# Wednesday, March 23 Announcements

```Wednesday, March 23
Announcements
If you submitted your Quiz #4 for a regrade, come pick it up
from me (before or after class)
WeBWorK #10 is due at 9pm
Covers material from “Week 10”
No class on Friday March 25 or Monday March 28
WeBWorK #11 is due next Wednesday (March 30) at 9pm
Covers material from “Week 11”—see syllabus on course
web page
Solutions to (all versions of) Quiz #5 are online
Quiz #5 papers will be available in the MLC starting
tomorrow
Grades posted in Connect by Friday
Wednesday, March 23
Clicker Questions
Clicker Question 1
Suppose the power series
∞
X
cn (x − 3)n converges when
n=0
x = −2 and diverges when x = −5. Of the values
x = −6, x = −4, x = 1, x = 4, x = 7, x = 9, x = 12, where can
we be sure that the series converges?
A. at x = 1, x = 4, and x = 7
Close enough to 3
B. at x = −4, x = 1, and x = 4
The series converges at
x = −2, so the radius of
convergence is at least
|(−2) − 3| = 5. The series
diverges at x = −5, so the
most |(−5) − 3| = 8. . . .
C. at x = −4, x = 1, x = 4,
x = 7, and x = 9 (maybe)
D. only at x = 1
E. none of the above
Clicker Question 2
Power series representation
x5
Find a power series that represents 2
on the interval
x +3
√ √
(− 3, 3).
A.
B.
C.
D.
∞
X
(−1)2n+5
n=0
∞
X
n=0
∞
X
n=0
∞
X
n=0
32n+6
xn
(−1)n 5n
√
x
( 3)n+1
(−1)n n
√
x
( 3)n+1
(−1)n
3n+1
x2n+5
E. none of the
above
Using a previous example
∞
X (−1)n
1
=
xn
x+3
3n+1
(|x| &lt; 3)
n=0
1
=
x2 + 3
x5
x2 + 3
=
∞
X
(−1)n
n=0
3n+1
∞
X
(−1)n
n=0
3n+1
(x2 )n
(x2 )n x5
(|x2 | &lt; 3)
(|x| &lt;
√
3)
```