a_ne^{inx}</math>
and<br />
:<math>B(x)=\sum_{n=-\infty}^\infty b_ne^{inx}</math>

respectively. Then

{{Equation box 1
|indent =
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|equation = {{NumBlk|:|<math>\sum_{n=-\infty}^\infty a_n\overline{b_n} = \frac{1}{2\pi} \int_{-\pi}^\pi A(x)\overline{B(x)} \, \mathrm{d}x,</math>|{{EquationRef|Eq.1}}}}
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where <math>i</math> is the [[imaginary unit]] and horizontal bars indicate [[complex conjugation]]. Substituting <math>A(x)</math> and <math>\overline{B(x)}</math>:

:<math>
\begin{align}
\sum_{n=-\infty}^\infty a_n\overline{b_n}
&= \frac{1}{2\pi} \int_{-\pi}^\pi \biggl( \sum_{n=-\infty}^\infty a_ne^{inx} \biggr) \biggl( \sum_{n=-\infty}^\infty \overline{b_n}e^{-inx} \biggr) \, \mathrm{d}x \\[6pt]
&= \frac{1}{2\pi} \int_{-\pi}^\pi \Bigl(a_1e^{i1x} + a_2e^{i2x} + \cdots\Bigr) \Bigl(\overline{b_1}e^{-i1x} + \overline{b_2}e^{-i2x} + \cdots\Bigr) \, \mathrm{d}x \\[6pt]
&= \frac{1}{2\pi} \int_{-\pi}^\pi \left(a_1e^{i1x} \overline{b_1}e^{-i1x} + a_1e^{i1x} \overline{b_2}e^{-i2x} + a_2e^{i2x} \overline{b_1}e^{-i1x} + a_2e^{i2x} \overline{b_2}e^{-i2x} + \cdots \right) \mathrm{d}x \\[6pt]
&= \frac{1}{2\pi} \int_{-\pi}^\pi \left(a_1 \overline{b_1} + a_1 \overline{b_2}e^{-ix} + a_2 \overline{b_1}e^{ix} + a_2 \overline{b_2} + \cdots\right) \mathrm{d}x
\end{align}
</math>
As is the case with the middle terms in this example, many terms will integrate to <math>0</math> over a full [[Periodic_function|period]] of length <math>2\pi</math> (see [[harmonic|harmonics]]):
:<math>
\begin{align}
\sum_{n=-\infty}^\infty a_n\overline{b_n} &= \frac{1}{2\pi} \left[a_1 \overline{b_1} x + i a_1 \overline{b_2}e^{-ix} - i a_2 \overline{b_1}e^{ix} + a_2 \overline{b_2} x + \cdots\right] _{-\pi} ^{+\pi} \\[6pt]
&= \frac{1}{2\pi} \left(2\pi a_1 \overline{b_1} + 0 + 0 + 2\pi a_2 \overline{b_2} + \cdots\right) \\[6pt]
&= a_1 \overline{b_1} + a_2 \overline{b_2} + \cdots \\[6pt]
\end{align}</math>

More generally, if <math>A(x)</math> and <math>B(x)</math> are instead two complex-valued functions on <math>\mathbb{R}</math> of period <math>P</math> that are [[square integrable]] (with respect to the [[Lebesgue measure]]) over intervals of period length, with [[Fourier series]]

:<math>A(x)=\sum_{n=-\infty}^\infty a_ne^{2\pi ni\left(\frac{x}{P}\right)}</math>
and<br />
:<math>B(x)=\sum_{n=-\infty}^\infty b_ne^{2\pi ni\left(\frac{x}{P}\right)}</math>

respectively. Then

{{Equation box 1
|indent =
|title=
|equation = {{NumBlk|:|<math>\sum_{n=-\infty}^\infty a_n\overline{b_n} = \frac{1}{P} \int_{-P/2}^{P/2} A(x)\overline{B(x)} \, \mathrm{d}x,</math>|{{EquationRef|Eq.2}}}}
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|border colour = #0073CF
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Even more generally, given an [[Locally compact abelian group|abelian locally compact group]] ''G'' with [[Pontryagin dual]] ''G^'', Parseval's theorem says the Pontryagin–Fourier transform is a unitary operator between [[Hilbert spaces]] ''L''<sup>2</sup>(''G'') and ''L''<sup>2</sup>(''G^'') (with integration being against the appropriately scaled [[Haar measure|Haar measures]] on the two groups.) When ''G'' is the [[unit circle]] '''T''', ''G^'' is the integers and this is the case discussed above. When ''G'' is the real line <math>\mathbb{R}</math>, ''G^'' is also <math>\mathbb{R}</math> and the unitary transform is the [[Fourier transform]] on the real line. When ''G'' is the [[cyclic group]] '''Z'''<sub>n</sub>, again it is self-dual and the Pontryagin–Fourier transform is what is called [[discrete Fourier transform]] in applied contexts.

Parseval's theorem can also be expressed as follows:

Suppose <math>f(x)</math> is a square-integrable function over <math>[-\pi, \pi]</math> (i.e., <math>f(x)</math> and <math>f^2(x)</math> are integrable on that interval), with the Fourier series
:<math>f(x) \simeq \tfrac12 a_0 + \sum_{n=1}^{\infty} \bigl(a_n \cos(nx) + b_n \sin(nx)\bigr).</math>
Then<ref>{{cite book | page=439 | author=Arthur E. Danese | title=Advanced Calculus | volume=1 | publisher=Allyn and Bacon, Inc. | location=Boston, MA | year=1965 }}</ref><ref>{{cite book |page=[https://archive.org/details/advancedcalculus00kapl_841/page/n533 519] | author=Wilfred Kaplan | title=Advanced Calculus |url=https://archive.org/details/advancedcalculus00kapl_841 |url-access=limited | publisher=Addison Wesley | location=Reading, MA | year=1991 | edition=4th | isbn=0-201-57888-3 |author-link=Wilfred Kaplan }}</ref><ref>{{cite book | page=[https://archive.org/details/fourierseries00tols/page/119 119] | author=Georgi P. Tolstov | title=Fourier Series | url=https://archive.org/details/fourierseries00tols | url-access=registration | translator-last=Silverman | translator-first=Richard | publisher=Prentice-Hall, Inc. | location=Englewood Cliffs, NJ | year=1962 }}</ref>

:<math>\frac{1}{\pi} \int_{-\pi}^{\pi} f^2(x) \,\mathrm{d}x = \tfrac12 a_0^2 + \sum_{n=1}^{\infty} \left(a_n^2 + b_n^2 \right).</math>