13
$\begingroup$

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?

$\endgroup$
2
  • 2
    $\begingroup$ What is $\mathbb{B}$? $\endgroup$ May 31 '12 at 18:14
  • 23
    $\begingroup$ A letter on my keyboard that's very close to N. I fixed the typo above. $\endgroup$
    – Potato
    May 31 '12 at 18:17
32
$\begingroup$

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.

$\endgroup$
3
  • 2
    $\begingroup$ Beat me by 24 seconds! $\endgroup$ May 31 '12 at 18:22
  • 1
    $\begingroup$ Nice, this proof also generalizes for any function that takes infinitely many values. $\endgroup$ May 31 '12 at 18:34
  • $\begingroup$ For the step from "$f(y) = 0$ for infinitely many values of $y$" to "$f$ is the zero polynomial," is there any way of further explaining / proving this? is it a theorem? $\endgroup$
    – Rax Adaam
    Oct 13 '15 at 18:24
5
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Could you explain further? I'm not familiar with the theory of Wronskians. $\endgroup$
    – Potato
    May 31 '12 at 18:25
  • 1
    $\begingroup$ 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. $\endgroup$
    – GEdgar
    May 31 '12 at 18:27
  • $\begingroup$ Is the calculation trivial? $\endgroup$
    – Pedro Tamaroff
    May 31 '12 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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