POWER

SERIES
TOPIC OUTLINE:

• • •
• • • • •
•

• •
•
1.0 INTRODUCTION
A power series is an infinite series in the form:
∞
𝒏
𝒇 𝒙 =෍ 𝒂𝒏 𝒙 − 𝒄
𝒏=𝟎

where 𝒂𝒏 are constants called the coefficients of the series, 𝒄 is a fixed number called the center
of the series, and 𝒙 is a variable. Power series are often used in mathematics to represent
functions as an infinite sum of terms. They are particularly useful in calculus, where they can be
used to represent functions that may be difficult to express otherwise.

Power series are fundamental in many areas of mathematics, including calculus, differential
equations, and complex analysis, and they have wide-ranging applications in physics,
engineering, and other scientific fields.
1.0 INTRODUCTION
When you encounter a differential equation that cannot be solved using traditional methods,
you may resort to power series methods as an alternative approach. The idea is to express the
solution of the differential equation as a power series, where the coefficients of the series are
determined by the differential equation itself. This is often done for equations with variable
coefficients, singular points, or nonlinear terms.
ODEs Solvable by Specific Methods: ODEs Requiring Power Series Methods:

1. Linear First-Order ODE w/ constant Coefficient 1. Non Linear ODE

𝑑𝑦
+ 2𝑦 = 3𝑥 𝑦 ′′ + 𝑥𝑦 = 0
𝑑𝑥

(Here, the nonlinear term xy makes it difficult to solve using traditional methods.
(This equation can be solved using an integrating factor method or by
Power series methods can be applied to express the solution in series)
variable separable)
2. Variable Coefficient
2. Variable Coefficient

′′ 1 + 𝑥 𝑦 ′′ − 2𝑥𝑦 ′ + 𝑦 = 0
𝑦 − 3𝑦 + 2𝑦 = 0
(The varying coefficients 1+x and -2x make this equation challenging for the
(This equation can be solved using homogenous equation approach)
specific methods of DE)
3. Singular Points
3. Singular Points
2𝑥 + 1 𝑑𝑥 + 2𝑦 − 𝑥 𝑑𝑦 = 0 𝑦
𝑦 ′′ + =0
𝑥
(This equation can be solved using exact differential equation)
(This equation has a singularity at x=0, making it unsuitable for standard
methods. Power series centers at x=0 can provide a solution in the vicinity of this
singularity)
1.0 INTRODUCTION
The general form of a power series expansion around a center 𝒄 is:

𝒇 𝒙 = σ∞
𝒏=𝟎 𝒂𝒏 𝒙 − 𝒄
𝒏

𝒇 𝒙 = 𝒂𝟎 + 𝒂𝟏 𝒙 − 𝒄 +𝒂𝟐 (𝒙 − 𝒄)𝟐 +𝒂𝟑 (𝒙 − 𝒄)𝟑 …

Here 𝑎0 , 𝑎1 , 𝑎2 , … are constants, and each term involves powers of (𝑥 − 𝑐)

For example, 𝒄 = 0 if, the expansion simplifies to:

𝒇 𝒙 = 𝒂𝟎 + 𝒂𝟏 𝒙 +𝒂𝟐 𝒙𝟐 +𝒂𝟑 𝒙𝟑 …

Power series can converge (meaning they approach a finite limit) or diverge (meaning they
don't approach any finite limit) for different values of 𝒙. The interval of convergence is the set of
all 𝒙 values for which the power series converges. This interval may include or exclude the center
𝒄, and its radius of convergence is a measure of how wide the interval is.
1.0 INTRODUCTION
When a series converges, it means that as you add more and more terms, the sum of those terms approaches a
finite value. In practical terms, if you substitute values of 𝒙 within the interval of convergence into the series, you'll
get a meaningful result. If the series is divergent for a certain value of 𝒙, it means that the series fails to converge
to a finite value for that specific 𝒙. In many applications, having a divergent series is not acceptable because it
implies that the mathematical model represented by the series doesn't behave as expected for that value of 𝒙.
For example:

1. Loss of Predictive Power: If you're using the series to model some real-world phenomenon, a divergent series
might indicate that your model doesn't accurately represent the behavior of the system under consideration.
This loss of predictive power can be problematic if you're relying on the model to make predictions or decisions.

2. Invalid Results: In mathematical computations or scientific simulations, encountering a divergent series might
lead to invalid results. It indicates that the quantity being modeled behaves in a way that's not consistent with
the assumptions made in constructing the series.

3. Breakdown of Methodology: Divergent series can sometimes arise due to improper application of
mathematical techniques or assumptions. In such cases, it's crucial to revisit the methodology used and possibly
revise it to ensure that the results obtained are valid and meaningful.
1.1 RADII AND INTERVAL OF CONVERGENCE
Radii also referred to as the radius of convergence. In the context of power series, the "radius of convergence"
refers to a critical property that determines within which range of values the power series converges.

For a given power series:

∞
𝒇 𝒙 =෍ 𝒂𝒏 𝒙 − 𝒄 𝒏
𝒏=𝟎

the radius of convergence, denoted by 𝑹, is a non-negative real number or +∞ such that the series converges
for all 𝒙 within the interval of convergence (𝒄 − 𝑹, 𝒄 + 𝑹) and diverges for all 𝒙 outside this interval.

Test for convergence

1. Ratio Test
2. P-Series Test
3. Alternating Series Test
4. Integral Test
RADII AND INTERVAL OF CONVERGENCE
TWO STEP PROCESS:
1. Apply an infinite series convergence test. This gives us an open interval of convergence.
2. Check the two end points to determine if the series converges or diverges. This information will give us the
interval of convergence

RATIO TEST
𝑎𝑛+1
𝐿 = lim
𝑛→∞ 𝑎𝑛

Then:

If 𝑳 < 𝟏 , the series converges

If 𝑳 > 𝟏 or 𝑳 = +∞ , the series diverges
If 𝑳 = 𝟏, the test is inconclusive

The ratio test is used to determine the radius of convergence.

RADII AND INTERVAL OF CONVERGENCE
P-SERIES TEST ALTERNATING SERIES TEST INTEGRAL TEST
∞
(AST) 1 1 𝑒 −𝑛 𝒍𝒏 (𝒏) 𝒄𝒐𝒔 (𝒏) 𝒔𝒊𝒏 (𝒏)
1 ∞
σ𝑛=1 , , 𝟐 , 𝟐 , 𝟐
෍ 𝑝 𝑛 𝑛+1 𝑛2 𝒏 𝒏 𝒏
𝑛=1 𝑛 ∞
෍ (−1)𝑛 … ∞
𝑛=1
Where 𝑝 is a real number න 𝑓 𝑥 𝑑𝑥
𝑛−1
(−1) 1
The p-series states: Supposed you have σ∞
𝑛=1 𝑛
Convergent: If the integral
∞
If 𝒑 > 𝟏 , the series converges Conditions: ‫׬‬1 𝑓 𝑥 𝑑𝑥 converges to a finite
If 𝒑 ≤ 𝟏 , the series diverges value
1. lim 𝑎𝑛 = 0
𝑛→∞
Divergent: If the integral
∞
2. 0 < 𝑎𝑛+1 ≤ 𝑎𝑛 ‫׬‬1 𝑓 𝑥 𝑑𝑥 approaches to ±∞

This two condition must be

satisfied to consider the series to
be convergent
PRACTICE PROBLEMS
Examples:
Find the radius and interval of convergence of the following power series.

∞ ∞ 2𝑛
𝑛 𝑛 𝑥+1
෍ −1 𝑛𝑥 ෍
1. 𝑛=1 4. 𝑛=1 𝑛!

∞ −1 𝑛 𝑥 𝑛 ∞ 5𝑥 𝑛
෍ ෍
2. 𝑛=1 𝑛 5. 𝑛=1 𝑛8

∞ 𝑥−2 𝑛
෍ 𝑛
3. 𝑛=1 (𝑛 + 1)(3 )
PRACTICE PROBLEMS
Examples:
Find the radius and interval of convergence of the following power series.

∞ 2𝑛 𝑥 𝑛 (𝑛 + 1) 1 1 1 ∞ 𝑥−7 𝑛
෍ 𝑅= , 𝐼(− , ) ෍ 𝑅 = 7, 𝐼(0, 14)
1. 𝑛=1 𝑛+9 2 2 2 4. 𝑛=1 7𝑛

∞ −1 𝑛 𝑥 𝑛
∞ 𝑥−6 𝑛
෍ 𝑅 = 8, 𝐼[−8, 8 ] ෍ 𝑅 = 6, 𝐼(3, 9]
2. 𝑛 2 5. 𝑛 −3 𝑛
𝑛=1 8 (𝑛 + 9) 𝑛=1

∞ −1 𝑛 𝑥 𝑛 10𝑛 1 1 1
෍ 𝑅= , 𝐼(− , )
3. 𝑛=1 𝑛+2 10 10 10
POWER SERIES OPERATIONS
Let 𝑓 𝑥 = σ∞ 𝑎
𝑛=0 𝑛 𝑥 𝑛 and g 𝑥 = σ∞ 𝑏 𝑥 𝑛
𝑛=0 𝑛

1. 𝑓 𝑥 ± 𝑔 𝑥 = σ∞ 𝑎
𝑛=0 𝑛 ± 𝑏𝑛 𝑥 𝑛

2. 𝑓 𝑘𝑥 = σ∞ 𝑎
𝑛=0 𝑛 𝑘 𝑛𝑥𝑛

3. 𝑓 𝑥 𝑁 = σ∞
𝑛=0 𝑎𝑛 𝑥
𝑛𝑁

For addition and subtraction of series. All we need to remember is that the two series start at same place and
both have the same exponent (n) of the 𝑥 − 𝑥0 𝑛 .
POWER SERIES OPERATIONS
INDEX SHIFTING

Index Shifting concerned with the exponent and where the series starts.

Example 1: Perform Index Shifting to the following equations.

∞
𝑓 𝑥 = ෍ 𝑛2 𝑎𝑛−1 𝑥 − 4 𝑛+2
𝑛=3

first is we let 𝑖 = 𝑛 − 3
Such that when 𝑛 = 3, 𝑖 = 0
The0n we rewrite all terms instead of “n” we use “I”
Hence 𝑛 = 𝑖 + 3

𝑓 𝑥 = σ∞ 2
𝑖=0(𝑖 + 3) 𝑎(𝑖+3)−1 𝑥 − 4
(𝑖+3)+2 𝑓 𝑥 = σ∞ 2
𝑖=0(𝑖 + 3) 𝑎(𝑖+2) 𝑥 − 4
(𝑖+5)

𝑓 𝑥 = σ∞𝑛=0(𝑛 + 3) 2𝑎
(𝑛+2) 𝑥 − 4 (𝑛+5)

Note: The upper limit wont change in this

process since infinity minus a finite value is still
infinity
POWER SERIES OPERATIONS
INDEX SHIFTING

Example 2: Rewrite the series such that n starts at 5 instead of n=3

∞
𝑓 𝑥 =෍ 𝑛2 𝑎𝑛−1 𝑥 + 4 𝑛+2
𝑛=3

∞
𝑓 𝑥 =෍ (𝑛 − 2)2 𝑎(𝑛−2)−1 𝑥 + 4 (𝑛−2)+2 Since we add 2 to the initial start of the series we need to
𝑛=3+2 minus 2 in every n-term so we have (𝑛 − 2)

∞
𝑓 𝑥 =෍ (𝑛 − 2)2 𝑎𝑛−3 𝑥 + 4 𝑛
𝑛=5
POWER SERIES OPERATIONS
INDEX SHIFTING

Example 3: Rewrite the series as a single series in terms of 𝑥 − 𝑥0 𝑛

∞ ∞
2 𝑛−4 𝑛+1
𝑓 𝑥 = 𝑥+2 ෍ 𝑛𝑎𝑛 𝑥 + 2 − ෍ 𝑛𝑎𝑛 𝑥 + 2
𝑛=3 𝑛=1

𝑓 𝑥 = σ∞
𝑛=3 𝑛𝑎𝑛 𝑥 + 2
𝑛−4+2 − σ∞
𝑛=1 𝑛𝑎𝑛 𝑥 + 2
𝑛+1

∞ ∞
𝑓 𝑥 =෍ 𝑛𝑎𝑛 𝑥 + 2 𝑛−2 − ෍ 𝑛𝑎𝑛 𝑥 + 2 𝑛+1 As specified in the problem the series should contain
𝑛=3 𝑛=1 𝑥 − 𝑥0 𝑛

∞ ∞
𝑓 𝑥 =෍ (𝑛 + 2)𝑎𝑛+2 𝑥 + 2 𝑛−2+2 − ෍ (𝑛 − 1)𝑎𝑛−1 𝑥 + 2 𝑛+1−1
𝑛=3−2 𝑛=1+1

∞ ∞
𝑓 𝑥 =෍ (𝑛 + 2)𝑎𝑛+2 𝑥 + 2 𝑛 − ෍ (𝑛 − 1)𝑎𝑛−1 𝑥 + 2 𝑛
𝑛=1 𝑛=2
∞ ∞ ∞
𝑛 𝑛 𝑓 𝑥 =෍ [ 𝑛 + 2 𝑎𝑛+2 − 𝑛 − 1 𝑎𝑛−1 ] 𝑥 + 2 𝑛
𝑓 𝑥 =෍ (𝑛 + 2)𝑎𝑛+2 𝑥 + 2 − ෍ (𝑛 − 1)𝑎𝑛−1 𝑥 + 2
𝑛=1 𝑛=1 𝑛=1
GEOMETRIC SERIES
A Geometric series in summation represents the sum of the terms of a geometric sequence. It has the form:

∞ 𝑎
෍ 𝑎𝑟 𝑛 =
𝑛=0 1−𝑟

where 𝑟 < 1 for convergence

𝑎
1. To write the given function in the form 𝑓 𝑥 = to find the value of “a” and “r”
1−𝑟
2. Write the function as a power series. If the power series is not centered at zero (0), replace 𝑥 with (𝑥 − 𝑐) and
then undo the addition or subtraction.
GEOMETRIC SERIES
Example 1: Determine a power series for the given function. Also determine the open interval of convergence.

1
𝑓 𝑥 =
1−𝑥

Centered at 0
𝟏 ∞
𝑎 𝒇 𝒙 = 𝒇 𝒙 =෍ 𝑥𝑛
Remember the general equation of geometric series σ∞ 𝑛
𝑛=0 𝑎𝑟 = 𝟏−𝒙 𝑛=0
1−𝑟

Hence a=1 and r=x Interval of Convergence

1
𝑓 𝑥 = =σ∞
𝑛=0 1 𝑥
𝑛
−1 < 𝑥 < 1
1−𝑥

∞ 0−1< 𝑥 <0+1
𝑓 𝑥 =෍ 𝑥𝑛
𝑛=0
−1 < 𝑥 < 1 I (−1,1)
GEOMETRIC SERIES
Example 2: Determine a power series for the given function. Also determine the open interval of convergence.

2
𝑓 𝑥 =
𝑥+4

Centered at 0
𝑎
Remember the general equation of geometric series σ∞
𝑛=0 𝑎𝑟 𝑛 =
1−𝑟

Rewrite your function 𝟐 ∞ −1 𝑛 𝑥 𝑛

𝒇 𝒙 = 𝒇 𝒙 =෍
𝑿−𝟒 𝑛=0 2 (4𝑛 )
2
𝑓 𝑥 = Hence Interval of Convergence
4 − (−𝑥)

2
1 𝑥 𝑥
𝑎= 𝑟= − −1 < <1
2 4
4 4
𝑓 𝑥 =
4 −𝑥 𝑥 𝑛
−( ) 𝑓 𝑥 = σ∞
1
− 0−4< 𝑥 <0+4
4 4 𝑛=0 2 4

1 −4 < 𝑥 < 4 I (−4,4)

∞ −1 𝑛 𝑥 𝑛
𝑓 𝑥 = 2 𝑓 𝑥 = ෍
−𝑥 𝑛=0 2 (4𝑛 )
1−( )
4
GEOMETRIC SERIES
Example 3: Determine a power series for the given function. Also determine the open interval of convergence.

6
𝑓 𝑥 =
2𝑥 + 1

Centered at 1
𝑎
Remember the general equation of geometric series σ∞
𝑛=0 𝑎𝑟 𝑛 =
Interval of Convergence
1−𝑟

Rewrite your function 2 2

𝑓 𝑥 = − (𝑥 − 1) < 1
2 𝑥−1 3
6 1 − [− ]
3
𝑓 𝑥 =
2 𝑥−1 −1+2 2
Hence − (𝑥 − 1) < 1
3
6
𝑓 𝑥 = 2
3 3
3 − [−2 𝑥 − 1 ] 𝑎=2 𝑟 = − (𝑥 − 1)
3 1− < − 𝑥−1 < 1+
2 2
6 2
𝑓 𝑥 = σ∞
𝑛=0 2 [− (𝑥 − 1)] 𝑛
1 5
𝑓 𝑥 = 3 3
3 2 𝑥−1 − < (𝑥 − 1) <
− [− ] 2 2
3 3 ∞ −1 𝑛 2𝑛+1 (𝑥 − 1)𝑛 1 5
I (− , )
𝑓 𝑥 = ෍ 2 2
𝑛=0 3𝑛
GEOMETRIC SERIES
Example 4: Determine a power series for the given function. Also determine the open interval of convergence.

𝑥+3
𝑓 𝑥 =
1 − 𝑥2

Centered at 0
𝑎
Remember the general equation of geometric series σ∞
𝑛=0 𝑎𝑟
𝑛 =
1−𝑟

Rewrite your function

2 1
𝑓 𝑥 = + Interval of Convergence
𝑥+3 𝐴 𝐵 1 − 𝑥 1 − (−𝑥)
𝑓 𝑥 = = +
(1 − 𝑥)(1 + 𝑥) (1−𝑥) (1+𝑥) For both 𝑥 < 1 and −𝑥 < 1
1𝑠𝑡 𝑇𝑒𝑟𝑚: 𝑎 = 2 ; 𝑟 = 𝑥
2𝑛𝑑 𝑇𝑒𝑟𝑚: 𝑎 = 1 ; 𝑟 = −𝑥 you get 𝐼(−1,1)
𝑥 + 3 = 𝐴 1 + 𝑥 + 𝐵(1 − 𝑥)
When x= -1 ∞ ∞
𝑓 𝑥 = ෍ 2𝑥 𝑛 + ෍ 1(−𝑥)𝑛
𝑛=0 𝑛=0
−1 + 3 = 𝐴 1 + (−1) + 𝐵(1 − (−1))
∞ ∞
𝐵=1 𝑓 𝑥 = ෍ 2𝑥 𝑛 + ෍ −1 𝑛 (𝑥)𝑛
When x= 1 𝑛=0 𝑛=0
∞
1 + 3 = 𝐴 1 + 1 + 𝐵(1 − 1) 𝑓 𝑥 = ෍ [2 + −1 𝑛 ] 𝑥 𝑛
𝐴=2 𝑛=0
TAYLOR SERIES AND MACLAURIN SERIES
In the Previous Section we started looking at writing down a power series representation of a function. The
problem with the approach in that section is that everything came down to satisfying or relating the function to
𝑎
. In P-series will assume that 𝑥 = 𝑥0 exist.
1−𝑟

P-Series is given by:

𝑓 𝑥 = σ∞
𝑛=0 𝑎𝑛 𝑥 − 𝑥0
𝑛
= 𝒂𝟎 + 𝒂𝟏 𝒙 − 𝒙𝟎 + 𝒂𝟐 𝒙 − 𝒙𝟎 𝟐
+ 𝒂𝟑 𝒙 − 𝒙𝟎 𝟑
+ 𝑎4 𝑥 − 𝑥0 4 …
𝑓′ 𝑥 = 𝒂𝟏 +2𝑎2 𝑥 − 𝑥0 + 3𝑎3 𝑥 − 𝑥0 2 + 4𝑎4 𝑥 − 𝑥0 3…

𝑓 ′′ (𝑥) = 2(1)𝑎2 +3 2 𝑎3 x − 𝑥0 + 4(3)𝑎4 𝑥 − 𝑥0 2…

𝑓 ′′′ (𝑥) = 3 2 (1)𝑎3 + 4 3 (2)𝑎4 (x − 𝑥0 ) …

𝑓 ′′′′ (𝑥) = 4 3 (2)(1)𝑎4

𝐻𝑒𝑛𝑐𝑒 𝑓 𝑥0 = 𝒂𝟎
𝑓′ 𝑥0 = 𝒂𝟏 𝒂𝟏 = 𝑓′ 𝑥0
𝑓 ′′ (𝑥0 )
𝑓′′ 𝑥0 = 𝟐(𝟏)𝒂𝟐 𝒂𝟐 =
2(1)
𝑓 ′′′ (𝑥0 )
𝑓′′′ 𝑥0 = 𝟑(𝟐)(𝟏)𝒂𝟑 𝒂𝟑 =
3(2)(1)

𝑓′′′′ 𝑥0 = 4(𝟑)(𝟐)(𝟏)𝒂𝟒 𝑓 ′′′′ (𝑥0 )

𝒂𝟒 =
4(3)(2)(1)
TAYLOR SERIES AND MACLAURIN SERIES
From this we can draw out the Taylor Series General Form:

∞ 𝒇𝒏 𝒙𝟎 𝒏
𝒇 𝒙 = σ𝒏=𝟎 𝒙 − 𝒙𝟎
𝒏!

𝒇′′ 𝒙𝟎 𝒙−𝒙𝟎 𝟐 𝒇′′′ 𝒙𝟎 𝒙−𝒙𝟎 𝟑

= 𝒇 𝒙𝟎 + 𝒇′(𝒙𝟎 )(𝒙 − 𝒙𝟎 ) + + +⋯
𝟐! 𝟑!

On the other hand the Maclaurin Series for 𝑓 𝑥 we let 𝑥0 = 0

∞ 𝒇𝒏 𝟎 𝒏
𝒇 𝒙 = σ𝒏=𝟎 𝒙
𝒏!

𝒇′′ 𝟎 𝒙 𝟐 𝒇′′′ 𝟎 𝒙 𝟑
= 𝒇 𝟎 + 𝒇′(𝟎)(𝒙) + + +⋯
𝟐! 𝟑!

To determine a condition that must be true in order for a Taylor Series to exist for a function lets first define the
nth degree Taylor Polynomial of 𝑓 𝑥

𝒇𝒊 𝒙𝟎
∞
𝒊
𝑻𝒏 𝒙 = ෍ 𝒙 − 𝒙𝟎
𝒊=𝟎 𝒊!
TAYLOR SERIES AND MACLAURIN SERIES
Table of Power Series Expansion
TAYLOR SERIES AND MACLAURIN SERIES
ACTIVITY: Using the Taylor Series or Maclaurin Series derived the Summation
and expansion of the following functions.

1. 𝑓 𝑥 = 𝑒 𝑥

2. 𝑓 𝑥 = cos 𝑥

3. 𝑓 𝑥 = sin 𝑥

1
4. 𝑓 𝑥 =
1+𝑥

5. 𝑓 𝑥 = 𝑡𝑎𝑛−1 𝑥
SOLVING ODE USING POWER SERIES
Examples: Use a power series to solve the differential equation.

1. 𝒚′′ + 𝒚 = 𝟎 ; 𝑥0 = 0

2. 𝒚′′ − 𝒙𝒚 = 𝟎 ; 𝑥0 = 0

3. 𝒚′′ − 𝒙𝒚 = 𝟎 ; 𝑥0 = −2

4. 𝒚′′ − 𝟐𝒙𝒚′ + 𝒚 = 𝟎 ; 𝑥0 = 0

5. 𝑥 2 + 1 𝑦 ′′ − 4𝑥𝑦 ′ + 6𝑦 = 0 ; 𝑥0 = 0
PERFORMANCE INNOVATIVE TASK
Make a Instructional Manual for the following Topics

1. Concept, Theorem, or other necessary discussion.

2. Examples with detailed solution (showcasing different scenarios in a problem). At least 10
Examples.
3. Practice Problems with Solutions (at least 15 Practice Problems)

Each Group must consist of 10-12 students

Ideal Topic Delegation

Laplace Transform- 3 students

Complex Numbers- 2 students
Vector and Vector Spaces- 1 students
Power Series- 3 students
Fourier Series- 2 Students
Matrices- 2 students