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Re: Re: Wrong limit?
I find it curious (and perhaps slightly disappointing) that replacing
Reduce by Resolve seems to fail (or at least it outruns my patience
while Reduce does not). This is puzzling because Resolve and not Reduce
is supposed to be "the principal tool" for quantifier elimination and,
according to the documentation "Resolve is in effect automatically
applied by Reduce." Presumably Reduce applies some additional
transformations not available to Resolve, which make it possible for it
to succeed where Resolve fails (or takes far too long), but what could
they be?
Andrzej Kozlowski
On 12 Nov 2009, at 19:59, Maxim wrote:
> On Nov 10, 5:03 am, DrMajorBob <btre... at austin.rr.com> wrote:
>> One could hope that Mathematica covered all cases in every symbolic
>> calculation, but it's not possible, and if it were, it would frequently be
>> inefficient.
>>
>> Consider the simple quadratic:
>>
>> Solve[a x^2 + b x + c == 0, x]
>>
>> {{x -> (-b - Sqrt[b^2 - 4 a c])/(
>> 2 a)}, {x -> (-b + Sqrt[b^2 - 4 a c])/(2 a)}}
>>
>> That's wrong when a == 0, just as in your Limit problem.
>>
>> Reduce is more complete, if that's what we want... but do we really want
>> every computation cluttered to the maximum degree?
>>
>> Reduce[a x^2 + b x + c == 0, x]
>>
>> (a != 0 && (x == (-b - Sqrt[b^2 - 4 a c])/(2 a) ||
>> x == (-b + Sqrt[b^2 - 4 a c])/(2 a))) || (a == 0 && b
> != 0 &&
>> x == -(c/b)) || (c == 0 && b == 0 && a == 0)
>>
>> Bobby
>>
>> On Fri, 06 Nov 2009 04:14:48 -0600, wiso <gtu2... at alice.it> wrote:
>>> Look at this:
>>
>>> Limit[(x^2 - a^2)/(5 x^2 - 4 a x - a^2), x -> a]
>>> Mathematica answer = 1/3
>>
>>> this is ok for a !=0, but if a = 0 the value is
>>
>>> Limit[(x^2 - a^2)/(5 x^2 - 4 a x - a^2) /. a -> 0, x -> 0]
>>> 1/5
>>
>> --
>> DrMajor... at yahoo.com
>
> Interestingly, the original problem can actually be solved using
> Reduce:
>
> In[1]:= Reduce[
> ForAll[eps, eps > 0, Exists[del, del > 0, ForAll[x,
> Implies[0 < Abs[x - a] < del,
> Abs[(x^2 - a^2)/(5 x^2 - 4 a x - a^2) - M] < eps]]]], Reals]
>
> Out[1]= (a < 0 && M == 1/3) || (a == 0 && M == 1/5) || =
(a > 0=
> && M ==
> 1/3)
>
> One just has to be careful to make sure the expressions do not become
> indeterminate, otherwise the meaning of the quantifiers is not really
> well defined. E.g.,
>
> In[2]:= Reduce[ForAll[x, x >= 0, 1/x > a]]
>
> Out[2]= False
>
> In[3]:= Reduce[ForAll[x, x >= 0, 1/x > 0]]
>
> Out[3]= True
>
> Maxim Rytin
> m.r at inbox.ru
>
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