Question

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

Theorem 5 states that if u1,..., uk are any (k-1) times differentiable functions which are linearly dependent on I, then W(u1(x),...,uk(x) = 0 on I. . The converse of theorem 5 is false for arbitrary functions.

Answer 1

the derivatives of x^3 must be 3x^2 6x 6 and 0 if it's a Wronskian

Answer 2

@wio

Answer 3

@dan815

Answer 4

the interval inclued a point where abs x^3 is undefined

Answer 5

where the derivative* is undefined

Answer 6

how do I take the derivative of an absolute value? just do it as normal with the x^3

Answer 7

@zzr0ck3r

Answer 8

:/

Answer 9

I'm doomed. not good with proof writing I swear

Answer 10

unless well everyone knows that abs(-1) is positive 1 .same for if there's a negative in the determinant force it to be positive.

Answer 11

and there's a second part to this as well...
are x^2 and abs(x)^3 solutions on [-1,1] of any second order homogeneous linear equation? any third order homogeneous linear equation? hint: what is the third derivative of abs(x)^3 at 0

Answer 12

second order homogeneous like y''+y'+8y
third y'''+y''+y'+9y

Answer 13

as examples

Answer 14

y have to break it up into parts

Answer 15

and u have to label the cusp points, where it is not differntiable

Answer 16

well absolute value...OH THAT GRAPH!

Answer 17

https://www.google.com/search?q=graph+abs%28x%29&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a