Dear All,
I am again puzzled at the output I get from maxima, but I may be
misunderstanding something.
Consider the very simple code below
assume(A_par>0);
assume(B_par>0);
assume(n>1);
beta_par:A_par*(n^2-1)/((2*(n^2-1)+1)/sqrt(n^2-1)*log(n+sqrt(n^2-1))+B_par*n)/n/(6*%pi);
limit(beta_par,n, infinity);
When I run it this is what I get
(%i23) batch("friction.max");
batching #p/home/iselllo/temp/friction.max
(%i24) assume(A_par > 0)
(%o24) [redundant]
(%i25) assume(B_par > 0)
(%o25) [redundant]
(%i26) assume(n > 1)
(%o26) [n > 1]
2
A_par (n - 1)
------------------------------------------------
2 2
(1 + 2 (n - 1)) log(sqrt(n - 1) + n)
B_par n + --------------------------------------
2
sqrt(n - 1)
------------------------------------------------
n
(%i27) beta_par : ------------------------------------------------
6 %pi
2
A_par (n - 1)
(%o27) ----------------------------------------------------------
2 2
(2 (n - 1) + 1) log(sqrt(n - 1) + n)
6 %pi n (-------------------------------------- + B_par n)
2
sqrt(n - 1)
(%i28) limit(beta_par, n, infinity)
Is (n - 1) (n + 1) positive, negative, or zero?
positive;
2
Is sqrt(n - 1) + n positive or negative?
positive;
Is n - 1 positive, negative, or zero?
positive;
Maxima encountered a Lisp error:
Error in $LIMIT [or a callee]: Bind stack overflow.
Automatically continuing.
To reenable the Lisp debugger set *debugger-hook* to nil.
(%o29) friction.max
First, I do not understand why I am asked if e.g. (n-1) is positive or
negative, having already assumed that n>1.
But the final message sounds Greek to me; I used the same expression
elsewhere and it is a perfectly reasonable function for n>=2.
Can anyone enlighten me here?
Many thanks
Lorenzo
--
Life is what happens to you while you're busy making other plans.