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Parseval's Theorem — Definition, Formula & Examples

Parseval's Theorem states that the total energy of a periodic function (computed by integrating its square over one period) equals the sum of the squares of its Fourier coefficients. In other words, you can calculate a function's energy either in the time domain or the frequency domain and get the same result.

If ff is a square-integrable function on [−π,π][-\pi, \pi] with Fourier coefficients a0,an,bna_0, a_n, b_n, then 1π∫−ππ[f(x)]2 dx=a022+∑n=1∞(an2+bn2).\frac{1}{\pi}\int_{-\pi}^{\pi} [f(x)]^2\, dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty}\left(a_n^2 + b_n^2\right). Equivalently, using the complex Fourier coefficients cnc_n, the theorem states 12π∫−ππ∣f(x)∣2 dx=∑n=−∞∞∣cn∣2\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|^2\,dx = \sum_{n=-\infty}^{\infty}|c_n|^2.

Key Formula

1π∫−ππ[f(x)]2 dx=a022+∑n=1∞(an2+bn2)\frac{1}{\pi}\int_{-\pi}^{\pi} [f(x)]^2\, dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty}\left(a_n^2 + b_n^2\right)
Where:
  • f(x)f(x) = A square-integrable periodic function on $[-\pi, \pi]$
  • a0a_0 = The zeroth (constant) Fourier cosine coefficient
  • ana_n = The $n$th Fourier cosine coefficient
  • bnb_n = The $n$th Fourier sine coefficient

How It Works

To apply Parseval's Theorem, first compute the Fourier coefficients of your function. Then square each coefficient, form the appropriate series, and sum. The result equals the average squared value of the function (scaled by the period). This is powerful because it converts a difficult integral into an infinite series, or vice versa — whichever is easier to evaluate. Parseval's Theorem also provides a convergence check: if the series of squared coefficients diverges, the function is not square-integrable on that interval.

Worked Example

Problem: Use Parseval's Theorem to evaluate ∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^2}, given that the Fourier series of f(x)=xf(x) = x on [−π,π][-\pi, \pi] has coefficients a0=0a_0 = 0, an=0a_n = 0, and bn=2(−1)n+1nb_n = \frac{2(-1)^{n+1}}{n}.
Compute the left side: Evaluate the integral of [f(x)]2=x2[f(x)]^2 = x^2 over [−π,π][-\pi, \pi].
1π∫−ππx2 dx=1π⋅2π33=2π23\frac{1}{\pi}\int_{-\pi}^{\pi} x^2\, dx = \frac{1}{\pi} \cdot \frac{2\pi^3}{3} = \frac{2\pi^2}{3}
Compute the right side: Since a0=0a_0 = 0 and an=0a_n = 0, the right side reduces to the sum of bn2b_n^2.
∑n=1∞bn2=∑n=1∞4n2\sum_{n=1}^{\infty} b_n^2 = \sum_{n=1}^{\infty} \frac{4}{n^2}
Equate and solve: Set the two sides equal and solve for the desired sum.
2π23=4∑n=1∞1n2  ⟹  ∑n=1∞1n2=π26\frac{2\pi^2}{3} = 4\sum_{n=1}^{\infty} \frac{1}{n^2} \implies \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}
Answer: ∑n=1∞1n2=π26\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}, which is the famous Basel problem result.

Why It Matters

Parseval's Theorem is central in signal processing, where it guarantees that energy computed in the time domain matches energy in the frequency domain. It also provides elegant proofs of exact values of series like ∑1/n2\sum 1/n^2 and ∑1/n4\sum 1/n^4, connecting Fourier analysis directly to number theory.

Common Mistakes

Mistake: Forgetting the factor of 12\frac{1}{2} on the a02a_0^2 term.
Correction: The constant term in Parseval's identity is a022\frac{a_0^2}{2}, not a02a_0^2. This factor arises because the standard definition of a0a_0 already includes an averaging factor different from ana_n for n≥1n \geq 1.