5.7 Parseval’s Identity
The coefficients carry the whole function, and they carry its size as well.
Theorem 5.7.1 (Parseval). If \(f\) is square-integrable on \([-\pi ,\pi ]\) then \[\frac {1}{\pi }\int ^{\pi }_{-\pi }\left [f(x)\right ]^{2}dx = \frac {a_0^{2}}{2} + \sum ^{\infty }_{n=1}\left (a_n^{2}+b_n^{2}\right ).\]
Note 5.7.2. Parseval’s identity is Pythagoras’ theorem in a space of functions. The trigonometric system is orthogonal, so it plays the role of a set of perpendicular axes; the coefficients are the components of \(f\) along those axes; and the identity says the squared length of \(f\) is the sum of the squares of its components. That it holds with equality — rather than merely \(\geq \), which orthogonality alone would give — is precisely the completeness of Theorem 4.5.1. Nothing of \(f\) has been missed.
Example 5.7.3. The odd extension of \(f(x)=x\) on \([-\pi ,\pi ]\) has \(b_n = 2(-1)^{n+1}/n\) and all \(a_n=0\). Parseval then gives \[\frac {1}{\pi }\int ^{\pi }_{-\pi }x^{2}\,dx = \frac {2\pi ^{2}}{3} = \sum ^{\infty }_{n=1}\frac {4}{n^{2}},\] whence \[\sum ^{\infty }_{n=1}\frac {1}{n^{2}} = \frac {\pi ^{2}}{6}.\] This is Euler’s evaluation of the Basel problem, obtained here as a by-product of expanding the function \(f(x)=x\).
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.