Square root of a square
Mariano Suárez-Alvarez
Mathematica evaluates FullSimplify[Sqrt[x^2]] to Sqrt[x^2], while it
tells me that FullSimplify[Sqrt[x^2] == x] is True.
Are these the expected answers?
-- m
Adam Strzebonski
>
> FindInstance[Sqrt[x^2] != x, x]
>
> {{x -> -(109/5) + (11*I)/5}}
>
>
> Andrzej Kozlowski
>
It is a bug in V7.0. The function used by FullSimplify to
simplify equations involving holonomic functions does not
handle branch cuts correctly. It can be disabled with
In[1]:= Unprotect[Holonomic`HolonomicFullSimplify];
Clear[Holonomic`HolonomicFullSimplify];
In[2]:= FullSimplify[Sqrt[x^2] == x]
2
Out[2]= Sqrt[x ] == x
Best Regards,
Adam Strzebonski
Wolfram Research
Leonid Shifrin
The first result (that Sqrt[x^2] does not simplify further without
additional assumptions) was expected. However, the second result (that
Mathematica simplifies Sqrt[x^2] == x to True) looks like a bug to me.
Regards,
Leonid
Bob Hanlon
FullSimplify[Sqrt[x^2] == x]
True
This is wrong since for real x, Sqrt is always non-negative and x could be =
negative.
x = -3;
Sqrt[x^2] == x
False
Bob Hanlon
---- "Mariano Su=C3=A1rez-Alvarez" <mariano.su...@gmail.com> wrote:=
=============
Andrzej Kozlowski
On 27 Feb 2010, at 09:13, Mariano Su=E1rez-Alvarez wrote:
> Hi all,
>
> Mathematica evaluates FullSimplify[Sqrt[x^2]] to Sqrt[x^2], while it
> tells me that FullSimplify[Sqrt[x^2] == x] is True.
> Are these the expected answers?
>
> -- m
>
The second certainly not the answer I would expect, in fact this looks
Christoph Lhotka
1) Simplify[Sqrt[x^2] == x, x \[Element] Reals]
2) Simplify[Sqrt[x^2] == x, x \[Element] Complexes]
3) Simplify[Sqrt[x^2] == x, x < 0]
works fine, while
1) FullSimplify[Sqrt[x^2] == x, x \[Element] Reals]
2) FullSimplify[Sqrt[x^2] == x, x \[Element] Complexes]
3) FullSimplify[Sqrt[x^2] == x, x < 0]
will return a wrong result in the 2nd case.
The problem seems to be only in version 7, not in version 6 or version 5.
chr
> :
Andrzej Kozlowski
Quote:
> It is a bug in V7.0. The function used by FullSimplify to
> simplify equations involving holonomic functions does not
> handle branch cuts correctly. It can be disabled with
>
> In[1]:= Unprotect[Holonomic`HolonomicFullSimplify];
> Clear[Holonomic`HolonomicFullSimplify];
>
> In[2]:= FullSimplify[Sqrt[x^2] == x]
>
> 2
> Out[2]= Sqrt[x ] == x
>
> Best Regards,
>
> Adam Strzebonski
> Wolfram Research
Andrzej Kozlowski
On 1 Mar 2010, at 09:41, Christoph Lhotka wrote:
> hi, this is something, which is really needed to be discussed:
>
> 1) Simplify[Sqrt[x^2] == x, x \[Element] Reals]
> 2) Simplify[Sqrt[x^2] == x, x \[Element] Complexes]
> 3) Simplify[Sqrt[x^2] == x, x < 0]
>
> works fine, while
>
> 1) FullSimplify[Sqrt[x^2] == x, x \[Element] Reals]
> 2) FullSimplify[Sqrt[x^2] == x, x \[Element] Complexes]
> 3) FullSimplify[Sqrt[x^2] == x, x < 0]
>
> will return a wrong result in the 2nd case.
>
> The problem seems to be only in version 7, not in version 6 or version =
5.
>
> chr
>
>
> Andrzej Kozlowski wrote:
>> On 27 Feb 2010, at 09:13, Mariano Su=E1rez-Alvarez wrote:
>>
>>
>>> Hi all,
>>>
>>> Mathematica evaluates FullSimplify[Sqrt[x^2]] to Sqrt[x^2], while it
>>> tells me that FullSimplify[Sqrt[x^2] == x] is True.
>>> Are these the expected answers?
>>>
>>> -- m
>>>
>>>
>>
>> The second certainly not the answer I would expect, in fact this =
looks to me like a serious (perhaps even "shocking") bug. It certainly =
janj...@gmail.com
> Andrzej Kozlowski wrote:
> > On 27 Feb 2010, at 09:13, Mariano Su=E1rez-Alvarez wrote:
>
> >> Hi all,
>
> >> Mathematica evaluates FullSimplify[Sqrt[x^2]] to Sqrt[x^2], while it
> >> tells me that FullSimplify[Sqrt[x^2] == x] is True.
> >> Are these the expected answers?
>
> >> -- m
>
>
> > FindInstance[Sqrt[x^2] != x, x]
>
> > {{x -> -(109/5) + (11*I)/5}}
>
> > Andrzej Kozlowski
>
> simplify equations involving holonomic functions does not
> handle branch cuts correctly. It can be disabled with
>
> In[1]:= Unprotect[Holonomic`HolonomicFullSimplify];
> Clear[Holonomic`HolonomicFullSimplify];
>
> In[2]:= FullSimplify[Sqrt[x^2] == x]
>
> 2
> Out[2]= Sqrt[x ] == x
>
> Best Regards,
>
> Adam Strzebonski
> Wolfram Research
This type of error seems to be more wide-spread in Mathematica.
The following Mathematica code gives two answers, one of which is
incorrect, even after entering the unprotect and clear commands above:
DSolve[{y'[x] == 2 y[x] (x Sqrt[y[x]] - 1), y[0] == 1}, y[x], x]
{{y[x] -> 1/(-1 + 2 E^x - x)^2}, {y[x] -> 1/(1 + x)^2}}
The error appears for me in all Mathematica versions I could test,
that is 5.2, 6.0 and 7.0. The first solution can only be imagined to
be correct if you assume Sqrt[x^2] ==x.
Regards,
Jan Jitse Venselaar
Andrzej Kozlowski
I think this is a different issue and may not involve any bug. This looks to me just the "usual" problem with parasite solutions that inevitably appear in functions like Solve when equations involving parameters are involved. I do not think DSolve attempts to verify the correctness of the solutions it returns (just as Solve does not in parametric cases), so there is no alternative to doing it by hand. However, when doing so its certainly a good idea to clear Holonomic`HolonomicFullSimplify, although in this particular case it does not matter.
In[1]:= sols =
DSolve[{Derivative[1][y][x] == 2*y[x]*(x*Sqrt[y[x]] - 1),
y[0] == 1}, y, x];
In[2]:= (FullSimplify[#1, x > 0] & )[
{Derivative[1][y][x] == 2*y[x]*(x*Sqrt[y[x]] - 1), y[0] == 1} /.
sols]
Out[2]= {{False, True}, {True, True}}
Andrzej Kozlowski=