Math 2260 Written HW #13 Solutions
1. Consider the power series
∞
X
4n+1 (x − 2)n .
n=0
(a) What is the interval of convergence of this series?
Answer: Using the Ratio Test on the series of absolute values yields
n+2
4
(x − 2)n+1 4n+2 (x − 2)n+1 = lim n+1 · lim
n→∞ 4
n→∞ |4n+1 (x − 2)n |
(x − 2)n = lim 4 · |x − 2|
n→∞
= 4|x − 2|.
Therefore the series definitely converges absolutely when 4|x − 2| < 1 or, equivalently,
when |x − 2| < 14 .
Now, we can check the endpoints: when x − 2 = 14 , the series becomes
n X
∞
∞
∞
X
4n+1 X
n+1 1
=
=
4 = 4 + 4 + 4 + 4 + ...,
4
4
4n
n=0
n=0
n=0
which obviously diverges (for example, it fails the nth term test).
When x − 2 = − 14 , the series becomes
n X
∞
∞
∞
n+1
X
X
1
n+1
n4
4
−
=
(−1)
=
(−1)n 4 = 4 − 4 + 4 − 4 + . . . ,
4
4n
n=0
n=0
n=0
which also diverges (again, it fails the nth term test).
Therefore, we see that the series converges for
1
1
− <x−2<
4
4
or
7
9
<x< ,
4
4
so the interval of convergence is (7/4, 9/4).
(b) On this interval of convergence, what does the series converge to?
Answer: Notice that
∞
X
4n+1 (x − 2)n = 4 + 4 · 4(x − 2) + 4 · [4(x − 2)]2 + 4 · [4(x − 2)]3 + . . .
n=0
is a geometric series with a = 4 and r = 4(x − 2). Therefore, by the geometric series test,
we know that it converges when |r| < 1, meaning when |4(x − 2)| < 1 or |x − 2| < 1/4
and diverges everywhere else (notice that this gives an alternative approach to solving
part (a) above!). Moreover, the geometric series test says that on this interval, which is
the same as we determined in part (a), the series must converge to
a
4
4
=
=
.
1−r
1 − 4(x − 2)
9 − 4x
1
2. The 0th order Bessel Function is defined to be
∞
X
(−1)k x2k
.
J0 (x) =
22k (k!)2
k=0
(a) For what values of x is this function defined? (In other words, where does the series
converge?)
Answer: The function is defined where the series converges. As always when trying to
determine where a power series converges, we use the Ratio Test on the series of absolute
values:
(−1)k+1 x2(k+1) 22(k+1) ((k+1)!)2 (−1)k+1 x2k+2 22k
(k!)2
·
lim
·
=
lim
·
(−1)k x2k k→∞
k→∞ (−1)k x2k 22k+2 ((k + 1)!)2
22k (k!)2 = lim 1 · x2 ·
k→∞
1
1
·
4 (k + 1)2
x2
k→∞ 4(k + 1)2
= 0.
= lim
Therefore, the function J0 (x) is defined for all x.
(b) Show that the series is a solution to the differential equation
x2 J000 + xJ00 + x2 J0 = 0.
Answer: First, I’ll compute the first and second derivatives of J0 (x):
J00 (x)
∞
∞
X
X
(−1)k 2k 2k−1
(−1)k
2k−1
2kx
=
x
=
22k (k!)2
22k (k!)2
k=1
k=1
J000 (x) =
∞
X
(−1)k 2k
k=1
(2k − 1)x2k−2 =
22k (k!)2
∞
X
(−1)k 2k(2k − 1)
22k (k!)2
k=1
x2k−2 .
Therefore, the terms in the above differential equation are
x2 J000 (x)
=x
2
∞
X
(−1)k 2k(2k − 1)
k=1
xJ00 (x) = x
x2 J0 (x) = x
22k (k!)2
∞
X
(−1)k 2k
22k (k!)2
k=1
∞
X
2
k=0
22k (k!)2
∞
X
(−1)k 2k
k=1
x2k
=
∞
X
(−1)k 2k(2k − 1)
k=1
x2k−1 =
(−1)k
x
2k−2
22k (k!)2
22k (k!)2
x2k
x2k
∞
X
(−1)k 2k+2
=
x
.
22k (k!)2
k=0
In order to combine terms, it would be best to have the sums all start at the same index
and (more importantly) for the power of x to be the same in each. Therefore, I will
re-index the x2 J0 (x) sum to start at k = 1 rather than k = 0:
∞
X
∞
X
(−1)k−1
(−1)k−1
2k
x J0 (x) =
x
=
x2k .
2(k−1) ((k − 1)!)2
22k−2 ((k − 1)!)2
2
k=1
k=1
2
2
Therefore,
∞
X
(−1)k 2k(2k − 1)
∞
X
(−1)k 2k
∞
X
(−1)k−1
x2k
22k (k!)2
22k−2 ((k − 1)!)2
k=1
k=1
k=1
∞ k
k
k−1
X
(−1) 2k(2k − 1) (−1) 2k
(−1)
=
+
+
22k (k!)2
22k (k!)2
22k−2 ((k − 1)!)2
k=1
∞
X
2k
1
k 2k(2k − 1)
+ 2k
−
=
(−1)
22k (k!)2
2 (k!)2 22k−2 ((k − 1)!)2
k=1
∞
X
1
k 2k(2k − 1 + 1)
=
− 2k−2
(−1)
22k (k!)2
2
((k − 1)!)2
k=1
∞
X
1
22 k 2
k
−
=
(−1)
22k k! · k! 22k−2 (k − 1)! · (k − 1)!
k=1
∞
X
1
1
k
=
(−1)
−
22k−2 (k − 1)! · (k − 1)! 22k−2 (k − 1)! · (k − 1)!
x2 J000 (x) + xJ00 (x) + x2 J0 (x) =
x2k +
x2k +
22k (k!)2
k=1
= 0,
as desired.
3. Find the Taylor series centered at x = π/4 of the function f (x) = cos x.
Answer: We need to compute a bunch of derivatives of f (x) = cos(x):
f 0 (x) = − sin(x)
f 00 (x) = − cos(x)
f 000 (x) = sin(x)
f (4) (x) = cos(x)
and then the pattern repeats every four terms thereafter. Hence,
1
f (π/4) = cos(π/4) = √
2
1
f 0 (π/4) = − sin(π/4) = − √
2
1
f 00 (π/4) = − cos(π/4) = − √
2
1
f 000 (π/4) = sin(π/4) = √
2
1
f (4) (π/4) = cos(π/4) = √
2
3
and so on. Therefore, the Taylor series for f (x) = cos(x) centered at x = π/4 is
1
1
1
1
1
√ − √ (x − π/4) − √
(x − π/4)2 + √
(x − π/4)3 + √
(x − π/4)4 − . . .
2
2
2 · 2!
2 · 3!
2 · 4!
∞
X
(−1)n(n+1)/2
√
=
(x − π/4)n .
2 · n!
n=0
4