How to evaluate A Taylor series at a specific value

[Albert Chan]

(08-22-2018 08:50 PM)Han Wrote:  Why would the evaluate of a function that contains an error term magically drop the error term?

Exactly !

Taylor series that carried the error term should not be evaluable, at any x.
The fact that taylor(sin(x)) | x=0.5 can return anything is not right. Unless ... (see below)

Quote:(x/x) | x=0 is indeterminate. The expression x/x is undefined at 0.
Even if one were to "evaluate first", the expression x/x only evaluates to 1 when x is non-zero.

It were meant to test the CAS order of evaluation.
If CAS does evaluation before substitution, x/x simplied as 1

But, if CAS return 0/0, taylor(sin(x)) | x=0.5 return sin(0.5) make sense.