My Math Forum Help with simplification

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 October 8th, 2016, 04:41 AM #1 Member   Joined: Sep 2013 Posts: 83 Thanks: 0 Help with simplification Can someone show me how the author arrived at the following results?
 October 8th, 2016, 06:32 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 First, this is almost impossible to read: it appears to be a full page of text in one jpg file. Second, and more important, virtually nothing is defined. I am guessing that m = 2. I am guessing that $\phi (\bar z)$ has $\alpha$ and $\beta$ as arguments. I suspect that the production function is in Cobb-Douglas form. I do know that the CES function is the result of Problem 2.6, but I do not know what Problem 2.6 is. Third, you have shown nothing of how far you have got in this problem Do you expect someone to work the entire thing out for you?
 October 10th, 2016, 07:28 AM #3 Member   Joined: Sep 2013 Posts: 83 Thanks: 0 take a look at the last re-arranging, supposed to be easy. i dont understand how a2 can appear twice, when its just one time in the first equation you dont have to know what each letter means, its just a "simple " rearranging. Last edited by Ku5htr1m; October 10th, 2016 at 08:28 AM.
 October 10th, 2016, 08:30 AM #4 Member   Joined: Sep 2013 Posts: 83 Thanks: 0 i began by simplifying to this expression $\displaystyle \left( a_{{1}}{z}^{b-1}[{a_{{1}}}^{-{b}^{-1}}[w_{{1}}]+{a_ {{2}}}^{-{b}^{-1}}[w_{{2}}]]{q}^{1-b}=w_{{1}} \right)$
October 12th, 2016, 10:35 AM   #5
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Quote:
 Originally Posted by Ku5htr1m take a look at the last re-arranging, supposed to be easy. i dont understand how a2 can appear twice, when its just one time in the first equation you dont have to know what each letter means, its just a "simple " rearranging.
It is not a simple re-arrangement, but it is correct.

Let's start by simplifying the notation with:

$t = z_1^*,\ \gamma = \beta - 1,\ \delta = \dfrac{\beta}{\gamma},\ and\ s = \alpha_1^{-(1/ \gamma)} * w_1^{\delta} + \alpha_2^{-(1/ \gamma)} * w_2^{\delta}.$

$Note\ 1:\ \delta * \gamma = \dfrac{\beta}{\gamma} * \gamma = \beta.$

$Note\ 2:\ \delta - \dfrac{1}{\gamma} = \dfrac{\beta}{\gamma} - \dfrac{1}{\gamma} = \dfrac{\beta - 1}{\beta - 1} = 1.$

$Note\ 3:Obviously,\ \delta - \dfrac{1}{\gamma} = 1 \implies 1 - \delta = -\ \dfrac{1}{\gamma}.$

$Note\ 4:\ \dfrac{\alpha_1^{\delta}}{w_1^{\delta}} * s = \left ( \dfrac{\alpha_1^{\delta}}{\cancel{w_1^{\delta}}} * \alpha_1^{-(1/ \gamma)} * \cancel{w_1^{\delta}} \right )+ \left ( \dfrac{\alpha_1^{\delta}}{w_1^{\delta}} * \alpha_2^{-(1/ \gamma)} * w_2^{\delta} \right) \implies$

$\dfrac{\alpha_1^{\delta}}{w_1^{\delta}} * s = \alpha_1^{\{\delta - (1 / \gamma)\}} + \alpha_2^{-(1/ \gamma )} * \left ( \dfrac{\alpha_1 * w_2}{w_1} \right)^{\delta} = \alpha_1^1 + \alpha_2^1 * \alpha_2^{- \delta} * \left ( \dfrac{\alpha_1 * w_2}{w_1} \right)^{\delta} \implies$

$\dfrac{\alpha_1^{\delta}}{w_1^{\delta}} * s = \alpha_1 + \alpha_2 \left (\dfrac{\alpha_1}{\alpha_2} * \dfrac{w_2}{w_1} \right)^{\delta}.$

$w_1 = \alpha_1 * t^{\gamma} * s^{(1 / \delta )} * q^{- \gamma} \implies log(w_1) = log(\alpha_1 * t^{\gamma} * s^{1 / \delta )} * q^{- \gamma}) \implies$

$log(w_1) = log(\alpha_1) + \gamma * log(t) + \dfrac{1}{\delta} * log(s) - \gamma * log(q) \implies$

$\delta * log(w_1) = \delta * log(\alpha_1) + \beta * log(t) + log(s) - \beta * log(q) \implies$

$\beta * log(t) = \delta * \{log(w_1 - \alpha_1)\} - log(s) + \beta * log(q) = log \left ( \dfrac{w_1^{\delta}}{\alpha_1^{\delta}} \right ) + log \left ( \dfrac{1}{s} \right) + \beta * log(q) \implies$

$log(t) = \dfrac{1}{\beta} * log \left ( \dfrac{w_1^{\delta}}{\alpha_1^{\delta}} * \dfrac{1}{s} \right) + log(q) \implies$

$t = \left ( \dfrac{w_1^{\delta}}{\alpha_1^{\delta}} * \dfrac{1}{s} \right)^{(1/ \beta)}q = \left ( \dfrac{\alpha_1^{\delta}}{w_1^{\delta}} * s \right)^{-(1/ \beta)} q \implies$

$t = \left \{ \alpha_1 + \alpha_2 \left (\dfrac{\alpha_1}{\alpha_2} * \dfrac{w_2}{w_1} \right)^{\delta} \right \}^{-(1/ \beta )}q.$

Quite ugly.

 October 12th, 2016, 02:38 PM #6 Member   Joined: Sep 2013 Posts: 83 Thanks: 0 Greatly appreciated the answer!! The only part i did not get is in at the ende where you take $\displaystyle \delta (ln(w_{1}-a_{1})) \rightarrow ln(w^{\delta}/a^{\delta})$. PsWritibg on my ipad so i cant write it out the entire expression/step) Last edited by Ku5htr1m; October 12th, 2016 at 02:41 PM.
October 12th, 2016, 04:36 PM   #7
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Quote:
 Originally Posted by Ku5htr1m Greatly appreciated the answer!! The only part i did not get is in at the ende where you take $\displaystyle \delta (ln(w_{1}-a_{1})) \rightarrow ln(w^{\delta}/a^{\delta})$. PsWritibg on my ipad so i cant write it out the entire expression/step)
$-\ log(x) = (-\ 1) log(x) = log(x^{-1}) = log \left( \dfrac{1}{x} \right )\ for\ all\ real\ x > 0.$

$So\ \delta * \{ log(w_1) - log( \alpha_1 ) \} = \{ \delta * log(w_1)\} - \{\delta * log (a_1)\} \implies$

$\delta * \{ log(w_1) - log( \alpha_1 ) \} = log(w_1^{\delta}) - log(\alpha_1^{\delta}) = log(w_1^{\delta}) + log \left ( \dfrac{1}{\alpha_1^{\delta}} \right ) \implies$

$\delta * \{ log(w_1) - log( \alpha_1 ) \} = log \left (w_1^{\delta} * \dfrac{1}{\alpha_1^{\delta}} \right ) = log \left ( \dfrac{w_1^{\delta}}{\alpha_1^{\delta}} \right ).$

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