- #1

kiwifruit

- 8

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## Homework Statement

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)

f'''=-3sin(x) -xcos(x)

f''''=-4cos(x) +xsin(x)

f'''''=5sin(x) +xcos(x)

with the formula for taylor polynomial

Pn(x) = n

Ʃ [f^k (a) (x-a)^k] / k!

k=0

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??