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Thread: Trouble with exponents and differentiation

  1. #1
    Member garymarkhov's Avatar
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    Trouble with exponents and differentiation

    I think I have an error in my calculation, because I should get $\displaystyle s= \alpha$ at the end but I do not. Can someone please check to see where I've gone wrong?

    Suppose you have: $\displaystyle c=(1-s) \left[ \frac{s}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}$

    Rewrite as

    $\displaystyle
    c=(1-s)s^\frac{\alpha}{1-\alpha} \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}
    $

    Expand the brackets:

    $\displaystyle
    c=(s^\frac{\alpha}{1-\alpha}-s^\frac{1}{1-\alpha}) \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}
    $

    To find golden rule consumption, take the partial:

    $\displaystyle
    \frac{\partial c}{\partial s}=\left( \frac{\alpha}{1-\alpha}s^\frac{2 \alpha -1}{1-\alpha}-\frac{1}{1-\alpha}s^\frac{\alpha}{1-\alpha} \right) \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}
    $

    Set that beastie equal to zero so we can find the local max:

    $\displaystyle
    \frac{\partial c}{\partial s}=\left( \frac{\alpha}{1-\alpha}s^\frac{2 \alpha -1}{1-\alpha}-\frac{1}{1-\alpha}s^\frac{\alpha}{1-\alpha} \right) \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha} = 0
    $

    Divide through by $\displaystyle \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}$ to get

    $\displaystyle
    \left( \frac{\alpha}{1-\alpha}s^\frac{2 \alpha -1}{1-\alpha}-\frac{1}{1-\alpha}s^\frac{\alpha}{1-\alpha} \right)= 0
    $


    $\displaystyle
    \frac{\alpha}{1-\alpha}s^\frac{2 \alpha -1}{1-\alpha} = \frac{1}{1-\alpha}s^\frac{\alpha}{1-\alpha}
    $

    Multiply both sides by $\displaystyle 1-\alpha$

    $\displaystyle
    \alpha s^\frac{2 \alpha -1}{1-\alpha} = s^\frac{\alpha}{1-\alpha}
    $

    Raise both sides to the power of $\displaystyle 1-\alpha$

    $\displaystyle
    \alpha s^{2 \alpha -1} = s^{\alpha}
    $

    Multiply both sides by $\displaystyle s^{-\alpha}$

    $\displaystyle
    \alpha s^{\alpha -1} = 1
    $

    $\displaystyle
    s^{\alpha -1} = \frac{1}{\alpha }
    $

    And finish by raising both sides by $\displaystyle \frac{1}{\alpha -1}$


    $\displaystyle
    s = \left( \frac{1}{\alpha } \right)^\frac{1}{\alpha -1}
    $

    We can rewrite as

    $\displaystyle
    s = \alpha^\frac{1}{1- \alpha}
    $

    But that hardly makes it look closer to $\displaystyle s=\alpha$, which I think is the right answer. What have I done wrong?
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  2. #2
    MHF Contributor ebaines's Avatar
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    Your problem is here:

    Quote Originally Posted by garymarkhov View Post
    Raise both sides to the power of $\displaystyle 1-\alpha$

    $\displaystyle
    \alpha s^{2 \alpha -1} = s^{\alpha}
    $
    When you raise a function ab^c to the power of d, you should get:
    $\displaystyle
    (\alpha b^c)^d = \alpha ^d*b^{cd}
    $
    but you have:
    $\displaystyle
    (\alpha b^c)^d = \alpha *b^{cd}
    $
    Note that the $\displaystyle \alpha $ term is missing an exponent.
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  3. #3
    Member garymarkhov's Avatar
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    Dear Lord,

    Thank you for Mr. ebaines. Amen.

    Love,

    Gary
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