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last edited 9 years ago by test1 |
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Editor:
Time: 2007/11/17 22:22:56 GMT-8 |
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changed: - For the following limit: \begin{axiom} limit(exp(exp(2*log(x^5+x)*log(log(x))))/exp(exp(10*log(x)*log(log(x)))), x = %plusInfinity) \end{axiom} the correct answer is +infinity. Simpler version is: \begin{axiom} limit(exp(2*log(x^5+x)*log(log(x)))-exp(10*log(x)*log(log(x))), x = %plusInfinity) \end{axiom} (again the correct answer is +infinity). Another problematic limit is: \begin{axiom} limit(max(x, exp(x))/log(min(exp(-x), exp(-exp(x)))), x = %plusInfinity) \end{axiom} where the correct answer is -1. BTW both examples are taken from Dominik Gruntz thesis form 1996 Waldek Hebisch From kratt6 Mon Jun 5 02:10:40 -0500 2006 From: kratt6 Date: Mon, 05 Jun 2006 02:10:40 -0500 Subject: Message-ID: <20060605021040-0500@wiki.axiom-developer.org> Yes, somebody should reorganize the limit package and implement Gruntz algorithm. This is on the WishList, by the way. I believe this is also related to asymptotic expansions as provided by Gdev from Maple, but I'm not sure. Could you do this? Martin
For the following limit:
limit(exp(exp(2*log(x^5+x)*log(log(x))))/exp(exp(10*log(x)*log(log(x)))),x = %plusInfinity)
(1) |
the correct answer is +infinity. Simpler version is:
limit(exp(2*log(x^5+x)*log(log(x)))-exp(10*log(x)*log(log(x))),x = %plusInfinity)
(2) |
(again the correct answer is +infinity).
Another problematic limit is:
limit(max(x,exp(x))/log(min(exp(-x), exp(-exp(x)))), x = %plusInfinity)
There are 1 exposed and 2 unexposed library operations named max having 2 argument(s) but none was determined to be applicable. Use HyperDoc Browse,or issue )display op max to learn more about the available operations. Perhaps package-calling the operation or using coercions on the arguments will allow you to apply the operation.
Cannot find a definition or applicable library operation named max with argument type(s) Variable(x) Expression(Integer)
Perhaps you should use "@" to indicate the required return type,or "$" to specify which version of the function you need.
where the correct answer is -1.
BTW both examples are taken from Dominik Gruntz thesis form 1996
Waldek Hebisch
Could you do this?
Martin