User Name Remember Me? Password

 Calculus Calculus Math Forum

 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
Senior Member

Joined: May 2016
From: USA

Posts: 1,310
Thanks: 551

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 $\displaystyle \delta (ln(w_{1}-a_{1})) \rightarrow ln(w^{\delta}/a^{\delta})$. Ps Writibg 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
Senior Member

Joined: May 2016
From: USA

Posts: 1,310
Thanks: 551

Quote:
 Originally Posted by Ku5htr1m Greatly appreciated the answer!! The only part i did not get is in at the ende where you take $\displaystyle \delta (ln(w_{1}-a_{1})) \rightarrow ln(w^{\delta}/a^{\delta})$. Ps Writibg on my ipad so i cant write it out the entire expression/step)
$-\ log(x) = (-\ 1) log(x) = log(x^{-1}) = log \left( \dfrac{1}{x} \right )\ for\ all\ real\ x > 0.$

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

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

$\delta * \{ log(w_1) - log( \alpha_1 ) \} = log \left (w_1^{\delta} * \dfrac{1}{\alpha_1^{\delta}} \right ) = log \left ( \dfrac{w_1^{\delta}}{\alpha_1^{\delta}} \right ).$ Tags simplification Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Chikis Elementary Math 2 September 11th, 2013 10:11 AM arron1990 Algebra 3 August 14th, 2012 04:50 AM Jakarta Algebra 3 August 4th, 2012 10:45 AM Salient Algebra 3 October 21st, 2008 01:17 PM arron1990 Calculus 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      