- #36
Jeff12341234
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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.
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?
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?
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)
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'?
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.
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.
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
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.
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 :/
Jeff12341234 said:So that's 2 votes for #1 and two votes for #2 so far. That's not good :(
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.
Jeff12341234 said: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..