14.3. Derivation of Parseval’s Theorem

We start with the Fourier series of f(x):

$f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos \left( \frac{n\pi x}{L} \right) + b_n \sin \left( \frac{n\pi x}{L} \right) \right)$

In deriving Parseval's Theorem, we begin with the Fourier series representation of a function f(x). This series expresses f(x) as a sum of sine and cosine functions, weighted by coefficients a_n and b_n. The constant term a_0 accounts for the average value of the function over the interval, while the sine and cosine terms capture the oscillatory nature of the function at various frequencies.

Think of the Fourier series as a musical chord, where each sine and cosine function is like an individual note. Just as different notes combine to create harmony in music, these mathematical components work together to reconstruct the original function.

To compute the integral $\frac{1}{L} \int_{-L}^{L} f(x)^2 \, dx$, we substitute the Fourier series expansion:

$\frac{1}{L} \int_{-L}^{L} f(x)^2 \, dx = \frac{1}{L} \int_{-L}^{L} \left( a_0 + \sum_{n=1}^{\infty} \left( a_n \cos \left( \frac{n\pi x}{L} \right) + b_n \sin \left( \frac{n\pi x}{L} \right) \right) \right)^2 \, dx$

Next, we need to evaluate the energy of the function f(x) over the interval [-L, L], which is done by computing the integral of its square. We substitute the Fourier series expression for f(x) into this integral. This step sets the stage for examining how the energy represented in the time domain correlates with the coefficients in the frequency domain.

Imagine measuring the total energy of a moving object by tracking its speed over time. The integral of the squared function gives us an aggregate measure of energy, much like how speed multiplied by time gives us distance traveled.

Now expanding the square and using the orthogonality of sine and cosine functions:

$\int_{-L}^{L} \cos(n\pi x/L) \cos(m\pi x/L) \, dx = \begin{cases} L, & n = m \neq 0 \ 0, & n \neq m \end{cases}$.

Similarly for sine functions and cross-terms.

Upon expanding the square, we analyze cross-terms involving both sine and cosine functions. The orthogonality property tells us that the integral of the product of two different sine or cosine functions over a complete period yields zero. This means that most mixed terms in the expansion will cancel out, simplifying our calculations.

This is like two musicians playing different instruments. If they play at the same time but in different keys, their sounds blend beautifully, but if they play two different notes, the blend cancels out or becomes silent, which reflects the phenomenon of orthogonality.

After evaluating the integrals and simplifying, all cross terms vanish due to orthogonality, and we are left with:

$\frac{1}{L} \int_{-L}^{L} f(x)^2 \, dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty} \left( \frac{a_n^2 + b_n^2}{2} \right)$

After simplifying the expression using orthogonality, we arrive at the final form of Parseval's Theorem. This shows that the mean square value (or energy) of the function over the time domain is equal to the sum of the squares of the Fourier coefficients. This relationship indicates that the 'energy' present within the function can be directly computed through its frequency components.

Think of transforming your favorite recipe from cooking (time domain) to a list of ingredients (frequency domain). All the different parts of the recipe contribute to the final dish, just as all frequency components of a function contribute to its overall energy.

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