asympt(ln(x+a),x);
gives a correct asymptotic series for large x. However,
asympt(ln(ln(x)+a),x);
gives only the trivial answer, i.e. the same. How do I convince Maple that \(\ln (x)\) also becomes large if \(x\) is large?
As far as I know, asympt(f(x), x, n), which is essentially the same as
subs(t=1/x, series(f(1/t), t, n)), is supposed to do its calculations using all
terms that are O(1/x^(n-epsilon)) for epsilon > 0 (i.e. it won’t include terms
like ln(x)/x^n), but will include x^(-n+1/10)). The asymptotic series you want
is
2 3 4 a a a a ln(ln(x)) + ----- - 1/2 ------ + 1/3 ------ - 1/4 ------ + ... ln(x) 2 3 4 ln(x) ln(x) ln(x)
which would have infinitely many terms that are not O(1/x^p) for any p > 0. So that does
not really fit the job description for "asympt". Anyway, you can get your series
by
subs(u = ln(x), asympt(ln(u+a), u));
What would you like the answer to be? Presumably you do not object if
asympt(x^2,x) returns x^2 ... Similarly, ln(ln(x)+a) is one of the basic types that Maple
uses. Examples:
> asympt(ln(ln(x)+a),x); ln(ln(x) + a) > asympt(ln(ln(2*x)),x); ln(ln(x) + ln(2)) > asympt(ln(ln(x^2)),x); ln(2 ln(x)) > asympt(ln(ln(3*x^2+2*x-2)),x,2); 2 1 ln(ln(3) + 2 ln(x)) + --------------------- + O(----) 3 (ln(3) + 2 ln(x)) x 2 x
I would like to thank the MUG contributors Robert Israel and Gerald Edgar for their responses on my question. They explain what happens inside Maple. Nevertheless, I’m not fully convinced that Maple’s results are consistent, and are what they should be.
The argument that Maple only expands in powers of x is probably true in most cases, but not in general. For example,
asympt(ln(exp(x)+a),x);
gives an asymptotic series in powers of exp(x).
Also:
asympt(ln(ln(x)+x+exp(x)),x);
gives powers of exp(x), an asymptotic expansion in x in the coefficients of each exp(x), and a
seemingly random "order" of the error O([x/exp(x)]^p).
On the other hand, the similar command
asympt(1/(exp(x)+a),x);
does not do a thing, apart from the error message
Error, (in asympt) unable to compute series At the same time,
asympt(1/(x+ln(x)+a),x);
gives a series in powers of x with ln(x) considered as a constant.
Apparently: in general the asymptotic gauge functions are x^p, and sometimes exp(p*x).
Mixed expressions of x and exp(x) can be handled under a logarithm, but not
under another power, and ln(x) is ignored as a function of x, and treated as a
constant.
I think a more consistent approach would be to consider
(negative powers of ...),..1, ...,ln(ln(x))^p, ln(x)^p, x^p, exp(x)^p, exp(x^q)^p, exp(exp(x)^q)^p, ....
as gauge functions. Maybe this is a bit too ambitious, but then at least
exp(-px), x^(-p), ln(x)^(-p), 1, ln(x)^p, x^p, exp(px).
Otherwise, the name "asymptotic expansion" is a bit misleading, and should be something like "Laurent series expansion around infinity".
There is a package (gdev) written by a third party that does certain expansions in Maple.
Get info from Bruno Salvy, Bruno.Salvy@inria.fr, or perhaps even on the web
at
ftp://ftp.inria.fr/INRIA/Projects/algo/www/intro.html
Here is your example...
> readlib(gdev); proc(fct:algebraic) ... end > readlib(glimit); proc(fct:algebraic) ... end > gdev(log(log(x)+a),x=infinity,4); 2 3 a a a 1 ln(ln(x)) + ----- - 1/2 ------ + 1/3 ------ + O(------) ln(x) 2 3 4 ln(x) ln(x) ln(x)