Re: Solving inequalities.
- To: mathgroup at smc.vnet.net
- Subject: [mg45209] Re: Solving inequalities.
- From: bobhanlon at aol.com (Bob Hanlon)
- Date: Sat, 20 Dec 2003 05:56:37 -0500 (EST)
- References: <bruqak$sue$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
$Version
4.2 for Mac OS X (August 22, 2002)
Off[Solve::ifun];
eqn = Simplify[x^(a+1)*(1-x)^(b-1) == (x^a)*(1-x)^b];
Simplify[Solve[eqn, x], 0<x<1 && 1<b<a]
{{x -> 1/2}, {x -> 0}, {x -> 1}}
f[p_] := p^k * (1-p)^(n-k);
eqn = Simplify[f'[p] == 0];
Simplify[Solve[eqn, p], 0<p<1 && 1<k<n-1]
{{p -> 0}, {p -> 1}, {p -> k/n}}
Bob Hanlon
In article <bruqak$sue$1 at smc.vnet.net>, gtsavdar at auth.gr (George) wrote:
<< I want to solve this equation: f'(p)=0.
With the restrictions 0<p<1 and 2<k<n and
f(p) = (p^k) * (1-p)^(n-k).
How this can be done inside Mathematica 5.0 or 4.2?
The following procedure doesn't work.
I define the f[p_] and then i enter:
Needs["Algebra`InequalitySolve`"]
InequalitySolve[f'[p] == 0 && 1 > p > 0 && k > 2 && n > k, p]
but no solution is found.
--------------------------------------------------------------------
The above problem is "the same" with the equation:
x^(a+1) * (1-x)^(b-1) = (x^a) * (1-x)^b with the restrictions
0<x<1 and 1<b<a.
So this equation is the same with the: x = 1-x <=> x=0.5
But how can i solve this in Mathematica 5.0 or 4.2?
The:
Needs["Algebra`InequalitySolve`"]
InequalitySolve[x^(a+1) * (1-x)^(b-1) == (x^a) * (1-x)^b && 1 > x > 0
&& a > b > 1, x]
doesn't work.
>><BR><BR>