1

Showing independence of $\{1,\cos x, \sin x, \ldots, \cos nx, \sin nx\,,\ldots\}$

There's infinitely many terms so I'm not sure how to do the definition I'm familiar with, like $\alpha(1)+\beta\cos x+\gamma \sin x+\ldots+\psi\cos nx +\mu\sin nx=0\implies \alpha=\beta=\ldots=\psi=\mu=0$

Any hints to get started?

Timbuc
  • 34,191
Jon
  • 25
  • 1
    Are you sure there's an infinite number of terms? – GFauxPas Nov 14 '14 at 16:23
  • Its the same, just pick a finite number of terms. – copper.hat Nov 14 '14 at 16:23
  • Easier perhaps to show that they are all orthogonal, from which linear indep follows. What's the underlying vector space, $C[-\pi, \pi]$? If so, integrate each pair over the interval $[-\pi, \pi]$ – Simon S Nov 14 '14 at 16:24
  • 2
    An infinite set is linearly independent if every finite subset of it is linear independendent – Learnmore Nov 14 '14 at 16:25
  • I think the "infinite number of terms" refers to $x$. – Suzu Hirose Nov 14 '14 at 16:26
  • 1
    Orthogonal makes no sense unless you have some added structure. You can add the structure, of course... – copper.hat Nov 14 '14 at 16:27
  • Related: http://math.stackexchange.com/questions/422347/how-to-prove-that-the-set-sinx-sin2x-sinmx-is-linearly-indepe – rschwieb Nov 14 '14 at 16:28
  • @rschwieb: That's very interesting! I didn't know I could prove independence with integrals. – Jon Nov 14 '14 at 16:32
  • @Jon: You can use any tricks you want. – copper.hat Nov 14 '14 at 16:37
  • @Jon It might be better to think of it as "proving independence with an inner product." and "Learning there is an inner product defined by an integral." Otherwise it sounds a bit like a weird trick that makes a huge jump :) – rschwieb Nov 14 '14 at 17:29

2 Answers2

3

The key results are $\int_0^{2 \pi} \sin (mt) \sin (nt)dt = \begin{cases} 0, & n \neq m \\ \pi, & n=m\end{cases}$, $\int_0^{2 \pi} \sin (mt) \cos (nt) dt = 0$ and $\int_0^{2 \pi} \cos (mt) \cos (nt)dt = \begin{cases} 0, & n \neq m \\ \pi, & n=m\end{cases}$.

Now suppose you have constants $\alpha_k, \beta_k$ such that $f(x) = \alpha_0 + \sum_{k=1}^m \alpha_k \cos (kx) + \sum_{k=1}^n \beta_k \sin (kx) = 0$.

Then $\int_0^{2\pi} f(x) dx = 2 \pi \alpha_0 = 0$, $\int_0^{2\pi} f(x) \cos(lx) dx = \pi \alpha_l = 0$ and $\int_0^{2\pi} f(x) \sin(lx) dx = \pi \beta_l = 0$. Hence the set is linearly independent.

copper.hat
  • 172,524
  • What are the $l$'s in the last passage of text? Are they arbitrary integers? – Jon Nov 14 '14 at 17:00
  • Arbitrary non-zero integers. By integrating $f$ multiplied by the appropriate function over $[0,2 \pi]$ you can 'pick out' the relevant multiplier. – copper.hat Nov 14 '14 at 17:02
0

Showing that they are orthogonal by integrating would be sufficient. Then a linear combination cannot be 0 because taking the integral with a product of one of the terms is nonzero.

Matt Samuel
  • 58,164