SECTION 5.4 Polynomial Approximations 293
Example 2. (A handy trick)If we wish to compute the Maclaurin
series for cosx, we could certainly follow the same procedure as for the
sinx in the above example. However, since cosx = dxd sinx, we can
likewise obtain the Maclaurin series for the cosxby differentiating that
for the sinx, this yields the series
cosx∼d
dx∑∞
n=0(−1)nx^2 n+1
(2n+ 1)!=
∑∞
n=0(−1)nx^2 n
(2n)!.
Example 3. We we wanted to compute the Maclaurin series for sinx^2 ,
then computing the higher-order derivatives of sinx^2 would be ex-
tremely tedious! A more sensible alternative would be to simply replace
xbyx^2 in the Maclaurin series for sinx:
sinx^2 ∼x^2 −x^6
3!+
x^10
5!−
x^14
7!+···=
∑∞
n=0(−1)nx^4 n+2
(2n+ 1)!.
Example 4.(A handy trick)Since ln(1 +x) =
∫ dx
1 +xwe may startwith the geometric series
1 −x+x^2 −···=∑∞
n=0(−1)nxn=1
1 +x,
and then integrate each term to obtain the Maclaurin series for ln(1 +
x) :
ln(1 +x)∼x−
x^2
2+
x^3
3−···=
∑∞
n=0(−1)n∫
xndx=∑∞
n=0xn+1
n+ 1.
(Note that there is no constant occuring in the above integrations since
whenx= 0, ln(1 +x) = ln 1 = 0.)
Example 5. Since
dn
dxnex=ex= 1 atx= 0, we immediately have theMaclaurin series expansion forex:
ex∼ 1 +x+
x^2
2!+
x^3
3!+··· =
∑∞
n=0xn
n!