Polynomial and Non-Polynomial Terminating Series Solutions to the Associated Laguerre Differential Equation

Abstract

Using the Frobenius method, we find all polynomial and non-polynomial terminating series solutions of the associated Laguerre differential equation and its special case, the Laguerre differential equation.

Full text

1 4 December 2018
Polynomial and Non-Polynomial Terminating Series Solutions
to the Associated Laguerre Differential Equation
Spiros Konstantogiannis
spiroskonstantogiannis@gmail.com
4 December 2018
Abstract
Using the Frobenius method, we find all polynomial and non-polynomial
terminating series solutions of the associated Laguerre differential
equation and its special case, the Laguerre differential equation.
Keywords: associated Laguerre equation, confluent hypergeometric equation,
associated Laguerre polynomials, non-polynomial terminating series solutions

4 December 2018
2
1. Terminating series solutions to the Laguerre equation
We’ll first seek terminating series solutions to the Laguerre equation
(
)
(
)
(
)
(
)
1 0
xy x x y x y x
l
¢¢ ¢
+ - + =
(1)
where
l
is a real parameter [1].
The Laguerre equation is written in standard form as
( ) ( ) ( )
1
0
x
y x y x y x
x x
l
-
¢¢ ¢
+ + =
We see that zero is a singular point and the functions
1
1
x
x x
x
-
= -
and
2
x x
x
l
l
=
are analytic at zero, thus zero is a regular singular point of the Laguerre equation, and
applying the Frobenius method [2], we can find a series solution – a so called
Frobenius solution – to (1) near zero. The indicial equation of (1) for its regular
singular point at zero is
(
)
1 0
r r r
- + =
,
thus it has two equal roots
1 2
0
r r
= =
Then (1) has, near zero, one Frobenius solution, which has the form
( )
10
n
n
n
y x c x
¥
=
=
å
The second linearly independent solution has a logarithmic part, thus it is not a
Frobenius series, and particularly it has the form
( ) ( )
2 1 1
ln
n
n
n
y x y x x d x
¥
=
= +
å
Since it has a logarithmic part, the second solution is not a terminating series, and thus
only the first solution may be a terminating series.
Differentiating the series
(
)
1
y x
term by term and shifting appropriately the
summation index, we obtain

4 December 2018
3
( ) ( )
1
1 1
1 0
1
n n
n n
n n
y x nc x n c x
¥ ¥
-+
= =
¢= = +
å å
and
( ) ( ) ( )( )
1
1 1 2
1 0
1 1 2
n n
n n
n n
y x n n c x n n c x
¥ ¥
-
+ +
= =
¢¢ = + = + +
å å
Substituting the expressions of
(
)
1
y x
and its first and second derivative into (1)
yields
( )( ) ( ) ( )
2 1
0 0 0
1 2 1 1 0
n n n
n n n
n n n
x n n c x x n c x c x
l
¥ ¥ ¥
+ +
= = =
+ + + - + + = Þ
å å å
( )( ) ( ) ( )
1 1
2 1 1
0 0 0 0
1 2 1 1 0
n n n n
n n n n
n n n n
n n c x n c x n c x c x
l
¥ ¥ ¥ ¥
+ +
+ + +
= = = =
Þ + + + + - + + = Þ
å å å å
( ) ( )
1 1
1 0 1 0
1 1 0
n n n n
n n n n
n n n n
n n c x n c x nc x c x
l
¥ ¥ ¥ ¥
+ +
= = = =
Þ + + + - + = Þ
å å å å
( ) ( )
1 1 1 0
1 1 1 0
1 1 0
n n n n
n n n n
n n n n
n n c x c n c x nc x c c x
l l
¥ ¥ ¥ ¥
+ +
= = = =
Þ + + + + - + + = Þ
å å å å
( ) ( ) ( )
( )
1 0 1 1
1
1 1 0
n
n n n n
n
c c n n c n c nc c x
l l
¥
+ +
=
Þ + + + + + - + = Þ
å
( ) ( ) ( )
(
)
2
1 0 1
1
1 0
n
n n
n
c c n c n c x
l l
¥
+
=
Þ + + + - - =
å
Since the powers of
x
are linearly independent, the last equation gives
1 0
0
c c
l
+ =
and
( ) ( )
2
1
1 0
n n
n c n c
l
+
+ - - =
,
1,2,...
n
=
The recurrence relation incorporates the first equation for
0
n
=
, thus it also holds for
0
n
=
, and then solving it for
1
n
c
+
, since
1
n
+
is non-zero, we end up to the recurrence
relation
( )
12
1
n n
n
c c
n
l
+
-
=+
with
0,1,...
n
=

4 December 2018
4
If
0,1,...
l
¹
, then
1
0
n
c
+
¹
if (and only if)
0
n
c
¹
, for every
0,1,...
n
=
Then, since
0
0
c
¹
, all coefficients are non-zero and thus the series does not terminate. On the
contrary, if
0,1,...
l
=
, i.e. if
l
is a non-negative integer, then
1
0
c
l
+
=
, and then all
subsequent coefficients also vanish, and thus the series terminates to a polynomial of
degree
l
. Therefore, the Laguerre equation has a terminating series solution if (and
only if)
0,1,...
l
=
, and then the solution is an
l
degree polynomial given by
( )
0
n
n
n
y x c x
l
=
=
å,
where the coefficients of the series are determined by the recurrence relation
( )
12
1
n n
n
c c
n
l
+
-
=+
(2)
with
0,1,..., 1
n
l
= -
.
Writing (2) as
( )
12
1
n n
n
c c
n
l
+
-
= - +
,
and applying it successively yields
( ) ( ) ( )
(
)
( )
(
)
( ) ( )
(
)
( )
( )
1 0
2 2 2 22
1 2
1 1 ... 1
1 1 1 1
n n
n n n n
n n
c c c
n
n n n n n
l l l
l l
+
- - - - - -
- -
= - = - - - - =
+ + - - -
( )
(
)
(
)
(
)
( )
( )
( )
(
)
(
)
( )
( )
1 1
0 0
2 2
1 ... 1 ...
1 1
1 ...1 1 !
n n
n n n n n
c c
n n n
l l l l l l
+ +
- - + - + - -
= - = - =
+ +
}
( ) ( )
( )
( )
( )
( )
( ) ( ) ( )
( )
1
11
0 0
2
1
! !
11 ! 1 ! 1 !
1 ! 1 !
n
nn
c c
n n n
n n
l
l l
l
l
+
³ + +
-
= - = =
+ + - +
+ - +
( )
( )
1
0
1
1
1 !
n
c
nn
l
+
-
æ ö
=
ç ÷
++
è ø
,
where in the last equality, we introduced the binomial coefficient
( ) ( )
( )
!
1
1 ! 1 !
nn n
ll
l
æ ö =
ç ÷
++ - +
è ø
The coefficient
1
n
c
+
is then

4 December 2018
5
( )
( )
1
1 0
1
1
1 !
n
n
c c
nn
l
+
+
-
æ ö
=
ç ÷
++
è ø
,
and obviously
( )
0
1
!
n
n
c c
n
n
l
-
æ ö
=
ç ÷
è ø
,
with
0
c
being a real constant.
Then, the polynomial solution to the Laguerre equation takes the form
( ) ( )
00
1
!
n
n
n
y x c x
n
n
l
l
l
=
-
æ ö
=
ç ÷
è ø
å
Note that in the previous series, the index
n
can be equal to
l
, while in the
recurrence relation (2) it cannot. In both cases, the index
n
is dummy.
The polynomial
( )
0
1
!
n
n
n
x
n
n
l
l
=
-
æ ö
ç ÷
è ø
å is defined as the Laguerre polynomial of degree
l
and it is denoted by
(
)
L x
l
[3-5], i.e.
( ) ( )
0
1
!
n
n
n
L x x
n
n
l
l
l
=
-
æ ö
=
ç ÷
è ø
å (3)
Thus, up to a multiplicative real constant, the Laguerre polynomial
(
)
L x
l
of degree
l
is the only polynomial solution to the Laguerre equation with parameter
l
.
2. Terminating series solutions to the associated Laguerre
equation
We’ll now seek terminating series solutions to the associated Laguerre equation
(
)
(
)
(
)
(
)
1 0
xy x x y x y x
n l
¢¢ ¢
+ + - + =
(4)
where
,
n l
are real parameters [1]. Writing (4) in standard form as
( ) ( ) ( )
1
1 0
y x y x y x
x x
n l
+
æ ö
¢¢ ¢
+ - + =
ç ÷
è ø
,
we see that zero is a singular point, and since the functions

4 December 2018
6
11 1
x x
x
nn
+
æ ö
- = + -
ç ÷
è ø
and
2
x x
x
l
l
=
are analytic at zero, zero is a regular singular point of the associated Laguerre
equation. Then, applying the Frobenius method, we can find a series solution of (4)
near zero. The indicial equation for the singularity at zero reads
(
)
(
)
(
)
2
1 1 0 0 0
r r r r r r r r r
n n n
- + + = Þ - + + = Þ + =
The roots of the indicial equation are then
1
0
r
=
and
2
r
n
= -
If
0
n
=
, the associated Laguerre equation reduces to the Laguerre equation, which we
examined earlier. If
0
n
¹
, the two roots are different and applying the Frobenius
method, we have the following cases
i. If
n
is not an integer, the two linearly independent solutions of (4) are Frobenius
series of the form
( )
10
n
n
n
y x c x
¥
=
=
å and
( )
20
n
n
n
y x x d x
n
¥
-
=
=
å
Since
n
-
is not an integer either,
(
)
2
y x
is not a polynomial, while
(
)
1
y x
is a
polynomial if (and only if) the first series terminates. Although
(
)
2
y x
is not a
polynomial, if the series of
(
)
2
y x
terminates,
(
)
2
y x
is written as the product of the
non-polynomial factor
x
n
-
with a polynomial.
ii. If
n
is a positive integer, then
0
n
> -
, and thus the first linearly independent
solution of (4) has again the form of
(
)
1
y x
, which is a polynomial if (and only if) the
first series terminates. The second linearly independent solution has the form
( ) ( )
3 1 0
ln
n
n
n
y x cy x x x h x
n
¥
-
=
= +
å
Since the series on the right-hand side begins at
0
n
=
, it has negative powers of
x
,
thus
(
)
3
y x
is not a polynomial. However, if the series terminates and
c
vanishes,
then
(
)
3
y x
is the sum of a polynomial and a finite number of negative powers of
x
,
i.e. it is a terminating series solution, but, as we’ll see, this does not happen here.

4 December 2018
7
iii. If
n
is a negative integer, then
0
n
- >
, and thus the first linearly independent
solution of (4) has the form of
(
)
2
y x
, which, in this case, is a polynomial if (and only
if) the series
0
n
n
n
d x
¥
=
å terminates, while the second linearly independent solution has
now the form
( ) ( )
4 2 0
ln
n
n
n
y x cy x x h x
¥
=
= +
å
and thus it is a polynomial if (and only if)
c
vanishes and the series on the right-hand
side terminates.
Let us examine analytically each of the previous cases.
i.
n
is not an integer
In this case, we have two Frobenius series solutions, which are
( )
10
n
n
n
y x c x
¥
=
=
å and
( )
20
n
n
n
y x x d x
n
¥
-
=
=
å
For the first solution, we have, differentiating term by term and shifting appropriately
the summation index,
( ) ( )
1
1 1
1 0
1
n n
n n
n n
y x nc x n c x
¥ ¥
-+
= =
¢= = +
å å
( ) ( ) ( )( )
1
1 1 2
1 0
1 1 2
n n
n n
n n
y x n n c x n n c x
¥ ¥
-
+ +
= =
¢¢ = + = + +
å å
Then, substituting into (4) yields
( )( ) ( ) ( )
2 1
0 0 0
1 2 1 1 0
n n n
n n n
n n n
x n n c x x n c x c x
n l
¥ ¥ ¥
+ +
= = =
+ + + + - + + = Þ
å å å
( )( ) ( )( ) ( )
1 1
2 1 1
0 0 0 0
1 2 1 1 1 0
n n n n
n n n n
n n n n
n n c x n c x n c x c x
n l
¥ ¥ ¥ ¥
+ +
+ + +
= = = =
Þ + + + + + - + + = Þ
å å å å
( ) ( )( )
1 1
1 0 1 0
1 1 1 0
n n n n
n n n n
n n n n
n n c x n c x nc x c x
n l
¥ ¥ ¥ ¥
+ +
= = = =
Þ + + + + - + = Þ
å å å å
( ) ( ) ( )( )
1 1 1 0
1 1 1 1
1 1 1 1 0
n n n n
n n n n
n n n n
n n c x c n c x nc x c c x
n n l l
¥ ¥ ¥ ¥
+ +
= = = =
Þ + + + + + + - + + = Þ
å å å å
( ) ( ) ( )( )
( )
1 0 1 1
1
1 1 1 1 0
n
n n n n
n
c c n n c n c nc c x
n l n l
¥
+ +
=
Þ + + + + + + + - + = Þ
å
( ) ( )( ) ( )
( )
1 0 1
1
1 1 1 0
n
n n
n
c c n n c n c x
n l n l
¥
+
=
Þ + + + + + + + - =
å

4 December 2018
8
Since all powers of
x
are linearly independent, the last equation gives
(
)
1 0
1 0
c c
n l
+ + =
and
(
)
(
)
(
)
1
1 1 0
n n
n n c n c
n l
+
+ + + + - =
,
1,2,...
n
=
The recurrence relation incorporates the first equation for
0
n
=
, thus it also holds for
0
n
=
, i.e.
(
)
(
)
(
)
1
1 1 0
n n
n n c n c
n l
+
+ + + + - =
,
0,1,...
n
=
Since
n
is not an integer,
(
)
1
n
n
¹ - +
, as
(
)
1
n
- +
is an integer for every
0,1,...
n
=
,
and thus the coefficient of
1
n
c
+
does not vanish. Then, solving the previous recurrence
relation for
1
n
c
+
yields
( )( )
1
1 1
n n
n
c c
n n
l
n
+
-
=+ + +
(5)
with
0,1,...
n
=
The series of
(
)
1
y x
then terminates if (and only if)
l
is a non-negative integer, i.e. if
(and only if)
0,1,...
l
=
, since then
1
0
c
l
+
=
, and then all subsequent coefficients
vanish, as seen from (5). In this case, the Frobenius solution
(
)
1
y x
is an
l
degree
polynomial.
Let us now examine if/when the series of
(
)
2
y x
terminates. In the region
0
x
>
,
(
)
2
y x
is written as
( )
20
n
n
n
y x d x
n
¥
-
=
=
å
Differentiating term by term yields
( ) ( )
1
20
n
n
n
y x n d x
n
n
¥
- -
=
¢= -
å
Note that, since
0
n
¹
, the last series begins at
0
n
=
.

4 December 2018
9
Differentiating the last series once again yields
( ) ( )( )
2
20
1
n
n
n
y x n n d x
n
n n
¥
- -
=
¢¢ = - - -
å
Since
0,1
n
¹
, the last series also begins at
0
n
=
.
Substituting the expressions of
(
)
2
y x
and its first and second derivative into (4)
yields
( )( ) ( ) ( )
2 1
0 0 0
1 1 0
n n n
n n n
n n n
x n n d x x n d x d x
n n n
n n n n l
¥ ¥ ¥
- - - - -
= = =
- - - + + - - + = Þ
å å å
( )( ) ( ) ( ) ( )
1 1
0 0 0 0
1 1 0
n n n n
n n n n
n n n n
n n d x n d x n d x d x
n n n n
n n n n n l
¥ ¥ ¥ ¥
- - - - - -
= = = =
Þ - - - + + - - - + = Þ
å å å å
( )( ) ( ) ( ) ( )
1 1
1 1 0 0
1 1 1 0
n n n n
n n n n
n n n n
n n d x n d x n d x d x
n n n n
n n n n n l
¥ ¥ ¥ ¥
- - - -
+ +
=- =- = =
Þ - - + + + - + - - + = Þ
å å å å
( )( ) ( )( ) ( )( )
1 1
0 1 0
0
1 1 1
n
n
n
d x n n d x d x
n n n
n n n n n n
¥
- - - - -
+
=
Þ - - - + - - + + + - +
å
( ) ( ) ( )
1
0 0 0
1 1 0
n n n
n n n
n n n
n d x n d x d x
n n n
n n n l
¥ ¥ ¥
- - -
+
= = =
+ + - + - - + = Þ
å å å
( )
1
0
0
1 1 d x
n
n n n
- -
æ ö
Þ - - + + - +
ç ÷
è ø
1442443
( )( ) ( )( ) ( )
1 1
0 0 0 0
1 1 1 0
n n n n
n n n n
n n n n
n n d x n d x n d x d x
n n n n
n n n n n l
¥ ¥ ¥ ¥
- - - -
+ +
= = = =
+ - - + + + - + - - + = Þ
å å å å
( )( ) ( )
( )
( )
1
0
1 1 0
n
n n
n
n n d n d x
n
n n n l n
¥-
+
=
Þ - + + - + + - - =
å
Thus, we end up to the equation
( )( ) ( )
( )
1
0
1 1 0
n
n n
n
n n d n d x
n
n l n
¥-
+
=
+ - + + - + =
å,
and since all powers of
x
are linearly independent,
(
)
(
)
(
)
1
1 1 0
n n
n n d n d
n l n
+
+ - + + - + =
,
0,1,...
n
=
Since
n
is not an integer,
1
n
n
¹ +
, as
1
n
+
is an integer for every
0,1,...
n
=
, and
thus the coefficient of
1
n
d
+
does not vanish. Then solving the previous recurrence
relation for
1
n
d
+
yields

4 December 2018
10
(
)
( )( )
1
1 1
n n
n
d d
n n
l n
n
+
- +
=+ - +
(6)
with
0,1,...
n
=
The series of
(
)
2
y x
then terminates if (and only if)
l n
+
is a non-negative integer,
i.e. if (and only if)
, 1,...
l n n
= - - +
Then, since
n
is not an integer,
l
is not an
integer either. The series
0
n
n
n
d x
¥
=
å is then a
l n
+
degree polynomial and the
Frobenius solution
(
)
2
y x
is then the product of the non-polynomial factor
x
n
-
and an
l n
+
degree polynomial.
ii.
n
is a positive integer
In this case, the first linearly independent solution of (4) has the same form as in case
i, i.e.
( )
10
n
n
n
y x c x
¥
=
=
å,
and thus, repeating what we did in the previous case, we end up to the recurrence
relation
(
)
(
)
(
)
1
1 1 0
n n
n n c n c
n l
+
+ + + + - =
,
0,1,...
n
=
Since
n
is a positive integer,
1 0
n
n
+ + ¹
, thus solving the previous recurrence
relation for
1
n
c
+
yields
( )( )
1
1 1
n n
n
c c
n n
l
n
+
-
=+ + +
,
0,1,...
n
=
,
which is the same recurrence relation as (5), but here
n
is a positive integer. As in
case i, the series
0
n
n
n
c x
¥
=
å terminates if (and only if)
l
is a non-negative integer.
The second linearly independent solution has the form
( ) ( )
3 1 0
ln
n
n
n
y x cy x x x h x
n
¥
-
=
= +
å
If
c
vanishes,
(
)
3
y x
takes the form of
(
)
2
y x
, and thus following the same steps as in
case i, we end up to the recurrence relation

4 December 2018
11
(
)
(
)
(
)
1
1 1 0
n n
n n h n h
n l n
+
+ - + + - + =
,
0,1,...
n
=
,
which holds in the region
0
x
>
. Since
n
is a positive integer, the factor
1
n
n
- +
vanishes for
1
n
n
= -
and
1
n
h
+
becomes undetermined, and then all subsequent
coefficients, i.e.
2 3
, ,...
n n
h h
+ +
, become also undetermined. This means that the
assumption that
c
vanishes does not hold, i.e.
0
c
¹
, and thus
(
)
3
y x
has a
logarithmic part, and then
(
)
3
y x
is not a terminating series.
Therefore, if
n
is a positive integer, the associated Laguerre equation has polynomial
solution if (and only if)
l
is a non-negative integer, and then the polynomial solution
is
( )
0
n
n
n
y x c x
l
=
=
å,
with the coefficients of the polynomial being determined by the recurrence relation
( )( )
1
1 1
n n
n
c c
n n
l
n
+
-
=+ + +
(7)
with
0,1,..., 1
n
l
= -
. If
0
l
=
, the polynomial is a constant. Setting
0
n
=
in (7), we
obtain the recurrence relation (2), which determines the Laguerre polynomials. Thus,
(7) also holds for
0
n
=
.
Writing (7) as
( )( )
1
1 1
n n
n
c c
n n
l
n
+
-
= - + + +
,
and applying it successively yields
( )( )
1
1 1
n n
n
c c
n n
l
n
+
-
= - =
+ + +
( )( ) ( )
(
)
( ) ( )
(
)
( )( ) ( )
(
)
( )
( )
( )
( )
0
1 2
1 1 ... 1
1 1 1 1 1 1
n n n n
n
c
n n n n n n n n n n
l l l
l
n n n n
- - - - - -
-
= - - - - =
+ + + + + - - + - - - -
( )
(
)
(
)
(
)
(
)
(
)
( )( ) ( )
( )
( ) ( )
( )
1
0
1 ...
11 ... 1 1 1
n
n n n n
c
n n n n n n n n
l l l
n n n
+
- - - - -
= - =
+ + + + - - + - -

4 December 2018
12
( ) ( ) ( )
( )( ) ( )( )
}
( ) ( )
( )
( ) ( )
1
1 1
0 0
!
1 !
1 ...
1 1 1 !
1 ... 1 1 ! 1 !
!
n
n n
n
n
c c
n
n n n n
l
l
l
l l l
n
n n n
n
³ +
+ +
- +
- -
= - = - =
+ +
+ + + + + +
( ) ( ) ( ) ( )
( )
1
0
! !
1
1 ! 1 ! 1 !
n
c
n n n
l n
n l
+
= - + + + - +
That is
( ) ( ) ( ) ( )
( )
1
1 0
! !
1
1 ! 1 ! 1 !
n
n
c c
n n n
l n
n l
+
+
= - + + + - + ,
and obviously
( ) ( ) ( )
0
! !
1
! ! !
n
n
c c
n n n
l n
n l
= - + -
,
with
0
c
being a real constant. Then, the polynomial solution is written as
( ) ( ) ( ) ( ) ( ) ( )
(
)
( ) ( )
0 0
0 0
!
! ! ! !
1 1
! ! ! ! ! ! !
n n
n n
n n
v
y x c x c x
n n n v n n n
l l
nl
l
l n l n
n l l n l
= =
+
= - = - =
+ - + + -
å å
( ) ( )
(
)
( ) ( )
0
0
!
! ! 1
! ! ! !
n
n
n
v
c
x
v n n n
l
l
l n
l n l
=
+
= -
+ + -
å
Setting
(
)
0 0
! ! !
c v c
l n l
+ ®
,
0
c
remains an arbitrary real constant, and we end up to
( ) ( )
(
)
( ) ( )
00
!
1
! ! !
n
n
n
v
y x c x
n n n
l
nl
l
n l
=
+
= - + -
å
Note that in the previous series, the index
n
can be equal to
l
, while in the
recurrence relation (7) it cannot. In both cases, the index
n
is dummy.
The polynomial
( )
(
)
( ) ( )
0
!
1
! ! !
n
n
n
v
x
n n n
l
l
n l
=
+
-+ -
å
is defined as the associated Laguerre
polynomial of degree
l
with parameter
v
, and it is denoted by
(
)
L x
nl
[6, 7], i.e.
( ) ( )
(
)
( ) ( )
0
!
1
! ! !
n
n
n
v
L x x
n n n
l
nl
l
n l
=
+
= - + -
å
(8)
with
, 0,1,...
n l
=

4 December 2018
13
Using that
(
)
( ) ( )
!
! !
v
v
n
n n
l
l
ln l
+
+
æ ö
=
ç ÷
-
+ -
è ø
,
the relation (8) is also written as
( ) ( )
0
1
!
n
n
n
v
L x x
n
n
l
nl
l
l
=
+-
æ ö
=
ç ÷
-
è ø
å (9)
with
, 0,1,...
n l
=
For
0
n
=
, (9) takes the form
( ) ( )
}
( ) ( )
0
0 0
1 1
! !
n n
n n
n n
n n
L x x x L x
n n
n n
l l
l
l l
l l
l l
l
æ ö æ ö
=
ç ÷ ç ÷
-
è ø è ø
= =
- -
æ ö æ ö
= = =
ç ÷ ç ÷
-
è ø è ø
å å ,
where, in the last equality, we used (3). Thus, for
0
n
=
, the associated Laguerre
polynomials reduce to the Laguerre polynomials, as expected, since the recurrence
relation (7) reduces to the recurrence relation (2) if
0
n
=
.
The associated Laguerre polynomials are also related to the Laguerre polynomials by
the relation
( ) ( ) ( )
1d
L x L x
dx
n
n
n
l l n
n
+
= -
(10)
The relation (10) can also be used to define the associated Laguerre polynomials [8,
9].
Indeed, the Laguerre polynomial of degree
l n
+
is, by means of (3),
( ) ( )
0
1
!
n
n
n
L x x
n
n
l n
l n
l n
+
+=
+-
æ ö
=
ç ÷
è ø
å
Differentiating the previous series term by term, we obtain
( ) ( ) ( ) ( )
( )
1
1
1
1 0
1 1
11! 1 !
n n
n n
n n
d
L x n x n x
n ndx n n
l n l n
l n
l n l n
+
+ + -
-
+= =
+ +
- -
æ ö æ ö
= = + =
ç ÷ ç ÷
++
è ø è ø
å å
( ) ( )
1
0
1
11
!
n
n
n
x
n
n
l n
l n
+ -
=
+-
æ ö
= -
ç ÷
+
è ø
å

4 December 2018
14
That is
( ) ( ) ( )
1
0
1
11
!
n
n
n
d
L x x
n
dx n
l n
l n
l n
+ -
+=
+-
æ ö
= -
ç ÷
+
è ø
å
Differentiating term by term again yields
( ) ( ) ( ) ( ) ( ) ( )
( )
1
21 2
1
21 0
1 1
1 1 1
1 2! 1 !
n n
n n
n n
d
L x n x n x
n ndx n n
l n l n
l n
l n l n
+
+ - + -
-
+= =
+ +
- -
æ ö æ ö
= - = - + =
ç ÷ ç ÷
+ ++
è ø è ø
å å
( ) ( )
2
2
0
1
12
!
n
n
n
x
n
n
l n
l n
+ -
=
+-
æ ö
= -
ç ÷
+
è ø
å
That is
( ) ( ) ( )
22
2
20
1
12
!
n
n
n
d
L x x
n
dx n
l n
l n
l n
+ -
+=
+-
æ ö
= -
ç ÷
+
è ø
å
Then
( ) ( ) ( )
0
1
1!
n
n
n
d
L x x
n
dx n
nl
n
l n
n
l n
n
+=
+-
æ ö
= -
ç ÷
+
è ø
å (11)
Since
(
)
( ) ( )
!
! !
v
n n
n n
l n l
l n
n ln l
+ +
+
æ ö æ ö
= =
ç ÷ ç ÷
+ -
+ -
è ø è ø
,
(11) is written as
( ) ( ) ( )
( )
( ) ( )
0
1
1 1
!
n
n
n
L x
v
d
L x x L x
n
dx n
nl
nl
n n n
l n l
n
l
l
+=
+
-æ ö
= - = -
ç ÷
-
è ø
å
144424443
That is
( ) ( ) ( ) ( ) ( ) ( )
( )
( )
2
1
1 1 1
d d
L x L x L x L x
dx dx
n n
n n n
n n
l n l l n l
n n
+ +
= - Þ - = -
14243
,
and we end up to
( ) ( ) ( )
1d
L x L x
dx
n
n
n
l l n
n
+
= -

4 December 2018
15
iii.
n
is a negative integer
In this case, the first linearly independent solution of (4) has the same form as
(
)
2
y x
of case i and thus, repeating the same steps, we end up to the recurrence relation
(
)
(
)
(
)
1
1 1 0
n n
n n d n d
n l n
+
+ - + + - + =
,
0,1,...
n
=
Since
0
n
- >
, then
1 0
n
n
- + ¹
and solving the previous recurrence relation for
1
n
d
+
yields
(
)
( )( )
1
1 1
n n
n
d d
n n
l n
n
+
- +
=+ - +
,
0,1,...
n
=
,
which holds in the region
0
x
>
. Then, the series
0
n
n
n
d x
¥
=
å terminates if (and only if)
l n
+
is a non-negative integer, i.e. if
, 1,...
l n n
= - - +
Since
n
is a negative integer,
l
is a positive integer, equal to or greater than
n
-
. Making the substitution
n n
® -
and
l n l
+ ®
, the last recurrence relation reads
( )( )
1
1 1
n n
n
d d
n n
l
n
+
-
=+ + +
(12)
where now
n
is a positive integer,
l
is a non-negative integer, and
0,1,..., 1
n
l
= -
, if
1
l
³
, or if
0
l
=
, the series is a constant. The recurrence relation (12) is the same as
the recurrence relation (7) of the previous case, with
,
n l
in (12) taking the same
values as the respective
,
n l
in (7). Thus, the series
0
n
n
n
d x
¥
=
å terminates, up to a
multiplicative real constant (as it happens in the previous case), to the associated
Laguerre polynomial
(
)
L x
n
l n
-+
, and then the polynomial solution
(
)
2
y x
is, again up to
a multiplicative real constant, equal to
(
)
x L x
n n
l n
- - +
(in the region
0
x
>
).
The second linearly independent solution of (4) has the form
( ) ( )
4 2 0
ln
n
n
n
y x cy x x h x
¥
=
= +
å,
and it is a terminating series – in this case a polynomial – if (and only if)
c
vanishes
and the series on the right-hand side terminates. If
c
vanishes, the solution
(
)
4
y x

4 December 2018
16
takes the form of the solution
(
)
1
y x
of case i, and thus, repeating the same steps, we
end up to the recurrence relation
(
)
(
)
(
)
1
1 1 0
n n
n n h n h
n l
+
+ + + + - =
,
0,1,...
n
=
Since
n
is a negative integer, the factor
1
n
n
+ +
vanishes for
(
)
1
n
n
= - +
, which is
non-negative, and
1
n
h
+
becomes undetermined, and then all subsequent coefficients,
i.e.
2 3
, ,...
n n
h h
+ +
, become also undetermined. This means that
c
does not vanish, and
thus
(
)
4
y x
has a logarithmic part, and then it is not a terminating series.
3. Summary and examples
To summarize the cases where the associated Laguerre equation and its special case,
the Laguerre equation, have terminating series solution, we have
1. If
n
is not an integer and
0,1,...
l
=
, the associated Laguerre equation has a
polynomial solution
( )
0
n
n
n
y x c x
l
=
=
å, with
( )( )
1
1 1
n n
n
c c
n n
l
n
+
-
=+ + +
,
0,1,..., 1
n
l
= -
If
0
l
=
, the polynomial is a constant.
2. If
n
is not an integer and
, 1,...
l n n
= - - +
, the associated Laguerre equation has a
terminating series solution, which is not a polynomial and has the form
( )
0
n
n
n
y x c x
l n
n
+
-
=
=
å, with
(
)
( )( )
1
1 1
n n
n
c c
n n
l n
n
+
- +
=- + +
,
0,1,..., 1
n
l n
= + -
If
l n
= -
,
(
)
y x x
n
-
:
.
The solution holds in the region
0
x
>
, while in the region
0
x
<
, it is derived
similarly. Observe that if
0
n
>
, then
(
)
(
)
0
0
lim sgn
x
y x c
+
®
= ¥
, and since
0
0
c
¹
,
(
)
y x
diverges at
+
0
.
3. If
, 0,1,...
n l
=
, the associated Laguerre equation has a polynomial solution, which
is, up to multiplicative real constant, the associated Laguerre polynomial
(
)
L x
nl
of
degree
l
with parameter
n
, which is given by

4 December 2018
17
( ) ( )
0
1
!
n
n
n
v
L x x
n
n
l
nl
l
l
=
+-
æ ö
=
ç ÷
-
è ø
å,
with
v
n
l
l
+
æ ö
ç ÷
-
è ø
being the binomial coefficient.
In the special case where
0
n
=
, the associated Laguerre equation reduces to the
Laguerre equation and the associated Laguerre polynomials reduce to the Laguerre
polynomials, i.e.
(
)
(
)
0
L x L x
l l
=
,
with
( ) ( )
0
1
!
n
n
n
L x x
n
n
l
l
l
=
-
æ ö
=
ç ÷
è ø
å,
since
n n
l l
l
æ ö æ ö
=
ç ÷ ç ÷
-
è ø è ø
.
As a consequence, the Laguerre equation has polynomial solution if (and only if)
0,1,...
l
=
, and the polynomial solution is, up to a multiplicative real constant, the
Laguerre polynomial
(
)
L x
l
of degree
l
.
The associated Laguerre polynomials and the Laguerre polynomials are related by the
relation
( ) ( ) ( )
1d
L x L x
dx
n
n
n
l l n
n
+
= -
,
which can also be used as the definition relation of the associated Laguerre
polynomials.
4. If
n
is a negative integer and
, 1,...
l n n
= - - +
, the associated Laguerre equation
has a polynomial solution, which is, up to a multiplicative real constant, equal to
(
)
x L x
n n
l n
- - +
. The solution holds in the region
0
x
>
, while in the region
0
x
<
, it is
derived similarly. The associated Laguerre polynomial
(
)
L x
n
l n
-+
is of degree
l n
+
,
thus the polynomial solution is a polynomial of degree
l
. If
l n
= -
, then
0
l n
+ =
,
and then the respective associated Laguerre polynomial is of zero degree, i.e. it is a
non-zero constant, and thus the polynomial solution is, up to a multiplicative real

4 December 2018
18
constant,
x
n
-
. We also note that the polynomial solution has at zero, a zero of
multiplicity
n
-
.
3.1. Examples
a.
1 2
n
=
,
2
l
=
The polynomial solution to the associated Laguerre equation is then
( )
2
0
n
n
n
y x c x
=
=
å, with
( )
1
2
3
1
2
n n
n
c c
n n
+
-
=æ ö
+ +
ç ÷
è ø
,
0,1
n
=
Thus
1 0 0
2 4
3
3
2
c c c
-
= = -
2 1 1 0
1 1 4
5
5 15
2
2
c c c c
-
= = - =
Then
( )
2 2 2
0 0 0
4 4 4 15 15
1 5 5
3 15 15 4 4
y x c x x c x x c x x
æ ö æ ö æ ö
¢
= - + = - + = - +
ç ÷ ç ÷ ç ÷
è ø è ø è ø
,
with
0 0
4 15
c c
¢
=,
0
c
a real constant. We thus obtain the monic polynomial
( )
2
15
5
4
p x x x
= - +
,
which is a solution to the associated Laguerre equation (4) for
1 2
n
=
,
2
l
=
, since
( ) ( ) ( ) ( )
2
1 3 15
1 2 2 2 5 2 5
2 2 4
xp x x p x p x x x x x x
æ ö æ ö æ ö
¢¢ ¢
+ + - + = + - - + - + =
ç ÷ ç ÷ ç ÷
è ø è ø è ø
2 2 2 2
0 0 0
15 15 15 15
2 3 2 5 2 10 2 2 5 5 10 0
2 2 2 2
x x x x x x x x x x x
= + - - + + - + = - + + + - - + =
14243 1442443 14243
b.
1 2
n
=
,
3 2
l
=
Then, the associated Laguerre equation has a non-polynomial terminating series
solution, which, in the region
0
x
>
, is

4 December 2018
19
( )
2
1 2
0
n
n
n
y x c x
-
=
=
å, with
( )
1
2
1
12
n n
n
c c
n n
+
-
=
æ ö
+ +
ç ÷
è ø
,
0,1
n
=
Thus
1 0 0
2
4
1
2
c c c
-
= = -
2 1 1 0
1 1 4
3
3 3
2
2
c c c c
-
= = - =
Then
( )
1 2 1 2 3 2
0
4
43
y x c x x x
-
æ ö
= - +
ç ÷
è ø
,
with
0
c
a real constant.
For
1 2
n
=
,
3 2
l
=
the associated Laguerre equation (4) reads
( ) ( ) ( )
3 3
0
2 2
xy x x y x y x
æ ö
¢¢ ¢
+ - + =
ç ÷
è ø
,
and it is satisfied by the function
1 2 1 2 3 2
4
4
3
x x x
-
- +
, since
1 2 1 2 3 2 1 2 1 2 3 2 1 2 1 2 3 2
4 3 4 3 4
4 4 4
3 2 3 2 3
x x x x x x x x x x x
- - -
¢¢ ¢
æ ö æ öæ ö æ ö
- + + - - + + - + =
ç ÷ ç ÷ç ÷ ç ÷
è ø è øè ø è ø
3 2 1 2 1 2 3 2 1 2 1 2 1 2 1 2 3 2
1 3 1 3
2 2 2 2 6 2
2 2 2 2
x x x x x x x x x x x
- - - - -
¢
æ ö æ öæ ö
= - - + + - - - + + - + =
ç ÷ ç ÷ç ÷
è ø è øè ø
5 2 3 2 1 2 3 2 1 2 1 2 1 2 1 2 3 2 1 2 1 2 3 2
3 3 1 3
3 3 2 2 6 2
4 4 2 2
x x x x x x x x x x x x x
- - - - - - -
æ ö
= + + - - + + + - + - + =
ç ÷
è ø
3 2 1 2 1 2 3 2 1 2 1 2 1 2 1 2 3 2 1 2 1 2 3 2
3 3 1 3
3 3 2 2 6 2
4 4 2 2
x x x x x x x x x x x x
- - - - - -
= + + - - + + + - + - + =
3 2 3 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 3 2 3 2
0 0
0 0
3 3 1 3
3 3 2 6 2 2 0
4 4 2 2
x x x x x x x x x x x x
- - - - - -
= - + - + + + + + - - + =
14444244443 1442443
1442443 1444442444443
c.
1
n
=
,
2
l
=

4 December 2018
20
In this case, the polynomial solution to the associated Laguerre equation is, up to a
multiplicative real constant, the associated Laguerre polynomial
(
)
1
2
L x
of degree 2
with parameter 1, which is
( ) ( )
2
1 2 2
20
3 3 3 31 1 3! 3! 1 3!
2 2 1 0
! 2 2!1! 1!2! 2 0!3!
n
n
n
L x x x x x x
n
n
=
-æ ö æ ö æ ö æ ö
= = - + = - + =
ç ÷ ç ÷ ç ÷ ç ÷
-
è ø è ø è ø è ø
å
2
1
3 3
2
x x
= - +
That is
( )
1 2
2
1
3 3
2
L x x x
= - +
Then
(
)
1 2
2
2 6 6
L x x x
= - +
The polynomial
(
)
1
2
2
L x
is a solution to the associated Laguerre equation (4) for
1
n
=
,
2
l
=
, since
( )
( )
( ) ( )
( )( )
2 2 2 2
6 6 1 1 6 6 2 6 6 2 2 2 6 2 12 12
x x x x x x x x x x x x x
¢¢ ¢
- + + + - - + + - + = + - - + - + =
2 2 2 2
0
0 0
2 4 12 2 6 2 12 12 2 2 2 4 6 12 12 12 0
x x x x x x x x x x x x
= + - - + + - + = - + + + + - - + =
14243
14243 144424443
Since the polynomial
(
)
1
2
2
L x
is a solution, the polynomial
(
)
1
2
L x
is also a solution,
as the associated Laguerre equation is linear and homogeneous.
d.
1
v
= -
,
3
l
=
In this case, the associated Laguerre equation has a polynomial solution, which, in the
region
0
x
>
, has the form
( )
2
1
0
n
n
n
y x c x
+
=
=
å, with
( )( )
1
2
1 2
n n
n
c c
n n
+
-
=+ +
,
0,1
n
=
Thus
1 0
c c
= -
2 1 0
1 1
6 6
c c c
-
= =
Then

4 December 2018
21
( )
( ) ( )
2 3 3 2 3 2
0
0 0
1
6 6 6 6
6 6
c
y x c x x x x x x c x x x
æ ö ¢
= - + = - + = - +
ç ÷
è ø
,
with
0 0
6
c c
¢
=,
0
c
a real constant. We thus obtain the monic polynomial
(
)
3 2
6 6
p x x x x
= - +
,
which is a solution to the associated Laguerre equation (4) for
1
n
= -
,
3
l
=
, since
( )
( )
( ) ( )
( )
( )
3 2 3 2 3 2 2
6 6 1 1 6 6 3 6 6 6 12 3 12 6
x x x x x x x x x x x x x x x x
¢¢ ¢
- + + - + - - + + - + = - - - + +
3 2 2 3 2 3 2
3 18 18 6 12 3 12 6 3 18 18
x x x x x x x x x x x
+ - + = - - + - + - + =
3 3 2 2 2
0 0 0
3 3 6 12 18 12 6 18 0
x x x x x x x x
= - + + + - - - + =
14243 144424443 1442443
4. Conclusions
We’ve found all polynomial and non-polynomial terminating series solutions to the
associated Laguerre equation and its special case, the Laguerre equation, and we’ve
shown that the associated Laguerre polynomials are not the only polynomial solutions
to the associated Laguerre equation, which has also non-polynomial terminating series
solutions when
n
is non-integer and
, 1,...
l n n
= - - +
Since the trivial substitution
1
c
n
= +
and
a
l
= -
converts the associated Laguerre
equation (4) to the confluent hypergeometric equation [10, 11], we’ve also found all
polynomial and non-polynomial terminating series solutions to the confluent
hypergeometric equation.
5. References
[1] http://mathworld.wolfram.com/LaguerreDifferentialEquation.html.
[2] Ravi P. Agarwal and Donal O’Regan, Ordinary and Partial Differential Equations
(Springer Science+Business Media, LLC, 2009), Lecture 5.
[3] http://mathworld.wolfram.com/LaguerrePolynomial.html.
[4] Arfken, Weber, and Harris, Mathematical Methods for Physicists (Elsevier Inc.,
Seventh Edition, 2013), p. 891.
[5] Mary L. Boas, Mathematical Methods in the Physical Sciences (John Wiley &
Sons, Inc., Third Edition, 2006), p. 609.
[6] Arfken, Weber, and Harris, Mathematical Methods for Physicists (Elsevier Inc.,
Seventh Edition, 2013), p. 892.

4 December 2018
22
[7] http://mathworld.wolfram.com/AssociatedLaguerrePolynomial.html.
[8] Arfken, Weber, and Harris, Mathematical Methods for Physicists (Elsevier Inc.,
Seventh Edition, 2013), p. 892.
[9] Mary L. Boas, Mathematical Methods in the Physical Sciences (John Wiley &
Sons, Inc., Third Edition, 2006), p. 610.
[10] Arfken, Weber, and Harris, Mathematical Methods for Physicists (Elsevier Inc.,
Seventh Edition, 2013), p. 345.
[11]
http://mathworld.wolfram.com/ConfluentHypergeometricDifferentialEquation.html.