Orthogonal polynomials
We start with
Definition 1. A sequence of polynomials {pn (x)}∞
n=0 with degree[pn (x)] = n for each n is
called orthogonal with respect to the weight function w(x) on the interval (a, b) with a < b if
(
Z b
0, m 6= n
w(x)pm (x)pn (x) dx = hn δmn with δmn :=
1, m = n.
a
The weight function w(x) should be continuous and positive on (a, b) such that the moments
Z b
µn :=
w(x) xn dx, n = 0, 1, 2, . . .
a
exist. Then the integral
Z
b
hf, gi :=
w(x)f (x)g(x) dx
a
denotes an inner product of the polynomials f and g. The interval (a, b) is called the interval
of orthogonality. This interval needs not to be finite.
If hn = 1 for each n ∈ {0, 1, 2, . . .} the sequence of polynomials is called orthonormal, and if
pn (x) = kn xn + lower order terms
with kn = 1
for each n ∈ {0, 1, 2, . . .} the polynomials are called monic.
Example. As an example we take w(x) = 1 and (a, b) = (0, 1). Using the Gram-Schmidt
process the orthogonal polynomials can be constructed as follows. Start with the sequence
{1, x, x2 , . . .}. Choose p0 (x) = 1. Then we have
hx, 1i
1
hx, p0 (x)i
p0 (x) = x −
=x− ,
hp0 (x), p0 (x)i
h1, 1i
2
p1 (x) = x −
since
Z
Z
1
h1, 1i =
dx = 1
1
and hx, 1i =
0
0
1
x dx = .
2
Further we have
hx2 , p0 (x)i
hx2 , p1 (x)i
p0 (x) −
p1 (x)
hp0 (x), p0 (x)i
hp1 (x), p1 (x)i
µ
¶
µ
¶
hx2 , x − 12 i
hx2 , 1i
1
1
1
1
2
2
= x −
−
x−
=x − − x−
= x2 − x + ,
1
1
h1, 1i
2
3
2
6
hx − 2 , x − 2 i
p2 (x) = x2 −
since
Z
2
1
hx , 1i =
0
1
x dx = ,
3
and
2
Z
hx −
1
2,
x−
1
2i
=
0
1µ
1
x−
2
Z
2
hx , x −
¶2
1
2i
=
0
1µ
Z
dx =
0
1
2
1
µ
¶
1
1 1
1
x x−
dx = − =
2
4 6
12
2
1
x −x+
4
2
¶
dx =
1 1 1
1
− + = .
3 2 4
12
The polynomials p0 (x) = 1, p1 (x) = x − and p2 (x) = x2 − x + 16 are the first three monic
orthogonal polynomials on the interval (0, 1) with respect to the weight function w(x) = 1.
1
Repeating this process we obtain
3
3
1
9
2
1
p3 (x) = x3 − x2 + x − , p4 (x) = x4 − 2x3 + x2 − x + ,
2
5
20
7
7
70
5
20
5
5
1
p5 (x) = x5 − x4 + x3 − x2 + x −
,
2
9
6
42
252
and so on.
√
The orthonormal polynomials would be q0 (x) = p0 (x)/ h0 = 1,
¶
µ
p
√
√
p2 (x)
1
2
=6 5 x −x+
,
q1 (x) = p1 (x)/ h1 = 2 3(x − 1/2), q2 (x) = √
6
h2
µ
¶
√
p3 (x)
3 2 3
1
3
q3 (x) = √
= 20 7 x − x + x −
,
2
5
20
h3
etcetera.
All sequences of orthogonal polynomials satisfy a three term recurrence relation:
Theorem 1. A sequence of orthogonal polynomials {pn (x)}∞
n=0 satisfies
pn+1 (x) = (An x + Bn )pn (x) + Cn pn−1 (x),
n = 1, 2, 3, . . . ,
where
An =
kn+1
,
kn
n = 0, 1, 2, . . .
and
Cn = −
An
hn
·
,
An−1 hn−1
n = 1, 2, 3, . . . .
Proof. Since degree [pn (x)] = n for each n ∈ {0, 1, 2, . . .} the sequence of orthogonal polynomials {pn (x)}∞
n=0 is linearly independent. Let An = kn+1 /kn . Then pn+1 (x) − An xpn (x) is a
polynomial of degree ≤ n. Hence
pn+1 (x) − An xpn (x) =
n
X
ck pk (x).
k=0
The orthogonality property now gives
hpn+1 (x) − An xpn (x), pk (x)i =
n
X
cm hpm (x), pk (x)i = ck hpk (x), pk (x)i = ck hk .
m=0
This implies
hk ck = hpn+1 (x) − An xpn (x), pk (x)i
= hpn+1 (x), pk (x)i − An hxpn (x), pk (x)i = −An hpn (x), xpk (x)i.
For k < n − 1 we have degree [xpk (x)] < n which implies that hpn (x), xpk (x)i = 0. Hence:
ck = 0 for k < n − 1. This proves that the polynomials satisfy the three term recurrence
relation
pn+1 (x) − An xpn (x) = cn pn (x) + cn−1 pn−1 (x), n = 1, 2, 3, . . . .
Further we have
hn−1 cn−1 = −An hpn (x), xpn−1 (x)i = −An
This proves the theorem.
2
kn−1
hn
kn
=⇒
cn−1 = −
An
hn
·
.
An−1 hn−1
Note that the three term recurrence relation for a sequence of monic (kn = 1) orthogonal
polynomials {pn (x)}∞
n=0 has the form
pn+1 (x) = xpn (x) + Bn pn (x) + Cn pn−1 (x) with Cn = −
hn
,
hn−1
n = 1, 2, 3, . . . .
A consequence of the three term recurrence relation is
Theorem 2. A sequence of orthogonal polynomials {pn (x)}∞
n=0 satisfies
n
X
pk (x)pk (y)
hk
k=0
and
n
X
{pk (x)}2
k=0
hk
=
kn
pn+1 (x)pn (y) − pn+1 (y)pn (x)
·
,
hn kn+1
x−y
n = 0, 1, 2, . . .
(1)
¡
¢
kn
· p0n+1 (x)pn (x) − pn+1 (x)p0n (x) ,
hn kn+1
n = 0, 1, 2, . . . .
(2)
=
Formula (1) is called the Christoffel-Darboux formula and (2) its confluent form.
Proof. The three term recurrence relation implies that
pn+1 (x)pn (y) = (An x + Bn )pn (x)pn (y) + Cn pn−1 (x)pn (y)
and
pn+1 (y)pn (x) = (An y + Bn )pn (y)pn (x) + Cn pn−1 (y)pn (x).
Subtraction gives
pn+1 (x)pn (y) − pn+1 (y)pn (x) = An (x − y)pn (x)pn (y) + Cn [pn−1 (x)pn (y) − pn−1 (y)pn (x)] .
This leads to the telescoping sum
n
n
X
X
pk (x)pk (y)
pk+1 (x)pk (y) − pk+1 (y)pk (x)
(x − y)
=
hk
Ak hk
k=1
k=1
−
n
X
pk (x)pk−1 (y) − pk (y)pk−1 (x)
k=1
=
Ak−1 hk−1
pn+1 (x)pn (y) − pn+1 (y)pn (x) k02 (x − y)
−
.
An hn
h0
This implies that
(x − y)
n
X
pk (x)pk (y)
k=0
hk
=
pn+1 (x)pn (y) − pn+1 (y)pn (x)
An hn
=
kn
(pn+1 (x)pn (y) − pn+1 (y)pn (x)) ,
hn kn+1
which proves (1). The confluent form (2) then follows by taking the limit y → x:
pn+1 (x)pn (y) − pn+1 (y)pn (x)
y→x
x−y
pn (x) (pn+1 (x) − pn+1 (y)) − pn+1 (x) (pn (x) − pn (y))
= lim
y→x
x−y
0
0
= pn (x)pn+1 (x) − pn+1 (x)pn (x).
lim
3
Zeros
Theorem 3. If {pn (x)}∞
n=0 is a sequence of orthogonal polynomials on the interval (a, b) with
respect to the weight function w(x), then the polynomial pn (x) has exactly n real simple zeros
in the interval (a, b).
Proof. Since degree[pn (x)] = n the polynomial has at most n real zeros. Suppose that pn (x)
has m ≤ n distinct real zeros x1 , x2 , . . . , xm in (a, b) of odd order (or multiplicity). Then the
polynomial
pn (x)(x − x1 )(x − x2 ) · · · (x − xm )
does not change sign on the interval (a, b). This implies that
Z b
w(x)pn (x)(x − x1 )(x − x2 ) · · · (x − xm ) dx 6= 0.
a
By orthogonality this integral equals zero if m < n. Hence: m = n, which implies that pn (x)
has n distinct real zeros of odd order in (a, b). This proves that all n zeros are distinct and
simple (have order or multiplicity equal to one).
Theorem 4. If {pn (x)}∞
n=0 is a sequence of orthogonal polynomials on the interval (a, b) with
respect to the weight function w(x), then the zeros of pn (x) and pn+1 (x) separate each other.
Proof. This follows from the confluent form (2) of the Christoffel-Darboux formula. Note
that
Z b
hn =
w(x) {pn (x)}2 dx > 0, n = 0, 1, 2, . . . .
a
This implies that
n
¡ 0
¢ X
kn
{pk (x)}2
0
· pn+1 (x)pn (x) − pn+1 (x)pn (x) =
> 0.
hn kn+1
hk
k=0
Hence
¢
kn ¡ 0
· pn+1 (x)pn (x) − pn+1 (x)p0n (x) > 0.
kn+1
Now suppose that xn,k and xn,k+1 are two consecutive zeros of pn (x) with xn,k < xn,k+1 .
Since all n zeros of pn (x) are real and simple p0n (xn,k ) and p0n (xn,k+1 ) should have opposite
signs. Hence we have
pn (xn,k ) = 0 = pn (xn,k+1 )
and p0n (xn,k ) · p0n (xn,k+1 ) < 0.
This implies that pn+1 (xn,k ) · pn+1 (xn,k+1 ) should be negative too. Then the continuity of
pn+1 (x) implies that there should be at least one zero of pn+1 (x) between xn,k and xn,k+1 .
However, this holds for each two consecutive zeros of pn (x). This proves the theorem.
Moreover, if {xn,k }nk=1 and {xn+1,k }n+1
k=1 denote the consecutive zeros of pn (x) and pn+1 (x)
respectively, then we have
a < xn+1,1 < xn,1 < xn+1,2 < xn,2 < · · · < xn+1,n < xn,n < xn+1,n+1 < b.
4
Gauss quadrature
If f is a continuous function on (a, b) and x1 < x2 < · · · < xn are n distinct points in (a, b),
then there exists exactly one polynomial P with degree ≤ n − 1 such that P (xi ) = f (xi ) for
all i = 1, 2, . . . , n. This polynomial P can easily be found by using Lagrange interpolation as
follows. Define
p(x) = (x − x1 )(x − x2 ) · · · (x − xn )
and consider
P (x) =
n
X
i=1
n
X
p(x)
(x − x1 ) · · · (x − xi−1 )(x − xi+1 ) · · · (x − xn )
f (xi )
=
f (xi )
.
0
(x − xi )p (xi )
(xi − x1 ) · · · (xi − xi−1 )(xi − xi+1 ) · · · (xi − xn )
i=1
Let {pn (x)}∞
n=0 be a sequence of orthogonal polynomials on the interval (a, b) with respect to
the weight function w(x). Then for x1 < x2 < · · · < xn we take the n distinct real zeros of the
polynomial pn (x). If f is a polynomial of degree ≤ 2n − 1, then f (x) − P (x) is a polynomial
of degree ≤ 2n − 1 with at least the zeros x1 < x2 < · · · < xn . Now we define
f (x) = P (x) + r(x)pn (x),
where r(x) is a polynomial of degree ≤ n − 1. This can also be written as
f (x) =
n
X
f (xi )
i=1
This implies that
Z b
Z
n
X
w(x)f (x) dx =
f (xi )
a
i=1
a
b
pn (x)
+ r(x)pn (x).
(x − xi )p0n (xi )
w(x)pn (x)
dx +
(x − xi )p0n (xi )
Z
a
b
w(x)r(x)pn (x) dx.
Since degree[r(x)] ≤ n − 1 the orthogonality property implies that the latter integral equals
zero. This implies that
Z b
Z b
n
X
w(x)pn (x)
dx, i = 1, 2, . . . , n.
w(x)f (x) dx =
λn,i f (xi ) with λn,i :=
0
a
a (x − xi )pn (xi )
i=1
This is the Gauss quadrature formula. This gives the value of the integral in the case that f is a
polynomial of degree ≤ 2n−1 if the value of f (xi ) is known for the n zeros x1 < x2 < · · · < xn
of the polynomial pn (x).
If f is not a polynomial of degree ≤ 2n − 1 this leads to an approximation of the integral:
Z b
Z b
n
X
w(x)pn (x)
λn,i f (xi ) with λn,i :=
w(x)f (x) dx ≈
dx, i = 1, 2, . . . , n.
(x
− xi )p0n (xi )
a
a
i=1
The coefficients {λn,i }ni=1 are called Christoffel numbers. Note that these numbers do not
depend on the function f . These Christoffel numbers are all positive. This can be shown as
follows. We have
Z b
pn (x)
λn,i =
w(x)`n,i (x) dx with `n,i (x) :=
, i = 1, 2, . . . , n.
(x − xi )p0n (xi )
a
5
Then `2n,i (x) − `n,i (x) is a polynomial of degree ≤ 2n − 2 which vanishes at the zeros {xn,k }nk=1
of pn (x). Hence
`2n,i (x) − `n,i (x) = pn (x)q(x) for some polynomial q of degree ≤ n − 2.
This implies that
Z
a
b
¡
¢
w(x) `2n,i (x) − `n,i (x) dx =
Z
a
b
w(x)pn (x)q(x) dx = 0
by orthogonality. Hence we have
Z b
Z b
λn,i =
w(x)`n,i (x) dx =
w(x) {`n,i (x)}2 dx > 0.
a
a
Now we can also prove
Theorem 5. Let {pn (x)}∞
n=0 be a sequence of orthogonal polynomials on the interval (a, b)
with respect to the weight function w(x) and let m < n. Then we have: between any two zeros
of pm (x) there is at least one zero of pn (x).
Proof. Suppose that xm,k and xm,k+1 are two consecutive zeros of pm (x) and that there is no
zero of pn (x) in (xm,k , xm,k+1 ). Then consider the polynomial
g(x) =
pm (x)
.
(x − xm,k )(x − xm,k+1 )
Then we have
g(x)pm (x) ≥ 0
for x ∈
/ (xm,k , xm,k+1 ).
Now the Gauss quadrature formula gives
Z b
n
X
w(x)g(x)pm (x) dx =
λn,i g(xn,i )pm (xn,i ),
a
i=1
{xn,i }ni=1
where
are the zeros of pn (x). Since there are no zeros of pn (x) in (xm,k , xm,k+1 )
we conclude that g(xn,i )pm (xn,i ) ≥ 0 for all i = 1, 2, . . . , n. Further we have λn,i > 0 for all
i = 1, 2, . . . , n which implies that the sum at the right-hand side cannot vanish. However, the
integral at the left-hand side is zero by orthogonality. This contradiction proves that there
should be at least one zero of pn (x) between the two consecutive zeros of pm (x).
2
y
1
0
0.2
0.4
0.6
0.8
1
x
–1
–2
The polynomials q2 (x), q3 (x) and q4 (x)
6
Classical orthogonal polynomials
The classical orthogonal polynomials are named after Hermite, Laguerre and Jacobi:
name
pn (x)
w(x)
Hermite
Hn (x)
e−x
(−∞, ∞)
Laguerre
Ln (x)
e−x xα
(0, ∞)
(1 − x)α (1 + x)β
(−1, 1)
1
(−1, 1)
Jacobi
Legendre
2
(α)
(α,β)
Pn
(x)
Pn (x)
(a, b)
The Hermite polynomials are orthogonal on the interval (−∞, ∞) with respect to the normal
2
distribution w(x) = e−x , the Laguerre polynomials are orthogonal on the interval (0, ∞) with
respect to the gamma distribution w(x) = e−x xα and the Jacobi polynomials are orthogonal
on the interval (−1, 1) with respect to the beta distribution w(x) = (1 − x)α (1 + x)β .
The Legendre polynomials form a special case (α = β = 0) of the Jacobi polynomials.
These classical orthogonal polynomials satisfy an orthogonality relation, a three term recurrence relation, a second order linear differential equation and a so-called Rodrigues formula.
Moreover, for each family of classical orthogonal polynomials we have a generating function.
In the sequel we will often use the Kronecker delta which is defined by
(
0, m 6= n
δmn :=
1, m = n
for m, n ∈ {0, 1, 2, . . .} and the notation
D=
d
dx
(3)
for the differentiation operator. Then we have Leibniz’ rule
n µ ¶
X
n
D [f (x)g(x)] =
Dk f (x)Dn−k g(x),
k
n
n = 0, 1, 2, . . .
(4)
k=0
which is a generalization of the product rule. The proof is by mathematical induction and by
use of Pascal’s triangle identity
µ ¶ µ
¶ µ
¶
n
n
n+1
+
=
, k = 1, 2, . . . , n.
k
k−1
k
7
Hermite
The Hermite polynomials are orthogonal on the interval (−∞, ∞) with respect to the normal
2
distribution w(x) = e−x . They can be defined by means of their Rodrigues formula:
Hn (x) =
(−1)n n
2
2
D w(x) = (−1)n ex Dn e−x ,
w(x)
n = 0, 1, 2, . . . ,
(5)
where the differentiation operator D is defined by (3). Since Dn+1 = D Dn , we obtain
£
¤
Dn+1 w(x) = D [Dn w(x)] = (−1)n D [w(x)Hn (x)] = (−1)n w0 (x)Hn (x) + w(x)Hn0 (x)
£
¤
= (−1)n+1 w(x) 2xHn (x) − Hn0 (x) , n = 0, 1, 2, . . . ,
which implies that
Hn+1 (x) = 2xHn (x) − Hn0 (x),
n = 0, 1, 2, . . . .
(6)
The definition (5) implies that H0 (x) = 1. Then (6) implies by induction that Hn (x) is a
polynomial of degree n. Further we have that H2n (x) is even and H2n+1 (x) is odd and that
the leading coefficient of the polynomial Hn (x) equals kn = 2n .
The Hermite polynomials satisfy the orthogonality relation
Z ∞
1
2
√
e−x Hm (x)Hn (x) dx = 2n n! δmn , m, n ∈ {0, 1, 2, . . .}.
π −∞
To prove this we use the definition (5) to obtain
Z ∞
Z
−x2
n
e Hm (x)Hn (x) dx = (−1)
−∞
∞
−∞
(7)
2
Hm (x)Dn e−x dx.
Now we use integration by parts n times to conclude that the integral vanishes for m < n.
For m = n we have using integration by parts
Z ∞
Z ∞
Z ∞
2
2
2
e−x Hn (x)Hn (x) dx = (−1)n
Hn (x)Dn e−x dx =
Dn Hn (x) · e−x dx
−∞
−∞
Z −∞
∞
√
2
= kn n!
e−x dx = 2n n! π.
−∞
This proves the orthogonality relation (7).
In order to find the three term recurrence relation we start with
2
w(x) = e−x
w0 (x) = −2xw(x).
=⇒
Then we have by using Leibniz’ rule (4)
Dn+1 w(x) = Dn w0 (x) = Dn [−2xw(x)] = −2xDn w(x) − 2nDn−1 w(x),
which implies that
Hn+1 (x) = 2xHn (x) − 2nHn−1 (x),
8
n = 1, 2, 3, . . . .
(8)
Combining (6) and (8) we find that
Hn0 (x) = 2nHn−1 (x),
n = 1, 2, 3, . . . .
(9)
Differentiation of (6) gives
0
Hn+1
(x) = 2xHn0 (x) + 2Hn (x) − Hn00 (x),
n = 0, 1, 2, . . . .
Now we use (9) to conclude that
2(n + 1)Hn (x) = 2xHn0 (x) + 2Hn (x) − Hn00 (x),
n = 0, 1, 2, . . . ,
which implies that Hn (x) satisfies the second order linear differential equation
y 00 (x) − 2xy 0 (x) + 2ny(x) = 0,
n ∈ {0, 1, 2, . . .}.
Finally we will prove the generating function
e
2xt−t2
=
∞
X
Hn (x)
n=0
We start with
n!
2
tn .
2
(10)
2
f (t) = e−(x−t) = e−x · e2xt−t .
The Taylor series for f (t) is
f (t) =
∞
X
f (n) (0)
n=0
n!
tn
with, by using the substitution x − t = u,
· n
¸
· n
¸
d −(x−t)2
d
2
2
(n)
n
−u2
f (0) =
e
= (−1)
e
= (−1)n Dn e−x = e−x Hn (x)
n
n
dt
du
t=0
u=x
for n = 0, 1, 2, . . .. Hence we have
2
2
2
e−x · e2xt−t = e−(x−t) = f (t) =
∞
X
f (n) (0)
n=0
This proves the generating function (10).
9
n!
tn = e−x
2
∞
X
Hn (x)
n=0
n!
tn .
Laguerre
The Laguerre polynomials are orthogonal on the interval (0, ∞) with respect to the gamma
distribution w(x) = e−x xα . They can be defined by means of their Rodrigues formula:
L(α)
n (x) =
1 1
1 x −α n £ −x n+α ¤
Dn [w(x) xn ] =
e x D e x
,
n! w(x)
n!
n = 0, 1, 2, . . . .
(11)
By using Leibniz’ rule (4) we have
n µ ¶
X
£
¤
n
n −x n+α
D e x
=
Dk e−x Dn−k xn+α
k
k=0
n µ ¶
X
n
=
(−1)k e−x (n + α)(n + α − 1) · · · (α + k + 1)xα+k
k
k=0
µ ¶
n
X
−x α
k n Γ(n + α + 1) k
x .
= e x
(−1)
k Γ(k + α + 1)
k=0
Hence we have
L(α)
n (x)
µ
¶ k
n
X
k n+α x
=
(−1)
,
n − k k!
n = 0, 1, 2, . . . ,
(12)
k=0
where
µ
¶
n+α
Γ(n + α + 1)
(k + α + 1)n−k
=
=
,
n−k
(n − k)! Γ(k + α + 1)
(n − k)!
k = 0, 1, 2, . . . , n.
(α)
This proves that Ln (x) is a polynomial of degree n. Since
µ
¶
(−1)k (α + 1)n
(α + 1)n (−n)k
k n+α
(−1)
=
=
,
n−k
(n − k)! (α + 1)k
n!
(α + 1)k
k = 0, 1, 2, . . . , n
we also have for n = 0, 1, 2, . . .
L(α)
n (x)
µ
¶
µ
¶
n
(α + 1)n X (−n)k xk
n+α
−n
=
=
;x .
1 F1
n!
(α + 1)k k!
n
α+1
(13)
k=0
Note that we have
L(α)
n (0)
µ
¶
n+α
(α + 1)n
=
=
,
n
n!
n = 0, 1, 2, . . .
(α)
and that the leading coefficient of the polynomial Ln (x) equals
kn =
Further we have
(−1)n
,
n!
n = 0, 1, 2, . . . .
µ
¶ k
µ
¶ k−1
n
n
X
d X
x
k n+α x
k n+α
(−1)
=
(−1)
dx
n − k k!
n − k (k − 1)!
k=0
k=1
µ
¶ k
n−1
X
n+α
x
(α+1)
= −Ln−1 (x), n = 1, 2, 3, . . . .
=
(−1)k−1
n − k − 1 k!
d (α)
L (x) =
dx n
k=0
10
(14)
Now we can prove the orthogonality relation
Z ∞
Γ(n + α + 1)
(α)
e−x xα L(α)
δmn ,
m (x)Ln (x) dx =
n!
0
α > −1
(15)
for m, n ∈ {0, 1, 2, . . .}. First of all, the integral converges if the moments
Z ∞
µn =
e−x xn+α dx
0
exists for all n ∈ {0, 1, 2, . . .}. This leads to the condition α > −1. Note that µn = Γ(n+α+1).
Now we use (11) to obtain
Z ∞
Z ∞
£
¤
1
−x α (α)
(α)
n −x n+α
e x Lm (x)Ln (x) dx =
L(α)
e x
dx.
m (x)D
n! 0
0
We apply integration by parts to obtain
Z ∞
Z
£
¤
(α)
n −x n+α
n
Lm (x)D e x
dx = (−1)
0
∞
0
−x n+α
Dn L(α)
dx
m (x)e x
which equals zero for m < n. For m = n we find
Z ∞
Z ∞
−x n+α
Dn L(α)
(x)e
x
dx
=
k
n!
e−x xn+α dx = (−1)n Γ(n + α + 1).
n
n
0
0
This proves the orthogonality relation (15).
The Laguerre polynomials can also be defined by their generating function
µ
¶ X
∞
xt
−α−1
n
(1 − t)
exp −
=
L(α)
n (x)t .
1−t
(16)
n=0
The proof is straightforward. Start with (13) and change the order of summation to obtain
∞
X
n
L(α)
=
n (x)t
n=0
∞
X
(α + 1)n
n=0
∞ X
∞
X
n!
tn
∞
n
∞ X
X
X
(−n)k xk
(α + 1)n (−1)k xk tn
=
(α + 1)k k!
(α + 1)k k! (n − k)!
k=0
k=0 n=k
∞
X
∞
(−xt)k X (α + k + 1)n n
t
k! n!
k!
n!
n=0
k=0 n=0
k=0
µ
¶
∞
∞
k
X
X
(−xt)
1
xt k
−α−k−1
−α−1
=
(1 − t)
= (1 − t)
−
.
k!
k!
1−t
=
(α + 1)n+k
(α + 1)k
(−1)k xk tn+k
=
k=0
k=0
This proves (16).
If we define
−α−1
F (x, t) := (1 − t)
¶
µ
xt
,
exp −
1−t
then we easily obtain
µ
¶
∂
xt
−α−2
F (x, t) = −t(1 − t)
exp −
∂x
1−t
11
=⇒
(1 − t)
∂
F (x, t) + tF (x, t) = 0
∂x
and
½
¾
µ
¶
−x(1 − t) − xt
xt
(α + 1)(1 − t)−α−2 + (1 − t)−α−1 ·
exp
−
(1 − t)2
1−t
½
¾
µ
¶
x
xt
=
α+1−
(1 − t)−α−2 exp −
,
1−t
1−t
∂
F (x, t) =
∂t
which implies that
(1 − t)2
∂
F (x, t) + [x − (α + 1)(1 − t)] F (x, t) = 0.
∂t
The first relation leads to
∞
∞
X
X
d (α)
n
n
(1 − t)
L (x)t + t
L(α)
n (x)t = 0
dx n
n=0
n=0
or equivalently
∞
∞
∞
X
X
X
d (α)
d (α)
n+1
Ln (x)tn −
Ln (x)tn+1 +
L(α)
= 0.
n (x)t
dx
dx
n=0
n=0
n=0
Hence we have
d (α)
d (α)
Ln+1 (x) −
L (x) + L(α)
n (x) = 0,
dx
dx n
n = 0, 1, 2, . . .
(17)
or equivalently, by using (14),
d (α)
(α+1)
L (x) = L(α)
(x),
n (x) − Ln
dx n
n = 0, 1, 2, . . . .
The second relation leads to
2
(1 − t)
∞
X
n−1
nL(α)
n (x)t
+ [x − (α + 1)(1 − t)]
n=1
∞
X
n
L(α)
n (x)t = 0
n=0
or equivalently
∞
X
n−1
nL(α)
−2
n (x)t
n=1
∞
X
n
nL(α)
n (x)t +
n=1
+x
∞
X
n=0
n
L(α)
n (x)t − (α + 1)
∞
X
n+1
nL(α)
n (x)t
n=1
∞
X
n
L(α)
n (x)t + (α + 1)
n=0
∞
X
n+1
L(α)
= 0.
n (x)t
n=0
Equating the coefficients of equal powers of t we obtain the three term recurrence relation
(α)
(α)
(n + 1)Ln+1 (x) + (x − 2n − α − 1)L(α)
n (x) + (n + α)Ln−1 (x) = 0,
n = 1, 2, 3, . . . .
Note that this can also be written as
h
i
h
i
(α)
(α)
(α)
(α)
xL(α)
(x)
+
(n
+
1)
L
(x)
−
L
(x)
−
(n
+
α)
L
(x)
−
L
(x)
= 0,
n
n
n
n+1
n−1
12
(18)
n = 1, 2, 3, . . . .
Now we differentiate and use (17) to obtain
x
d (α)
(α)
(α)
Ln (x) + L(α)
n (x) − (n + 1)Ln (x) + (n + α)Ln−1 (x) = 0,
dx
n = 1, 2, 3, . . . .
This implies that
x
d (α)
(α)
Ln (x) = nL(α)
n (x) − (n + α)Ln−1 (x),
dx
n = 1, 2, 3, . . . .
(19)
Now we differentiate (19) and use (17) and (19) to find
x
d (α)
d
d (α)
d2 (α)
Ln (x) +
Ln (x) = n L(α)
L
(x)
n (x) − (n + α)
2
dx
dx
dx
dx n−1 ¸
·
d (α)
d (α)
d
= (n + α)
L (x) −
L
(x) − α L(α)
(x)
dx n
dx n−1
dx n
d
(α)
(x)
= −(n + α)Ln−1 (x) − α L(α)
dx n
d
d (α)
= x L(α)
(x) − nL(α)
L (x).
n (x) − α
dx n
dx n
(α)
This proves that the polynomial Ln (x) satisfies the second order linear differential equation
xy 00 (x) + (α + 1 − x)y 0 (x) + ny(x) = 0,
n ∈ {0, 1, 2, . . .}.
Finally, we use the generating function (16) to prove the addition formula
L(α+β+1)
(x
n
+ y) =
n
X
(α)
β)
Lk (x)Ln−k (y),
n = 0, 1, 2, . . . .
(20)
k=0
The generating function (16) implies that
∞
X
Ln(α+β+1) (x
n=0
+ y)t
n
µ
¶
(x + y)t
= (1 − t)
exp −
1−t
¶
µ
¶
µ
yt
xt
−β−1
−α−1
· (1 − t)
exp −
= (1 − t)
exp −
1−t
1−t
à n
!
∞
∞
∞
X (α)
X
X X (α)
(β)
k
(β)
m
=
Lk (x)t ·
Lm (y)t =
Lk (x)Ln−k (y) tn .
−α−β−2
k=0
m=0
This proves (20).
13
n=0
k=0
Jacobi
The Jacobi polynomials are orthogonal on the interval (−1, 1) with respect to the beta distribution w(x) = (1 − x)α (1 + x)β . They can be defined by means of their Rodrigues formula:
Pn(α,β) (x) =
=
£
¤
(−1)n 1
n
2 n
D
w(x)
(1
−
x
)
2n n! w(x)
h
i
(−1)n
−α
−β
n
n+α
n+β
(1
−
x)
(1
+
x)
D
(1
−
x)
(1
+
x)
2n n!
(21)
for n = 0, 1, 2, . . .. By using Leibniz’ rule (4) we have
n µ ¶
h
i X
n
Dn (1 − x)n+α (1 + x)n+β =
Dk (1 − x)n+α Dn−k (1 + x)n+β
k
k=0
n µ ¶
X
n
=
(−1)k (n + α)(n + α − 1) · · · (n + α − k + 1)(1 − x)n+α−k
k
k=0
× (n + β)(n + β − 1) · · · (β + k + 1)(1 + x)β+k
µ
¶µ
¶
n
X
n+β
k n+α
(−1)
(1 − x)n+α−k (1 + x)β+k , n = 0, 1, 2, . . . .
= n!
k
n−k
k=0
This implies that
Pn(α,β) (x)
µ
¶µ
¶
n
(−1)n X
n+β
k n+α
=
(−1)
(1 − x)n−k (1 + x)k ,
2n
k
n−k
n = 0, 1, 2, . . . .
(22)
k=0
(α,β)
This shows that Pn
(x) is a polynomial of degree n. Note that we have the symmetry
Pn(α,β) (−x) = (−1)n Pn(β,α) (x),
and
Pn(α,β) (1)
n = 0, 1, 2, . . .
µ
¶
µ
¶
n+α
(α,β)
n n+β
=
and Pn
(−1) = (−1)
,
n
n
(23)
n = 0, 1, 2, . . . .
In order to find a hypergeometric representation we write for x 6= 1
µ
¶ n µ
¶µ
¶µ
¶
x+1 k
x−1 nX n+α n+β
(α,β)
Pn
(x) =
, n = 0, 1, 2, . . . .
n
n−k
2
x−1
k=0
Now we have for x 6= 1
µ
¶
µ
¶k X
¶i
k µ ¶µ
x+1 k
k
2
2
= 1+
=
,
i
x−1
x−1
x−1
k = 0, 1, 2, . . . .
i=0
Now we obtain by changing the order of summations for x 6= 1 and n = 0, 1, 2, . . .
µ
¶µ
¶µ ¶ µ
¶ n n µ
¶i
x − 1 nXX n + α n + β
k
2
Pn(α,β) (x) =
k
n−k
i
2
x−1
i=0 k=i
¶ n n−i µ
¶µ
¶µ
¶µ
¶i
µ
n+β
i+k
2
x − 1 nXX n + α
.
=
2
i+k
n−i−k
i
x−1
i=0 k=0
14
Now we reverse the order in the first sum to find for x 6= 1 and n = 0, 1, 2, . . .
µ
¶ n n µ
¶µ
¶µ
¶µ
¶n−i
x − 1 nXX
n+α
n+β
n−i+k
2
(α,β)
Pn
(x) =
2
n−i+k
i−k
n−i
x−1
i=0 k=0
µ
¶µ
¶µ
¶
µ
¶
n X
n
X
n+α
n+β
n−i+k
x−1 i
=
n−i+k
i−k
n−i
2
i=0 k=0
=
n X
n
X
i=0 k=0
=
Γ(n + α + 1)
(n − i + k)! Γ(i − k + α + 1)
¶
x−1 i
2
µ
¶
1−x i
Γ(i + α + 1) i! Γ(n − i + β + 1)
2
Γ(n + β + 1)
(n − i + k)!
×
(i − k)! Γ(n − i + k + β + 1) (n − i)! k!
n
(−n)i
Γ(n + α + 1)Γ(n + β + 1) X
n!
×
i=0
n
X
(−i)k (−i − α − 1)k
k=0
(n − i + β + 1)k k!
µ
.
Since i ∈ {0, 1, 2, . . . , n} we have by using the Chu-Vandermonde summation formula
µ
¶
n
X
(−i)k (−i − α − 1)k
−i, −i − α − 1
(n + α + β + 1)i
= 2 F1
;1 =
.
(n − i + β + 1)k k!
n−i+β+1
(n − i + β + 1)i
k=0
Hence we have by using Γ(n − i + β + 1) (n − i + β + 1)i = Γ(n + β + 1)
µ
¶
n
Γ(n + α + 1) X (−n)i (n + α + β + 1)i 1 − x i
(α,β)
Pn
(x) =
n!
Γ(i + α + 1) i!
2
i=0
µ
¶
n
Γ(n + α + 1) X (−n)i (n + α + β + 1)i 1 − x i
=
, n = 0, 1, 2, . . . .
n! Γ(α + 1)
(α + 1)i i!
2
i=0
This proves the hypergeometric representation
µ
¶
µ
¶
n+α
−n, n + α + β + 1 1 − x
Pn(α,β) (x) =
F
;
,
2 1
n
α+1
2
n = 0, 1, 2, . . . .
Note that this result also holds for x = 1. In view of the symmetry (23) we also have
µ
¶
µ
¶
−n, n + α + β + 1 1 + x
(α,β)
n n+β
Pn
(x) = (−1)
;
, n = 0, 1, 2, . . . .
2 F1
n
β+1
2
Note that the hypergeometric representation implies that
µ
¶
µ ¶
d (α,β)
n + α (−n)(n + α + β + 1)
1
Pn
(x) =
· −
dx
n
α+1
2
µ
¶
−n + 1, n + α + β + 2 1 − x
× 2 F1
;
α+2
2
µ
¶
µ
¶
n+α+β+1 n+α
−n + 1, n + α + β + 2 1 − x
=
F
;
2 1
2
n−1
α+2
2
n + α + β + 1 (α+1,β+1)
=
Pn−1
(x), n = 1, 2, 3, . . . .
2
15
(24)
Another consequence of the hypergeometric representation is that the leading coefficient of
(α,β)
the polynomial Pn
(x) equals
µ
¶
n + α (−n)n (n + α + β + 1)n (−1)n
(n + α + β + 1)n
kn =
=
, n = 0, 1, 2, . . . .
n
(α + 1)n n!
2n
2n n!
Now it can be shown that the Jacobi polynomials satisfy the orthogonality relation
Z 1
2α+β+1 Γ(n + α + 1) Γ(n + β + 1)
(α,β)
(1 − x)α (1 + x)β Pm
(x)Pn(α,β) (x) dx =
δmn
(2n + α + β + 1) Γ(n + α + β + 1) n!
−1
for α > −1, β > −1 and m, n ∈ {0, 1, 2, . . .}. This can be shown by using the definition (21)
and integration by parts. The value of the integral in the case m = n can be computed by
using the leading coefficient and then writing the integral in terms of a beta integral:
Z 1
n
o2
(1 − x)α (1 + x)β Pn(α,β) (x) dx
−1
=
=
=
=
Z
h
i
(−1)n 1 (α,β)
n
n+α
n+β
P
(x)D
(1
−
x)
(1
+
x)
dx
n
2n n! −1
Z 1
1
Dn Pn(α,β) (x)(1 − x)n+α (1 + x)n+β dx
2n n! −1
Z
(n + α + β + 1)n 1
(1 − x)n+α (1 + x)n+β dx
22n n!
−1
Z 1
Γ(2n + α + β + 1)
(1 − x)n+α (1 + x)n+β dx,
Γ(n + α + β + 1) 22n n! −1
n = 0, 1, 2, . . .
and by using the substitution 1 − x = 2t
Z 1
Z 1
n+α
n+β
(1 − x)
(1 + x)
dx =
(2t)n+α (2 − 2t)n+β 2 dt
−1
0
Z
1
= 22n+α+β+1
tn+α (1 − t)n+β dt = 22n+α+β+1 B(n + α + 1, n + β + 1)
0
= 22n+α+β+1
= 22n+α+β+1
Γ(n + α + 1)Γ(n + β + 1)
Γ(2n + α + β + 2)
Γ(n + α + 1)Γ(n + β + 1)
,
(2n + α + β + 1) Γ(2n + α + β + 1)
(α,β)
The Jacobi polynomials Pn
n = 0, 1, 2, . . . .
(x) satisfy the second order linear differential equation
(1 − x2 )y 00 (x) + [β − α − (α + β + 2)x] y 0 (x) + n(n + α + β + 1)y(x) = 0,
n = 0, 1, 2, . . . .
A generating function for the Jacobi polynomials is given by
∞
X
2α+β
=
Pn(α,β) (x)tn ,
α
β
R(1 + R − t) (1 + R + t)
n=0
16
R=
p
1 − 2xt + t2 .
Legendre
The Legendre polynomials are orthogonal on the interval (−1, 1) with respect to the weight
function w(x) = 1. They can be defined by means of their Rodrigues formula:
Pn (x) =
£
¤ (−1)n n £
¤
(−1)n 1
n
2 n
D
w(x)
(1
−
x
)
= n
D (1 − x2 )n ,
n
2 n! w(x)
2 n!
This is the special case α = β = 0 of the Jacobi polynomials:
¶
µ
−n, n + 1 1 − x
(0,0)
Pn (x) = Pn (x) = 2 F1
,
;
2
1
n = 0, 1, 2, . . . .
n = 0, 1, 2, . . . .
(25)
(26)
Further we have
Pn (−x) = (−1)n Pn (x),
Pn (1) = 1 and Pn (−1) = (−1)n ,
n = 0, 1, 2, . . . .
The leading coefficient of the polynomial Pn (x) equals
kn =
(−n)n (n + 1)n (−1)n
(2n)!
= n
,
n
(1)n n!
2
2 (n!)2
The orthogonality relation is
Z 1
Pm (x)Pn (x) dx =
−1
2
δmn ,
2n + 1
n = 0, 1, 2, . . . .
m, n ∈ {0, 1, 2, . . .}.
(27)
This can be shown by using the Rodrigues formula (25) and integration by parts
Z
1
(−1)n
Pm (x)Pn (x) dx = n
2 n!
−1
Z
1
−1
Pm (x)D
n
£
2 n
(1 − x )
¤
1
dx = n
2 n!
Z
1
−1
Dn Pm (x) (1 − x2 )n dx,
which vanishes for m < n. For m = n we have
Z 1
Z 1
Z
(2n)! 1
n
2 n
2 n
D Pn (x) (1 − x ) dx = kn n!
(1 − x ) dx = n
(1 − x2 )n dx.
2 n! −1
−1
−1
Finally we have by using the substitution 1 − x = 2t for n = 0, 1, 2, . . .
Z 1
Z 1
Z 1
2 n
n
n
(1 − x ) dx =
(1 − x) (1 + x) dx =
(2t)n (2 − 2t)n 2 dx
−1
−1
0
= 22n+1 B(n + 1, n + 1) = 22n+1
Hence we have
Z 1
−1
{Pn (x)}2 dx =
Γ(n + 1) Γ(n + 1)
22n+1 (n!)2
=
.
Γ(2n + 2)
(2n + 1)!
(2n)! 22n+1 (n!)2
2
=
,
22n (n!)2 (2n + 1)!
2n + 1
This proves the orthogonality relation (27).
17
n = 0, 1, 2, . . . .
In order to find a generating function for the Legendre polynomials we use the hypergeometric
representation (26) to find
µ
¶
∞
∞ X
n
X
X
(−n)k (n + 1)k 1 − x k n
n
Pn (x)t =
t
(1)k k!
2
n=0
n=0 k=0
µ
¶
∞ X
∞
X
(−n)k (n + 1)k 1 − x k n
=
t
k! k!
2
k=0 n=k
µ
¶
∞ X
∞
X
(−n − k)k (n + k + 1)k 1 − x k n+k
=
t
k! k!
2
k=0 n=0
µ
¶
∞
∞
X
(2k)! x − 1 k k X (2k + 1)n n
t
t
=
k! k!
2
n!
n=0
k=0
µ
¶k
∞
X
(2k)! x − 1
tk (1 − t)−2k−1
=
k! k!
2
k=0
·
¸
∞
X
(1/2)k
2(x − 1)t −1/2
k
−2k−1
−1
=
[2(x − 1)t] (1 − t)
= (1 − t)
1−
k!
(1 − t)2
k=0
=
£
(1 − t)2 − 2(x − 1)t
¤−1/2
1
.
= (1 − 2xt + t2 )−1/2 = √
1 − 2xt + t2
This proves the generating function
∞
X
1
√
=
Pn (x)tn .
1 − 2xt + t2 n=0
(28)
If we define F (x, t) = (1 − 2xt + t2 )−1/2 , then we have
∂
1
x−t
F (x, t) = − (1 − 2xt + t2 )−3/2 (−2x + 2t) =
.
∂t
2
(1 − 2xt + t2 )3/2
This implies that
(1 − 2xt + t2 )
∂
F (x, t) = (x − t)F (x, t).
∂t
Now we use (28) to obtain
2
(1 − 2xt + t )
∞
X
nPn (x)t
n−1
= (x − t)
n=1
∞
X
Pn (x)tn .
n=0
This can also be written as
∞
∞
∞
∞
∞
X
X
X
X
X
nPn (x)tn−1 − 2x
nPn (x)tn +
nPn (x)tn+1 = x
Pn (x)tn −
Pn (x)tn+1
n=1
n=1
n=1
n=0
n=0
or equivalently
∞
X
(n + 1)Pn+1 (x)tn − x
n=0
∞
X
(2n + 1)Pn (x)tn +
n=0
∞
X
(n + 1)Pn (x)tn+1 = 0.
n=0
This leads to P1 (x) = xP0 (x) and the three term recurrence relation
(n + 1)Pn+1 (x) − (2n + 1)xPn (x) + nPn−1 (x) = 0,
18
n = 1, 2, 3, . . . .
(29)
Chebyshev
For x ∈ [−1, 1] the Chebyshev polynomials Tn (x) of the first kind and the Chebyshev polynomials Un (x) of the second kind can be defined by
sin(n + 1)θ
,
sin θ
Tn (x) = cos(nθ) and Un (x) =
The orthogonality property is given by
Z 1
Z
2 −1/2
(1 − x )
Tm (x)Tn (x) dx =
−1
x = cos θ,
n = 0, 1, 2, . . . .
(30)
π
cos(mθ) cos(nθ) dθ = 0,
m 6= n
0
and
Z
Z
1
2 1/2
(1 − x )
−1
Um (x)Un (x) dx =
π
sin(m + 1)θ sin(n + 1)θ dθ = 0,
m 6= n.
0
Both families of orthogonal polynomials satisfy the three term recurrence relation
Pn+1 (x) = 2xPn (x) − Pn−1 (x),
n = 1, 2, 3, . . . ,
since we have
Tn+1 (x) + Tn−1 (x) = cos(n + 1)θ + cos(n − 1)θ = 2 cos θ cos(nθ) = 2xTn (x)
and
Un+1 (x) + Un−1 (x) =
2 cos θ sin(n + 1)θ
sin(n + 2)θ + sin(nθ)
=
= 2xUn (x).
sin θ
sin θ
Note that
T0 (x) = U0 (x) = 1,
T1 (x) = x and U1 (x) = 2x.
We also have
Tn (x) = Un (x) − xUn−1 (x),
n = 1, 2, 3, . . . ,
since
Un (x) − xUn−1 (x) =
sin(n + 1)θ − cos θ sin(nθ)
sin θ cos(nθ)
=
= cos(nθ) = Tn (x).
sin θ
sin θ
In order to find a generating function for the Chebyshev polynomials Tn (x) of the first kind,
we multiply the recurrence relation by tn+1 and take the sum to obtain
∞
X
Tn+1 (x)t
n+1
= 2x
n=1
∞
X
Tn (x)t
n+1
n=1
If we define
F (x, t) =
∞
X
−
∞
X
Tn−1 (x)tn+1 .
n=1
Tn (x)tn ,
|t| < 1,
n=0
then we have
F (x, t) − T1 (x)t − T0 (x) = 2xt [F (x, t) − T0 (x)] − t2 F (x, t).
19
This implies that
(1 − 2xt + t2 )F (x, t) = T0 (x) + T1 (x)t − 2xtT0 (x) = 1 + xt − 2xt = 1 − xt.
Hence we have the generating function
∞
X
Tn (x)tn = F (x, t) =
n=0
1 − xt
,
1 − 2xt + t2
|t| < 1.
In the same way we have for the Chebyshev polynomials Un (x) of the second kind:
G(x, t) =
∞
X
Un (x)tn ,
|t| < 1
n=0
where
(1 − 2xt + t2 )G(x, t) = U0 (x) + U1 (x)t − 2xtU0 (x) = 1 + 2xt − 2xt = 1.
Hence we have
∞
X
1
,
1 − 2xt + t2
Un (x)tn = G(x, t) =
n=0
|t| < 1.
This can be used, for instance, to prove that
n
X
Tk (x)xn−k = Un (x),
n = 0, 1, 2, . . . .
k=0
In fact, we have for |t| < 1
!
à n
∞
∞ X
∞
∞ X
∞
X
X
X
X
Tk (x)xn tn+k
Tk (x)xn−k tn =
Tk (x)xn−k tn =
n=0
k=0
=
k=0 n=0
k=0 n=k
∞
X
∞
X
k=0
n=0
Tk (x)tk ·
(xt)n =
1 − xt
1
·
2
1 − 2xt + t 1 − xt
∞
=
X
1
=
Un (x)tn .
2
1 − 2xt + t
n=0
In a similar way we can prove that
n
X
Pk (x)Pn−k (x) = Un (x),
n = 0, 1, 2, . . . ,
k=0
where Pn (x) denotes the Legendre polynomial. In fact, we have for |t| < 1
à n
!
∞ X
∞
∞ X
∞
∞
X
X
X
X
n
n
Pk (x)Pn−k (x) t =
Pk (x)Pn−k (x)t =
Pk (x)Pn (x)tn+k
n=0
k=0
k=0 n=0
k=0 n=k
=
∞
X
k=0
Pk (x)tk ·
∞
X
1
1
Pn (x)tn = √
·√
2
1 − 2xt + t
1 − 2xt + t2
n=0
∞
=
X
1
=
Un (x)tn .
1 − 2xt + t2
n=0
20