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Linear independent or dependent

  1. Jul 27, 2005 #1
    Given y1=3t, y2=|t|, determine whether the given pair of functions is linearly independent or linearly dependent.

    I can't do the Wronskian on this one since y2 is not differentiable at 0, is there any other way to do this?
     
  2. jcsd
  3. Jul 27, 2005 #2

    lurflurf

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    It is easy to see that they are linearly independent, as one is even and one is odd (I presume you mean for all real numbers).
    A similar approch would be to notice (ie with wronskian) that they are dependent on (-infinity,0] and [0,infinity) then show that sets of coefficients that make them linearly dependent in each region have nothing in common.
    A third way is show that if
    u*y1+y2=0
    u is not constant
    All three of these are basically the same.
    say
    f(x)=f(-x)
    g(x)=-g(-x)
    if
    a f(x)+b g(x)=0
    then
    a f(-x)+b g(x)=0
    so
    a f(x)-b g(x)=0
    thus
    a f(x)=0
    b g(x)=0
    This holds for all x
    f(x),g(x) cannot be zero every where
    thus a=b=0 and f and g are linearly independent
     
    Last edited: Jul 27, 2005
  4. Jul 27, 2005 #3
    In the book, they did something the likes of:
    suppose k1 and k2 are two constants, and k1(3t)+k2(|t|)=0, then choose two points t0 and t1 where t1 is not equal to t0, if the determinant of the coefficients is not zero than the two functions are independent.

    but when I did it out, they are zero for t>1 and t<1. I'm confused, I guess, since I don't understand why the determinant thing is true...

    (See the picture for what I did)
     

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    Last edited: Jul 27, 2005
  5. Jul 27, 2005 #4

    lurflurf

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    A fourth way.
    The determinant tell us if the linear system
    3k1*t0+k2*|t0|=0
    3k1*t1+k2*|t1|=0
    has a unique solution
    we know k1=k2=0 is a solution
    If the determinat is zero every where the system is linearly dependent.
    If the determinat is ever nonzero the system is linearly independent.
    you found
    3(t0|t1|-t1|t0|)=3|t0|*|t1|(sign(t0)-sign(t1))
    so the determinant will be nonzero if t0*t1<0
    for example if t0=3 t1=-7
    the determinant is 42
    so linearly independant
    I find this method a bit odd for this problem.
    It was probably used because it is more general.
    Did you like my suggestions?
    In particular if
    u*y1+y2=0
    3u*t+|t|=0
    u=-1/3 t<0
    u=1/3 t>0
    Hence
    Linearly independent
     
  6. Jul 27, 2005 #5

    cronxeh

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    W=det(3t abs(t); 3 abs(t)/t) =

    3*t*abs(1,t)-3*abs(t) != 0
     
  7. Jul 27, 2005 #6
    Can you tell me how that makes it linearly independent cuz I don't quite get it...
     
  8. Jul 27, 2005 #7
    what does abs(1,t) mean? And can't you cancel out the t's when you do 3*t*|t|/t?
     
  9. Jul 27, 2005 #8

    lurflurf

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    Things are simpler since n=2
    we want to study
    a y1+b y2=0
    3at+b|t|=0
    A nice way to see this is to concider (-infinity,0] and [0,infinity)
    (-infinity,0]
    3at+b|t|=0
    3at-bt=0
    (3a-b)t=0
    (3a-b)=0
    [0,infinity)
    3at+b|t|=0
    3at+bt=0
    (3a+b)t=0
    (3a+b)=0
    so if this holds for the whole real line we have a system
    3a-b=0
    3a+b=0
    This can only be true if a=b=0
    Linearly independent
    Another view if they were dependent
    3at+b|t|=0
    b is not zero (if it were a would need to be since t is not always 0)
    (1/b)(3at+b|t|)=0
    3(a/b)t+|t|=0
    call a/b u and let it depend on t
    3ut+|t|=0
    solve for u
    u=-sign(t)/3
    u is clearly not constant so they must be linearly independent.
    Also check out the above deal where an even function and an odd function must be linearly independent.
     
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