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Original Problem: Determine if the set of functions $$\{ y_1(x),y_2(x),y_3(x) \} = \{x^2, \sin x, \cos x \}$$ is linearly independent.

I understand I have to use the Wronskian method, but how would it work for three functions with sine and cosine? Can someone help me give a brief overview of what I need to do and does the terms actually cancel?

$$W(y_1,y_2,y_3)(x) = \det \begin{pmatrix} x^2 & \sin x & \cos x \\ 2x & \cos x & -\sin x \\ 2 & -\sin x & -\cos x \end{pmatrix}$$

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    $\begingroup$ Compute the determinant now. $\endgroup$
    – user7440
    Aug 14, 2019 at 22:56
  • $\begingroup$ If there is a constant linear dependence between the functions, it also occurs between their derivatives and between their second derivatives. So assume a constant linear dependence, and evaluate the functions, their derivatives and second derivatives, at $0$. $\endgroup$
    – Aphelli
    Aug 14, 2019 at 23:02
  • $\begingroup$ Check the section of linear independence here: en.wikipedia.org/wiki/Wronskian $\endgroup$
    – NoChance
    Aug 14, 2019 at 23:47

3 Answers 3

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I think you can use a more elementary method. $\{x^2,\sin x, \cos x\}$ are linearly independent iff $ax^2+b\sin x+c \cos x=0 \implies a,b,c=0$.

But, if $ax^2+b\sin x+c \cos x=0$, then, since $x^2$ is unlimited while $\sin x$ and $\cos x$ are not, $a=0$.

Also $\sin (0)=0, \cos(0)=1 \implies c=0$. So, $a,b,c=0$.

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  • $\begingroup$ Agreed. Whilst the Wronskian can not determine whether ${x, |x|}$ are linear independent (for $x \in \mathbb{R}$), your approach can. $\endgroup$
    – user7440
    Aug 15, 2019 at 2:32
  • $\begingroup$ I would add that your propositions are all all-quantified over $x$. Linear independence means $(\forall x. ax^2 + b\sin(x) + c\cos(x) = 0) \Rightarrow a=b=c=0$. Only this all-quantification justified the conclusion of $a=0$. $\endgroup$
    – ComFreek
    Aug 16, 2019 at 8:15
  • $\begingroup$ oh yes, I meant the equality as an equality between functions $\endgroup$ Aug 16, 2019 at 13:46
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$ax^2+b\cos x+c\sin x=0\,,\forall x\implies a(0)^2+b\cos0+c\sin0=0\implies b=0$.

So we have $ax^2+c\sin x=0\,,\forall x\implies a( \pi)^2+c\sin (\pi)=0\implies a(\pi)^2=0\implies a=0$.

So we have $c\sin x=0\,,\forall x\implies c=0$.

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$$ *** $$

Let the set of functions Y = $\{ y_1(x),y_2(x),y_3(x) \} = \{x^2, \sin x, \cos x \}$, be linearly dependent.

Since you did not mention the anything about where the function is coming from and going to and about the field. I will suppose that, Y: $\mathbb{R}$ -> $\mathbb{R}$

Then by definition of linear dependence, there exists scalars (not all zero), such that, $$\ c_1.\mathbf{y_1(x)} + c_2.\mathbf{y_2(x)} + c_3.\mathbf{y_3(x)} = \mathbf{0(x)},\ \ \ \ x \in \mathbb{R},c_1,c_2,c_3 \in \mathbb{N} $$

$$ c_1.x^2 + c_2.\sin x + c_3.\cos x = 0 ,\ \ \ \ x \in \mathbb{R},c_1,c_2,c_3 \in \mathbb{N}$$

$$ c_2.\sin x + c_3.\cos x = -c_1.x^2 $$

$$ c_2.\sin x + c_3.\cos x = c_4.x^2, \ \ \ \ c_4 = -c_1 $$

Is there a possibility that the manipulation of just the scalars $c_1, c_2, c_3$, we can equate the above equation?

Let $c_2$ = 0, then,

$$ c_3. \cos x = c_4.x^2 \ is \ only \ possible \ when \ c_3 = c_4 = 0$$

Let $c_3$ = 0, then,

$$ c_2. \sin x = c_4.x^2 \ is \ only \ possible \ when \ c_2 = c_4 = 0$$

Therefore, $ c_2.\sin x + c_3.\cos x = c_4.x^2$ is only possible when $c_2 = c_3 = c_4 = 0$

Therefore, the set of function forms a linearly independent set.

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Disclaimer: I don't have a definite reason why $ c_3. \cos x = c_4.x^2 \ is \ only \ possible \ when \ c_3 = c_4 = 0$, yet.

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