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List Archives >  Maple User Group List Archive >  Archive by date >  This Month By Date >  This Month By Topic

[MUG] solving inequality

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[MUG] solving inequality
Author: M A Suzen    Posted: Mon, 28 Oct 2002 14:36:30 +0100

>> From: M A Suzen "mehmet.suzen"

Hello All,
Are there any workaround to get a solution for below in equality
in terms of n? solve seems is not able to solve it.

Cheers,

Mehmet
> restart:
>
ineqn:=n*ln(n)-(1-alpha)*n*ln((1-alpha)*n)-n*alpha*ln(n*alpha)+ln(B)>=-4(1-
alpha)*n*ln(2):
> B:=2^((3-2*alpha)*n)-2^(4*(1-alpha)*n):
> e:=evala(ineqn):
> c:=simplify(e,symbolic):
> assume(alpha,real):solve(c,n);

[MUG] Re: solving inequality
Author: Maple User Group    Posted: Fri, 1 Nov 2002 10:17:54 -0500 (

>> From: Maple User Group "maple_gr"

| >> From: M A Suzen "mehmet.suzen"
| Are there any workaround to get a solution for below in equality
| in terms of n? solve seems is not able to solve it.
| ineqn:=n*ln(n)-(1-alpha)*n*ln((1-alpha)*n)-n*alpha*ln(n*alpha)+ln(B)>=-4(1-
| alpha)*n*ln(2):
| > B:=2^((3-2*alpha)*n)-2^(4*(1-alpha)*n):
| > e:=evala(ineqn):
| > c:=simplify(e,symbolic):
| > assume(alpha,real):solve(c,n);

-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-

Date: Tue, 29 Oct 2002 21:23:39 -0800 (PST)
From: Robert Israel "israel"
To: "maple-list"
Subject: solving inequality

I suspect you want n > 0, and also to assume 1/2 < alpha < 1, not just
alpha real, in order to make ln((1-alpha)*n), ln(n*alpha) and ln(B)
real when n > 0. It's easy to prove that rhs(c) - lhs(c) is an increasing
function of n for such alpha, with limits -infinity and +infinity
as n -> 0+ and infinity respectively, so there is exactly one n (say
n_0) that makes the inequality an equality, and the solution of the
inequality is n = [n_0, infinity).

Not surprisingly, Maple doesn't find a closed-form solution for n_0; I
doubt there is such a closed-form solution. For particular values of
alpha in the interval 1/2 < alpha < 1, you can use fsolve.

Robert Israel "israel"
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2



-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-

Date: Wed, 30 Oct 2002 09:35:11 -0500
To: "maple-list"
From: "Gerald A. Edgar" "edgar"
Subject: solving inequality

Why do you think there is a formula for the solution? Subtract to
get an inequality (formula) >= 0, then exponentiate to get
(formula)>=1. Simplifying the formula, we get something
of the form U^n-V^n >=1, where U and V depend on alpha.
If we solve U^n-V^n=1, then we can solve the inequality,
since U^n-V^n is a continuous function of the real variable n.
But I see no reason to think there is a nice formula
for solutions of U^n-V^n=1.

If I am not mistaken,

1
U = 128 --------------------------------------------
(1 - alpha) alpha (2 alpha)
(1 - alpha) alpha 2



1
V = 256 --------------------------------------------
(1 - alpha) alpha (4 alpha)
(1 - alpha) alpha 2



--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/


-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-

From: "Daniel Willard" "willardd"
To: "maple-list"
Subject: solving inequality
Date: Wed, 30 Oct 2002 10:49:24 -0500

I think you are going to find this is in the "too hard" box for Maple. Your
"c" can be recast into n*(Complex number in alpha) <= ln( sum of
exponentials in n) and this seems to say that for some range(s) of alpha n
will be complex. Try first plotting something.

Previous by date: [MUG] Set And List Problem,  UKC Account
Next by date: [MUG] filter transformations -- basic algebraic manipulations, Ben Nevile
Previous thread: [MUG] Maple 8 under Debian GNU/Linux, Curtis Vinson
Next thread: [MUG] filter transformations -- basic algebraic manipulations, Ben Nevile



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