Question

Determine the scalar product of the functions \(f(x)=\cos (2 n \pi x / L)\) and \(g(x)=\cos (2 p \pi x / L)\) on the domain \(0 \leq x \leq L\), where \(n \neq p\) are positive integers greater than zero.

Short answer

Answer: The scalar product of the given functions on the domain \(0\leq x\leq L\) is 0.

Step-by-step solution

Find the product of \(f(x)\) and \(g(x)\)

To calculate the product, simply multiply both functions: \( f(x)g(x)=\cos(2 n \pi x / L)\cos(2 p \pi x / L)\)

Integrate the product over the given domain

Now we need to integrate this product on the interval \(0\leq x\leq L\). We will use the trigonometric identity: \(\cos(a) \cos(b)=\frac{1}{2}\left[\cos(a-b)+\cos(a+b)\right]\) Applying this identity to the product \(f(x)g(x)\) with \(a=2n\pi x/L\) and \(b=2p\pi x/L\), we get: \( f(x)g(x) = \frac{1}{2}\left[\cos\left(\frac{2(n-p)\pi x}{L}\right) + \cos\left(\frac{2(n+p)\pi x}{L}\right)\right]\) Now we will integrate the product over the domain \(0\leq x\leq L\): \(\int_0^L f(x)g(x)dx = \frac{1}{2}\int_0^L \left[\cos\left(\frac{2(n-p)\pi x}{L}\right) + \cos\left(\frac{2(n+p)\pi x}{L}\right)\right]dx\) Integration of both cosine terms, we get: \(= \frac{1}{2}\left[\frac{L}{2(n-p)\pi}\sin\left(\frac{2(n-p)\pi x}{L}\right) + \frac{L}{2(n+p)\pi}\sin\left(\frac{2(n+p)\pi x}{L}\right)\right] \bigg|_0^L\) We can now evaluate the expression at the limits: \(= \frac{1}{2}\left[\frac{L}{2(n-p)\pi}\sin\left(\frac{2(n-p)\pi L}{L}\right) + \frac{L}{2(n+p)\pi}\sin\left(\frac{2(n+p)\pi L}{L}\right) - \left(\frac{L}{2(n-p)\pi}\sin\left(\frac{2(n-p)\pi (0)}{L}\right) + \frac{L}{2(n+p)\pi}\sin\left(\frac{2(n+p)\pi (0)}{L}\right)\right)\right]\) Notice that all sine terms with integer multiples of \(\pi\) are zero. Therefore: \(\int_0^L f(x)g(x)dx = 0\) So, the scalar product of the given functions \(f(x)\) and \(g(x)\) on the domain \(0\leq x\leq L\) is \(\boxed{0}\).