use the wronskian to show that the given function are linearly independent on the given interval

a. {1,x,x^2} on (negative infinity, positive infinity)
b. {sinx,cosx,tanx} on ( negative pi over two, pi over two)

Question

use the wronskian to show that the given function are linearly independent on the given interval 

a. {1,x,x^2} on (negative infinity, positive infinity)
b. {sinx,cosx,tanx} on ( negative pi over two, pi over two)

turingtest · Answer

the wronskian is$$W=\left|\begin{matrix}u&v&w\u'&v'&w'\u''&v''&w''\end{matrix}\right|$$which if it is nonzero, proves the linear independence of the solutions u, v, and w

anonymous · Answer

i tried the first one and i got a zero. so how can i prove its linearly independent?

turingtest · Answer

it isn't zero
u=1
v=x
w=x^2
so$$W=\left|\begin{matrix}u&v&w\u'&v'&w'\u''&v''&w''\end{matrix}\right|=\left|\begin{matrix}1&x&x^2\0&1&2x\0&0&2\end{matrix}\right|\neq0$$

anonymous · Answer

you are right sorry i missed up lol thank you so much. how about for the second one i get a very complex trig function. do i have to plug in the values of the interval ?

turingtest · Answer

you should get something that can be simplified, though it may be a bit tricky, to something that is never zero in that interval
the way you wrote it the interval is open, so would not include those endpoints
I haven't tried it...

turingtest · Answer

$$W=\left|\begin{matrix}u&v&w\u'&v'&w'\u''&v''&w''\end{matrix}\right|=\left|\begin{matrix}\sin x&\cos x&\tan x\\cos x&-\sin x&\sec^2 x\-\sin x&-\cos x&2\sec^2x\tan x\end{matrix}\right|$$$$=-\sin^2x(2\sec^2x\tan x)-\sec x\sin x-\cos^2x\tan x-\sin^2x\tan x...$$$$...+\cos x\sec^2 x\sin x-2\cos^2 x\sec^2x\tan x$$let's see if we can simplify....

turingtest · Answer

multiple typos :S$$=-\sin^2x(2\sec^2x\tan x)-\sec^2x\sin x\cos x-\cos^2x\tan x-\sin^2x\tan x...$$$$...+\cos x\sec^2 x\sin x-2\cos^2 x\sec^2x\tan x$$I hope it's right now...

turingtest · Answer

$$=-2\tan^3x-\tan x-\cos^2x\tan x-\sin^2x\tan x+\tan x-2\tan^3x$$hope I didn't mess up yet...

turingtest · Answer

$$W=-4\tan^3x-\tan x$$hmm... that has a point where it is zero
I think I made some mistake somewhere :/

anonymous · Answer

first determinant has a sign error i think...+sec^2(x)sin(x)cos(x)

anonymous · Answer

$$i got -2\tan ^{3}x +\tan x -\cos x \sin x \sec ^{2}x-\tan x-\sin x \cos x+ \tan x \sin ^{2}x$$

turingtest · Answer

@mathmuse mine has that sign

turingtest · Answer

oh well, good luck finding my mistake
I'm out !

anonymous · Answer

haha thanks anyways

anonymous · Answer

hmmm
$$-2\tan^3x-2tanx-sinxcosx-\sin^2x$$

anonymous · Answer

either it tends to zero at x = 0.