How to evaluate A Taylor series at a specific value

[Mark Hardman]

Bernard's example used an extra, undocumented 4th paramter: polynom. This appears to return the expansion without the O term. If we use his syntax, it works:

taylor(sin(x),x,5,polynom) = x-1/6*x^3+1/120*x^5

taylor(sin(x),x,5,polynom)|x=(1/2) = 1841/3840 = 0.479427083333