I am confused about how my professor resolved a fourier approximation. The question was: Find the fourth-order fourier approximation of: $$f(x) = \sin (5x)$$ on the interval: $$0\le x \le 2\pi$$ He said something about the given sine function being an orthonormal basis in this situation What he specifically wrote is that:
$$\{\cos (nx), \sin (nx)\} , n \in \mathbb{Z}$$ is an othonormal basis for $$C[0,2\pi]$$
He said that since the function in question is $\sin (5x)$ then that implies that it is a higher order (5) than the order of the approximation (4) which means that the approximation of the function along the interval is $0$. I don't really understand this. Why does the function's fourier approximation go to $0$ on the interval?
Is it because the given function is a sine function and the fourier approximation is composed of summations of sine and cosine functions? Or is it because the function is of $5x$ and thus the order n = 4 of the approximation is less than the order of the function being approximated? $\implies$ n < 5
Since $f(x)=\sin(5 x)$ is an odd function of $x$, the Fourier series for $f(x)$ only consists of $\sin$ terms, and since the Fourier sin series for $f(x)$ on the interval $(0, 2 \pi)$ is
$$f(x)=\sum\limits_{n=1}^\infty a_n\, \sin(n x)$$
its not too hard to see that
$$a_n=\left\{\begin{array}{cc} 1 & n=5 \\ 0 & n\ne 5 \\ \end{array}\right..$$