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It seems rather obvious to me that $\sin(x)$, $\sin(2x)$, $\sin(3x)$ are linearly independent in $\operatorname{Map}(\mathbb{R},\mathbb{R})$, but I'm not sure how to prove it (or disprove it if I'm wrong?).
I know that for linear independence it must
$$ a\sin(x)+b\sin(2x)+c\sin(3x)=0 $$ $$ \implies a=b=c=0$$ but how can I show this is true for any $x \in \mathbb{R}$?

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    $\begingroup$ The language needs some precision: essentially an order-of-quantifiers thing: it's not that for given $x$ there are no such $a,b,c$, but that there are no $a,b,c$ such that the relation would hold for all $x$. Thus, @mookid's computation uses the assumption that the relation holds for all $x\in [0,2\pi]$ (or even "almost all"). The other answers are more economical in the number of different $x$ values needed. One imagines that a clever choice of three would suffice. $\endgroup$ – paul garrett May 14 '15 at 21:42
  • $\begingroup$ @paulgarrett the question states the space in which the functions are independent. $\endgroup$ – mookid May 14 '15 at 23:01
  • $\begingroup$ @mookid, yes, indeed, and/but the most primitive sort of linear algebra often fails to give prototypes for proofs of linear independence in spaces of functions on uncountable physical spaces, etc. My remark about "economical" was not at all meant to deprecate your solution... which, in fact, connects better with the larger sense of the situation. But the literal question (not so reasonable, in fact) can be addressed by shortcuts, obviously. $\endgroup$ – paul garrett May 14 '15 at 23:04
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Hint: if $f(x) = a\sin (x) + b\sin(2x) + c\sin(3x) = 0$ then compute $$ \int _0^{2\pi} f(x) \sin(x)dx\\ \int _0^{2\pi} f(x) \sin(2x)dx\\ \int _0^{2\pi} f(x) \sin(3x)dx\\ $$

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  • $\begingroup$ I don't quite understand this approach. Care to explain why this works? $\endgroup$ – soohyung May 14 '15 at 21:45
  • $\begingroup$ @soohyung first of all, since $f(x)=0$, all the integrals will evaluate to $0$. The idea is that if you insert $a\sin x +b \sin 2x +c\sin 3x$ into the integrals and calculate, you will see that the first integral is equal to $a\pi$, the second one is equal to $b\pi$, and the third one is equal to $c\pi$. $\endgroup$ – Arthur May 14 '15 at 22:00
  • $\begingroup$ mookid, can we use the basic definition of linear independence and prove the result via the Period of sin function concept? Let $\sin kx=\alpha_1 \sin x+.....\alpha_n\sin nx.$ and can we show that since l.h.S is a $2\pi/k$ periodic, R.H.S is not ?? $\endgroup$ – Upstart Jan 29 '18 at 8:38
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Differentiate the LHS of the equality and evaluate at $0$ we get

$$a+2b+3c=0\tag1$$ Now differentiate twice we get $$a+2^3b+3^3c=0\tag2$$ and again we get

$$a+2^5b+3^5c=0\tag3$$ and finally $(1),(2),(3)$ give the desired result.

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    $\begingroup$ I like this approach a lot, thank you. $\endgroup$ – soohyung May 14 '15 at 21:44
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Pick three values for $x$, for example $\pi/2,\pi/3,\pi/4$. The first gives $a = c$. The last gives you $b = 0$. The second will give you $a=c=0$.

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  • $\begingroup$ I can't prove it for $x \in \mathbb{R}$ this way though, can I now? That's why I thought it was obvious but wasn't sure how to prove it for any x. $\endgroup$ – soohyung May 14 '15 at 21:43
  • $\begingroup$ Yes, but in your case, the family is not independent. Assume that there are $a,b,c$, such that the relation you wrote is true for all $x \in \Bbb{R}$. Now you choose three values for $x$, in order to find a system of equations satisfied by $a,b,c$. You can choose whatever values you like, since the relation is valid for all $x \in \Bbb{R}$. If the family is not dependent, you'll find, after a few tries, that $a=b=c=0$, and this means that you have independence. $\endgroup$ – Beni Bogosel May 14 '15 at 21:50
  • $\begingroup$ Ah, thanks for the clarification! $\endgroup$ – soohyung May 14 '15 at 21:54
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The hypothesis is that $a\sin(x)+b\sin(2x)+c\sin(3x)=0$ for all $x\in\mathbb{R}$, because this should be the zero function.

Now you can evaluate the expression for any particular value of $x$; if we do it for $x=\pi/2$ we get $$ a\sin\frac{\pi}{2}+b\sin\pi+c\sin\frac{3\pi}{2}=0 $$ that is, $a-c=0$.

Now we can try $x=\pi/4$, that gives $$ a\sin\frac{\pi}{4}+b\sin\frac{\pi}{2}+c\sin\frac{3\pi}{4}=0 $$ or $a/\sqrt{2}+b+c/\sqrt{2}=0$.

Plug in another suitable value and you'll be done.

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Let $a,b,c\in\mathbb R$ such that $a\sin(x)+b\sin(2x)+c\sin(3x)=0$ for all $x\in\mathbb R$. Denote $f(x)=a\sin x+b\sin 2x+c\sin 3x$. Then $f'(x)=0$ and $f''(x)=0$ for all $x\in\mathbb R$. Hence, for any $x\in\mathbb R$ for $a,b,c$ we have a homogeneous system of linear equations ($a,b,c$ - unknowns) \begin{cases} a\sin(x)+b\sin(2x)+c\sin(3x)=0\\ a\cos(x)+b 2\cos(2x)+c 3\cos(3x)=0\\ a\sin(x)+b 4\sin(2x)+c 9\sin(3x)=0 \end{cases} with a matrix $$ W(x)= \begin{pmatrix} \sin(x) & \sin(2x) & \sin(3x)\\ \cos(x) & 2\cos(2x) & 3\cos(3x)\\ \sin(x) & 4\sin(2x) & 9\sin(3x) \end{pmatrix}. $$ Obviously, $$ \det(W(\frac{\pi}{2}))= \begin{vmatrix} 1 & 0 & -1\\ 0 & -2 & 0\\ 1 & 0 & -9 \end{vmatrix}\neq 0, $$ therefore our system has only zero solution, i.e. $a=b=c=0$.

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