Wronskian and linear dependence

[Sarahisme]

Hey everyone,

 

i am having a bit of trouble trying to do this question.

 

 

how do i show that they are linearly dependent ? i tried using the wronskian but this doesnt work does it? (because just because the wroskian is 0 on the interval it doesnt neccessarily mean that the two functions are dependent, right?)

 

this is what i have done so far:

 

(a)

for 0<t<1

|t| = t

so f(t) = t^2.t = t^3 = g(t)

 

therefore f(t) and g(t) are linearly dependent on 0<t<1

 

 

for -1<t<0

|t| = -t

so f(t) = t^2.-t = -t^3 = -g(t)

 

therefore f(t) and g(t) are linearly dependent on -1<t<0

 

 

for -1<t<1

because f(t) is a different multiple of g(t) on 0<t<1 than on for -1<t<0, therefore f(t) & g(t) are linearly independent on for -1<t<1

 

 

also for part (b) i get W(f,g) = {t^6}/|t| which is obviously not zero for all -1<t<1 ?? :S

=========================

does that makes sense to you guys?? because i am not to sure it makes sense to me!

 

 

 

any help would be lovely

 

-Sarah

[Tom Mattson]

how do i show that they are linearly dependent ?

 

You've already done it.

 

i tried using the wronskian but this doesnt work does it? (because just because the wroskian is 0 on the interval it doesnt neccessarily mean that the two functions are dependent' date=' right?)

[/quote']

 

That's right.

 

this is what i have done so far:

 

(a)

for 0<t<1

|t| = t

so f(t) = t^2.t = t^3 = g(t)

 

therefore f(t) and g(t) are linearly dependent on 0<t<1

 

 

for -1<t<0

|t| = -t

so f(t) = t^2.-t = -t^3 = -g(t)

 

therefore f(t) and g(t) are linearly dependent on -1<t<0

 

That's just what I would have done.

 

for -1<t<1

because f(t) is a different multiple of g(t) on 0<t<1 than on for -1<t<0, therefore f(t) & g(t) are linearly independent on for -1<t<1

 

 

also for part (b) i get W(f,g) = {t^6}/|t| which is obviously not zero for all -1<t<1 ?? :S

 

How did you get that for [imath]W(f,g)[/imath]? The Wronskian clearly vanishes on [imath]-1<x<0[/imath] and on [imath]0<x<1[/imath]. Having said that, the Wronskian doesn't exist at [imath]x=0[/imath] because [imath]f[/imath] is not differentiable there. So the problem statement is wrong when it claims that [imath]W(f,g)\equiv 0[/imath] everywhere in [imath][-1,1][/imath].

[Sarahisme]

ok, thanks for that Tom!