How to evaluate A Taylor series at a specific value

[Komanguy]

(08-21-2018 04:47 PM)parisse Wrote:  But a Taylor expansion has two parts: the polynomial part and the remainder term. Unfortunately students tend to forget the remainder term, and that will not help if the calculator ignores it as well. That's why you have taylor(.) and taylor(.,polynom) on the HP. And if you don't remind the polynom trick, you can still run f:=taylor(sin(x),x=0,5) then f|x=0.5 and ignore the order_size term.
:-)

Best answer.

Thanks a lot.