[Date Index]
[Thread Index]
[Author Index]
Re: Number of roots from Solve?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg48444] Re: [mg48423] Number of roots from Solve?
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Sun, 30 May 2004 06:12:05 -0400 (EDT)
*References*: <200405290706.DAA20383@smc.vnet.net> <DCCD3B24-B159-11D8-9808-000A95B4967A@mimuw.edu.pl>
*Sender*: owner-wri-mathgroup at wolfram.com
I should have added that you can indeed predict the number of roots.
In[1]:=
e := (1/z)*((2 + z^2*(b - 1))/(b + 1))^
((1 + b)/(2*(b - 1)))
In[2]:=
a = 10; b = Rationalize[1.4];
In[3]:=
Numerator[Together[e - 10]]
Out[3]=
z^6 + 15*z^4 + 75*z^2 - 2160*z + 125
Hence you will get 6 roots.
In[4]:=
a = 10; b = Rationalize[1.05];
In[5]:=
Limit[Numerator[Together[e - 10]]/z^41, z -> Infinity]
Out[5]=
Sqrt[41]
So this time you will indeed get 41 roots.
Andrzej
On 29 May 2004, at 19:20, Andrzej Kozlowski wrote:
> 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
>>
>> --
>> This message has been scanned for viruses and
>> dangerous content by the Cranfield MailScanner, and is
>> believed to be clean.
>>
>>
>
Prev by Date:
**Re: how can I solve a function Erfc**
Next by Date:
**How to get the real part of an integral?**
Previous by thread:
**Re: Number of roots from Solve?**
Next by thread:
**Re: Number of roots from Solve?**
| |