Re: Generated parameters

• To: mathgroup at smc.vnet.net
• Subject: [mg116282] Re: Generated parameters
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Wed, 9 Feb 2011 02:12:46 -0500 (EST)

```Because there is more than one solution.

sol = Reduce[v/d^(i/-2) == vP/d^(iP/-2) && (iP == -1 || iP == 0),
{iP, vP}, Reals, Backsubstitution -> True];

Simplify[sol, {C[1] == 0, d > 0}]

(iP == 0 && d^(i/2)*v == vP) || (iP == -1 && d^((1 + i)/2)*v == vP)

Or use Solve

Simplify[Solve[v/d^(i/-2) == vP/d^(iP/-2) && (iP == -1 || iP == 0),
{iP, vP}, Reals], d > 0]

{{iP -> -1, vP -> d^((1 + i)/2)*v}, {iP -> 0, vP -> d^(i/2)*v}}

Bob Hanlon

---- olfa <olfa.mraihi at yahoo.fr> wrote:

=============
Hi Mathematica community,
For this system:
Reduce[v/d^(i/-2) == vP/d^(iP/-2) &&
(iP == -1 || iP == 0), {iP, vP}, Reals, Backsubstitution -> True]

the output is:
(iP == 0 && C[1] \[Element] Integers && d < 0 && i == -2 C[1] &&
vP == (-d)^-C[1] v)
||
(iP == 0 && C[1] \[Element] Integers &&
C[1] <= -1 && d < 0 && i == -2 C[1] &&
vP == (-d)^-C[1] v)
||
(iP == 0 && C[1] \[Element] Integers &&
d < 0 && i == -2 C[1] && vP == -(-d)^-C[1] v)
||
(iP == 0 &&
C[1] \[Element] Integers && C[1] <= -1 && d < 0 && i == -2 C[1] &&
vP == -(-d)^-C[1] v)
||
(d > 0 && iP == 0 &&
vP == d^(i/2) v)
||
(d > 0 && iP == -1 && vP == d^(1/2 + i/2) v)

1)Why Reduce generates parameters for this example?
2) and how to avoid them knowing that for this system the solution
that I hope to get is
( iP == 0 && vP == d^(i/2) v)
||
( iP == -1 && vP == d^(1/2 + i/2) v)

Thank you.

```

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