MA1S12: SOLUTIONS TO TUTORIAL 9
1. Find the radius of convergence for the power series
∞
X
(2x)k
k=0
∞
X
(−1)k xk
k!
k=1
k
Solution: Recall that the Maclaurin series for ex is
∞
X
xk
k=0
k!
and this series converges for all values of x. The first series is the
Maclaurin series for e2x (just substitute 2x for x). Again this series
converges for all values of x and so the radius of convergence is
R = ∞.
To find the radius of convergence for the second series we apply
the ratio test to the series of non-negative terms
∞ ∞
X
(−1)k xk X
|x|k
=
k
k
k=1
Letting uk =
|x|k
k
k=1
we have
ρ =
=
uk+1
k→∞ uk
lim
|x|k+1 k
k→∞ (k + 1) |x|k
lim
k|x|
k→∞ k + 1
|x|
= lim
k→∞ 1 + 1
k
=
lim
= |x|
1
2
MA1S12: SOLUTIONS TO TUTORIAL 9
The series converges when ρ < 1, so for all values of x ∈ (−1, 1).
The series diverges when ρ > 1, so for all values of x ∈
/ [−1, 1].
We conclude that the radius of convergence for the power series is
R = 1.
2. Find the first three non-zero terms of the Maclaurin series for
f (x) = x sin(3x)
Solution: The Maclaurin series for sin x is
sin x =
∞
X
(−1)k
k=0
x2k+1
(2k + 1)!
Substituting 3x for x we get the Maclaurin series for sin(3x),
∞
X
(3x)2k+1
sin(3x) =
(−1)
(2k + 1)!
k=0
k
Now multiplying by x we get,
x sin(3x) =
∞
X
(−1)k
k=0
32k+1 x2k+2
(2k + 1)!
The first three non-zero terms are,
2
X
k=0
(−1)k
32k+1 x2k+2
9x4 81x6
= 3x2 −
+
(2k + 1)!
2
40
3. Find the Taylor series for
f (x) =
1
x
at the point a = 1 and use it to approximate f (x) at the point 1.1.
Solution: The Taylor series for f (x) at a = 1 is
∞
X
f (k) (1)
k=0
k!
(x − 1)k = f (1) + f 0 (1)(x − 1) +
f 00 (1)
(x − 1)2 + · · ·
2!
MA1S12: SOLUTIONS TO TUTORIAL 9
3
We have
k f k (x)
f k (1)
1
x
0 f (x) =
f (1) = 1
1 f 0 (x) = − x12
f 0 (1) = −1
2 f 00 (x) =
f 00 (1) = 2
2
x3
3 f (3) (x) = − x3!4 f (3) (1) = −(3!)
The general term is
f (k) (x) = (−1)k
k!
=⇒ f (k) (1) = (−1)k k!
xk+1
Thus the Taylor series for f (x) at a = 1 is
∞
X
(−1)k (x − 1)k
k=0
We can approximate f (x) at points near 1 using Taylor polynomials
(i.e. partial sums),
sn (x) =
n
X
(−1)k (x − 1)k
k=0
Consider the Taylor polynomials of degree 1, 2 and 3,
s1 (x) = 1 − (x − 1)
s2 (x) = 1 − (x − 1) + (x − 1)2
s3 (x) = 1 − (x − 1) + (x − 1)2 − (x − 1)3
At the point 1.1 we have the approximations
s1 (1.1) = 1 − (0.1) = 0.9
s2 (1.1) = 1 − (0.1) + (0.1)2 = 0.91
s3 (1.1) = 1 − (0.1) + (0.1)2 − (0.1)3 = 0.909
4
MA1S12: SOLUTIONS TO TUTORIAL 9
The exact value is a rational number with a repeating decimal
f (1.1) =
1
10
=
= 0.90909090
1.1
11
Alternatively, notice that we can represent f (x) by a geometric
series,
∞
X
1
1
=
=
(1 − x)k
x
1 − (1 − x) k=0
provided |1 − x| < 1. We can express this geometric series in the
form of a power series centred at a = 1,
∞
1 X
=
(−1)k (x − 1)k
x k=0
Again this is the Taylor series for f (x) at the point a = 1.
4. Approximate the following integral by representing the integrand as
a power series.
Z
1
sin(x2 ) dx
0
Solution: The Maclaurin series for sin x is
∞
X
k=0
(−1)k
x2k+1
(2k + 1)!
Substituting x2 for x we find that the Maclaurin series for sin(x2 ) is
∞
X
k=0
∞
(−1)k
X
(x2 )2k+1
x4k+2
=
(−1)k
(2k + 1)! k=0
(2k + 1)!
MA1S12: SOLUTIONS TO TUTORIAL 9
5
Integrating term by term gives
Z
0
1
=
1
∞
X
k=0
"
∞
X
#
x4k+2
sin(x ) dx =
(−1)
dx
(2k + 1)!
0
k=0
Z 1
∞
X
(−1)k
=
x4k+2 dx
(2k
+
1)!
0
k=0
1
∞
X
(−1)k
x4k+3
=
(2k + 1)! 4k + 3 0
k=0
Z
2
k
(−1)k
(2k + 1)!(4k + 3)
We can approximate the integral by considering partial sums of this
series. The first three partial sums correct to two decimal places are
s0 =
1
= 0.33
3
1
1
−
= 0.31
3 42
1
1
1
s2 = −
+
= 0.31
3 42 5!(11)
s1 =