Franck MICHEL on Thu, 12 Dec 2002 12:26:44 +0100

 Re: How to eliminate the big-oh

```>As Michael say, x has an higher priority than t and truncate only truncate
with
>respect of the highest priority variable.
>
>This is a problem here.
>Assuming z has higher priority that x, you can do
>
>subst(truncate(subst(taylor((1-x)/(1-t),t),x,z)),z,x)

Thank you, it works. Certainly a naive question, but: how can we know the
level of priority of variables? Is it possible to change it?

>fortunately there is a  simpler solution, convert your polynomial to a
vector:
>
>Pol(truncate(Vec(taylor((1-x)/(1-t),t))))

It's a nicest solution, unfortunately it does not work for other expressions.
For example, if we slightly modify (1-x)/(1-t) and consider

truncate(taylor((1-x)/(x-t),t)), we get -1+O(t^16)

but subst(truncate(subst(taylor((1-x)/(x-t),t),x,z)),z,x) gives the good
expansion.

If we ask for the type of taylor(1/(1-t),t), we get "SER"; for
taylor((1-x)/(1-t),t) we get "POL"; and for taylor((1-x)/(x-t),t) we get
"RFRAC". When we have x and t, the type is determined with respect to x. It
is determined with respect to t after changing z in x as you have
suggested. If we have to deal with a quotient with multiple variables (x,
y, z, etc.., and t), I presume all the variables above the level of
priority of t would have to be substituted by variables below the level of
priority of t. But it is not very convenient, the determination of levels
of priority of variables is a little bit mysterious for me and I am
wondering if there would be a better solution than using substitutions with
low-level variables.