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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?

Thanks.

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5 Answers 5

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, $\pi\neq 0$.

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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.

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1  
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.

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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).

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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 $S^1$.

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