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Math Help - Stuck on transformation question

  1. #1
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    Stuck on transformation question

    f(x) = \frac{e^{2x}}{x^2}

    Question is, show that, for a certain value of k,

    f(x - 1) = k(\frac{1}{x-1})^2 f(x)

    My effort:

    f(x-1) = \frac{e^{2(x-1)}}{(x-1)^2}

    = \frac{e^{2x-2}}{(x-1)^2}

    = \frac{e^{2x}}{e^2(x-1)^2} = \frac{x^2}{e^2} (\frac{1}{x-1})^2 \frac{e^{2x}}{x^2}

    Therefore k = \frac{x^2}{e^2}

    But apparently the answer should be \frac{1}{e^2}

    What have I done wrong?
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  2. #2
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    e^(i*pi)'s Avatar
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    Re: Stuck on transformation question

    Quote Originally Posted by angypangy View Post
    f(x) = \frac{e^{2x}}{x^2}

    Question is, show that, for a certain value of k,

    f(x - 1) = k(\frac{1}{x-1})^2 f(x)

    My effort:

    f(x-1) = \frac{e^{2(x-1)}}{(x-1)^2}

    = \frac{e^{2x-2}}{(x-1)^2}

    = \frac{e^{2x}}{e^2(x-1)^2}

     = \frac{x^2}{e^2} (\frac{1}{x-1})^2 \frac{e^{2x}}{x^2} .

    Therefore k = \frac{x^2}{e^2}

    But apparently the answer should be \frac{1}{e^2}

    What have I done wrong?
    I get the same answer as you, I have put my working in a spoiler.

    Spoiler:
    f(x-1) = \dfrac{e^{2x}}{e^2(x-1)^2} = \dfrac{e^{2x}}{e^2} \cdot \left(\dfrac{1}{x-1}\right)^2


    From the question: f(x-1) = k \left(\dfrac{1}{x-1}\right)^2 f(x)

     \dfrac{e^{2x}}{e^2} \cdot \left(\dfrac{1}{x-1}\right)^2 = k \left(\dfrac{1}{x-1}\right)^2 \cdot \dfrac{e^{2x}}{x^2}

    Immediately we can see that e^{2x} \text{  and  } \left(\dfrac{1}{x-1}\right)^2 cancel:

    \dfrac{1}{e^2} = \dfrac{k}{x^2}

    k = \dfrac{x^2}{e^2}
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  3. #3
    Member vernal's Avatar
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    Re: Stuck on transformation question

    Quote Originally Posted by angypangy View Post
    f(x) = \frac{e^{2x}}{x^2}

    Question is, show that, for a certain value of k,

    f(x - 1) = k(\frac{1}{x-1})^2 f(x)



    My effort:

    f(x-1) = \frac{e^{2(x-1)}}{(x-1)^2}

    = \frac{e^{2x-2}}{(x-1)^2}

    = \frac{e^{2x}}{e^2(x-1)^2} = \frac{x^2}{e^2} (\frac{1}{x-1})^2 \frac{e^{2x}}{x^2}

    Therefore k = \frac{x^2}{e^2}

    But apparently the answer should be \frac{1}{e^2}

    What have I done wrong?

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  4. #4
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    Re: Stuck on transformation question

    vernal,

    \frac{e^{2x}}{(x-1)^2} \neq f(x) = \frac{e^{2x}}{x^2}
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  5. #5
    Member vernal's Avatar
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    Re: Stuck on transformation question

    Quote Originally Posted by Deveno View Post
    vernal,

    \frac{e^{2x}}{(x-1)^2} \neq f(x) = \frac{e^{2x}}{x^2}
    oh sory

    Last edited by vernal; December 22nd 2011 at 08:10 AM.
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  6. #6
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    Re: Stuck on transformation question

    Wow, I beat the book That feels good!
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  7. #7
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    Re: Stuck on transformation question

    yes, and that leads to the same result as the OP and e^(i*pi) obtained, that k = \frac{x^2}{e^2}
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