1132
Calculus II
uiz 4
Names:
Date: 3/30/09
Question:
1
2
3
4
5
Total
Points:
2
2
2
2
2
10
Score:
You are allowed to work in groups of no more than five (5) people. This quiz is open book
and open notes. Each group only has to hand in one (1) quiz, so please make sure it is
written in a neat and concise manner.
1. (2 points) Find the radius of convergence and interval of convergence of
∞
X
(−1)n
n=1
xn
.
n2 5 n
Solution: The radius of convergence is given by the ratio test.
xn2 |x|
(−1)n+1 xn+1
n2
n2 5n |x|
=
=
lim
·
=
lim
.
lim 2
n+1
n
n
2
n→∞ 5(n + 1)
n→∞ (n + 1) 5
(−1) x
5 n→∞ (n + 1)2
5
By the ratio test, this series converges for all |x|/5 < 1, so |x| < 5 and the radius of convergence is
R = 5. To find the interval of convergence, we need to investigate the behavior at the endpoints:
x = −5 and x = 5. At x = −5, we have
∞
X
(−1)n
n=1
∞
∞
X
X
(−5)n
5n
1
=
=
,
2 5n
2
n2 5n
n
n
n=1
n=1
which converges since it is a p-series with p = 2 > 1. When x = 5, we have
∞
X
(−1)n
n=1
∞
X
5n
(−1)n
=
.
n2 5n
n2
n=1
We can P
see this series converges by two methods. First, notice that the series converges absolutely
1
1
because
n2 converges. Alternatively, we can use the alternating series test. Notice that bn = n2 is
positive, decreasing and tends to zero as n → ∞, so it converges by the alternating series test. Thus
the interval of convergence is [−5, 5].
2. (2 points) Find the Maclaurin series and the radius of convergence for
f (x) =
x2
.
1+x
Solution: Using the geometric series, we get
∞
∞
X
X
1
1
n
=
=
(−x) =
(−1)n xn .
1+x
1 − (−x) n=0
n=0
Then multiplying through by x2 gives
∞
∞
X
X
x2
= x2
(−1)n xn =
(−1)n xn+2 .
1+x
n=0
n=0
To find the radius of convergence, we use the ratio test.
n+3 (−1)n+1 xn+3 = lim x
= lim |x| = |x|.
lim n→∞
n
n+2
n+2
n→∞ x
n→∞
(−1) x
Since it converges for all |x| < 1, the radius of convergence is R = 1.
Page 2
3. (2 points) Use series to evaluate
lim
x→0
sin x − x
.
x3
Solution: Recall the Maclaurin series expansion for sin x is
sin x =
∞
X
(−1)n
n=0
x3
x5
x7
x2n+1
=x−
+
−
+ ···
(2n + 1)!
3!
5!
7!
Subtracting x gives
∞
X
x3
x5
x7
x2n+3
sin x − x = − +
−
+ ··· =
.
(−1)n+1
3!
5!
7!
(2n + 3)!
n=0
Finally, dividing everything by x3 gives
∞
sin x − x X
1
x2
x4
x2n
n+1
=
=
−
+
−
+ ··· .
(−1)
x3
(2n + 3)!
3!
5!
7!
n=0
Taking the limit at x → 0 we have
sin x − x
1
x2
x4
1
1
=
lim
−
+
−
+
·
·
·
=− =− .
x→0
x→0
x3
3!
5!
7!
3!
6
lim
Page 3
4. (2 points) Find the Taylor series for f (x) = x−x3 centered at a = −2. What is its radius of convergence?
Solution: We use the formula for a Taylor series at a. We have
f (x) =
∞
∞
X
X
f (n) (a)
f (n) (−2)
(x − a)n =
(x + 2)n .
n!
n!
n=0
n=0
So we need to calculate the derivatives at x = −2.
f (x) = x − x3
f 0 (x) = 1 − 3x2
f 00 (x) = −6x
f 000 (x) = −6
f (4) (x) = 0
..
.
f (−2) = −2 − (−2)3 = −2 + 8 = 6
f 0 (−2) = 1 − 3(−2)2 = 1 − 12 = −11
f 00 (−2) = −6(−2) = 12
f 000 (−2) = −6
f (4) (−2) = 0
..
.
Hence the Taylor series centered at a = −2 is
f (x) = 6 − 11(x + 2) +
12
6
(x + 2)2 − (x + 2)3 = 6 − 11(x + 2) + 6(x + 2)2 − (x + 2)3 .
2!
3!
This series is finite so it converges for all x ∈ R, so R = ∞.
Page 4
5. (2 points) Show that the function defined by
(
f (x) =
2
e−1/x ,
0,
x 6= 0
x=0
is not equal to its Maclaurin series.
Solution: We must first calculate some derivatives, and unfortunately the best way to find the derivative of this function is to use the definition of the derivative along with l’Hôpital’s rule (otherwise, when
you evaluate at zero you’ll get a function divided by zero). We have
2
−1/x2
x
e−1/x
1/x
f (x) − f (0)
= lim 1/x2 = 0.
= lim
= lim 1/x2 = lim
f (0) = lim
2
x→0 −2e1/x /x3
x→0 2e
x→0
x→0 e
x→0
x−0
x
0
If you continue to use l’Hôpital’s rule to find f 00 (0), f 000 (0), and so on, you’ll see that f (n) (0) = 0 for
all n. This implies
∞
X
f (n) (0) n
f (x) =
x = 0 + 0 + ···
n!
n=0
But f (x) is only zero when x = 0, so this function does not equal its power series.
Page 5