MATH 106A - CALCULUS II
FALL 2005
QUIZ 6
NAME:
Show ALL your work CAREFULLY.
(a) Find the second-order Maclaurin’s polynomial M2(x) for the function
f(x) = sin(x2).
Maclaurin’s polynomials are simply Taylor’s polynomials with
x0=0. To find M2(x), we first evaluate the derivatives of f(x)=
sin(x2). By the chain rule, f0(x) = cos(x2)·2x. Together with the
product rule,
(1) f00(x)=(−sin(x2)·2x)·2x+ cos(x2)·2
= 2cos(x2)−4x2sin(x2).
It follows that f(0) = 0,f0(0) = 0,and f00(0) = 2 cos(0)−0=2. Thus,
M2(x)=0+0x+2
2! x2=x2.
(b) Use the Taylor’s theorem to give an upper bound for the error commit-
ted by using M2(x) to estimate f(x) for −1≤x≤1. [Do your differentiation
carefully.]
Taylor’s theorem asserts that |f(x)−M2(x)|≤ K3
3! |x|3where K3is
a constant such that |f000(x)|≤K3. By differentiating f00(x)from
(1),weget
(2) f000(x)=−2 sin(x2)·2x−4[2x·sin(x2)+x2cos(x2)·2x]
=−[12xsin(x2)+8x3cos(x2)].
Since |sin(x2)|≤1,|cos(x2)|≤1, and |x|≤1for −1≤x≤1, it follows
that |f000(x)|≤12+8 = 20 so that we let K3=20. Hence, we conclude
that |f(x)−M2(x)|≤ 20
3! =10
3.
Date: October 28, 2005.
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