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August 4th, 2011, 02:39 PM   #1
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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
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August 4th, 2011, 06:29 PM   #2
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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.
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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"?
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August 8th, 2011, 01:20 PM   #4
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Re: is there an analytical solution to (x+a)^g-(x-a)^g=p*a*x

http://en.wikipedia.org/wiki/Taylor_series
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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?
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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.
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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
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August 9th, 2011, 11:37 AM   #8
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Re: is there an analytical solution to (x+a)^g-(x-a)^g=p*a*x

Wish I could be more helpful...!
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