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Decide whether or not the set of functions

$$S= \left\{\cos^2(x),\cos(2x),\pi \right\}.$$

is linearly dependent or independent. Defend your decision.

I know that in order to solve this I need to test for dependence and check if there are any non-trivial solutions that will give me a $0$, hence linearly independent. So I started off by saying that in order for $\pi$ to be $0$, its coefficient (say $c_3$) has to equal $0$, and if that is true I can continue to solve for the other two by plugging in specific $x$'s and solving for $c_1$ and $c_2$. My question is if this process is correct, and is this really linearly independent or dependent?

Dan Rust
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user145592
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3 Answers3

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Hint:

$$ \cos(2x) = 2\cos^2(x)-1. $$

Check the definitions of linear dependence and independence. See another technique.

Community
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Mhenni Benghorbal
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1

Hint: The Wronskian, $W(\cos^2(x),\cos(2x),\pi)=0$

Sujaan Kunalan
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  • Note, $W(f,g,h)=0$ does not in itself imply that $f,g,h$ are a dependent system - you need further conditions such as analyticity, or not all pairs have zero Wronksian for some point $x$. – Dan Rust Apr 25 '14 at 13:16
  • @DanielRust: Oh, ok, I wasn't aware of that. Thanks. But in this case, all the functions are analytic, yes? – Sujaan Kunalan Apr 25 '14 at 13:43
  • Yeah we're fine here, I was just adding it in case the OP forgot to check if he decides to use the Wronksian method :). It's a good hint. – Dan Rust Apr 25 '14 at 13:56
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For the sake of simplicity, consider $S'=\{\cos^2(x), \cos(2x), 1\}$ instead. Let's look at some values of $x$ where those functions are easy to evaluate to see about possible dependencies, say $x\in\{0,\frac\pi 2,\pi\}$. The values at these points are $$ v_1=\pmatrix{1\\0\\1},\ v_2= \pmatrix{1\\-1\\1}, v_3=\pmatrix{1\\1\\1}. $$ Any linear dependence $\lambda_1 \cos^2(x)+\lambda_2\cos(2x)+\lambda_3=0$ would give us a linear dependence $\lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3 = 0$. The solutions of the latter equation are exactly $(\lambda_1, \lambda_2, \lambda_3)\in\operatorname{span}\big((-2,1,1)\big)$. So the only non-trivial combination to consider is $$ -2\cos^2(x) + \cos(2x) + 1 = 0. $$ Thus, your set $S$ is linearly dependent if and only if $\cos(2x)=2\cos^2(x)-1$.

Christoph
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