Show that $~x^3~$ and $~|x|^3~$ are not linearly dependent on $~[-1,1]~$, but that Wronskian,$$W(x^3, |x|^3) = 0~.$$ This shows that the converse of Theorem $5$ is false.

Are $~x^3~$ and $~|x|^3~$ solutions on $~[-1,1]~$ on any second order homogeneous linear equation?

Any third order homogeneous linear equation?

Theorem 5 : If $~u_1,\cdots,u_k~$ are any $~(k-1)~$ times differentiable functions which are linearly dependent on $~I~$, then Wronskian, $~W(u_1(x),\cdots,u_k(x) = 0~$ on $~I~$.

  • 1
    $\begingroup$ Do you know what it would mean for $x^3$ and $|x|^3$ to be linearly dependent? Do you know the definition of the Wronskian? What have you tried, and what's giving you trouble? $\endgroup$
    – user61527
    Commented Oct 20, 2013 at 1:16
  • $\begingroup$ ummm...well when the Wronskian is 0, it just vanishes and the results become inconclusive. Also I have taken the first and second derivatives of x^3 and the absolute value of x^3 and had a determinant of 0. $\endgroup$
    – usukidoll
    Commented Oct 20, 2013 at 1:22
  • $\begingroup$ Seriously guys...this question is insanely old and I'm getting downvoted. lol! hysterical rolls eyes $\endgroup$
    – usukidoll
    Commented Feb 8, 2014 at 9:57

1 Answer 1


Consider what $|x|^3$ really is: a piecewise function. You can split it up around zero. Are there a single set of scalars you can multiply by these two functions over $[-1,1]$ that shows they are linearly dependent?

Recall what it means to be linearly (in)dependent. I gave you a hint above.

  • $\begingroup$ alright... yes I can graph it as a piece wise function it is a cusp for the absolute value and I know there is symmetry around the origin since the function is odd for x^3. I do know that linearly independent is when all values are 0 which is c1=c2=c3=cr=0 and for dependency occurs when not all of the values are 0 but the result is still going to end up as 0 like 4c1+c2+0c3=0 $\endgroup$
    – usukidoll
    Commented Oct 20, 2013 at 1:40
  • $\begingroup$ See, you want to consider each part of the function separately. How would you write (or graph) the function on the interval $[-1,0]$? Here's a hint: for the interval $[0,1]$, $|x|^3 \equiv x^3.$ So consider each sub-interval separately and find your constants (there should be two). Are they the same over both sub-intervals? $\endgroup$
    – Nico
    Commented Oct 20, 2013 at 1:48
  • $\begingroup$ well I do know that anything negative in the absolute value turns positive. $\endgroup$
    – usukidoll
    Commented Oct 20, 2013 at 2:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .