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Math Help - [SOLVED] linear independence exponential functions

  1. #1
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    [SOLVED] linear independence exponential functions

    Given three exponential functions e^x , e^2x , e^3x , show they are linearly independent.
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    Quote Originally Posted by rick81 View Post
    Given three exponential functions e^x , e^2x , e^3x , show they are linearly independent.
    Do you know the polynomials  y,y^2 and  y^3 are linearly independent? Hmm.. why am I talking about that?
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    Quote Originally Posted by rick81 View Post
    Given three exponential functions e^x , e^2x , e^3x , show they are linearly independent.
    Say that ae^x + be^{2x}+ce^{3x} = 0 for all x\in \in \mathbb{R}.
    Then taking the derivative, ae^x + 2be^{2x} + 3ce^{3x} = 0.
    Take the derivative again, ae^x + 4be^{2x}+9ce^{3x}=0.

    For any value of x this gives us a 3\times 3 system of equation.
    The determinant of this system (for this value of x) is:
    \left| \begin{array}{ccc} e^x & e^{2x} & e^{3x} \\ e^x & 2e^{2x} & 3e^{3x} \\ e^x & 4e^{2x} & 9e^{3x} \end{array} \right|

    Show this determinat is not the zero function. Therefore, it would imply that for some x we have a homogenous system with a non-vanishing determinant and hence the trivial solution a=b=c=0 is the only such solution.
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    that's a pretty nifty trick differentiating it to get 3 equations so you have a 3x3 matrix! How's this one:

    a e^x+be^{2x}+ce^{3x}=0

    e^x(a+be^x+ce^{2x})=0

    a+be^x+ce^{2x}=0

    a=-be^x-ce^{2x}

    a+0e^x+0e^{2x}=-be^x-ce^{2x}

    Comparing coefficients on both sides gives a=b=c=0.

    ?

    P.S: 600th post!
    Last edited by Showcase_22; May 26th 2009 at 02:04 AM. Reason: 600th post!!
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    Quote Originally Posted by Showcase_22 View Post
    that's a pretty nifty trick differentiating it to get 3 equations so you have a 3x3 matrix! How's this one:

    a e^x+be^{2x}+ce^{3x}=0

    e^x(a+be^x+ce^{2x})=0

    a+be^x+ce^{2x}=0

    a=-be^x-ce^{2x}

    a+0e^x+0e^{2x}=-be^x-ce^{2x}

    Comparing coefficients on both sides gives a=b=c=0.
    "Comparing coefficients" like that only works if the functions are independent. You are assuming what you want to prove!

    ?

    P.S: 600th post!
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    Quote Originally Posted by Showcase_22 View Post
    that's a pretty nifty trick differentiating it to get 3 equations so you have a 3x3 matrix! How's this one:

    a e^x+be^{2x}+ce^{3x}=0

    e^x(a+be^x+ce^{2x})=0

    a+be^x+ce^{2x}=0

    a=-be^x-ce^{2x}

    a+0e^x+0e^{2x}=-be^x-ce^{2x}

    Comparing coefficients on both sides gives a=b=c=0.

    ?

    P.S: 600th post!

    Showcase ,i will to extent your proof and try to solve the problem without using determinants.

    So you proved :

     a +be^x + ce^{2x} = 0

    Now differentiate both sides and you get:

     be^x + 2ce^{2x}=0\Longrightarrow e^x( b + 2ce^x) =0\Longrightarrow b + 2ce^x = 0

    Differentiate again both sides and you get:

     2ce^x =0\Longrightarrow c=0

    From the equation :  b + 2ce^x=0 you get b=0

    AND from the equation :  a + be^x + ce^{2x} =0 you get a=0
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    You could, of course, do this without differentiating at all (although that is easier!).

    You want to find ae^x+ be^{2x}+ ce^{3x}= 0. Take x= 0 so you have a+ b+ c= 0. Take x= 1 so you have ae+ be^2+ ce^3= 0. Take x= 2 so you have ae^2+ be^4+ ce^6= 0. You now have three linear equations to solve for a, b, and c. Do that and show that a= b= c= 0 is the only solution.
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