CHS Calc BC: Number 10 from Set #3

Friday, April 11, 2008

Can anyone explain how to do this problem? Thanks!

Anonymous said...: Well I haven't really done this, but I'm going to take a shot here and hope I'm doing the right thing. Uh well you know that the Taylor polynomial for f(x)=e^-x is Tn(x)=1-x+(x²/2!)-(x³/3!)+....-(x^7/7!)+...+(-1)ⁿ (xⁿ/n!)+... right? So I guess you would just plug in 2 to Tn(x) find what the values are. Like for T6(2) you would plug 2 as x into T(x) until the sixth degree term, ie T6(2)=1-2+(2²/2!)-(2³/3!)+(2^4/4!)-(2^5/5!)+(2^6/6!). So I guess you'll just have to go through the whole list of choices. You'll be using f(2)=e^-2, but you'll only need T(x) for T5(2), T6(2), and T7(2) so it shouldn't be too bad. I hope I didn't lose you. :|; April 11, 2008 at 10:29 PM