Hello John

thank you for this.  I tried the same thing on mathematica,
which managed to simplify 'c' back to 'a'.

I don't quite understand the culture of sage-support yet.
Is commenting on mathematica's ability to do a particular
task a useful thing to say?  Or does it just annoy everyone?

best wishes

Robin



On Thu, Jul 7, 2011 at 8:55 PM, John Cremona <john.crem...@gmail.com> wrote:
> With d=c-a, not even d.simplify_radical() gives 0.
>
> Simplifying "nested radicals" is a notoriously hard problem in
> symbolic computer algebra.  As this example shows (unless there are
> other tricks to try which I do not know about), Sage's symbolic system
> is not up to examples like this.
>
> As an algebraist I would use a different approach.  Note that
> sage: type(a)
> <type 'sage.symbolic.expression.Expression'>
> but as these quantities are all algebraic one can also use
> sage: QQbar
> Algebraic Field
>
> for example
>
> sage: QQbar(a) == QQbar(c)
> True
>
> Behind the scenes, this is checking first that a and c have the same
> minimal polynomial:
> sage: QQbar(a).minpoly()
> x^4 - 4*x^3 - 4*x^2 + 16*x - 8
> sage: QQbar(c).minpoly()
> x^4 - 4*x^3 - 4*x^2 + 16*x - 8
> and also (by using numerical approximations to whatever precision is
> necessary) that they are the same root of that poly.
>
> John Cremona
>
>
> On Jul 5, 11:38 pm, robin hankin <hankin.ro...@gmail.com> wrote:
>> Hi.  I am having difficulty using sage to manipulate surds.
>>
>> Consider:
>>
>> a = 1 + sqrt(2) + sqrt(3)
>> b= (a^2).expand()
>> c = sqrt(b)
>>
>> Then 'y' should be equal to 'a'.
>>
>> But, given 'y' I cannot make sage return the simple form.
>> Trying
>>
>> y.simplify_full()
>>
>> doesn't do what I want.
>>
>> How do I make sage recognize that a=c, other than using n()?
>>
>> --
>> Robin Hankin
>> Uncertainty Analyst
>> hankin.ro...@gmail.com
>
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-- 
Robin Hankin
Uncertainty Analyst
hankin.ro...@gmail.com

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