MATH 101 V2A
March 27th – Practice problems
Practice Problems:
1. Find the Maclaurin series for the following functions. What is the radius of convergence? (Hint: For
(d)-(g) you should not have to calculate f (n) (0) – you can use power series methods to find the series).
(a) ex
Solution: For all x,
ex =
∞
X
x2
x3
xn
=1+x+
+
+ ...
n!
2
6
n=0
(b) cos(x)
Solution: For all x,
cos(x) =
∞
X
(−1)n x2n
x2
x4
=1−
+
− ...
(2n)!
2
24
n=0
(c) sin(x)
Solution: For all x,
sin(x) =
∞
X
(−1)n x2n+1
x3
x5
=x−
+
− ...
(2n + 1)!
6
120
n=0
(d) log(1 + x)
Solution: For |x| < 1,
Z
x
log(1 + x) =
0
Z
x
=
0
1
dt
1+t
!
∞
X
n n
(−1) t
dt
n=0
∞
X
(−1)n xn+1
.
=
n+1
n=0
(e)
1
1−x
Solution: For |x| < 1,
∞
X
1
=
xn .
1 − x n=0
(f)
1
(1+x)2
Solution: For |x| < 1,
d
1
=
(1 + x)2
dx
(g)
−1
1+x
=−
∞
X
(−1)n xn =
n=0
∞
X
(−1)n+1 xn .
n=0
1
6+x2
Solution: For |x| < 6,
2 n X
∞
∞
1
1X
1
x
x2n
1
n
=
=
(−1)
=
(−1)n n+1 .
·
2
2
x
6+x
6 1+ 6
6 n=0
6
6
n=0
2. Find the Maclaurin series for sin(x2 ). Where does it converge? What about sin(x2 )/x?
Hint: For all x,
sin(x2 ) =
∞
∞
X
X
(−1)n (x2 )2n+1
(−1)n x4n+2
=
.
(2n + 1)!
(2n + 1)!
n=0
n=0
Note that the series obtained by dividing the Maclaurin series for sin(x2 ) by x is NOT the Maclaurin
series for f (x) = sin(x2 )/x, since f (0), f 0 (0), f 00 (0), . . . are not defined. Instead, one can show that it
is the Maclaurin series for the function
(
sin(x2 )
, if x 6= 0
x
.
g(x) =
0,
if x = 0
3. If f (x) = cos(3x), what is f (105) (0)? What about f (106) (0)?
Solution: We know that, for all x,
cos(3x) =
∞
∞
X
X
(−1)n (3x)2n
(−1)n 32n x2n
=
,
(2n)!
(2n)!
n=0
n=0
and that this is the Maclaurin series for f (x) = cos(3x), i.e.
∞
∞
X
f (n) (0) n X (−1)n 32n x2n
x =
.
n!
(2n)!
n=0
n=0
Hence f (105) (0) = 0, and f (106) (0) = 106! ·
(−1)53 3106
106!
2
= −3106 .
4. Without using l’Hospital’s Rule, evaluate
(e2x − 1) log(1 + x2 )
.
x→0
(1 − cos(3x))2
lim
Hint: Expand the functions using Taylor series and show that the limit is equal to 0.
3