Hopefully someone can help me out with this. I have a question by my prof that goes like:
Prove that the following function is orthogonal to each other (i.e., they form a basis in the space of square integrable functions):
$$\sin(kt)\text{ and }\cos(mt)$$
($k\neq m$ natural numbers, $-\pi \leq t \leq \pi$.) and he gave me a hint, which is a definition: Dot product
$$\langle f, g\rangle = \int_{-\pi}^{\pi} f(x)g(x) dx$$
and if the dot product is $0$, then the functions are orthogonal.
Could anyone please explain me this definition and give me an example on how I am supposed to use this?
$\endgroup$ 32 Answers
$\begingroup$Two vectors $x,y \in \mathbb{R}^n$ are called orthogonal iff there is a $90^\circ$ angle between them, which happens iff $\left<x,y\right> = 0$. Generalizing this to some space $S$ where elements are functions, we define an inner product on two functions $f,g \in S$ as$$ \left<f,g\right> = \int_{-\pi}^\pi f(x)g(x)dx $$and your question asks to prove that $\sin(kt),\cos(mt)$ are orthogonal. To do that, you must show that $\left<\sin(kt),\cos(mt)\right> = 0$ under your assumptions. Can you finish?
$\endgroup$ $\begingroup$If $f = \sin(kt)$ and $g = \cos(kt)$, then
$$ \left<f,g \right> = \left<\sin(kt),\cos(kt)\right> = \int_{-\pi}^{\pi}\sin(kt)\cos(kt) dt. $$
But if $h(t) = \sin(kt)\cos(kt)$, then
$$ h(-t) = \sin(-kt)\cos(kt) = - \sin(kt)\cos(kt) = -h(t). $$So, $h$ is odd. Since $\left[-\pi,\pi \right]$ is a symmetrical interval around $0$,
$$ \int_{-\pi}^{\pi}\sin(kt)\cos(kt) dt = 0. $$You can calculate this integral aswell by substitution, calling $kt = u$ and $s = \cos(u)$.
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