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Math Help - Induction Proof

  1. #1
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    Induction Proof

    Hi,

    Is this correct?

    Prove by induction that \frac{d^n}{dx^n}(xe^{-x}) = (-1)^n(x - n)e^{-x} for all positive integers n.

    \frac{d^n}{dx^n}(xe^{-x})  =  (-1)^n(x - n)e^{-x}

    \frac{d}{dx}(xe^{-x}) = -xe^{-x} + e^{-x}

    = e^{-x(1 - x) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\text{...1}

    Prove true for n = 1

    (-1)^n(x - n)e^{-x}

    (-1)^1(x - 1)e^{-x}

    (-1)(x - 1)e^{-x}

    (-x + 1)e^{-x} \,\,\,\,\,\,\,\,\,\,\text{...1}

    Prove true for n = k

    (-1)^n(x - n)e^{-x}   =  (-1)^k(x - k)e^{-x}

    Prove true for n = k + 1}

    (-1)^n\,(x - n)e^{-x}\,=\,(-1)^k\,+\,1[x - (k + 1)]e^{-x}

    = (-1)^k\,+\,1(x - k - 1)e^{-x}

    = (-1)^k\,(-1)^1\,(x - k - 1)e^{-x}

    (-1)^{n}\,(x - n)e^{-x} = (-1)^{k}\,(-x + k + 1)e^{-x}
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  2. #2
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    Quote Originally Posted by Hellbent View Post
    Hi,

    Is this correct?

    Prove by induction that \frac{d^n}{dx^n}(xe^{-x}) = (-1)^n(x - n)e^{-x} for all positive integers n.

    \frac{d^n}{dx^n}(xe^{-x})  =  (-1)^n(x - n)e^{-x}

    \frac{d}{dx}(xe^{-x}) = -xe^{-x} + e^{-x}

    = e^{-x(1 - x) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\text{...1}

    Prove true for n = 1

    (-1)^n(x - n)e^{-x}

    (-1)^1(x - 1)e^{-x}

    (-1)(x - 1)e^{-x}

    (-x + 1)e^{-x} \,\,\,\,\,\,\,\,\,\,\text{...1}

    Prove true for n = k

    (-1)^n(x - n)e^{-x}   =  (-1)^k(x - k)e^{-x}

    Prove true for n = k + 1}

    (-1)^n\,(x - n)e^{-x}\,=\,(-1)^k\,+\,1[x - (k + 1)]e^{-x} this and the following lines are expressed incorrectly

    = (-1)^k\,+\,1(x - k - 1)e^{-x}

    = (-1)^k\,(-1)^1\,(x - k - 1)e^{-x}

    (-1)^{n}\,(x - n)e^{-x} = (-1)^{k}\,(-x + k + 1)e^{-x}
    \displaystyle\frac{d^n}{dx^n}\left(xe^{-x}\right)=(-1)^n(x-n)e^{-x}


    Base Case

    n=1

    \displaystyle\frac{d}{dx}\left(xe^{-x}\right)=e^{-x}-xe^{-x}=(1-x)e^{-x}=(-1)^1(x-1)e^{-x}


    P(k)

    \displaystyle\frac{d^k}{dx^k}\left(xe^{-x}\right)=(-1)^k(x-k)e^{-x}


    P(k+1)

    \displaystyle\frac{d^{k+1}}{dx^{k+1}}\left(xe^{-x}\right)=(-1)^{k+1}\left[x-(k+1)\right]e^{-x}


    Proof that P(k+1) will be valid if P(k) is

    \displaystyle\frac{d}{dx}(-1)^k(x-k)e^{-x}=(-1)^k\frac{d}{dx}(x-k)e^{-x}=(-1)^k\left[-(x-k)e^{-x}+e^{-x}(1)\right]

    =(-1)^k(-1)\left[(x-k)e^{-x}-e^{-x}\right]=(-1)^{k+1}\left[x-(k+1)\right]e^{-x}
    Last edited by Archie Meade; November 24th 2010 at 10:09 AM. Reason: tidied notation
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