2. So, start asking yourself this question:
Can x be 0, 1
/2 , –1
/2, 3
/4, –3
/4, 1, –1, 3
/2, –3
/2, 2, –2, and so on?
Power Series—Introduction
A power series is a series of the form Σcnxn = c0 + c1x + c2x2 + c3x3 + ...n = 0
∞
Example 1:
Say cn = 1 for all n, we have Σxn
as our first power series.n = 0
∞
Well, what we really want to know is: for what values of x will
Σxn
converge?n = 0
∞
Yes, when x is any of
these numbers, Σxn
becomes convergent
geometric series
since |r| < 1
n = 0
∞
No, when x is any of
these numbers, Σxn
becomes divergent
geometric series
since |r| ≥ 1
n = 0
∞
Conclusion:
Σxn
converges when x is any number between –1 and 1, i.e., –1 < x < 1,
or the open interval: (–1, 1).
n = 0
∞
So?
3. Power Series—Interval and Radius of Convergence
Recall our power series Σxn
converges for x ∈ (–1, 1).
Therefore, we say (–1, 1) is the interval of convergence (IOC) and we
define the radius of convergence (ROC) as half of the length of IOC.
Since the length of the interval
(i.e., the distance from –1 to 1)
is 2, therefore, the ROC is 1.
n = 0
∞
–1 0 1
Interval of Convergence = (–1, 1)
Radius of Convergence = 1
Again, ask yourself this question:
Can x be 0, 1
/2 , –1
/2, 3
/4, –3
/4, 1, –1, 3
/2, –3
/2, 2, –2, and so on?
Yes, if x is 0, it
will make every
term 0
Example 2:
Σn!xn
n = 0
∞
No, if x is any of these
numbers, Σn!xn
will divergen = 0
∞
IOC = {0}
–1 0 1
ROC = 0
4. Power Series—The Three Types of IOCs
Once more, ask yourself this question:
Can x be 0, 1
/2 , –1
/2, 3
/4, –3
/4, 1, –1, 3
/2, –3
/2, 2, –2, and so on?
–2 0 2–1 1
Interval of Convergence = (–∞, ∞)
Radius of Convergence = ∞
Yes, if x is any of these numbers, Σxn
/n! will convergen = 0
∞
he Three Types of IOCs:
or any given power series Σcnxn
, there are only three possibilities:
1) The series converges on an interval with a finite length (Example 1)
2) The series converges at only one number (Example 2)
3) The series converges for all real numbers (Example 3)
n = 0
∞
Example 3:
Σxn
/n!n = 0
∞
5. Power Series—More on IOCs and ROCs
IOC ROC
[1, 4) ½(3) = 1.5
(–2, 2] ½(4) = 2
Examples:
1)
2)
3)
0 2 41 3
–2 1 3–1 20
–2 1 3–1 20
{½} 0
6. Power Series—How to Guess the IOC?
One number, in particular, will obviously make the series converge.
What is this number? [Hint: this number will make every term = 0.]
Example 1:
1
( 3)n
n
x
n
∞
=
−
∑
x = 3
Of course, x may be other numbers too. If so, on the number line,
we begin at 3 and move to the right and left to obtain our interval of
convergence.
0 2 41 3
Therefore, the IOC is [2, 4), and ROC = 1.
∑
∞
=1
1
n n
If x = 4, then the series
becomes which is
a divergent series
1
( 1)n
n n
∞
=
−
∑
If x = 2, then the series
becomes which
is a convergent series
7. Power Series—How to Guess the IOC? (cont’d)
What is the number that makes every term = 0?
x = 1
Again, on the number line, we begin at 1 and move to the right and
left to obtain our interval of convergence.
0 2 41 3
Therefore, the IOC is [0, 2], and ROC = 1.
If x = 2, then the series
becomes which is
a convergent series
2
1
1
n n
∞
=
∑
If x = 0, then the series
becomes which
is a convergent series
2
1
( 1)n
n n
∞
=
−
∑
Example 2:
2
1
( 1)n
n
x
n
∞
=
−
∑
8. Power Series—How to Find the IOC in General?
Answer: Kind of like doing the Ratio Test
Example 1:
1
( 3)n
n
x
n
∞
=
−
∑ n
x
a
n
n
)3( −
=
1
1
( 3)
1
n
n
x
a
n
+
+
−
=
+
( )
1
1
1
31
1 ( 3)
n
n
n n
n n
xa n
a
a a n x
+
+
+
−
= × = ×
+ −
( 3)
lim lim 3 1 3 3
1 1n n
n x n
x x x
n n→∞ →∞
−
= × − = × − = −
+ +
|x – 3| < 1
–1 < x – 3 < 1
2 < x < 4
+3 + 3 +3
and
Step 1: Find :1n
n
a
a
+
Step 2: Find :1
lim n
n
n
a
a
+
→∞
Step 3: Set above limit < 1 and solve for x:
Step 4: Check whether endpoints work or not:
x – 3
1
( 1)n
n n
∞
=
−
∑If x = 2, the series becomes which is a convergent series, so 2 is included.
∑
∞
=1
1
n n
If x = 4, the series becomes which is a divergent series, so 4 should be excluded.
∴IOC is [2, 4), and ROC = 1.
( 3)
1
n x
n
−
=
+
9. Power Series—How to Find the IOC in General? (cont’d)
Recap:
2) We then check the endpoints (to see whether they will make the
power series converges or not) by substituting each endpoint into
the series. (Step 4)
1
lim n
n
n
a
a
+
→∞
1) We find and set it < 1, and solve for x. (Steps 1–3)
Example 2:
2
1
(2 1)n
n
x
n
∞
=
−
∑
Example 3:
2
0
( 1) (2 )
!
n n
n
x
n
∞
=
−
∑
Example 4:
1
0
( 2)
3
n
n
n
n x∞
+
=
+
∑
10. Power Series—How to Find the IOC in General? (cont’d)
Recap:
2) We then check the endpoints (to see whether they will make the
power series converges or not) by substituting each endpoint into
the series. (Step 4)
1
lim n
n
n
a
a
+
→∞
1) We find and set it < 1, and solve for x. (Steps 1–3)
Example 2:
2
1
(2 1)n
n
x
n
∞
=
−
∑
Example 3:
2
0
( 1) (2 )
!
n n
n
x
n
∞
=
−
∑
Example 4:
1
0
( 2)
3
n
n
n
n x∞
+
=
+
∑