# The Linear Independence of {$\sin(2^m x), \sin(2^{m-1}x), \ldots, \sin(2x), \sin(x)$}.

Let $$m \in \mathbb{N}$$. I wish to prove the linear independence of the set of vectors {$$\sin(2^m x), \sin(2^{m-1}x), \ldots, \sin(2x), \sin(x)$$} $$\subset F(\mathbb{R})$$. I had attempted to prove the property by induction upon $$m$$: the base case, quite evidently, was trivial to prove (insofar as $$\sin(x)≠0$$), but I could not proceed very well with the induction hypothesis. How might I do so? Or is there perhaps a better way by which to show the property holds?

• Here's a nice trick: They are all eigenvectors of the linear operator $\frac{d^2}{dx^2}$ with distinct eigenvalues Aug 23, 2020 at 17:45

We can prove a more general result. Let $$n \ge 1.$$ We show that the set $$\{\sin(x), \sin(2x), \ldots, \sin(nx)\} \subset F(\Bbb R)$$ is linearly independent.

Indeed, suppose that $$a_1, \ldots, a_n$$ are reals such that $$a_1\sin(x) + \cdots + a_n\sin(nx) = 0$$ for all $$x \in \Bbb R.$$

Fix $$k$$ such that $$1 \le k \le n$$. Multiplying both sides with $$\sin(kx),$$ we get $$a_1\sin(x)\sin(kx) + \cdots + a_n\sin(nx)\sin(kx) = 0$$ for all $$x\in\Bbb R$$.

Now, integrate both sides from $$0$$ to $$2\pi$$ and note that $$\int_0^{2\pi}\sin(ix)\sin(kx){\mathrm d}x = 0 \iff i \neq k.$$

Thus, we get that $$a_k\int_0^{2\pi}(\sin(kx))^2{\mathrm d}x = 0$$ or $$a_k = 0.$$

As $$k$$ was arbitrary, we get that each $$a_i$$ is zero, proving linear independence.

EDIT: Note that if we consider the vector space $$\mathcal{C}[0, 2\pi]$$, the set of real-valued continuous functions on $$[0, 2\pi],$$ we can define an inner product on it as $$\langle f, g\rangle := \int_0^{2\pi}f(t)g(t){\mathrm d}t.$$ The above shows that $$\{\sin(x), \ldots, \sin(nx)\}$$ is an orthogonal subset of $$\mathcal{C}[0, 2\pi]$$ under this inner product. In particular, it is linearly independent. (Since each vector is nonzero.)

From this, it is easy to see that the functions considered as elements of $$F(\Bbb R)$$ must also be linearly independent.

• Not mentioning orthogonal bases and inner products seems like a missed opportunity.
– Pedro
Aug 23, 2020 at 18:05
• I did wish to mention inner products but the ambient vector space is $F(\Bbb R)$ and not $\cal C[0, 2\pi]$. Though you are right, I still should mention it. I'll add a note to this effect. Aug 23, 2020 at 18:07