Parseval's Theorem is an interesting result of applied mathematics that occurs in several different guises. In many problems we wish to break down a given function into linear combinations of orthogonal eigenfunctions. Parseval's theorem expresses the notion that the total energy of the original function is the sum of the energies of the different eigenfunctions.

Although Parseval's Theorem can be applied to any orthogonal basis of eigenfunctions it is most often discussed when applied to the Fourier Series or the Fourier Transform, and these two cases are discussed in more detail below.

#### Applied to the Fourier Transform

For a complex function f(x) with Fourier Transform F(k), Parseval's Theorem states that

```integral(|f(x)|2dx, x=-inf...inf) = integral(|F(k)|2dk, k=-inf...inf)
```

We prove this result as follows, using * to denote conjugation. Starting from the Fourier inversion formula,

```f(x) = 1/sqrt(2π) integral(exp(ikx)F(k)dk, k=-inf...inf)
f*(x) = 1/sqrt(2π) integral(exp(-ikx)F*(k)dk, k=-inf...inf)

LHS = integral(f(x)f*(x)dx, x=-inf...inf)
= 1/sqrt(2π) integral(f(x)F*(k)exp(-ikx)dxdk, x,k=-inf...inf)
= integral((1/sqrt(2π) f(x)exp(-ikx)dx) F*(k)dk, x,k=-inf...inf)
= integral(F(k)F*(k)dk, k=-inf...inf)
= RHS, as required.
```

#### Applied to the Fourier Series

Let f(x) be a real periodic square integrable function with period 2L. The Fourier Series is defined as

```            inf               inf
f(x) = a0 +  Σ ancos(nπx/L) +  Σ bnsin (nπx/L).
n=1               n=1
```

Parseval's Theorem states that

```                                           inf
integral(|f(x)|2dx, x=0...2L) = L(1/2 a02 + Σ (an2+bn2))
n=1
```

The proof of the result is similar to the Fourier Transform case and follows from application of the orthogonality relations of sine and cosine.

#### Example

In the Fourier Series node, we showed that the sawtooth wave f(x)=x on the open interval (-L,L) can be written as

```      inf
f(x) = Σ 2L/(nπ) (-1)n+1 sin (nπx/L).
n=1
```

Parseval's Theorem states that

``` inf
L Σ (2L/(nπ))2 = integral(|f(x)|2dx, x=-L...L)
n=1           = integral(x2dx, x=-L...L)
= 2L3/3.
```

Rearranging this gives us

```inf
Σ 1/n2 = π2/6.
n=1
```

Parseval's Theorem has allowed us to prove a well-known and interesting mathematical result about the transcendental number pi.

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