Co-21C Problem Solving #5T (Answers)
1) Write each series in sigma notation, then find the interval of convergence. Check endpoints!
"
(!1)n +1 (x ! 1)n
(x !1) 2 (x !1) 3 (x !1) 4
a) ln x = (x !1) !
Converges on ( 0, 2 ]
+
!
+ ... = #
n
2
3
4
n =1
b) cos x = 1!
c)
"
(!1)n x 2n
Converges on ( !", " )
(2n)!
n=0
x2 x4 x6 x8
+
! +
+ ... =
2! 4! 6! 8!
1
= 1! x + x 2 ! x 3 + x 4 ! ... =
1+ x
d) tan!1 x = x !
x3 x5 x7
+
! + ... =
3
5
7
#
"
# (!x)
Converges on ( !1,1)
n
n=0
(!1)n x 2n +1
n = 0 (2n + 1)
"
#
Converges on [ !1,1]
2) Find the radius of convergence. For what x values does the series converge conditionally? Absolutely?
"
(x ! 5) n
a) # 2 n
Radius = 3. Converges absolutely on [ 2, 8 ]
n=1 n 3
!
b)
(x + 3) n
n 4 n +1
n=1
"
Radius = 4. Converges conditionally at x = -7; absolutely on ( !7,1)
"
c)
n!(x ! 4) n
2n n 3
n=1
#
!
Radius = 0. Converges absolutely only at x = 4.
d)
"
(x + 2) n
6n
n=1
Radius = 6. Converges absolutely on ( !8, 4 )
e)
" x 2 ! 4 %n
)$# 5 '&
n=1
Radius = 3. Converges absolutely on ( !3, 3)
f)
"
(
!
ln n n
x
n=1 n
3) Suppose we know
Radius = 1. Converges conditionally at x = -1;, absolutely on ( !1,1)
"
# a (x ! 2)
n
n
converges absolutely at x = -3 and diverges at x = 10. What, if anything, can be said about:
n= 0
a) Convergence at x = 8?
From the given information, we know the radius of convergence is at least 5 from the
center of 2, but 8 is 6 units away from the center, so we can’t tell.
b) Absolute convergence at x = 5?
c) Convergence at x = 7?
Since 5 is within the radius of convergence, we know it converges absolutely.
If the radius of convergence is 5 (based on the fact that it converges at -3), then it still may or
may not converge at the other endpoint of 7, so again we can’t tell.
4) Use the formula for the sum of a convergent geometric series to find a power series for the following functions.
Also find the interval of convergence for each one.
!
1
a)
= 1 + x 3 + x 6 + x 9 + ... = " x 3n ; Converges on (-1, 1)
3
1! x
n=0
b)
"
1
2
4
6
1
!
4x
+
16x
!
64x
+
...
=
=
# !4x 2
1+ 4 x 2
n=0
c)
!
x2
2
6
10
14
x
+
x
+
x
+
x
+
...
=
x2 x4
=
"
1! x 4
n=0
(
( )
n
)
n
; Converges on ( ! 12 , 12 )
; Converges on ( !1,1)
5a) Find a power series for f (x) =
f (x) =
1
and find the interval of convergence.
1+ x 2
"
1
= 1 ! x 2 + x 4 ! x 6 + ... = # !x 2
2
1+ x
n=0
(
)
n
, which converges on ( !1,1)
b) Integrate the power series you found in part a term by term to find a new power series.
What function does your new power series represent? Does the interval of convergence change?
1
! 1+ x
2
dx = ! 1 " x 2 + x 4 " x 6 + ... dx
tan !1 x = x !
x3 x5 x7
+
!
+ ... =
3
5
7
(!1)n x 2n +1
, which still converges on ( !1,1) , but also converges
n = 0 (2n + 1)
conditionally at the endpoints x = -1 and 1.
"
#
"
"
x 3 2x 5 17x 7
+
+
+ ... , which converges for ! < x < , to find the first 4 terms of
2
2
3 15
315
a power series for sec 2 x . For what values of x should the full series converge?
6) Use the series tan x = x +
$
2x 4 17x 6
d
d !
x 3 2x 5 17x 7
+
(first 4 terms)
+
+ ...& = 1 + x 2 +
( tan x ) = # x + +
3
45
dx
dx "
3 15
315
%
The full series should also converge on !
"
"
< x < according to the Term-by-Term Differentiation Theorem.
2
2