Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Could you help me to show that the functions $\sin(x),\sin(2x),...,\sin(mx)\in V$ are linearly independent, where $V$ is the space of real functions?


share|cite|improve this question
up vote 10 down vote accepted

Suppose that, for every $x\in\Bbb R$ we have $$a_1\sin x+a_2\sin 2x+\cdots+a_m\sin mx=0$$

Take $i\in \{1,\dots,m\}$, and consider $\sin ix$. Multiply throughout and integrate from $x=0$ to $x=2\pi$. Do this for $i=1,\dots,m$. Use that $$\int_0^{2\pi} \sin mx\sin nxdx=\begin{cases}0& m\neq n\\ \pi &m=n\end{cases}$$

ADD If the above wasn't entirely clear, for each $1\leq k\leq m$

$$\begin{align}\sum_{j=1}^m a_j\sin jx&=0\\ \sum_{j=1}^m a_j\sin kx\sin jx&=0\\ \sum_{j=1}^m a_j\int_0^{2\pi}\sin kx\sin jxdx&=0\\ a_k \pi&=0\\ {}&{}\\ a_k&=0\end{align}$$

since, of course, $\pi\neq 0$.

share|cite|improve this answer

For variety....

Let $D$ be the differentiation operator. It is a linear transformation.

$D^2 \sin(kx) = -k^2 \sin(kx)$, so $\sin(kx)$ is an eigenfunction (i.e. an eigenvector when the vector space is a space of functions) of $D^2$ with eigenvalue $-k^2$.

Since each of the eigenvalues $-1, -4, -9, \cdots, -m^2$ are distinct, the eigenfunctions must be linearly independent.

share|cite|improve this answer
Very nice approach! – Tom Oldfield Jun 17 '13 at 0:49

Let $e_k(t) = \sin kt$, and consider the space $V=\operatorname{sp} \{ e_k \}_{k=1}^m$, with the inner product $\langle f_1, f_2 \rangle = \frac{1}{\pi} \int_{-\pi}^\pi f_1(t) f_2(t) dt$.

A quick computation shows that $\langle e_i, e_j \rangle = \delta_{ij}$, hence the $e_k$ are orthonormal.

If $\sum \alpha_k e_k = 0$, then since $\langle e_i, \sum \alpha_k e_k \rangle = \alpha_k = 0$, we see that the $e_k$ are linearly independent.

Alternative approach:

Let $f(t) = \sum_{k=1}^m \alpha_k \sin kt $, and suppose $f(t) = 0$ for all $t$. Let $D = \frac{d}{dt}$. Note that $(D^{2n+1} f)(t)= (-1)^n\sum_{k=1}^m \alpha_k k (k^2)^n \cos kt $ ($=0$ for all $t$, of course). In particular, $\lim_{n \to \infty} (-1)^n\frac{(D^{2n+1} f)(0)}{m^{2n+1}} = \alpha_m = 0$. Since $\alpha_m = 0$, we have $\lim_{n \to \infty} (-1)^n\frac{(D^{2n+1} f)(0)}{(m-1)^{2n+1}} = \alpha_{m-1} = 0$, and so on. Repeating this shows that $\alpha_k = 0$ for all $k$. Hence the functions $t \mapsto \sin kt$ are linearly independent.

share|cite|improve this answer
I guess you are a much faster typist... – copper.hat Jun 17 '13 at 0:25
I changed a $\pi$ to $-\pi$. – Pedro Tamaroff Jun 17 '13 at 0:25

Hint: Find $m$ values $a_1,...,a_m$, such that the matrix $$\pmatrix{\sin(a_1) & \sin(2a_1) & \dots & \sin (ma_1) \\ \sin(a_2) & \sin(2a_2) & \dots & \sin (ma_2) \\ \vdots && \ddots & \vdots \\ \sin(a_m) & \sin(2a_m) & \dots & \sin (ma_m) \\ }$$ is regular (has nonzero determinant).

share|cite|improve this answer

If $\{ \sin x, \sin 2x, \ldots, \sin mx\}$ is linear dependent, then for some $a_1,\ldots,a_m \in \mathbb{R}$, not all zero, we have:

$$\sum_{k=1}^m a_k \sin kx = 0, \text{ for all } x \in \mathbb{R}$$

This in turn implies for every $z \in S^1 = \{ \omega \in \mathbb{C} : |\omega| = 1\}$, if we write $z$ as $e^{ix}$, we have:

$$0 = \sum_{k=1}^m a_k \sin kx = \sum_{k=1}^m a_k \frac{z^k - z^{-k}}{2i} = \frac{z^{-m}}{2i}\sum_{k=1}^m a_k\left(z^{m+k}-z^{m-k}\right)$$

This contradicts with the fact the rightmost side of above expression is $\frac{z^{-m}}{2i}$ multiplied by a non-zero polynomial in $z$ and has at most finitely many roots on $S^1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.