Calculus  Taylor #2
posted by COFFEE on .
Find the Taylor series for f(x) centered at the given value of 'a'. (Assume that 'f' has a power series expansion. Do not show that Rn(x)>0.)
f(x) = x3, a = 1
and what i've done so far:
f (x) = x^3
f ' (x) = 3x^2
f '' (x) = 6x^1
f ''' (x) = 6x
f (1) = 1
f ' (1) = 3
f '' (1) = 6
f ''' (1) = 6
using taylor series equation.. my final answer that was wrong:
((1(x+1)^0)/(0!))+((3(x+1)^1)/(1!))+((6(x+1)^2)/(2!))+((6(x+1)^3)/(3!))
.. is this what the question was asking for? if not, what is it then? thank you very much for your assistance.
f ''' (x) = 6
f'''(1) = 6

Looks good, assuming your arithmetic is ok.