On 02/24/2013 02:54 PM, ?? ?? wrote:
> Thank you, Jaime.
>
> It works!
> How about this problem? (I change an equation just a little.)
>
> Executing
> float(rectform(solve(x^3-3*x^2-2*x+1,x)));
For your information this polynomial has discriminant D=257
there is no factorization of the polynomial, so we have casus
irreducibilis here.
see http://en.wikipedia.org/wiki/Casus_irreducibilis
Andre
> results
> [x=.3433795689528897-2.220446049250313*10^-16*%i,x=2.220446049250313*10^-16*%i-.8342431843139217,x=1.110223024625157* 10^-16*%i+3.490863615361032]
> though
> realroots(x^3-3*x^2-2*x+1);
> is
> [x=-27992557/33554432,x=11521907/33554432,x=117133945/33554432].
>
> I want to know general methods to avoid a small imaginary error.
>
> Thanks,
> -
> ICHIKAWA, Yuji
>
> On 2013/02/24, at 21:49, Jaime Villate <villate at fe.up.pt> wrote:
>
>> On 02/24/2013 11:24 AM, ?? ?? wrote:
>>> Executing
>>> solution: solve(x^3-x^2+1=0, x), rectform;
>>> ...
>>>
>>> The second one in solution is a real root.
>>> To know the concrete value, executing
>>> solution, float;
>>> results
>>> [x=0.744861766619749*%i+.8774388331233485,x=-.7548776662466969,x=.8774388331233485-0.744861766619749*%i].
>>>
>>> I want to get this result by single command.
>>> But executing
>>> solution: solve(x^3-x^2+1=0, x), rectform, float;
>>> results
>>> [x=.7448617666197254*%i+.8774388331233376,x=-8.480870326997723*10^-17*%i-.7548776662466756,x=.8774388331233378-
>>> .7448617666197253*%i].
>>>
>>> The second one includes a very small imaginary part.
>>>
>>> How can I avoid it?
>> Hi,
>> try
>> solution: float( rectform( solve( x^3-x^2+1=0, x)));
>>
>> which allows you to make the three things in the order you want them to be made. In your original command
>>
>> solution: solve(x^3-x^2+1=0, x), rectform, float;
>>
>> you used function ev(), which is more unpredictable in the way the modifiers rectform and float will be applied.
>>
>> Regards,
>> Jaime
>>
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