# Parseval's Theorem/Formulation 2

## Theorem

Let $f$ be a real function which is square-integrable over the interval $\left[{-\pi \,.\,.\, \pi}\right]$.

Let $f$ be expressed by the Fourier series:

$\displaystyle f \left({x}\right) = \sum_{n \mathop = -\infty}^\infty c_n e^{i n x}$

where:

$\displaystyle c_n = \frac 1 {2 \pi} \int_{-\pi}^\pi f \left({t}\right) e^{-i n t} \, \mathrm d t$

Then:

$\displaystyle \frac 1 {2 \pi} \int_{-\pi}^\pi \left\lvert{f \left({x}\right)}\right\rvert^2 \, \mathrm d x = \sum_{n \mathop = -\infty}^\infty \left\lvert{c_n}\right\rvert^2$

## Proof

 $\displaystyle \frac 1 {2 \pi} \int_{-\pi}^\pi \left\vert{f \left({x}\right)}\right\vert^2 \, \mathrm d x$ $=$ $\displaystyle \frac 1 {2 \pi} \int_{-\pi}^\pi f \left({x}\right) \overline {f \left({x}\right)} \, \mathrm d x$ Modulus in Terms of Conjugate $\displaystyle$ $=$ $\displaystyle \frac 1 {2 \pi} \int_{-\pi}^\pi \sum_{n \mathop = -\infty}^\infty c_n e^{i n x} \overline {\sum_{m \mathop = -\infty}^\infty c_m e^{i m x} } \, \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \frac 1 {2 \pi} \int_{-\pi}^\pi \sum_{n \mathop = -\infty}^\infty c_n e^{i n x} \sum_{m \mathop = -\infty}^\infty \overline {c_m} e^{-i m x} \, \mathrm d x$ Sum of Complex Conjugates $\displaystyle$ $=$ $\displaystyle \frac 1 {2 \pi} \int_{-\pi}^\pi \sum_{n \mathop = -\infty}^\infty \sum_{m \mathop = -\infty}^\infty c_n \overline {c_m} e^{i x \left({n - m}\right)} \, \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \frac 1 {2 \pi} \sum_{n \mathop = -\infty}^\infty \sum_{m \mathop = -\infty}^\infty c_n \overline {c_m} \int_{-\pi}^\pi e^{i x \left({n - m}\right)} \, \mathrm d x$ Fubini's Theorem $\displaystyle$ $=$ $\displaystyle \frac 1 {2 \pi} \sum_{n \mathop = -\infty}^\infty \sum_{m \mathop = -\infty}^\infty c_n \overline {c_m} 2 \pi \delta_{n m}$ Integral over 2 pi of Exponential of i by n x $\displaystyle$ $=$ $\displaystyle \frac {2 \pi} {2 \pi} \sum_{n \mathop = -\infty}^\infty c_n \overline {c_n}$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = -\infty}^\infty \left\vert{c_n}\right\vert^2$

$\blacksquare$

## Source of Name

This entry was named for Marc-Antoine Parseval.