Engineering Mathematics-I

Lecture 40 : Power Series

Panchatcharam Mariappan1

1 Assistant Professor

Department of Mathematics and Statistics

IIT Tirupati, Tirupati
Power Series

1
Power Series
When the linear ODE contains variable coefficients, then power series method
is very helpful to solve the problem. The solution of the ODE is in the form of
power series.

Definition 1 (Power Series)

A power series (in powers of (x − x0 )) is an infinite series of the form
∞
X
am (x − x0 )m = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + · · · (1)
m=0

Here x is a variable. a0 , a1 , a2 , · · · are constants, called the coefficient of the

series. x0 is a constant, called the center of the series.

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Power Series
If x0 = 0, then we obtain the power series in power of x.
∞
X
am xm = a0 + a1 x + a2 x2 + · · · (2)
m=0

The nth partial sum of (1) is

sn (x) = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + · · · + an (x − x0 )n

where n = 0, 1, · · · . The remainder of (1) is given by

Rn (x) = an+1 (x − x0 )n+1 + an+2 (x − x0 )n+2 + · · ·

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Power Series
If for some x1 , the sequence s1 (x), s2 (x), · · · converges, say

lim sn (x1 ) = s(x1 )

n→
−∞
then the series (1) is called convergent at x = x1 , the number s(x1 ) is called
the value or sum of (1) at x1 , and we write
∞
X
s(x1 ) = am (x − x0 )m
m=0

If the sequence s1 (x), s2 (x), · · · does not converge at x = x1 , we say (1)

diverges at x = x1

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Power Series

Remarks
• If we choose x = x0 , the series (1) reduces to a single term a0 , and hence
the series converges at x0
• If there are values other than x0 at which (1) converges, these values
form an interval. This interval is called the interval of convergence.
• If the interval of Convergence is I = |x − x0 | < R, the (1) converges for all
x ∈ I and diverges for all |x − x0 | > R.
• The quantity R is called the radius of convergence.
• If the series converges for all x, we set R = ∞

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Power Series
To find the radius of convergence, we use the following ratio and root tests
(provided these limits exist.)

1 an+1
= lim (3)
R n→
− ∞ an

1 p
= lim n |an | (4)
R n→−∞

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Power Series

Example 1
Find the radius and interval of convergence of the following power series
∞
X
(m + 1)mxm
m=0

1 am+1 (m + 2)(m + 1) 2
= lim = lim = lim 1 + =1
R m→ − ∞ am m→−∞ (m + 1)m m→−∞ m
=⇒ R = 1, I = (−1, 1)

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Power Series

Example 2
Find the radius and interval of convergence of the following power series
∞
X
xm
m=0

1 am+1 1
= lim = lim =1
R m→ − ∞ am m→−∞ 1
=⇒ R = 1, I = (−1, 1)

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Power Series

Example 3
Find the radius and interval of convergence of the following power series
∞
X (−1)m m
x
m
m=1

1 am+1 (−1)m+1 m m+1−1

= lim = lim = lim =1
R m→ − ∞ am m→− ∞ (−1)m (m + 1) m→−∞ m+1
=⇒ R = 1, I = (−1, 1]

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Power Series

Example 4
Find the radius and interval of convergence of the following power series
∞
X 5m m
x
m!
m=0

1 am+1 5m+1 m! 5
= lim = lim = lim =0
R m→ − ∞ am m→
− ∞ 5 (m + 1)!
m m→−∞ m+1
=⇒ R = ∞, I = (−∞, ∞)

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Power Series
Termwise Addition: Two power series may be added term by term. If
∞
X ∞
X
am (x − x0 )m and bm (x − x0 )m
m=0 m=0

have positive radii of convergence and their sums are f (x) and g(x), then
∞
X
(am + bm )(x − x0 )m
m=0

converges and represents f (x) + g(x) for each x that lies in the interior of the
convergence interval common to each of the two given series.

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Power Series
Termwise Multiplication: Two power series may be multiplied term by term. If
∞
X ∞
X
am (x − x0 )m and bm (x − x0 )m
m=0 m=0

have positive radii of convergence and their sums are f (x) and g(x), then
product of these two series
∞
! ∞ ! ∞
X X X
m m
(am )(x − x0 ) (bm )(x − x0 ) = (a0 bm +a1 bm−1 +· · · am b0 )(x−x0 )m
m=0 m=0 m=0

converges and represents f (x)g(x) for each x that lies in the interior of the
convergence interval common to each of the two given series.

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Power Series
Termwise Differentiation: A power series may be differentiated term by term.
If
∞
X
y(x) = am (x − x0 )m
m=0

converges for |x − x0 | < R, R > 0, then the series obtained by differentiating

term by term also converges for those x and represents the derivative y ′ of y
for those x
∞
X
y ′ (x) = mam (x − x0 )m−1
m=1

Vanishing of all coefficients: If a power series has a positive radius of

convergence and the sum is identically zero throughout its interval of
convergence, then each coefficient of the series must be zero.
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Power Series
Let us solve one of the following problem. Remaining are left for self-study
1. y ′ + y = 0
2. y ′ − 5y = 0
3. (1 + x)y ′ = y
4. y ′ = −2xy

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Power Series

Example 5
Find the power series solution for the following ODEs

y′ + y = 0

Consider
∞
X
y(x) = am xm
m=0
∞
X
y ′ (x) = mam xm−1
m=1

Substitute the values of y and y ′

in the respective equation and use the property
termwise addition and vanishing all coefficients of the series
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Power Series
∞
X ∞
X
m−1
mam x + am xm = 0
m=1 m=0
∞
X
=⇒ ((m + 1)am+1 + am )xm = 0
m=0
=⇒ am+1 = −am /(m + 1)
a0 −a0 (−1)m a0
=⇒ a1 = −a0 , a2 = , a3 = , am =
2! 3! m!
2 3 ∞ m
x x X (−1) m
y = a0 (1 − x + − + · · · ) = a0 x
2! 3! m!
m=0
−x
=⇒ y = a0 e

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Power Series

Example 6
Find the power series solution for the following ODEs

y ′ − 5y = 0

Consider
∞
X
y(x) = am xm
m=0
∞
X
y ′ (x) = mam xm−1
m=1

Substitute the values of y and y ′

in the respective equation and use the property
termwise addition and vanishing all coefficients of the series
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Power Series
∞
X ∞
X
m−1
mam x −5 am xm = 0
m=1 m=0
X∞
((m + 1)am+1 − 5am )xm = 0
m=0
=⇒ am+1 = 5am
52 a0 53 a0 5m a0
=⇒ a1 = 5a0 , a2 = , a3 = , am =
2! 3! m!
∞
52 x2 53 x3 X 5m m
y = a0 (1 + 5x + + + · · · ) = a0 x
2! 3! m!
m=0
5x
y = a0 e

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Power Series

Example 7
Find the power series solution for the following ODEs

(1 + x)y ′ = y

Consider
∞
X
y(x) = am xm
m=0
∞
X
y ′ (x) = mam xm−1
m=1

Substitute the values of y and y ′

in the respective equation and use the property
termwise addition and vanishing all coefficients of the series
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Power Series
∞
X ∞
X
m−1
(1 + x) mam x − am xm = 0
m=1 m=0
∞
X ∞
X ∞
X
m−1 m
mam x + mam x − am xm = 0
m=1 m=1 m=0
X∞ X∞
mam xm−1 + (m − 1)am xm = 0
m=1 m=0
X∞
((m + 1)am+1 + (m − 1)am )xm = 0
m=0
m−1
=⇒ am+1 = − am =⇒ a1 = a0 , a2 = 0, a3 = a4 = · · · = 0
m+1
y = a0 (1 + x) 20
Power Series

Example 8
Find the power series solution for the following ODEs

y ′ = −2xy

Consider
∞
X
y(x) = am xm
m=0
∞
X
y ′ (x) = mam xm−1
m=1

Substitute the values of y and y ′

in the respective equation and use the property
termwise addition and vanishing all coefficients of the series
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Power Series

∞
X ∞
X
m−1
mam x + 2x am xm = 0
m=1 m=0
X∞ X∞
(m + 1)am+1 xm + 2am xm+1 = 0
m=0 m=0
a2m−2
Equating coefficient of x2m−1 , 2ma2m + 2a2m−2 = 0 =⇒ a2m = −
m
Equating coefficient of x2m , 2m + 1a2m+1 + 2a2m−1 = 0 =⇒ a2m+1 = 0
1
=⇒ a2m = (−1)m
m!
x4 x6 x8
y = a0 (1 − x2 + + − + ···)
2! 3! 4!
2
y = a0 e−x 22
Analytic and Singular
Points

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Analytic

Definition 2 (Analytic)
A function f is said to be anaytic at a point x0 if it can be represented by a
power series
X∞
am (x − x0 )m (5)
m=0

with either a positive or infinite radius of convergence. If the function is ana-

lytic at all points, we say it is analytic. Here x is a variable. a0 , a1 , a2 , · · · are
constants, called the coefficient of the series. x0 is a constant, called the cen-
ter of the series.

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Ordinary and Singular Points

Definition 3 (Ordinary and Singular Points)

A point x = x0 is said to be an ordinary point of the differential equation

y ′′ + p(x)y ′ + q(x)y = r(x) (6)

if both coefficients p(x) and q(x) of (6) are analytic at x0 . A point that is not an
ordinary point of the differential equation (6) is said to be a singular point of
the ODE.

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Existence of Power Series Solution

Theorem 9 (Existence of Power Series Solution)

If p, q and r in (6) are analytic at x = x0 , then every solution of (6) is analytic
at x = x0 and can be represented by a power series in powers of x − x0 with
radius of convergence R > 0.

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Regular and irregular Singular Points

Definition 10 (Regular and irregular Singular Points)

A singular point x = x0 is said to be a regular singular point of the differential
equation (6) if the functions (x − x0 )p(x) and (x − x0 )2 q(x) are both (6) analytic
at x0 . A point that is not regular is said to be an irregular singular point of the
ODE.

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Regular and irregular Singular Points

Example 11
Identify ordinary, singular, regular singular and irregular singular points for the
following ODES.

(x2 − 4)2 y ′′ + 3(x − 2)y ′ + 5y = 0

x = 2 and x = −2 are singular points. Remaining points are ordinary points.

3(x − 2) 5 3(x − 2)2 3

p(x) = , q(x) = (x − 2)p(x) = = ,
(x2 − 4)2 (x2 − 4)2 (x2 − 4)2 (x + 2)2
5(x − 2)2 5
(x − 2)2 q(x) = 2 2
=
(x − 4) (x + 2)2

Both (x − 2)p(x) and (x − 2)2 q(x) are analytic at x = 2. Therefore, 2 is a regular

singular point
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Regular and irregular Singular Points

3(x2 − 4) 3 2 5(x + 2)2 5

(x + 2)p(x) = 2 2
= , (x + 2) q(x) = 2 2
=
(x − 4) (x − 2)(x + 2) (x − 4) (x − 2)2

Although (x + 2)2 q(x) is analytic at x = −2, (x + 2)p(x) is not analytic at

x = −2. Therefore, −2 is an irregular singular point.

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Regular and irregular Singular Points

Example 12
Identify ordinary, singular, regular singular and irregular singular points for the
following ODES.

x3 y ′′ + 4x2 y ′ + 3y = 0
x = 0 is a singular point. Remaining points are ordinary points.
4 3 3
p(x) = , q(x) = 3 =⇒ xp(x) = 4, x2 q(x) =
x x x
Although xp(x) is analytic at x = 0, x2 q(x) is not analytic at x = 0. Therefore, 0
is an irregular singular point.

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Regular and irregular Singular Points

Example 13
Identify ordinary, singular, regular singular and irregular singular points for the
following ODES.

3xy ′′ + y ′ − y = 0
x = 0 is a singular point. Remaining points are ordinary points.
1 −1 1 −x
p(x) = , q(x) = =⇒ xp(x) = , x2 q(x) =
3x 3x 3 3
Both xp(x) and x2 q(x) is analytic at x = 0. Therefore, 0 is a regular singular
point.

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Regular and irregular Singular Points

Example 14
Identify ordinary, singular, regular singular and irregular singular points for the
following ODES.

(x2 + x − 6)y ′′ + (x + 3)y ′ + (x − 2)y = 0

x = −3 and x = 2 are singular points. Remaining points are ordinary points.
1 1
p(x) = , q(x) =
x−2 x+3
x+3
(x + 3)p(x) = , (x + 3)2 q(x) = x + 3
x−2

Both (x + 3)p(x) and (x + 3)2 q(x) are analytic at x = −3. Therefore, −3 is a

regular singular point
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Regular and irregular Singular Points
Therefore, −3 is a regular singular point

(x − 2)2
(x − 2)p(x) = 1, (x − 2)2 q(x) =
x+3

Both (x − 2)p(x) and (x − 2)2 q(x) are analytic at x = 2. Therefore, 2 is a

regular singular point

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 Thanks
Doubts and Suggestions
panch.m@iittp.ac.in

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Engineering Mathematics-I
Lecture 40 : Power Series

Panchatcharam Mariappan1

1 Assistant Professor

Department of Mathematics and Statistics

IIT Tirupati, Tirupati

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