بسم الله الرحمن الرحيم
Power Series
We will see that several important types of functions, including f (x) = 𝑒 𝑥 , can be
represented exactly by an infinite series called a power series.
𝑥
𝑥2 𝑥3 𝑥𝑛
𝑒 = 1 + 𝑥 + + + ⋯+
2! 3! 𝑛!
For each real number x, it can be shown that the infinite series on the right
converges ( )تتقاربto the number 𝑒 𝑥 . Before doing this, however, some
preliminary results dealing with power series will be discussed. beginning with
the next definition.
Definition of Power Series
If x is a variable, then an infinite series of the form
∞
∑ 𝑎𝑛 𝑥 𝑛 = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3 + ⋯ + 𝑎𝑛 𝑥 𝑛 + ⋯
𝑛=0
is called a power series. More generally, an infinite series of the form
∞
∑ 𝑎𝑛 (𝑥 − 𝑐)𝑛 = 𝑎0 + 𝑎1 (𝑥 − 𝑐) + 𝑎2 (𝑥 − 𝑐)2 + ⋯ + 𝑎𝑛 (𝑥 − 𝑐)𝑛 + ⋯
𝑛=0
is called a power series centered at c, where c is a constant.
Radius and Interval of Convergence
A power series in x can be viewed as a function of x
∞
𝑓(𝑥) = ∑ 𝑎𝑛 (𝑥 − 𝑐)𝑛
𝑛=0
where the domain of f is the set of all x for which the power series converges.
Of course, every power series converges at its center c because:
𝑓(𝑐) = ∑∞ 𝑛 ∞ 𝑛
𝑛=0 𝑎𝑛 (𝑐 − 𝑐) = ∑𝑛=0 𝑎𝑛 ⋅ 0 = 𝑎0 ⋅ 1 + 𝑎1 ⋅ 0 + 𝑎2 ⋅ 0 + ⋯ = 𝑎0
THEOREM
For a power series centered at c, precisely
one of the following is true.
1. The series converges only at c.
2. There exists a real number R > 0 such
that the series converges
absolutely for
|𝑥 − 𝑐| < 𝑅
and diverges for
|𝑥 − 𝑐| > 𝑅
3. The series converges absolutely for all x.
* The number R is the radius of convergence of the power series.
To determine the radius of convergence of a power series, use the Ratio Test.
The Ratio Test
Let ∑ 𝑎𝑛 be a series with nonzero terms.
1. The series ∑ 𝑎𝑛 converges absolutely when
𝑎𝑛+1
lim | |<1
𝑛→∞ 𝑎𝑛
2. The series ∑ 𝑎𝑛 diverges when
𝑎𝑛+1 𝑎𝑛+1
lim | |>1 or lim | |=∞
𝑛→∞ 𝑎𝑛 𝑛→∞ 𝑎𝑛
3. The Ratio Test is inconclusive when
𝑎𝑛+1
lim | |=1
𝑛→∞ 𝑎𝑛
2𝑛
Example: Determine the convergence or divergence of ∑∞
𝑛=0 𝑛!
𝑎𝑛+1 2𝑛+1 𝑛! 2
lim = lim ⋅ = lim =0<1
𝑛→∞ 𝑎𝑛 𝑛→∞ (𝑛 + 1)! 2𝑛 𝑛→∞ 𝑛 + 1
Example(1): Finding the Radius of Convergence
Find the radius of convergence of ∑∞
𝑛=0 𝑛! 𝑥
𝑛
Solution
For x = 0, you obtain
∞
𝑓(0) = ∑ (𝑛! ∗ 0𝑛 ) = 1 + 0 + 0 + ⋯ = 1
𝑛=1
For any fixed value of x such that |x| > 0, let 𝑢𝑛 = 𝑛! 𝑥 𝑛 . Then
𝑢𝑛+1 (𝑛 + 1)! ⋅ 𝑥 𝑛+1
= lim = |𝑥| ⋅ lim (𝑛 + 1) = ∞
𝑢𝑛 𝑛→∞ 𝑛! ⋅ 𝑥 𝑛 𝑛→∞
Therefore, by the Ratio Test, the series diverges for |x| > 0 and converges only
at its center, 0. So, the radius of convergence is R = 0.
Example(2): Find the radius of convergence of ∑∞
𝑛=0 3(𝑥 − 2)
𝑛
Solution
𝑢𝑛+1 3(𝑥 − 2)𝑛+1
= lim = lim (𝑥 − 2) = |𝑥 − 2|
𝑢𝑛 𝑛→∞ 3(𝑥 − 2)𝑛 𝑛→∞
By the Ratio Test, the series converges for |x – 2| < 1 and diverges for
|𝑥– 2| > 1. Therefore, the radius of convergence of the series is R = 1.
Example(3): Find the radius of convergence of ∑∞
𝑛=0 3(𝑥 − 2)
𝑛
(−1)𝑛 𝑥 2𝑛+1
Find the radius of convergence of ∑∞
𝑛=0 (2𝑛+1)!
.
Solution
𝑢𝑛+1 (−1)𝑛+1 𝑥 2𝑛+3 (2𝑛 + 1)! 𝑥2
= ⋅ =−
𝑢𝑛 (2𝑛 + 3)! (−1)𝑛 𝑥 2𝑛+1 (2𝑛 + 3)(2𝑛 + 2)
For any fixed value of x, this limit is 0. So, by the Ratio Test, the series
converges for all x. Therefore, the radius of convergence is R = ∞.
Endpoint Convergence
the interval of convergence of a power series can take any one of the six
forms:
𝑥𝑛
Example: Find the interval of convergence of ∑∞
𝑛=1 𝑛2
Solution
𝑥𝑛 𝑥 𝑛+1
𝑢𝑛 = 2 , 𝑢𝑛+1 =
𝑛 (𝑛 + 1)2
𝑢𝑛+1 𝑥 𝑛+1 𝑛2 𝑛2
= ⋅ =𝑥⋅ = |𝑥|
𝑢𝑛 (𝑛 + 1)2 𝑥 𝑛 (𝑛 + 1)2
So, the radius of convergence is R = 1. Because the series is centered at x = 0, it
converges in the interval (−1, 1). When x = 1, you obtain the convergent p-
series
∞
1 1 1 1 1
∑ = + + + +⋯ This series converges when 𝑥 = 1.
𝑛 2 12 22 32 42
𝑛=1
When x = −1, you obtain the convergent alternating series
∞
(−1)𝑛 1 1 1 1
∑ = − + − + − ⋯ This series converges when 𝑥 = −1.
𝑛2 12 22 32 42
𝑛=1
Therefore, the interval of convergence is [−1, 1].
Differentiation and Integration of Power Series
Once you have defined a function with a power series, it is natural to wonder
how you can determine the characteristics of the function. Is it continuous?
Differentiable?
THEOREM: Properties of Functions Defined by Power Series
If the function
∞
𝑓(𝑥) = ∑ 𝑎𝑛 (𝑥 − 𝑐)𝑛 = 𝑎0 + 𝑎1 (𝑥 − 𝑐) + 𝑎2 (𝑥 − 𝑐)2 + 𝑎3 (𝑥 − 𝑐)3 + ⋯
𝑛=0
has a radius of convergence of R > 0, then, on the interval (c − R, c + R) then f is
differentiable (and therefore continuous). Moreover, the derivative and the
antiderivative of f are as follows.
1- 𝑓 ′ (𝑥) = ∑∞
𝑛=1 𝑛 𝑎𝑛 (𝑥 − 𝑐)
𝑛−1
= 𝑎1 + 2𝑎2 (𝑥 − 𝑐) + 3𝑎3 (𝑥 − 𝑐)2 + 4𝑎4 (𝑥 − 𝑐)3 +..
𝑎𝑛 (𝑥−𝑐)𝑛+1 𝑎1 (𝑥−𝑐)2 𝑎2 (𝑥−𝑐)3
2- ∫ 𝑓(𝑥) 𝑑𝑥 = 𝐶 + ∑∞
𝑛=0 = 𝐶 + 𝑎0 (𝑥 − 𝑐) + + +⋯
𝑛+1 2 3
• The radius of convergence of the series obtained by differentiating or
integrating a power series that of the original power
series.
• The interval of convergence, however, may differ as a result of the
behavior at the endpoints.
Example
∞
𝑥𝑛 𝑥2 𝑥3 𝑥4
𝑓(𝑥) = ∑ =1+𝑥+ + + +
𝑛! 2! 3! 4!
𝑛=0
′ (𝑥)
2𝑥 3𝑥 2 4𝑥 3 𝑥2 𝑥3 𝑥4
𝑓 =1+ + + + ⋯ = 1 + 𝑥 + + + + ⋯ = 𝑓(𝑥)
2 3! 4! 2 3! 4!
𝑥2 𝑥3 𝑥2 𝑥3 𝑥4 𝑥5
∫ 𝑓(𝑥) 𝑑𝑥 = ∫ (1 + 𝑥 + + + ⋯ ) 𝑑𝑥 = 𝐶 + 𝑥 + + + + +⋯
2! 3! 2 3 ⋅ 2! 4 ⋅ 3! 5 ⋅ 4!
𝑥2 𝑥3 𝑥4 𝑥5
= 𝐶 + 𝑥 + + + + + ⋯ = 𝐶 + 𝑓(𝑥)
2! ! 3 4! 5!
Do you recognize this function?
Intervals of Convergence for 𝒇(𝒙), 𝒇′ (𝒙), and ∫ 𝒇(𝒙) 𝒅𝒙
𝑥𝑛
Consider the function 𝑓(𝑥) = ∑∞
𝑛=1 Find the interval of convergence for
𝑛
each of the following. a- 𝑓(𝑥) b- 𝑓 ′ (𝑥) c- ∫ 𝑓(𝑥) 𝑑𝑥
Solution
a- For f (x), the series Interval of convergence: [−1, 1)
b- For f' (x), the series Interval of convergence: (−1, 1)
c- For ∫ 𝑓(𝑥), the series Interval of convergence: [−1, 1]
Operations with Power Series
The operations described above can change the interval of convergence for the
resulting series. For example, in the addition shown below, the interval of
convergence for the sum is the intersection of the intervals of convergence of
the two original series.
∞ ∞ ∞
𝑥 𝑛 1
∑ 𝑥 + ∑ ( ) = ∑ (1 + 𝑛 ) 𝑥 𝑛
𝑛
2 2
𝑛=0 𝑛=0 𝑛=0
Convergence: (−1,1) ∩ (−2,2) = (−1,1)
Identity Property If ∑∞ 𝑛
𝑛=0 𝑐𝑛 (𝑥 − 𝑎) = Analytic at a Point A function f is analytic at
0, 𝑅 > 0 for all numbers x in the interval a point a if it can be represented by a
of convergence, then 𝑐𝑛 = 0 for all n. power series in x - a with a positive or
infinite radius of convergence.