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Using the Wronskian for linear independence/dependence

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data

    f1 = 0 , f2 = x , f3 = e^x

    I am supposed to find out if these are linearly independent or dependent. Just by looking at it, I can't see a way to write one of the functions as a combination of the other two with constant multiples, so to make sure that it is linearly independent, I used the Wronskian

    f1' = 0 f2' =1 f3'= e^x
    f1" = 0 f2"=0 f3" = e^x

    So I evaluated the determinant of

    0 x e^x
    0 1 e^x
    0 0 e^x

    And that equals zero, which would mean to me that the functions are linearly dependent. I'm not sure how that could be. Any thoughts?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 9, 2008 #2

    Hurkyl

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    Then you're thinking too narrowly. I bet you've forgotten about the simplest linear combination of two functions.
     
  4. Nov 9, 2008 #3
    Well... I know that I can't use zero as a constant multiple I think I am overlooking something really simple, but I still can't see how to combine them.
     
  5. Nov 9, 2008 #4

    Hurkyl

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    Why not?
     
  6. Nov 9, 2008 #5

    gabbagabbahey

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    If the Wronskian is zero for all x, then the only thing you can conclude from that, is that the functions may or may not be linearly dependent .

    Just because linear dependence implies that the Wronskian is zero, does not mean that the Wronskian being zero implies linear dependence.
     
  7. Nov 9, 2008 #6
    So how would I go about showing that they are linearly independent if I can't use the Wronskian? Do I just say that there is no way to write one function as a linear combination of the other two?
     
  8. Nov 9, 2008 #7
    Let's go back to the definition of L.I and L.D.

    If I can write c1* f1 +c2* f2 +c3* f3 = 0 for some scalar c1 c2 c3, and not all of them are zero, then we found L.D. So, can you think of anything? Here's a hit, two of the scalars are the same number.
     
  9. Nov 9, 2008 #8

    Hurkyl

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    gabbagabbahey wasn't telling you that they're linearly independent. He merely said that a zero Wronskian is not proof of linear dependence.

    (At least, he better not have been telling you they're linearly independent -- because they really are linearly dependent!)
     
  10. Nov 9, 2008 #9

    gabbagabbahey

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    Well, for constants [itex]c_1[/itex], [itex]c_2[/itex] and [itex]c_3[/itex]...the only way that [itex]c_1 f_1+c_2 f_2 +c_3 f_3= c_2 x+ c_3 e^x=0[/itex] is if [itex]c_2=c_3=0[/itex], but that doesn't mean [itex]c_1[/itex] has to be zero does it?
     
  11. Nov 9, 2008 #10
    Ok... I understand. I just reread the definition for linear independence in book, and it all makes sense now. Thanks very much.
     
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