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Linear independence

  1. Jul 1, 2009 #1
    CAN somebody, please write down a formula defining the linear independence of the following functions??

    {[tex] e^x,e^{2x}[/tex]}
     
  2. jcsd
  3. Jul 1, 2009 #2
    If e^x and e^2x are linearly independent then the equation

    ae^x+be^2x=0 must have the only solution a,b=0.

    Differentiate to obtain,

    ae^x+2be^2x=0

    Subtract the first equation from the second to obtain,
    be^2x=0
    which is true if and only if b=0.
    Then it is obvious that a=0.

    Therefore, e^x and e^2x are linearly independent.
     
  4. Jul 1, 2009 #3
    The definition says: if [tex]a,b[/tex] are constants and [tex]a e^x + b e^{2x} = 0[/tex] for all [tex]x[/tex], then [tex]a = b = 0[/tex].
     
  5. Jul 1, 2009 #4
    Would you say that your definition is equivalent to:

    for all a,b,x and [tex]ae^x + be^{2x}=0[/tex] ,then a=b=0 ?

    or in a more combact form:

    for all a,b,x [tex][ae^x + be^{2x} =0\Longrightarrow ( a=b=0)][/tex]
     
  6. Jul 1, 2009 #5

    HallsofIvy

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    "For all a, b" makes no sense if "a= b= 0".

    Just "if, for all x, [tex]ae^x + be^{2x}=0[/tex] ,then a=b=0" or
    "(for all x [tex]ae^x + be^{2x} =0)\Longrightarrow ( a=b=0)[/tex]"
     
  7. Jul 2, 2009 #6
    You mean that a and ,b cannot have any other value apart from zero??

    In that case you do not have to prove anything because ae^x + be^2x =0

    OR given any a,b,x and if [tex] ae^x + be^{2x} =0[/tex] ,then you can prove that the only value a and b can take is zero??
     
  8. Jul 2, 2009 #7
    Incomplete. How about this...

    [tex]\text{for all } a,\; \text{for all } b\; [(\text{for all } x,\; ae^x + be^{2x} =0)\Longrightarrow ( a=b=0)][/tex]
     
  9. Jul 2, 2009 #8

    HallsofIvy

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    Given that [itex]ae^x+ be^{2x}= 0[/itex] for all x, then, taking x= 0, [itex]a+ b= 0[/itex]. Since [itex]ae^x+ be^{2x}= 0[/itex] for all x, differentiating with respect to x, [itex]ae^x+ 2be^{2x}= 0[/itex] for all x and, setting x= 0 again, [itex]a+ 2b= 0[/itex]. Subtracting the first equation from the second, (a+ 2b)- (a+ b)= b= 0. Putting b= 0 in either equation, a= 0. Is that what you are asking?

    NO!! It makes NO sense say "for all a, a= 0"! "If, for all x, ax= 0, then a= 0." It makes no sense to say "for all a" there.
     
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