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Math Help - Algebra with exponents

  1. #1
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    Algebra with exponents

    Hi all-

    I need to see the steps to get

    (1- 1/2 + 1/2*e^t)^25 * (1- 1/2 + 1/2*e^(-t))^25 * e^(25t)

    to be (1- 1/2 + 1/2*e^t)^50

    Can anyone help, i am having trouble seeing how?

    Brian
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  2. #2
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    Quote Originally Posted by B_Miner View Post
    Hi all-

    I need to see the steps to get

    (1- 1/2 + 1/2*e^t)^25 * (1- 1/2 + 1/2*e^(-t))^25 * e^(25t)

    to be (1- 1/2 + 1/2*e^t)^50

    Can anyone help, i am having trouble seeing how?

    Brian
    Probably there is a typo in your question because 1 - 1/2 = 1/2 . I don't understand why you didn't simplify this sum first.

    Nevertheless I show you what to do:

    \left(1-\frac12 + \frac12 \cdot e^t\right)^{25} \cdot \left(1-\frac12 + \frac12 \cdot e^{-t}\right)^{25} \cdot e^{25t} = \left(\frac12 + \frac12 \cdot e^t\right)^{25} \cdot \left(e^t \left(\frac12 + \frac12 \cdot e^{-t}\right) \right)^{25}   =  \left(\frac12 + \frac12 \cdot e^t\right)^{25} \cdot \left(\frac12 \cdot e^t+ \frac12  \right)^{25} = \left(\frac12 + \frac12 \cdot e^t\right)^{50}
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  3. #3
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    Thanks very much. I struggle trying to "see the end" and making the right moves.

    PS the reason it was not simplified is that the result should be the mgf of the binomial random variable.
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  4. #4
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    Quote Originally Posted by B_Miner View Post
    [snip]the result should be the mgf of the binomial random variable.
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