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Homework Help: Test the set of functions for linear independence in F

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Test the set of {1, ln(2x), ln(x^2)} for linear independence in F, the set of all functions.

    If it is linearly dependent, express one of the functions as a linear combination of the others.


    2. Relevant equations

    N/A

    3. The attempt at a solution

    I know if [ a(1) + b(ln(2x)) + c(ln(x^2)) = 0 ] implies [ a = b = c = 0 ], then the set is linearly independent. Otherwise, it is linearly dependent.

    The book gives an example problem where you can plug in 0 for x and solve for the coefficients, which are then shown to be 0. But that doesn't really work here.

    Plugged in x=1: a+b(ln2) + cln(1) =0 implies a + bln2 = 0.

    It works neatly in the book example (sinx, cosx) but doesn't really in this case. I can't think of any way to go about doing this...
     
  2. jcsd
  3. Feb 20, 2012 #2

    LCKurtz

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    Try simplifying your ln functions using properties of logarithms and see if that gives you any ideas.
     
  4. Feb 21, 2012 #3
    Thank you for the help. Apologies again if I'm missing something that's obvious, but I still cannot find the answer. This is my process:

    a(1) + b(ln2x) + c(ln(x^2))

    = a(1) + b(ln2 + lnx) + c(2lnx)

    = a + bln2 + blnx + 2clnx

    = a + b*ln2 + (b+2c)*lnx

    After trying to play around with it some more, set up some kind of system of equations, reason it out, etc. I still can't find some way to prove or disprove that a = b = c = 0. Any more suggestions? :\
     
    Last edited: Feb 21, 2012
  5. Feb 21, 2012 #4

    Dick

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    That's good. So you've got a+b*ln(2)=0 and b+2c=0. How about putting c=1 and b=(-2). Now what's a?
     
  6. Feb 21, 2012 #5
    Okay, to follow the above poster:

    I have a system of equations of

    a + bln2 = 0
    b + 2c = 0

    I set c = 1 and b = -2

    Then I get a = 2ln2

    Plug this back into the original equation to get

    2ln2 - 2ln2 - 2lnx + 2lnx = 0

    Which is always true, therefore the functions ARE linearly independent.


    My understanding of some of these concepts is probably shaky, since reading through this I still haven't managed to convince myself that this is true (like for instance, how did the values for b and c come about? guess and check?). But I will go over the concepts several times and see if it sticks, thanks everyone!
     
  7. Feb 21, 2012 #6

    Dick

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    No, no. Finding a nonzero a, b, c shows that they are linearly dependent. Not independent!
     
  8. Feb 22, 2012 #7
    Ahaha... no wonder it wasn't making sense to me. It's clearer now, much thanks for your patience!

    (To clarify for myself and maybe others: it was a matter of finding nonzero coefficients a, b, c and still have the linear combination of the functions equal to zero; if we found such a, b, and c, then the set of functions functions cannot be linearly independent.)

    So to finish up the problem, I would show linear dependence by writing one function in terms of the other:

    First plug in our values for a, b, c: (2ln2)(1) + (-2)(ln2x) + (ln(x^2)) = 0

    which equals:

    2ln2 + ln(x^2) = 2ln2x

    which we can confirm by playing with ln properties:

    2ln2 + 2lnx = 2ln2x
    2(ln2 + lnx) = 2ln2x
    2ln2x = 2ln2x

    Okay. Thanks again.
     
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