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A function is given as

$ln (y) = ln(\alpha)-\frac{\lambda}{\gamma}ln(\delta L^{-\gamma}+(1-\delta)K^{-\gamma})$

I need to find the second order Taylor $ln(y)$ around $\gamma=0$.

How can it be done since putting zero for gamma gives infinity?

Thank you

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  • $\begingroup$ How large are L and K, and is $\delta$ small? $\endgroup$ Dec 15 '14 at 7:07
  • $\begingroup$ $\alpha>0$, $0<\delta<1$ and $\gamma \geq -1$ $\endgroup$
    – Elekko
    Dec 15 '14 at 7:11
  • $\begingroup$ L and K are just some constant $\endgroup$
    – Elekko
    Dec 15 '14 at 7:11
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This is my take on a first-order approximation. If you want more, you can extend it to second order.

For small $g$ (instead of $\gamma$), $A^{-g} =e^{-g \ln A} \sim 1-g \ln A $.

Therefore, using $\ln(1-z) \sim -z$ (and $d$ instead of $\delta$),

$\begin{array}\\ \frac1{g}\ln(d L^{-g}+(1-d)K^{-g}) &\sim \frac1{g}\ln(d (1-g\ln L) + (1-d)(1-g \ln K))\\ &= \frac1{g}\ln(d -dg\ln L + (1-d)-(1-d)(g \ln K))\\ &= \frac1{g}\ln(1 -g(d\ln L +(1-d) \ln K))\\ &\sim \frac1{g}(-g(d\ln L +(1-d) \ln K)))\\ &\sim -(d\ln L +(1-d) \ln K)))\\ \end{array} $

I'll leave it at this.

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