Using the Wronskian to determine linear independence intervals

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w = wronskian (determinant result)

If w = 0
--function set is L.D.
else
--function set is L.I. everywhere

If w = some function
--set w = 0
--solve for x
--If you can solve for x
----function set is L.I. except at x value(s)
--else
----function set is L.I. everywhere

Is that right?

source: "Here W=0 only when x=0 . Therefore x^2 and x are independent except at x=0 . " http://planetmath.org/WronskianDeterminant.html
 
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Jeff12341234 said:
w = wronskian (determinant result)

If w = 0
--function set is L.D.

That is false in general. If the Wronskian of two functions is nonzero anywhere on an interval, the functions are linearly independent on that interval. But the Wronskian of two functions can be identically zero on an interval yet the functions are linearly independent on that interval. It makes a difference if the functions are solutions of a second order linear differential equation with certain hypotheses. Only then can you say that their Wronskian is either identically zero or never zero. Surely your text has a theorem about this.
 


Jeff12341234 said:
w = wronskian (determinant result)

If w = 0
--function set is L.D.
else
--function set is L.I. everywhere

If w = some function
--set w = 0
--solve for x
--If you can solve for x
----function set is L.I. except at x value(s)
--else
----function set is L.I. everywhere

Is that right?

source: "Here W=0 only when x=0 . Therefore x^2 and x are independent except at x=0 . " http://planetmath.org/WronskianDeterminant.html

Mostly correct. If you read the wikipedia article carefully, it will give you an example of two functions where w=0, but they are linearly independent. w=0 usually means they are linearly dependent, but there are exceptions. To show functions are linearly independent on an interval (a,b) you don't HAVE to solve for x. All you have to do is find a point x in (a,b) where w(x)≠0. If solving for x is the way you want to do it, that's fine.

Planetmath is usually a good source, but I'm going to have to take issue with their saying x, x^2 are LI except at 0. There are LI on (-inf,inf). No 0 exceptions. Remember all you need is ONE POINT in an interval where w(x)≠0 to show LI. The fact there are isolated points where w=0 doesn't change anything.

If you want to talk about linear independence on an interval consisting of a single point, like [0,0] then that is a whole different issue and doesn't have anything to do with wronskians. On a single point interval both functions are just numbers, and two numbers are always linearly dependent, whether the wronskian is zero or not.
 
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I looked back over that wiki article. I can't make sense of it enough to modify the "procedural code" that I previously posted.
 
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Jeff12341234 said:
I looked back over that wiki article. I can't make sense of it enough to modify the "procedural code" that I previously posted.

Beats me. I don't know what else to say that I haven't already repeated several times.
 


I need an actual procedure to perform to test for those rare exceptions you mentioned for instances where w=0 and the functions are still L.I.
 


Jeff12341234 said:
I need an actual procedure to perform to test for those rare exceptions you mentioned for instances where w=0 and the functions are still L.I.

Take the wikipedia example of f1(x)=x|x| and f2(x)=x^2. Go back to the definition of linear independence. It says try to find constants a and b, not both 0 such that a*f1(x)+b*f2(x)=0 for all x. Put in x=1, that gives a+b=0. Put in x=(-1), that gives -a+b=0. Solve those two equations for a and b and conclude the only possibility is a=b=0. So they are linearly independent.
 


so basically the wronskian fails in these circumstances since performing the test results in w=0 implying L.D.

Another procedural test needs to be used. Could that test be to solve a system of equations such that the first equation, a*f1(x)+b*f2(x)=0, x=1, and the second equation,a*f1(x)+b*f2(x)=0, x=-1. If a=b=0 the set of functions are L.I. Can I use that procedure and be confident that I've covered all of the possibilities now? If the test I just described can always be used, why not use it in place of the wronskian?
 


Jeff12341234 said:
so basically the wronskian fails in these circumstances since performing the test results in w=0 implying L.D.

Another procedural test needs to be used. Could that test be to solve a system of equations such that the first equation, a*f1(x)+b*f2(x)=0, x=1, and the second equation,a*f1(x)+b*f2(x)=0, x=-1. If a=b=0 the set of functions are L.I. Can I use that procedure and be confident that I've covered all of the possibilities now? If the test I just described can always be used, why not use it in place of the wronskian?

You can't always pick x=-1 and x=1. If you pick the two functions x^2 and x^4 then that procedure will give you a+b=0 and a+b=0 for both equations. But the wronskian will tell you they are linearly independent. Better there to pick x=1 and x=2. So it's not really a fixed procedure, you have to think about the specific functions and make sure that you choose good points. If you want to show functions are linearly independent that way, sure, you can use it instead of the wronskian.
 


ok. That's good to know.

Going back to the original question of this thread, I just ended up writing that the functions are L.I. everywhere except for x=0 and x=2/7 at which, on those points, they are L.D. Is there a better way to answer part A and and B that what I wrote?
 


Jeff12341234 said:
ok. That's good to know.

Going back to the original question of this thread, I just ended up writing that the functions are L.I. everywhere except for x=0 and x=2/7 at which, on those points, they are L.D. Is there a better way to answer part A and and B that what I wrote?

I already said I disagree with the Planetmath phrasing in post #33. But if want to ignore that and you like it, fine. But the question is asking for intervals, isn't it?
 


Well, you said that pseudo code I wrote earlier was right except for special cases when w=0, which we went over. So I'm guessing the pseudo code should really be:

--If you can solve for x
----function set is L.I. everywhere

instead of:

--If you can solve for x
----function set is L.I. except at x value(s)
 


Jeff12341234 said:
Well, you said that pseudo code I wrote earlier was right except for special cases when w=0, which we went over. So I'm guessing the pseudo code should really be:

--If you can solve for x
----function set is L.I. everywhere

instead of:

--If you can solve for x
----function set is L.I. except at x value(s)

If you solve for x and you only get a finite number of solutions, then the wronskian is still has a nonzero value on any interval whose length isn't zero. So, yes, they are linearly independent on any interval whose length isn't equal to zero.
 


So for this question, when asked where the function set is L.I., (-inf, inf) would be the best answer and for L.D., it would be 'none'?
 


Jeff12341234 said:
So for this question, when asked where the function set is L.I., (-inf, inf) would be the best answer and for L.D., it would be 'none'?

That is exactly what I would say. Though for B you might note if the interval contains only a single point then they are LD, for reasons that have nothing to do with wronskians. So I wouldn't consider those exceptions. I as said way back, that part of the question is kind of goofy.
 


I did some more research on this. Appearently you're allowed to plug in an x value first (typically x=0) to the square matrix BEFORE doing the determinant. After that the logic is simply:

If w = 0
--function set is L.D. and is NOT the whole solution
else
--function set is L.I. and is the general solution

When I do that for this problem, i get a matrix where the top row is 5 , 0 and the second row is 10 , 0. This results in a determinant that is 0... making it L.D. I really need a once-and-for-all, authoritative proof for what this should be because I'm getting all kinds of conflicting results. To top it all off, he's asking for intervals for each which isn't even mentioned in most resources I've read.
 
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Jeff12341234 said:
I did some more research on this. Appearently you're allowed to plug in an x value first (typically x=0) to the square matrix BEFORE doing the determinant. After that the logic is simply:

If w = 0
--function set is L.D. and is NOT the whole solution
else
--function set is L.I. and is the general solution

When I do that for this problem, i get a matrix where the top row is 5 , 0 and the second row is 10 , 0. This results in a determinant that is 0... making it L.D. I really need a once-and-for-all, authoritative proof for what this should be because I'm getting all kinds of conflicting results. To top it all off, he's asking for intervals for each which isn't even mentioned in most resources I've read.

That is incorrect. The wronskian is 0 at x=0. You knew that from the beginning, right? It's also zero at 2/7. It doesn't change the fact that the functions are linearly independent. Here's a definitive statement. f1 and f2 are linearly dependent on an interval if and only if there are constants not both zero such that a*f1(x)+b*f2(x)=0 on the interval. That's 'definitive' because that is the 'definition' of linear dependence. Here's another definitive statement. The wronskian is zero on an interval if the functions are linearly dependent. I know that because I know how to prove it. That means if the wronskian is nonzero anywhere on an interval then they are not linearly dependent. Hence they are linearly independent. That's definitive because it's just logic.

Here's another definitive statement. I am getting really sick of the sound of my own voice as I repeat these things over and over again. If your research is turning up conflicting things then you will have to decide what is correct. I've told you everything I know about the subject several times. You decide.
 
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And for my parting shot I will just observe that the two original functions can't be solutions of a second order linear DE with nonzero leading coefficient in the first place because their Wronskian is neither nonzero nor identically zero.

And Dick, you deserve a medal for your persistence. But wait..., you already have one.
 


For part A) L.I. on the intervals (-inf, 0) U (0, 2/7) U (2/7). Largest interval is (-inf, 0)
..-------B) L.D. at x=0 and x=2/7
 


LCKurtz said:
And for my parting shot I will just observe that the two original functions can't be solutions of a second order linear DE with nonzero leading coefficient in the first place because their Wronskian is neither nonzero nor identically zero.

And Dick, you deserve a medal for your persistence. But wait..., you already have one.

Thanks, LCKurtz. I decided to actually try to practice patience on this one, I thought it might work. Not sure it did. And sure, good point. I agree totally. Solutions to nice ODE's will allow you to say wronskian=0 means linearly dependent.
 


Jeff12341234 said:
For part A) L.I. on the intervals (-inf, 0) U (0, 2/7) U (2/7). Largest interval is (-inf, 0)
..-------B) L.D. at x=0 and x=2/7

Why, why, why? Give some sort of reason! Your best answer was in post 44. Yet you continue to yammer on about this for no reason I can think of.
 


The answer is either:

Possible answer #1
For part A) L.I. on the intervals (-inf, 0) U (0, 2/7) U (2/7). Largest interval is (-inf, 0)
..-------B) L.D. at x=0 and x=2/7

or:

Possible answer #2
For part A) (-inf,inf). Largest interval is (-inf, inf)
..-------B) none

The way the question is worded implies the first answer is more likely to be correct. I'll keep researching it.
 
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Jeff12341234 said:
The answer is either:

For part A) L.I. on the intervals (-inf, 0) U (0, 2/7) U (2/7). Largest interval is (-inf, 0)
..-------B) L.D. at x=0 and x=2/7

or:

For part A) (-inf,inf). Largest interval is (-inf, inf)
..-------B) none

The way the question is worded implies the first answer is more likely to be correct. I'll keep researching it.

Keep researching it. Because I have no idea what you are talking about.
 


Out of all of the examples I've looked at, I've yet to see one where they found two points for x. The wronskian either simply equals zero or it doesn't in all of the examples I've seen :/
 


Jeff12341234 said:
Out of all of the examples I've looked at, I've yet to see one where they found two points for x. The wronskian either simply equals zero or it doesn't in all of the examples I've seen :/

x and x^2 have a wronskian of 2x^2-x^2=x^2. They are NOT LINEARLY DEPENDENT even though the wronskian is zero at x=0. Do you think they are?
 


purplemath.com implies Possible answer #1 is correct: "Here W=0 only when x=0 . Therefore x^2 and x are independent except at x=0 . " http://planetmath.org/WronskianDeterminant.html

Some guy on chegg.com says that Possible answer #1 is correct: http://media.cheggcdn.com/media/c72/c72af4bb-22a6-4f97-8c51-f964eb461e24/phpdbpdV9.png

Paul's Online notes imply that Possible answer #2 is correct but maybe not since it doesn't go into details about intervals of the answer..
"Here we know that the two functions are linearly independent and so we should get a non-zero Wronskian... The Wronskian is non-zero as we expected provided . This is not a problem. As long as the Wronskian is not identically zero for all t we are okay." http://tutorial.math.lamar.edu/Classes/DE/Wronskian.aspx

You say Possible answer #2 is correct.

This implies that Possible answer #2 is correct. "Let f and g be differentiable on [a,b]. If Wronskian W(f,g)(t0) is nonzero for some t0 in [a,b] then f and g are linearly independent on [a,b]. If f and g are linearly dependent then the Wronskian is zero for all t in [a,b]." http://ltcconline.net/greenl/courses/204/ConstantCoeff/linearIndependence.htm So that's 2 votes for #1 and 3 votes for #2 so far.
 
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Jeff12341234 said:
So that's 2 votes for #1 and two votes for #2 so far. That's not good :(

I really don't care much what the polling is on this problem. Answer #1 is wrong. Answer #2 is correct. I think I've explained why I think so. Why do you keep asking? If you want a consensus by majority then you should keep researching to break the tie. If you can't think about this for yourself, then that seems the only option.
 


I keep asking because I don't fully understand it myself. I now can reliably determine if a set of functions are L.I. or not BUT, out of the 20+ examples that I've worked, none of them are like the problem that prompted this thread. none of them returned x values like this problem does. So the part that the instructor is asking about, the intervals, I don't get. The problem is either worded really retarded or answer #2 is wrong.
 


Jeff12341234 said:
I keep asking because I don't fully understand it myself. I now can reliably determine if a set of functions are L.I. or not BUT, out of the 20+ examples that I've worked, none of them are like the problem that prompted this thread. none of them returned x values like this problem does. So the part that the instructor is asking about, the intervals, I don't get. The problem is either worded really retarded or answer #2 is wrong.

The problem is certainly phrased oddly. Just believe the Paul's Online Notes version of the story.
 
Well we got our tests back. The correct answer for part A was (-inf, 0) but he said any of the three intervals could've been used. It was L.I. on the intervals (-inf, 0) U (0, 2/7) U (2/7, inf). I got part A correct on the test.

Part B was really weird. The answer was "any interval that contained the two points" so (-inf, inf) could've been used. I got this part wrong.

Weird stuff..
 

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