Re: Solve vs. nonlinearity
- To: mathgroup at smc.vnet.net
- Subject: [mg118307] Re: Solve vs. nonlinearity
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Thu, 21 Apr 2011 03:10:40 -0400 (EDT)
Good point! Try this, then:
myL = w1*n + w2*k + g*(q0^r - (a*n^r + (1 - a) k^r));
grad = D[myL, {{n, k, g}}];
s1 = First@Quiet@Solve[grad == 0, {n, k, g}];
grad /. % // PowerExpand // Simplify;
s2 = First@Quiet@Solve[% == 0, k];
s = Thread[{n, g, k} -> ({n /. s1 /. s2, g /. s1 /. s2, k /. s2})] //
PowerExpand // Simplify
grad /. % // PowerExpand // Simplify
{n -> (-1)^(-1/r) a^(1/(1 - r)) q0 (w1 - a w1)^(1/(-1 + r)) w2^(1/(
1 - r)) (-1 + a -
a^(1/(1 - r)) (w1 - a w1)^(r/(-1 + r)) w2^(-(r/(-1 + r))))^(-1/
r), g -> ((-1)^(-1/r) q0^(1 - r)
w2 (-1 + a -
a^(1/(1 - r)) (w1 - a w1)^(r/(-1 + r)) w2^(-(r/(-1 + r))))^((-1 +
r)/r))/((-1 + a) r),
k -> (-1)^(-1/r)
q0 (-1 + a -
a^(1/(1 - r)) (w1 - a w1)^(r/(-1 + r)) w2^(-(r/(-1 + r))))^(-1/r)}
{0, 0, 0}
No guarantees, of course! PowerExpand may be assuming too much for your
application.
Bobby
On Wed, 20 Apr 2011 03:26:35 -0500, Alan G Isaac <alan.isaac at gmail.com>
wrote:
>>> Is there a simple way to approach getting Mathematica to
>>> produce a solution in the following problem?
>>> (Without assigning to r.)
>>>
>>> myL = w1*n + w2*k + g*(q0^r - (a*n^r + (1-a) k^r));
>>> grad = D[myL, {{n, k, g}}]
>>> Solve[grad == 0, {n, k, g}]
>
>
> On 4/17/2011 8:00 PM, DrMajorBob wrote:
>> Umm... evaluate the code?
>
>> {{n -> ((((k^(1 - r))^(1/(1 - r)))^(-1 + r) (w1 - a w1))/(a w2))^(
>> 1/(-1 + r)), g -> -((k^(1 - r) w2)/((-1 + a) r))}}
>
>
>
> Note that this does not solve for k.
> (Or possibly I'm missing your point.)
>
> Thanks,
> Alan
>
--
DrMajorBob at yahoo.com