Hello,
1. I fully agree with you that
RJF> It is a difficult engineering question to figure
RJF> out how much work to do on simplifying input
RJF> without special requests, like trigsimp. If you
RJF> do too much, commands can take too much time.
I would even say, "It is a very difficult engineering question"
2. I could agree with you - if Maxima would return the
input unevaluated. Well... the system just cannot do
it.
However, in the given case, Maxima returns an answer
which is mathematically incorrect, namely, ind.
3. Some other system handles this successfully, say, any
version of Derive since 1980 to 2006, and Mathematica,
at least MMA 5.2. Here we are
LIM(SIN(SIN(z)^2+COS(z)^2),z,inf)
Limit[Sin[Sin[z]^2 + Cos[z]^2], z -> Infinity]
SIN(1)
Sin[1]
LIM(SIN(SIN(SIN(SIN(SIN(SIN(z)^2+COS(z)^2))))),z,inf)
Limit[Sin[Sin[Sin[Sin[Sin[Sin[z]^2 + Cos[z]^2]]]]], z -> Infinity]
SIN(SIN(SIN(SIN(SIN(1)))))
Sin[Sin[Sin[Sin[Sin[1]]]]]
So why Maxima cannot achieve what other have done OK?
Sincerely,
VB
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Sunday, August 27, 2006, 5:42:27 PM, you write:
RF> Yet again you conflate a problem with limit with a simplification problem.
RF> trigsimp( sin(sin(z)^2+cos(z)^2) ); gives sin(1).
RF> It is a difficult engineering question to figure out how much work to do on
RF> simplifying input without special requests, like trigsimp. If you do too
RF> much, commands can take too much time.
RF> RJF
>> -----Original Message-----
>> From: maxima-bounces at math.utexas.edu
>> [mailto:maxima-bounces at math.utexas.edu] On Behalf Of Vladimir
>> Bondarenko
>> Sent: Sunday, August 27, 2006 2:02 AM
>> To: maxima at math.utexas.edu
>> Subject: Maxima 5.9.3 bug in limit
>>
>> Hello,
>>
>> (%i1) limit(sin(sin(z)^2+cos(z)^2),z,inf);
>>
>> (%o1) ind
>>
>>
>> Of course, the correct answer is sin(1).