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How to show the functions $x^3$ and $x^2|x|$ are linearly independent on the real line ?

I know these are linearly independent because neither can be written as a constant multiple of the other for the whole real line. Is there any way to show their linear independence using Wronskian?

vishu
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1 Answers1

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Let $f_1(x) = x^3, f_2(x) = |x|x^2$.

Note that the Wronskian $W(f_1,f_2) = 0$, so one cannot use the Wronskian here.

Suppose $g=\lambda_1 f_1 + \lambda_2 f_2 = 0$ for some $\lambda_1,\lambda_2$.

Then $g(1) = \lambda_1+ \lambda_2, g(-1) = -\lambda_1+ \lambda_2$, and solving $\lambda_1+ \lambda_2= 0, -\lambda_1+ \lambda_2 = 0$ gives $\lambda_1 = \lambda_1 = 0$.

copper.hat
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  • @YvesDaoust: I updated my answer after Umberto's remark. – copper.hat Mar 20 '17 at 15:48
  • But these two ($x^3$ and $|x|x^2$) are solutions for the linear ODE $x^2\frac{d^2y}{dx^2}-4x\frac{dy}{dx}+6y=0$ on $\mathbb R$, right? Then how they can be linearly independent while Wronskian is $0$ on $\mathbb R$ ? – Messi Lio Sep 05 '23 at 17:49
  • @MessiLio As has been mentioned many times. linear independence does not imply that the Wronskian is non zero. See https://en.wikipedia.org/wiki/Wronskian#The_Wronskian_and_linear_independence for example. – copper.hat Sep 05 '23 at 22:46
  • Actually, I'd believed the extra condition ''being solutions of a second order linear differential equation'' will make equivalence between independence and non vanishing state of Wronskian. So this answer need correction, right? – Messi Lio Sep 06 '23 at 01:19
  • @MessiLio The ODE $ \frac{d^2y}{dx^2}-{4 \over x}\frac{dy}{dx}+{6 \over x^2}y=0 $ is not defined at $x=0$, so you cannot apply the result referenced. – copper.hat Sep 06 '23 at 02:10