Re: Number of roots from Solve?

• To: mathgroup at smc.vnet.net
• Subject: [mg48443] Re: [mg48423] Number of roots from Solve?
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Sun, 30 May 2004 06:12:04 -0400 (EDT)
• References: <200405290706.DAA20383@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```I think having inexact coefficients does not make any difference here
since I am pretty sure Solve first applies Rationalize to everything.
In fact I don&t think it would not make any sense to do otherwise,
given that Solve uses only algebraic and not numerical techniques.
Since the equations are non-polynomial I can see nothing at all strange
about them having very different numbers of roots.

Andrzej Kozlowski

Since the equation is non-polynomial I do not see an

On 29 May 2004, at 16:06, Goyder Dr HGD wrote:

> I am working with the expression
>
> a =.; b =.; e = (1/z)((2 + z^2 (b - 1))/(b + 1))^((1 + b)/(2(b - 1)))
>
> and was not surprised to find a number of complex and real roots (6)
> for
> the particular case
>
> a = 10.; b = 1.4; s = Solve[a == e]; {Length[s], s}
>
> I was surprised that the number of roots changed to 41 for the case
>
> a = 10.; b = 1.05; s = Solve[a == e]; {Length[s], s}
>
> With
>
> a = 10.; b = 1.03; s = Solve[a == e]; {Length[s], s}
>
> I never got an answer (or the calculation takes longer than my
> patience).
>
> Is it possible to predict the number of roots that an expression like e
> above will produce?
> Does Mathematica find all the roots, particularly as we are dealing
> with
> approximate numbers?
>
> Many thanks
>
> Hugh Goyder
>
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```

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