the question asks to find the quadratic maclaurin polynomial for f(x)
Given f(x) = x sin(x)
The Attempt at a Solution
i know that a maclaurin series is when a=0 in a taylor series.
i did the 1st-5th derivatives of f(x) and then used the formula for taylor polynomial and set a=0.
f'=sin(x) + x cos(x)
f''=2cos(x) -x sin(x)
with the formula for taylor polynomial
Pn(x) = n
Ʃ [f^k (a) (x-a)^k] / k!
by setting a=0 i got Pn(x)=0+0+x^2 -0- (x^4)/6............
however the answer seems to be P(x)=x^2
anyone able to explain to me how that is so?
i think i got it... is the quadratic maclaurin polynomial just the 4th order maclaurin polynomial??