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 August 4th, 2011, 02:39 PM #1 Newbie   Joined: Aug 2011 Posts: 4 Thanks: 0 is there an analytical solution to (x+a)^g-(x-a)^g=p*a*x Hello guys, I'm working on an economics model where a worker has a concave incentive contract and a production function y=x+mu where x is the effort provided and mu is an error term. Anyway, when I want to calculate the worker's optimal level of effort I end up with something like: (x+a)^g-(x-a)^g=p*a*x with g<1 and x>0, p>0, a>0. When I assign values to the parameters a,g and p I obtain a graphical solution. Nevertheless, I would need an analytical one to use in a further step of the model. I'm struggling with that since two days and I start to fear that there is no analytical solution. Am I right? Do you have any suggestion? Thanks, Giancarlo August 4th, 2011, 06:29 PM #2 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: is there an analytical solution to (x+a)^g-(x-a)^g=p*a*x I don't see an analytical solution either. I don't suppose an approximate solution would help? You could do a series expansion if it would. August 8th, 2011, 10:19 AM   #3
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Re: is there an analytical solution to (x+a)^g-(x-a)^g=p*a*x

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 Originally Posted by CRGreathouse I don't see an analytical solution either. I don't suppose an approximate solution would help? You could do a series expansion if it would.
Thank you for the answer. Nevertheless, could you be more precise? What do you mean with a "series expansion"? August 8th, 2011, 01:20 PM #4 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: is there an analytical solution to (x+a)^g-(x-a)^g=p*a*x August 8th, 2011, 02:12 PM   #5
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Re: is there an analytical solution to (x+a)^g-(x-a)^g=p*a*x

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 Originally Posted by CRGreathouse http://en.wikipedia.org/wiki/Taylor_series
An approximation would imply, to be precise enough, that I know all the parameters and then more or less where the x I'm looking for is located in order to approximate my function in the neighborhood of that point. Isn't it? August 8th, 2011, 03:23 PM   #6
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Re: is there an analytical solution to (x+a)^g-(x-a)^g=p*a*x

Quote:
 Originally Posted by sciacallojo2 An approximation would imply, to be precise enough, that I know all the parameters and then more or less where the x I'm looking for is located in order to approximate my function in the neighborhood of that point. Isn't it?
Depends on the function; some are very sensitive to those factors and others not as much. You may find techniques like the secant method, Newton's method, or bisection to be useful. August 9th, 2011, 10:20 AM   #7
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Re: is there an analytical solution to (x+a)^g-(x-a)^g=p*a*x

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Originally Posted by CRGreathouse
Quote:
 Originally Posted by sciacallojo2 An approximation would imply, to be precise enough, that I know all the parameters and then more or less where the x I'm looking for is located in order to approximate my function in the neighborhood of that point. Isn't it?
Depends on the function; some are very sensitive to those factors and others not as much. You may find techniques like the secant method, Newton's method, or bisection to be useful.
I will look it up. Thanks August 9th, 2011, 11:37 AM #8 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: is there an analytical solution to (x+a)^g-(x-a)^g=p*a*x Wish I could be more helpful...! Tags agxagpax, analytical, solution 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 munkifisht Calculus 2 June 13th, 2012 12:02 PM finfanillo Linear Algebra 0 March 21st, 2012 10:14 PM julian21 Geometry 2 July 17th, 2010 01:06 PM jangolobow Calculus 1 April 2nd, 2010 08:51 AM ask2 Calculus 2 November 13th, 2009 02:08 AM

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