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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 s= \alpha at the end but I do not. Can someone please check to see where I've gone wrong?

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

    Rewrite as

    <br />
        c=(1-s)s^\frac{\alpha}{1-\alpha} \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}<br />

    Expand the brackets:

    <br />
        c=(s^\frac{\alpha}{1-\alpha}-s^\frac{1}{1-\alpha}) \left[ \frac{1}{n+x+\delta} \right]^\frac{\alpha}{1-\alpha}<br />

    To find golden rule consumption, take the partial:

    <br />
        \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}<br />

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

    <br />
        \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<br />

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

    <br />
        \left( \frac{\alpha}{1-\alpha}s^\frac{2 \alpha -1}{1-\alpha}-\frac{1}{1-\alpha}s^\frac{\alpha}{1-\alpha} \right)= 0<br />


    <br />
        \frac{\alpha}{1-\alpha}s^\frac{2 \alpha -1}{1-\alpha} = \frac{1}{1-\alpha}s^\frac{\alpha}{1-\alpha}<br />

    Multiply both sides by 1-\alpha

    <br />
        \alpha s^\frac{2 \alpha -1}{1-\alpha} = s^\frac{\alpha}{1-\alpha}<br />

    Raise both sides to the power of 1-\alpha

    <br />
        \alpha s^{2 \alpha -1} = s^{\alpha}<br />

    Multiply both sides by s^{-\alpha}

    <br />
        \alpha s^{\alpha -1} = 1<br />

    <br />
        s^{\alpha -1} = \frac{1}{\alpha }<br />

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


    <br />
        s = \left( \frac{1}{\alpha } \right)^\frac{1}{\alpha -1} <br />

    We can rewrite as

    <br />
        s = \alpha^\frac{1}{1- \alpha} <br />

    But that hardly makes it look closer to 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 1-\alpha

    <br />
\alpha s^{2 \alpha -1} = s^{\alpha}<br />
    When you raise a function ab^c to the power of d, you should get:
    <br />
(\alpha b^c)^d = \alpha ^d*b^{cd}<br />
    but you have:
    <br />
(\alpha b^c)^d = \alpha *b^{cd}<br />
    Note that the  \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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