The expression y = (1-x)/(1-x*cos(t)) is, as given, undefined whenever
x*cos(t)=1, for example at (x,t)=(1,0).
When x=1 it simplifies to 0/(1-cos(t)), which equals 0 except where
cos(t)=1 where it is undefined but has a limiting value of 0.
When t=0 it simplfies to (1-x)/(1-x), which equals 1 except when x=1
where it is undefined, but has a limiting value of 1.
So you get different limits when first x -> 1 and then t->0 compared
with first t->0 and then x->1. The function has no continuous
extension to (x,t)=(1,0). Hence I would not expect a computer algebra
system to give the same answers with simple substitutions in the two
orders.
On 21 July 2014 15:14, kcrisman <kcris...@gmail.com> wrote:
>> Hi guys,
>>
>> This is so simple that probably someone else has already noticed it, but
>> just in case:
>>
>> sage: x,t = var('x,t')
>> sage: f = (1-x)/(1-x*cos(t))
>> sage: f(x=1)
>> 0
>> sage: f(t=0)(x=1)
>> 1
>>
>
> My guess is that this is more of a convention than anything else.
>
> sage: x/x
> 1
> sage: 0/x
> 0
>
> Maxima:
> (%i1) x/x;
> (%o1) 1
> (%i2) 0/x;
> (%o2) 0
>
> If Mma and Maple do it too, that would be my guess. In any case, it is
> 'known' and I bet you'll find other examples with a search of the email
> lists (though searching for x/x might be hard!). It's possible to not
> immediately do such reductions
>
> sage: x.mul(1/x,hold=True)
> x/x
>
> but I'm not sure how to combine that with the substitution that you are
> doing.
>
> - kcrisman
>
>> The second one is, of course, the correct answer. (FYI, Mathematica9
>> fails, too.)
>>
>> Wouldn't the first one return some sort of conditional expression: "if t=0
>> then 1, else 0"
>>
>> I would be happy to help in the debugging, if I can get some indication of
>> what is running in the background, i.e. what function is called when one
>> does the substitution "f(x=1)".
>>
>> Cheers,
>> Jesús Torrado
>
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