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The following problem is from Golan's linear algebra book. I have posted a proposed solution in the answers.

Problem: For $n\in \mathbb{N}$, consider the function $f_n(x)=\sin^n(x)$ as an element of the vector space $\mathbb{R}^\mathbb{R}$ over $\mathbb{R}$. Is the subset $\{f_n:\ n\in\mathbb{N}\}$ linearly independent?

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What is $\mathbb{B}$? –  Chris Eagle May 31 '12 at 18:14
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A letter on my keyboard that's very close to N. I fixed the typo above. –  Potato May 31 '12 at 18:17
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3 Answers

up vote 24 down vote accepted

Suppose we have real numbers $a_j$ such that $\sum_1^k a_j \sin^j(x)=0$ for every real $x$. Consider the polynomial $f(y)=\sum_1^k a_j y^j$. By assumption, we know that $f(\sin(x))=0$ for every $x$. Since $\sin(x)$ can take any value between $-1$ and $1$, we have that $f(y)=0$ for any $y$ between $-1$ and $1$. But then $f(y)=0$ for infinitely many values of $y$, and so $f$ is the zero polynomial, i.e. $a_j=0$ for all $j$. Thus the only linear dependence is the trivial one, and so our set is linearly independent.

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Beat me by 24 seconds! –  Qiaochu Yuan May 31 '12 at 18:22
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Nice, this proof also generalizes for any function that takes infinitely many values. –  Mikko Korhonen May 31 '12 at 18:34
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Here is a generalization.

Let $X$ be a set, let $F$ be a field, and let $h:X\to F$ be a function. Then $\{h,h^2,h^3,h^4,\ldots\}$ is linearly independent in the $F$-vector space $F^{X}$ if and only if $h(X)$ is infinite.

Proof can be found at the linked question, or using the same idea as in Chris Eagle's answer.

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The Wronskian of $\sin(t),...,\sin^n(t)$ is $$ 1! 2! 3! \dots (n-1)! \sin^n(t) \cos^{n(n-1)/2}(t) $$ not identically zero.

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Could you explain further? I'm not familiar with the theory of Wronskians. –  Potato May 31 '12 at 18:25
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When you take a course in differential equations, you will learn of the Wronskian. The Wronskian is a certain determinant calculation that you can do with a finite list of differentiable functions: if the functions are linearly dependent, then the Wronskian is identically zero. –  GEdgar May 31 '12 at 18:27
    
Is the calculation trivial? –  Pedro Tamaroff May 31 '12 at 18:50
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