Thanks to Emmanuel Charpentier for the solution.
But the trick is S1[0]^2 that needs human works.
Le samedi 18 octobre 2014 11:58:37 UTC+2, Emmanuel Charpentier a écrit :
>
> Well, after a bit of sleep, the solution was (semi-)obvious :
>
> sage: var("w,t")
> (w, t)
> sage: E1=-(1/2*sqrt((4*w+1)+1))*t+w==0
> sage: S1=E1.solve(w)
> sage: S1
> [w == 1/2*t*sqrt(4*w + 2)]
> sage: S2=((S1[0])^2).solve(w)
> sage: S2
> [w == 1/2*t^2 - 1/2*sqrt(t^2 + 2)*t, w == 1/2*t^2 + 1/2*sqrt(t^2 + 2)*t]
>
> A numerical check is enough to exclude S2[0] : it's a spurious "solution"
> introduced when we solved a^2==0 instead of a==0.
>
> Now, to check that S2[1] is indeed a solution of E1, Sage isn't enough :
>
> sage: E2=E1.subs(S2[1])
> sage: E2
> 1/2*t^2 - 1/2*sqrt(2*t^2 + 2*sqrt(t^2 + 2)*t + 2)*t + 1/2*sqrt(t^2 + 2)*t
> == 0
> sage: bool(E2)
> False
>
> The trouble is with the "biradical" term. Let's isolate it (by adding it
> to both sides of E2) :
>
> sage: E3=E2+1/2*sqrt(2*t^2 + 2*sqrt(t^2 + 2)*t + 2)*t
> sage: E3
> 1/2*t^2 + 1/2*sqrt(t^2 + 2)*t == 1/2*sqrt(2*t^2 + 2*sqrt(t^2 + 2)*t + 2)*t
> sage: bool(E3)
> False
> sage: bool(E3^2)
> True
>
> Again, we got Sage so check she equality of the squares of both sides of
> E3 (hence E1). But in this case, we didn't "introduce spurious solutions".
> So, QED (I think...).
>
> Bonus question : is there a way to get Sage (or Maxima) to recognize the
> existence of "biradical" terms and act accordingly ?
>
> HTH,
>
> --
> Emmanuel Charpentier
>
> Le vendredi 17 octobre 2014 23:00:28 UTC+2, Emmanuel Charpentier a écrit :
>>
>>
>>
>> Le vendredi 17 octobre 2014 16:37:55 UTC+2, vdelecroix a écrit :
>>>
>>> 2014-10-17 10:09 UTC, Emmanuel Charpentier <[email protected]>:
>>> > Ahem !
>>> >
>>> > On one machine :
>>> >
>>> > sage: sage.version.version
>>> > '6.4.beta4'
>>> > sage: var("w,t")
>>> > (w, t)
>>> > sage: solve(-(1/2*sqrt((4*w+1)+1))*t+w==0,w)
>>> > [w == 1/2*t*sqrt(4*w + 2)]
>>>
>>> This solution is implicit. This is the problem.
>>>
>>
>> Indeed. But one can get solutions for an equation close to this implicit
>> solution :
>>
>> sage: E1=-(1/2*sqrt((4*w+1)+1))*t+w==0
>> sage: S1=E1.solve(w)
>> sage: S1
>> [w == 1/2*t*sqrt(4*w + 2)]
>> sage: S2=(S1[0]^2).solve(w)
>> sage: S2
>> [w == 1/2*t^2 - 1/2*sqrt(t^2 + 2)*t, w == 1/2*t^2 + 1/2*sqrt(t^2 + 2)*t]
>>
>> A bit of numerics (plotting for t \in [-1,1]) shows that [S2[0] isn't
>> acceptable but that S2[1] might be. I didn't (yet) succeed in proving this.
>>
>> Any idea ?
>>
>> --
>> Emmanuel Charpentier
>>
>>
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