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Re: comparing implicit 0 with machine floats
*To*: mathgroup at smc.vnet.net
*Subject*: [mg71205] Re: [mg71108] comparing implicit 0 with machine floats
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Fri, 10 Nov 2006 06:38:09 -0500 (EST)
*References*: <11095316.1163039426494.JavaMail.root@eastrmwml01.mgt.cox.net> <F3C6DEC6-64C4-45D4-98CC-8A79F52F3B0A@mimuw.edu.pl>
In fact, things look even more surprising (to me anyway).
In[1]:=
Simplify[Sqrt[2] + Sqrt[3] - Sqrt[5 + 2*Sqrt[6]] == 0.]//Timing
Out[1]=
{0.046486 Second,True}
while
In[2]:=
Timing[Simplify[Sqrt[2] + Sqrt[3] - Sqrt[5 + 2*Sqrt[6]]]]
Out[2]=
{0.0008739999999998194*Second, Sqrt[2] + Sqrt[3] -
Sqrt[5 + 2*Sqrt[6]]}
Note that the first evaluation took a non-negligible amount of time.
One reason why I am surprised is that I would have thought one would
need to increase $MaxExtraPrecision to something like
Accuracy[0.]
307.653
Andrzej Kozlowski
On 9 Nov 2006, at 12:10, Andrzej Kozlowski wrote:
> Yes indeed, it does. This is rather curious because Simplify does
> not actually reduce Sqrt[2] + Sqrt[3] - Sqrt[5 + 2*Sqrt[6]] to 0
> (since Simplify does not make use of RootReduce) one needs
> FullSimplify to do that.
>
> In[51]:=
> Simplify[Sqrt[2] + Sqrt[3] - Sqrt[5 + 2*Sqrt[6]]]
>
> Out[51]=
> Sqrt[2] + Sqrt[3] - Sqrt[5 + 2*Sqrt[6]]
>
> In[52]:=
> FullSimplify[Sqrt[2] + Sqrt[3] - Sqrt[5 + 2*Sqrt[6]]]
>
> Out[52]=
> 0
>
> Andrzej Kozlowski
>
> On 9 Nov 2006, at 11:30, Bob Hanlon wrote:
>
>> Using Simplify will force the comparison
>>
>> Sqrt[2] + Sqrt[3] - Sqrt[5 + 2*Sqrt[6]] == 2.//Simplify
>>
>> False
>>
>> Sqrt[2] + Sqrt[3] - Sqrt[5 + 2*Sqrt[6]] == N[10^3]//Simplify
>>
>> False
>>
>>
>> Bob Hanlon
>>
>> ---- Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
>>> Consider the following:
>>>
>>>
>>> Sqrt[2] + Sqrt[3] - Sqrt[4 + 2*Sqrt[6]] == 2.
>>>
>>> False
>>>
>>> No problem here. Now let's make a small change
>>>
>>> In[17]:=
>>> Sqrt[2] + Sqrt[3] - Sqrt[5 + 2*Sqrt[6]] == 2.
>>>
>>> Out[17]=
>>>
>>> Sqrt[2] + Sqrt[3] - Sqrt[5 + 2*Sqrt[6]] == 2.
>>>
>>> In fact the expression on the LHS is exactly 0:
>>>
>>> In[19]:=
>>> RootReduce[Sqrt[2]+Sqrt[3]-Sqrt[5+2*Sqrt[6]]]
>>>
>>> Out[19]=
>>>
>>> The curious thing is that if you try a comparison between a zero of
>>> this kind and any machine float, however large, Mathematica 5.1 will
>>> return the original input:
>>>
>>> In[20]:=
>>> Sqrt[2] + Sqrt[3] - Sqrt[5 + 2*Sqrt[6]] == N[10^3]
>>>
>>> Out[20]=
>>> Sqrt[2] + Sqrt[3] - Sqrt[5 + 2*Sqrt[6]] == 1000.
>>>
>>> yet if the number of the left hand side is altered, however
>>> slightly,
>>> the comparison will be made:
>>>
>>>
>>> Sqrt[2] + Sqrt[3] - Sqrt[5 + 2*Sqrt[6+1/10^20]] == 2.
>>>
>>> False
>>>
>>> This suggests that Mathematica actually did perform a computation of
>>> the left hand side in the examples where it just returned the input
>>> and having discovered that it could not determine if the LHS is an
>>> exact zero decided "not to answer the question". But this seems
>>> quite
>>> unreasonable; after all it is not being asked if the LHS is an exact
>>> 0, or even an approximate 0, but if it is an approximate large
>>> number
>>> like 1000., and this it certainly can decide.
>>>
>>> I believe this used to be handled differently (better?) in older
>>> versions of Mathematica but I no longer have any installed to check.
>>>
>>> Andrzej Kozlowski
>>>
>>
>
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