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Math Help - Derivatives/Logarithms/Elasticity Question

  1. #1
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    Question Derivatives/Logarithms/Elasticity Question

    Hey all, I'm working on this problem for my economics. Although it's economics and not "math", the steps involved shouldn't require any knowledge of economics. It involves logarithms and elasticities and derivatives, but I'm having a bit of trouble getting the solution. We've been given the answer (Theta/1-Theta). The question is asking to "derive the elasticity of steady-state output per worker (y) with respect to the saving rate (s)". Now for the record, if it helps, that in the steady state, the growth rate of y is 0 (it is constant). This is the equation as presented.



    Now the first step I was told by my professor was to take the logs. So I did that, and I come up with. For the record, I replaced theta with "a" and the depreciation with "x", as I didn't have access to those at the moment.



    From here though, differentiating with respect to ln(s), I cannot for the life of me figure out where to get the (a/(1-a)) from.

    Also of note, the formula for elasticity for example with X and Y, is (X/Y) * (dY/dX) if it helps to work with the first equation.

    Any help is appreciated.
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  2. #2
    MHF Contributor ebaines's Avatar
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    I get an answer of  \frac { \theta } { 1 - \theta} (n+ \delta) . Here's how:

    You are looking for  \frac S Y \frac {dY} {dS} . Assuming that the quantities  n, \delta, \theta are all constants then:

     <br />
Y = \left(\frac S {n + \delta} \right)^{\frac {\theta} {1-\theta}}<br />
    <br />
\frac {dY} {dS} = \frac {\theta} { 1 - \theta} \left(\frac S {n + \delta} \right)^{\frac {\theta} {1-\theta}-1}<br />
= \frac {\theta} { 1 - \theta} \left( \frac S {n + \delta}\right)^{\frac {2\theta-1} {1-\theta}}<br />

    So:

    <br />
\frac S Y \frac {dY} {dS} = \frac S {(\frac S {n+\delta})^{\frac {\theta} { 1- \theta}} }\cdot \frac {\theta} { 1 - \theta} (\frac S {n + \delta})^{\frac {2\theta-1} {1-\theta}}<br />
= \frac {\theta} {1 - \theta} \left( \frac S {n + \delta}\right) ^ {\frac {\theta - 1} {1 - \theta}} S<br />
= \frac {\theta} {1 - \theta} \left( \frac S {n + \delta}\right) ^ {-1} S<br />
    <br />
= \frac {\theta} {1- \theta} (n+ \delta)<br />

    Are you sure that there isn't supposed to be that  (n + \delta) term in the answer?
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  3. #3
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    Thank you for your work. And for the record, I am 100% positive it should be theta/1-theta. I've found powerpoints and such with that figure given, it's just that I've had a tough time finding derivations of it. They usually just give it to their students as given. I asked the professor for more guidance, but he insists on never giving out answers. This is what he had to say...

    "Hi,

    Your first steps are correct: you could just use (S/Y) (dY/dS). Note that the derivative of Y with respect to S could be written, in this case, as (theta/(1-theta))*Y/S. And then the result you are looking for follows pretty much immediately.

    Jon"

    I'm still just totally dumbfounded on what he means though. I'm thinking this is one of those things that once figured out will be so readily apparent I'll *facepalm*
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  4. #4
    MHF Contributor ebaines's Avatar
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    I realize now that I made an error in my post - I forget that the derivative should inlcude the term  (n + \delta ) - Doh! So it should be like this:

     <br />
\frac {dY} {dS} =\left( \frac {\theta} { 1 - \theta} \right) \left(\frac S {n + \delta} \right)^{\frac {2 \theta -1} {1-\theta}} \frac 1 {(n+ \delta)}<br />

    So:

    <br />
\frac S Y \frac {dY} {dS} = S \left( \frac {n+\delta} S \right)^{\frac {\theta} { 1- \theta} } \left( \frac {\theta} { 1 - \theta} \right) \left( \frac S {n+ \delta} \right) ^{\frac {2\theta-1} {1-\theta}}\left( \frac 1 {n+ \delta} \right)<br />

    <br />
= \frac {\theta} {1- \theta}<br />

    Sorry for the error in my earlier post.
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  5. #5
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    Wow, thank you so much ebaines! I definitley look forward to using this forum in the future! Now in the second part he asks for elasticity of output with respect to n, but now that I've kind of seen the way to go about it, I think I may be able to figure it out!

    Thanks again
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  6. #6
    MHF Contributor ebaines's Avatar
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    Upon reflection, the solution would have been much easier to arrive at by simply considering the function:

    <br />
y = A x^n<br />

    Then

    <br />
\frac {dy} {dx} = nAx^{n-1} = \frac n x Ax^n = \frac {ny} x<br />

    So:

    <br />
\frac x y \frac {dy} {dx} = n<br />

    In your problem you have  n = \theta / (1 - \theta ) .
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  7. #7
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    A different example

    Let's say that you're given the function and you're asked to find the elasticity of t in respects to x.

    T=100x/20+x+49

    I know that it's a bit easier then the previous example, but I'm stuck on it.
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