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Math Help - Proof, linear independence

  1. #1
    Senior Member I-Think's Avatar
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    Proof, linear independence

    Let f,g,\in{F(R,R)} be the functions defined by f(t)=e^{rt} and g(t)=e^{st}, where r\neq{s}. Prove that f and g are linearly independent in F(R,R).

    Assume ae^{rt}+ be^{st}=0
    ae^{rt} = -be^{st}
    ln(a)+rt = ln(-b) +st
    ln(\frac{-a}{b})=t(s-r)

    For our assumption to be true, the values of a and b are dependent on the value of t, therefore there is no universal value of a or b, except 0, for which af+bg=0
    Thus, these functions are linearly independent

    This proof is correct?
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  2. #2
    Super Member girdav's Avatar
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    You don't know if a or b are positive. But you can take t=0 in the equation, go back to the equation and take the derivative on both side, then take t=0.
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  3. #3
    Senior Member I-Think's Avatar
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    Actually, we do know that one of the two constants is negative and the other is positive. t\in{R} and e^x is never negative, so for the equation to be true, one of the constants must be negative and the other positive.
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  4. #4
    Super Member girdav's Avatar
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    Yes, it would be perfect if you add "We assume a>0, if it's not the case we switch r and s.
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  5. #5
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    ae^{rt}+be^{st} is the 0 function, not just the number 0. this means this equation holds for any and all values of t.

    in particular, it holds for t = 0, whence b = -a.

    thus we have a(e^{rt} - e^{st}) = 0 for all t. if a is not 0, then

    e^{rt} = e^{st} for all real t, taking logs of both sides gives r = s, a contradiction, so a = 0 and b = -a = 0.

    EDIT: why do i do it this way? because in the original poster's proof, he takes the log of a number which he is trying to prove is 0.

    now, by the previous two comments, you can actually assume that a > 0, and b < 0, and then the original poster's argument works,

    but it is not easy to see that ln(a/-b) = t(s - r) immediately implies a contradiction. i feel it is more straight-forward to use the fact that R

    is a field (and thus has no zero divisors).
    Last edited by Deveno; June 4th 2011 at 01:03 PM.
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