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РАЗНОЕ
УДК 513.5
Y. Rouba, K. Smatrytski
GENERATING FUNCTION AND ITS APPLICATION FOR GENERALIZATION OF LEGENDRE POLYNOMIALS
Проводится обобщение полиномов Лежандра с помощью производящей функции. Найден явный вид искомых функций. Рассмотрены некоторые частные случаи. Особое внимание уделено случаю, когда параметры, определяющие изучаемые функции, являются различными, симметричными относительно нуля действительными числами. Изучены некоторые свойства этих функций. Основываясь на результатах численных экспериментов, выдвинута гипотеза о корнях исследуемых функций.
In the present paper the generalization of Legendre polynomials with the help of generating function is studied. The explicit form of considered functions is found. Some special cases are considered. Particular attention is paid to the case when the parameters defining the studied functions are different, symmetric about zero real numbers. Some properties of constructed functions are obtained. Based on the results of numerical experiment a hypothesis about zeroes of these functions is stated.
Ключевые слова: полиномы Лежандра, производящая функция, ортогональные системы.
Key words: Legendre polynomials, generating function, orthogonal systems.
Introduction
Generating function is one of the classical ways for construction of orthogonal systems. Application of generating functions for construction of orthogonal systems of algebraic polynomials is described in detail in the [1]. In the case of rational functions this problem is more complicated. In 1964 M. M. Dzhrbashyan and A. A. Kitbalyan applied generating function for construction of systems of rational functions which generalized Chebyshev polynomials of the first and the second type [2]. It should be mentioned that in the work [3] the construction of system of rational functions which generalized Jacobi polynomials with respect to the weight -y/(1 - x)/(1 + x) was described.
In the present paper the generalization of Legendre polynomials with the help of generating function is considered.
It's well known that the functions
1
F(x, z) = . , x e(-1,1), F(x, 0) = 1,
V1 - 2xz + z2
© Rouba Y., Smatrytski K., 2014
Вестник Балтийского федерального университета им. И. Канта. 2014. Вып. 10. С. 149-155.
150
is a generating function for algebraic Legendre polynomials, i. e.
F(x,z) = yz„ , |z|< 1.
„=0 n!
In other words, the Legendre polynomials are the Taylor coefficients of the function F(x, z),
P„( x) = 2- i F( x, z) 0 <P< 1.
2— 11 z
|z|=p
1. Main result
For the generalization of Legendre polynomials on rational case we consider the following rational functions. Let the sequence {ak} k=0 of complex numbers be such that a0 = 0, | ak | < 1, k e W0. Using this numbers we define the functions:
%(z) = 1, <pn (z) J la2 nf^, n e M0.
1 ~a„z k=0 1 -akz Lemma 1 [1]. The system {<pn (z)}™=0 is a orthogonal on the unit circle | z | = 1. Note, that if | z | = 1 then
_ ^¿jTRjini,!^ .^EEn1^- (1)
1 -anz k=0 1 ~akz z — an \=\ z-ak
Here and later the function mn (z) is defined by formula (1) also in the case
|z|*1.
The functions, which generalize the Legendre polynomials, we define as follows
Ln (x) = 2- i F(x, zfrM z, (2)
where r is a circle | z |= p, 0 < p < 1, such that | ak |< p, k = 1,2,..., n .
Remark. If a = 0, k = 1,2,..., n, then mn (z) = —, and Ln (x) = -1 i F(x, z)-^,
j z„' "w 2-|z|=p zn+1
i. e. in this case Ln (x) is a Legendre polynomials, orthogonal on the segment [-1,1] with respect to the weight 1.
Theorem 1. If ak ^ 0, k = 1,2,..., n, and ak k ^ j, k, j = 1,2,..., n,
then the following formula holds
(
Ln (x) = V1— | an
(-1)" 1-| Ok |2 1 j1 -ajak +
n a k=1 ak (ak -an ) ^ - 2xak +a2k j:1 ak - aj
V k=1
1 1 _ A
1 j-j 1 -ajan
a
ft-iaa j=1 a -a
Proof. Using formula (1) we obtain
Ln (x ) = -!_ J 1 J^2 n-
2ni rV 1 -2xz + z2 z-an k=1 z-ak z Then we apply the substitution
Vl - 2xz + z2 = 1 - zt or z = 2(t - x). (4)
In this case
t2 -1
dz = -2-—22tx + 1 dt, "K-lxz + z2 = -t 2tX +1
(t2 -1)2 t2 -1
1 -qz = t2 -1 -2ak(t-x) - = an(1 -12) + 2(t-x) z-ak _ ak(1 -12) + 2(t-x)' n = t2 -1 .
Therefore
L (x)= V 1-lan I2 r1 - 12 -1 -2ak(t-x) dt
2xi r t-x an(1 -12) + 2(t-x)k=1 ak(1 -12) + 2(t-x) ' where C is an image of r when mapping t(z).
Under the conditions of the theorem the integrand
F(t) = t-1 1 ff t2 -1 - 2q (t - x)
() t-x an(1 -12) + 2(t-x)-1=1 ak(1 -12) + 2(t-x) has simple poles at the point t = x and at the roots of the equations ak(1 -12) + 2(t-x)' k = 1,2'...'n '
i. e.
= 1 1 - 2xak +ak = 1 + V1 - 2xak +ak k = tk'1 = ' tk'2 = ' k = 1'2'...'n .
ak ak
It is not difficult to show that the points tk4' k = 1'2'...' n ' are inside the curve C' and the points tk 2' k = 1'2'...'n' are outside this curve. To prove it one needs to find the corresponding points zk^ ■ = z(tk^ ■)' k = 1'2'...' n' j = 1'2 using (4) and verify that zkAzk 2 = 1' | zk4|< 1' k = 1'2'...'n . By the Cauchy's residue theorem we get
Ln (x) = V 1-1 a„ |2 [Res(F' x) + jj Res(F' tw)j . (5)
It is easy to see that
(-1)"
Res(F' x) = VL. (6)
nak
k=1
Now we calculate residues at the points tk 4' k = 1' 2'...' n -1. We have
t2 -1-2ak(t-x)
t2 -1 Res(F, tu) =t
t - x
a„(1-t2) +2(t - x)
t2 -1-2oj(t-x)
41-
,=t a(1 -t2) +2(t -x)
pk
-ak(t tk,2)
757
Then we find
t2 -1 t - x
_2_
a,
an(1 -12) + 2(t - x)
an ^ + 2 | (t - x) t - x
152
-an — + 2
1 -yj 1 - 2xak + a,2 2 (a„-an akx-yjl - 2xa, + a,2 )
t2 -1 -2a-(t-x)
a, (1 -12) + 2(t - x)
--1 - 2a,
t - x 1
1 -12 + 2
t_tu a,-+ 2
' t - x
— - 2a —
\_'_ _1 -aa
a
-a - — + 2 a -a,
1 a,
t2 -1 - 2ak (t - x)
-a,(t - h ,2)
t2 -1 t - x
- 2a, | (t - x)
-a, (t - t, ,2 )
a - 2at
a,
1 -yj 1 - 2xak +a,2
1 -^/1-2x—+ af 1 + ^fl—2x——+—a
I1"^^^ V1 -2x— +aa )0H — |2)
yj1 - 2 xak +
Since that,
1-\a \2 Res(F, tk ,1) - 1 |ak 1
ak
k (ak - an )s]1 - 2xak +a2 j=1
And the residue at the point tn1
jjjj1 -a,ak
=1 a, -a,
1 * k
(7)
Res(F, tn,1) _
t2 -1
t - x
t_tn1 -an(t - tn,2 ) 1
jjj t2 -1 - 2a, (t - x)
t _t a, (1 -12) + 2(t - x)
- 2xan
jj1 -a,an
+ a,f ,_1 an -a,
(8)
"n ' 'n 3 n j
To complete the proof one must put (6), (7) and (8) into (5). □
Corollary 1. By the formula (3) the function Ln (x) can be written in the form
Ln (x)
c,
- 2xa
, +ak
where ck, k = 1,2,..., n, are some complex numbers. Its singular points are as follows
xk = 1 + ak , k = 1,2,...,n .
2ak
2
1
, _0
2. Some important cases
1) Obviously' L0( x) = 1.
2) Let a0 = 0' a1 = a' |a|< 1. Then the second summand in the right side of (3) vanishes' and the product in the third summand is equal to 1. Since that'
Vh
|2 (
L1(x ) =■
-1 +
Vl - 2 xa+a2
Note that
lim L1 (x ) = lim
. 1 -V1 - 2 xa + a
2 2 xa-a2 = lim—-,-. -
-11 + V 1 - 2xa + a2
= x.
a
Besides' the function L1(x) has the only zero at the point
153
_ a x = I-
3) Let the sequence {ak} 1=0 be as follows: a0 = 0' ' a2'...' a^n_1 (distinct
points)' an = 0 . In this case
cpn (z) ' V (z) = - n
1 "41 -akz
k=o 1 -akz
z k= 1 z —
and in a similar way
L ( ) = _Lr t2 -1 |4_t2 Ln (x) = 4m" J (t - x)2 U —
t2 -1 "4 t2 -1 - 2ak (t - x)
c (t - x)2 k=1 ak (1 -12 ) + 2(t - x)
dt
or
where
L„ (x) = 1 (Res(F, x) + g Res(F, tM) |,
F(t) = t2 -1 n_ t2 -1 - 2ak (t - x)
(t-x)2 k=1 ak(1 -12) + 2(t-x) tk 4 is the root of the equation ak (1 -12) + 2(t - x) = 0 which lies inside the curve C.
Now we find residues. The point t = x is a pole of the second order. Thus
Res(F,x) =
(t2 - 1)n
n-1 l2
t2 -1 - 2ak (t - x) 'i! ak (1 -12) + 2(t - x)
= 2xfl -Ot-1^ +(x2 - Din
n-1 n-1 x2 -1 f t2 -1 - 2ak (t - x) A
k=i ak (1 - x )
j =1 k=1 ak (1 x )
k * j
— (1 -12) + 2(t - x)
(-1)"-12x ( 2 - _(-1)n 2(1-1 -j |2 ) = (-l)n-12 f n- 1- I a
: "i +(x n-i
n-k ^
a-(1 - x2) "-1
n
x + X"
2
i=i a, ak v ' 1 j 7
k=1 k * j
a
154
The points tk A' k = 1' 2'...' n -1' are the simple poles. We have
( ____A
Res(Ft ) = t2 -1 t2 -1 - 2ak (t - x) ff t2 -1 - 2ak (t - x)
' W (t - x)2 -ak(t - tk 2) }=f ak(1 -12) + 2(t - x)
j*k
I1 -akx-Jl_2Xaa+aa)(1-|ak |2) n-11 -0"ak
—-L-n—~
1 - akx -*J 1 - 2xak + a,
a;
•^1 - 2 xa
k +ak2
j=1 ak -aj
j*k
ak
2(1-|ak |2) — -ajak
;•x/l_"2xa
k + ak j=1 ak -a,
j * k
Finally' L" (x) =
(-1)n
n
ak
x +
z
11- | a, |2 A n-1
a-
-z-
1- | a.
j /
- 2 xak
n4.1 -aIak
n-j . (9)
+ ak j=1 ak-aj
j *k
Now we consider the following sequence of parameters. Let P1' P2' ..v Pn be a sequence of distinct real positive numbers. Then the parameters {ak} 2n+1 are as follows: a0 = 0' a2k-1 = Pk' a2k = -Pk' k = 1'2'...'n ' a2n+1 = 0. In this case by the formula (9) we get
L2n+1(x) = "
n
-z-
1- | a.
ak
- 2xak
-r I"!—
+ ak2 j=1 ak-aj
j *k
1 1
-05 /
1 ; 0 8 0 6 0 4 0 2 \ 0 2 0 4 x 0 6 oy 1
/ -fl-5-
' -1-
These functions are odd' L2 n+1 (1) = 1. Besides using numerical experiment we can draw the graph of some of these functions. In the figures 1' 2' 3 we have the graphs of L3(x)' L5(x)' L7(x) for P1 = 0.5' P2 = 0.6' P3 = 0.7.
Fig. 1. The graph of L3( x) for P1 = 0.5
Fig. 2. The graph of L5(x) for P2 = 0.6 Fig. 3. The graph of L7(x) for P3 = 0.7
k=1
k=1 a
k=1
Concerning these results some questions can be raised:
(a) Does function L2n+1(x) has 2n + 1 distinct real zeroes in the interval (-1, 1)?
(b) Numerical results show that the system {Ln (x)} is not orthogonal. Is there a sequence {an} such that corresponding system {Ln (x)} is orthogonal?
Note, that if the answer to the first question is positive than we can use these zeroes as nodes for construction of interpolation polynomial.
References
1. Suetin P. K. Classical orthogonal polynomials. Moskow, 1978. (in Russian)
2. Dzhrbashyan M. M, Kitbalyan A. A. On one generalization of Chebyshev polynomials / / Reports of Armenian SSR Academy of science. 1964. Vol. 38, N 5. P. 263-270. (in Russian)
3. Rovba E. A. Orthogonal systems of rational functions on the segment and quadratures of Gauss-type // Mathematica Balkanica. 1999. Vol. 13, N 1-2. P. 187-198.
Об авторах
Евгений Алексеевич Ровба — д-р физ.-мат. наук, проф., Гродненский государственный университет им. Янки Купалы, Беларусь.
E-mail: [email protected]
Константин Анатольевич Смотрицкий — канд. физ.-мат. наук, доц., Гродненский государственный университет им. Янки Купалы, Беларусь.
E-mail: [email protected]
About the authors
Prof. Yauheni Rouba — Yanka Kupala State University of Grodno, Belarus.
E-mail: [email protected]
Dr Kanstantin Smatrytski — Ass. Prof., Yanka Kupala State University of Grodno, Belarus.
E-mail: [email protected]
