Yes, taylor is a good tool for this (though technically the result is not a
Taylor series), e.g.
(%i1) ex: sqrt(x^2+10);
(%o1) sqrt(x^2+10)
(%i2) taylor(ex,x,inf,3);
(%o2) x + 5/x - 25/(2*x^3) + ...
Perhaps more interesting:
(%i3) log(exp(x)+1);
(%o3) log(%e^x+1)
(%i4) taylor(%,x,inf,3);
(%o4) +x + %e^-x - (%e^-x)^2/2 + (%e^-x)^3/3 + ...
On Thu, Dec 23, 2010 at 13:40, Lorenzo Isella <lorenzo.isella at gmail.com>wrote:
> Dear All,
> I am not that much into maxima, so perhaps I am asking a trivial question.
> I know that maxima can be used e.g. to calculate the limits of a function
> e.g. lim f(x) when x--> infinity.
> However, sometimes you simply have a complicated function which can be
> expressed as something asymptotically much simpler when e.g. x is large.
> To fix the ideas, consider f(x)=sqrt(x^2+10).
> Now, I would simply be able to guess, using maxima, that f(x) goes like x
> for x>>1. How can I do that? Should I resort to a Taylor expansion in such
> cases?
> Many thanks
>
> Lorenzo
> _______________________________________________
> Maxima mailing list
> Maxima at math.utexas.edu
> http://www.math.utexas.edu/mailman/listinfo/maxima
>