 Open Access
 Total Downloads : 775
 Authors : B.M. Yisa, R.B. Adeniyi
 Paper ID : IJERTV1IS6013
 Volume & Issue : Volume 01, Issue 06 (August 2012)
 Published (First Online): 30082012
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
A Generalized Formula For Canonical Polynomials For mth Order NonOverdetermined Ordinary Differential Equations (ODES)
B.M. YISA AND R.B. ADENIYI
DEPARTMENT OF MATHEMATICS, UNIVERSITY OF ILORIN, ILORIN, NIGERIA
The canonical polynomials play a major role in the Recursive formulation of the Tau method of Lanczos and Ortiz. However,the construction of these polynomials, subsequent to its use, in the tau method is highly demanding and hence polynomial associated with individual DE are often constructed. In this paper, we shall present a derived formula which captures the poly nomials for a general class of problems involving nonoverdetermined mth order ODEs (m not fixed). The derivative of this polynomial will also be obtained. For the purpose of validating these results, they are cast as the orems for which mathematical induction principle is employed. We hope to incorporate the results obtained into the tau approximation process for purpose of deriving a general tau approximant of the solution of this class of problems in a future work.

Differential equations result from physical models of anything that varies
– whether in space, in time, in value, in cost, in colour, etc. For example, dif ferential equations exist for models of quantities such as: volume, pressure, temperature, density, magnetization, fracture Strength, dislocation density, chemical potential, etc. These differential equations take the general form
yn = f (x, y, yI , …, yn1) (1.1a)
dxi
where y(i) diy , i = 1, 2, …, n, is the ith derivative of y with respect to x. The above equation is referred to as nth order differential equation because the highest order derivative appearing in the equation is of order n. A unique
soluton y(x) of (1.1a) can be obtained when given supplementary conditions
y(i)(0) = i, for i = 0, 1, …n 1 (1.1b)
(1.1a) and (1.1b) is referred to as initial value problem (IVP). One of the methods that give an accurate approximate solution to (1.1) is the Tau Method.
The essential of the Tau Method (see Lanczos[11,12] and Ortiz[14]) is to perturb the given differential problem in such a way that its exact solution becomes a polynomial. This is achieved by using a polynomial perturbation term, added to the right hand side of the differential equation. The desired Tau approximation is written in terms of a special polynomial basis, called the canonical polynomial basis, uniquely associated with the given differen tial operator D (see Ortiz[19]) which defines the given problem. Such basis does not depend on the degree of approximation. The order of the approxi mation can be increased by just adding one or more canonical polynomials to those already generated and updating the coefficients affecting them.
To give more flexibility in computation of Tau solution, Lanczos[12] in 1956, introduced a systematic use of the socalled canonical polynomials in the Tau method. A recursive generation of Lanczos canonical polynomials was proposed by Ortiz[14]. The basic approach proposed by Ortiz will be restated briefly in this section as contained in Adeniyi et al[2].
Let y(x) be a known function which satisfies
Ly(x) = f (x) (1.2a)
where L is an mth order linear differential operator with polynomial coeffi cients and
f (x) = f0 + f1x + f2x2 + …. + fx
is a given polynomial of degree with real coefficients fi, i = 0(1).
In addition, we assume that y(x), a x b, satisfies the following general linear bondary conditions
tj m1
a(lkj)y(k)(xij) = j, j = 1(1)m (1.2b)
l=1 k=0
where y(k)(xij), k = 0(1)m 1, is the value of y(k)(x) at x = xij, l = 1(1)tj, j = 1(1)m; tj denotes the point of evaluation; alkj, xij and j are given real numbers. Uniquely associated with operator L in (1.2a) is a sequence
{Qr(x)}, r NoS, of canonical polynomials Qr(x) such that
LQr(x) = xr (1.3)
where s is a small finite or empty set of indices with cardinality s(s m+h); h is the maximum difference between the exponent r of x and the leading exponent of the generating polynomial Lxr, for r No. The construction of Qr(x) using Lxr is described exhaustively by Ortiz[14]. To apply the
constructed polynomials Qr(x), r N0, in the Tau method, we consider the perturbed equation
Lyn(x) = f (x) + Hn(x) (1.4)
of (1.2a), where
Hn(x) =
m+s1 i=0
m+s1Tnm+i+1(x) (1.5)
Adopt (1.3) in (1.4) to get
Lyn(x) = fiLQi(x) +
i=0
m+s1 i=0
m+s1
nm+i+1 r=0
r
C(nm+i+1)LQr(x) (1.6)
From the linearity of the operator L, we obtain
Lyn(x) = L
i=0
fiQi(x) +
m+s1 i=0
m+s1
nm+i+1
r=0
C(nm+i+1)Qr (x)
r
By assuming that L1 exists and is unique, we have
yn(x) = fiQi(x) +
i=0
m+s1 i=0
m+s1
nm+i+1 r=0
r
C(nm+i+1)Qr (x) (1.7)

In this paper, we intend to obtain a general formula for the canonical polynomials together with its derivatives for the initial value problem (IVPs) mth order ordinary differential equation (ODE)
Ly(x) :=
m
r=0
Nr
k=0
Prkxk
y(r)(x) =
F
r=0
frxr (2.1a)
(r)
m1
Ly(xrk) := arky (xrk) = k, k = 1(1)m (2.1b)
r=0
where Nr, F are given nonnegative integers and ark, xrk, k, fr, Prk are given real numbers by seeking an approximant
n
yn(x) = arxr, n < + (2.2)
r=0
which is the exact solution of the corresponding perturbed problem
where
F
Lyn(x) = frxr + Hn(x) (2.3a)
r=0
Lyn(xrk) = k, k = 1(1)m (2.3b)
m+s1
Hn(x) =
r=0
m+sr Tnm+r+1(x) (2.4)
is the perturbation term. The parameters r, r = 1(1)m + s, are to be determined,
Tr(x) = Cos
rCos1
2x a b b a
r
(
k=0
(r)
C
k
xk (2.5)
is the Chebyshev polynomial valid in the interval [a, b] (assuming that (2.1) is defined in this interval) and
s = max {Nr r 0 r m} (2.6)
The canonical polynomials for the initial value problems (2.1) will be obtained in this section for cases m=1,2,3, and 4 before the general formula for the case m=m is obtained. Since we are presently considering non overdetermined problems, s will be zero throughout this section.
The following individual cases are considered and from which the general case will be deduced.
Case m = 1 :
(P1,0 + P1,1x) yI (x) + P0,0y(x) = f (x) (2.7a)
y(0) = 0 (2.7b)
By Definition (see [1],[6],[7],[8]),
xr = LQr(x)
From (2.7a), the differential operator L is given by
d
L = (P1,0 + P1,1x) dx + P1,0
Lxr = rP1,0xr1 + (rP1,1 + P0,0) xr
r
Lx = rP1,0LQr1(x) + (rP1,1 + P0,0) LQr(x)
Lxr = L {rP1,0Qr1(x) + (rP1,1 + P0,0) Qr(x)}
And due to the existence of L1, we have
xr = rP1,0Qr1(x) + (rP1,1 + P0,0) Qr(x)
therefore
xr rP1,0Qr1(x)
Qr(x) =
rP1,1 + P0,0
, r 0
The results for r = 0, 1, 2, 3 are
1
Q0(x) = P
0,0
x P1,0
Q (x) =
1 P0,0 + P1,1 P0,0 (P0,0 + P1,1)
Q2(x) = P
x2
+ 2P
(P
2P1,0x
+ P ) (P
+ 2P )
0,0
1,1
0,0
1,1
0,0
1,1
2P 2
+ 1,0
P0,0 (P0,0 + P1,1) (P0,0 + 2P1,1)
x3 3P1,0x2
Q3(x) = P
0,0
+ 3P1,1 (P0,0 + 2P1,1) (P0,0
+ 3P
1,1)
6P 2 x
Case m = 2 :
+ 1,0
(P0,0 + P1,1) (P0,0 + 2P1,1) (P0,0 + 3P1,1)
1,0
6P 3
P0,0 (P0,0 + P1,1) (P0,0 + 2P1,1) (P0,0 + 3P1,1)
( )
P2,0 + P2,1x + P2,2×2 yII (x) + (P1,0 + P1,1x) yI (x) + P0,0y(x) = f (x)
(2.8a)
yI (0) = 1, y(0) = 2 (2.8b)
From (2.8a),
L = (P
2,0
+ P2,1
x + P
2,2
2
) d
x2 + (P
dx2
1,0
d
+ P1,1x) dx
+ P0,0
Passing through the same process as in the case m=0, we have
xr [r (r 1) P2,1 + rP1,0] Qr1(x) r (r 1) P2,0Qr2(x)
Qr(x) =
r (r 1) P2,2
+ rP1,1
+ P0,0
, r 0
From which we obtain Qr(x) for r = 0, 1, 2, 3 as
1
Q0(x) = P
0,0
x P1,0
Q (x) =
1 P0,0 + P1,1 P0,0 (P0,0 + P1,1)
Q2(x) = P
0,0
x2
+ 2P
1,1
2P1,0x
2
2P
1,0
(P0,0
+ P1,1
) (P
0,0
+ 2P1,1
+
) P0,0
(P0,0
+ P1,1
) (P0,0
+ 2P
1,1)
Q3(x) = P
0,0
x3
+ 3P
6P
2
1,1
3P1,0x2
1,0x
(P
0,0
+ 2P
1,1
) (P
0,0
+ 3P
1,1
+
) (P
0,0
+ P1,1
) (P0,0
6P 3
+ 2P
1,1
) (P0,0
+ 3P
1,1)
1,0
P0,0 (P0,0 + P1,1) (P0,0 + 2P1,1) (P0,0 + 3P1,1)
Case m = 3 :
( ) ( )
P3,0 + P3,1x + P3,2×2 + P3,3×3 yIII (x) + P2,0 + P2,1x + P2,2×2 yII (x)
+ (P1,0 + P1,1x) yI (x) + P0,0y(x) = f (x) (2.9a)
y(0) = 0, yI (0) = 1, yII (0) = 2 (2.9b)
From (2.9a), we obtain
xr r (r 1) (r 2) P3,0Qr3(x)
Qr (x) = r (r 1) (r 2) P
3,3
+ r (r 1) P
2,2
+ rP1,1
+ P0,0
[r (r 1) (r 2) P3,1 + r (r 1) P2,0] Qr2(x)
r (r 1) (r 2) P3,3 + r (r 1) P2,2 + rP1,1 + P0,0
[r (r 1) (r 2) P3,2 + r (r 1) P2,1 + rP1,0] Qr1(x) , r 0
r (r 1) (r 2) P3,3 + r (r 1) P2,2 + rP1,1 + P0,0
From which we obtain Qr(x) for r = 0, 1, 2, 3 as
1
Q0(x) = P
0,0
x P1,0
Q (x) =
1 P0,0 + P1,1 P0,0 (P0,0 + P1,1)
x2 (2P2,1 + 2P1,0) x
Q2(x) = P
0,0
+ 2P
1,1
+ 2P
2,2
(P
0,0
+ P1,1
) (P0,0
+ 2P
1,1
+ 2P
2,2)
P1,0 (2P2,1 + 2P1,0) 2P2,0
+
P0,0 (P0,0 + P1,1) (P0,0 + 2P1,1 + 2P2,2) P0,0 (P0,0 + 2P1,1 + 2P2,2)
Case m = 4 :
( )
( ) ( )
P4,0 + P4,1x + P4,2×2 + P4,3×3 + P4,4×4 y(iv)(x)
+ P3,0 + P3,1x + P3,2×2 + P3,3×3 ylll(x) + P2,0 + P2,1x + P2,2×2 yll(x)
+ (P1,0 + P1,1x) yl(x) + P0,0y(x) = f (x) (2.10a)
y(0) = 0, yl(0) = 1, yll(0) = 2, ylll(0) = 3 (2.10b)
This yields the canonical polynomial
Qr(x) =
xr
r(r 1)(r 2)(r 3)P4,4 + r(r 1)(r 2)P3,3 + r(r 1)P2,2 + rP1,1 + P0,0 [r(r 1)(r 2)(r 3)P4,3 + r(r 1)(r 2)P3,2 + r(r 1)P2,1 + rP1,0] Qr1(x) r(r 1)(r 2)(r 3)P4,4 + r(r 1)(r 2)P3,3 + r(r 1)P2,2 + rP1,1 + P0,0
[r(r 1)(r 2)(r 3)P4,2 + r(r 1)(r 2)P3,1 + r(r 1)P2,0] Qr2(x)
r(r 1)(r 2)(r 3)P4,4 + r(r 1)(r 2)P3,3 + r(r 1)P2,2 + rP1,1 + P0,0
[r(r 1)(r 2)(r 3)P4,1 + r(r 1)(r 2)P3,0] Qr3(x)
r(r 1)(r 2)(r 3)P4,4 + r(r 1)(r 2)P3,3 + r(r 1)P2,2 + rP1,1 + P0,0
r(r 1)(r 2)(r 3)P4,0Qr4(x)
r(r 1)(r 2)(r 3)P4,4 + r(r 1)(r 2)P3,3 + r(r 1)P2,2 + rP1,1 + P0,0
For r = 0, 1, 2,
1
Q0(x) = P
0,0
x P1,0
Q (x) =
1 P0,0 + P1,1 P0,0 (P0,0 + P1,1)
x2 2 (P1,0 + P2,1) x
Q2(x) = P
0,0
+ 2P
1,1
+ 2P
2,2
(P
0,0
+ P1,1
) (P0,0
+ 2P
1,1
+ 2P
2,2)
(2 P
+ 1,0 (P0,0 + P1,1) P2,0 + P1,0P2,1)
2
P0,0 (P0,0 + P1,1) (P0,0 + 2 (P1,1 + P2,2))
Hence , we obtained the canonical polynomials formula of the mth order ODEs as
Qr(x) =
Theorem
r m
x
k=1
m j=k
k=0
m
r j
( ( ) j! P Q (x)j,jk rk
k
k,k
k!(r)P
, r 0 (2.11)
Theorem
If the linear differential operator associated with the mth order linear ODE:
Ly(x) = P0(x)y(0)(x) + … + Pm(x)y(m)(x) = f (x)
where Pk(x) is a kth degree polynomials, y(k)(x) is the kth derivative of
y(x) and f (x) is a polynomial function, is given by
dm dm1
then
L = Pm(x) dxm + Pm1(x) dxm1 + … + P0(x)
Qr(x) =
r m
x
k=1
m j=k
m
r j
( ( ) j! P Q (x)j,jk rk
k!(r)P
, r 0 (2.12)
Proof
x
( ( ) j! P Q (x)j,jk 1k
We shall employ the principles of mathematical induction to establish the validity of our Qr(x). If we fix r = 1, then we can apply the principles of mathematical induction on m:
Q1(x) =
m k=1
m j=k
k=0
m
1
j
k
k,k
k!(1)P
(2.13)
1
We try for m = 1,
x 1!(1)P1,0Q0(x)
0
1
Q1(x) = 0!(1)P
0,0
+ 1!(1)P
1,1
Q (x) = x P1,0Q0(x)
(2.14)
1 P0,0 + P1,1
x
( ( ) j! P Q (x)j,jk 1k
and this is the same as our Q1(x) obtained in (2.7) above, confirming that it is true for m = 1.Now assume that (2.11) is true for m = n, thus (2.11) becomes
Q1(x) =
n k=1
n j=k
k=0
n
1
j
k
k,k
k!(1)P
(2.15)
The next thing is to show that (2.11) holds for m = n + 1. From our con struction of Q1(x) in (2.13) and (2.15) for m = n to m = n + 1, we have
k=1
k=0
k
x n
n j=k
k,k
(Pj,jk) j!(1) Q1k
n+1,n+1
Q (x) =
k!(1)P
+ (n + 1)!( 1 )P
j
n+1
(Pn+1,nk+1) (n + 1)!( 1
+
k,k
) Q1k
n+1,n+1
n k!(1)P
n+1
k=0
k
n+1
+ (n + 1)!( 1 )P
(2.16)
1 n
x n
n (Pj,jk) j!(1) + (Pn+1,nk+1) (n + 1)!( 1
k,k
) Q1k
Q1(x) =
k=1
j=k
j
k=0
k
n+1(P
)k!(1)
n+1
(2.17)
x n
n (Pj,jk) j!(1) + (Pn+1,nk+1) (n + 1)!( 1
) Q1k
Q1(x) =
k=1
j=k
j
n+1(P
)k!(1)
n+1
Thus, it is true for m = n + 1 also.
k=0
k,k k
(2.18)
If we choose to fix r = l, so that we can again apply the principle of mathematical induction on m,
Ql(x) =
l m
x
k=1
m j=k
m
1
( ( ) j! P Q (x)j,jk lk
j
k!(1)P
, r 0 (2.19)
We try for m = 1,
xl 1!(1)P1,0Ql1(x)
0
1
Ql(x) =
0!(1)P
0,0
+ 1!(1)P
1
1,1
(2.20)
xl lP1,0Ql1(x)
Ql(x) =
P0,0 + lP1,1
(2.21)
and this is in conformity with our earlier results, thus, it is true for m = 1.
Now, assume that (2.19) is true for m = n, thus (2.19) becomes
Ql(x) =
l n
x
k=1
n j=k
k=0
n
1
( ( ) j! P Q (x)j,jk lk
j
k
k,k
k!(1)P
(2.22)
The next thing is to prove that (2.19) holds for m = n + 1.
From our construction of Ql(x) for m = n up to m = n + 1, we have
k=1
k=0
k
n
xl n
n j=k
k,k
(Pj,jk) j!(1) Qlk(x)
n+1,n+1
p>Ql(x) =
k!(1)P + (n + 1)!( 1 )P
j
n+1
(Pn+1,nk+1) (n + 1)!( 1 ) Q1k(x)
+
k,k
n+1,n+1
n k!(1)P
n+1
k=0
k
n+1
+ (n + 1)!( 1 )P
(2.23)
xl n
n (Pj,jk) j!(1) + (Pn+1,nk+1) (n + 1)!( 1
k=1
j=k
j
k,k
) Qlk
Ql(x) =
k=1
j=k
j
k=0
k
xl n+1 n+1 (j!(1)Pj,jk Qlk(x)
n+1(P
)k!(1)
n+1
(2.24)
(2.24)
k=0
k
k,k
Ql(x) =
n+1 k!(1)P
(2.25)
Thus it is true for m = n + 1.
Finally, if we choose to fix r = l + 1, and we apply the principle of math ematical induction on m. With r = l + 1, (2.11) is now
Ql+1
(x) =
l+1 m
x
k=1
m j=k
m
(j!(l+1)Pj,jk Qlk+1(x)
j
k!(l+1)P
(2.26)
We try for m = 1
,
k=0
k k,k
Q (x) =
xl+1 1!(l+1)P1,0Ql(x)
0
1
(2.27)
l+1
0!(l+1)P
1
0,0
+ 1!(l+1)P
1,1
xl+1 (l + 1)Ql(x)
Ql+1(x) =
P0,0
+ (l + 1)P1,1
(2.28)
which shows it is true for m = 1.
j
We assume it is true for m = n so that (2.26), with r = l + 1, becomes
k=0
k
k,k
Ql+1
(x) =
l+1 n
x
k=1
n j=k
n
(j!(l+1)Pj,jk Qlk+1(x) k!(l+1)P
(2.29)
The next thing is to prove that (2.26) holds for m = n + 1,
Q (x) =
xl+1 n
n j=k
(Pj,jk) j!(l+1) Qlk+1(x)
k=1
l+1
n k=0
k!(l+1)P
k,k
+ (n + 1)!( l+1 )P
j
n+1
n+1,n+1
k
n+1
(Pn+1,nk+1) (n + 1)!( l+1 ) Q1k+1(x)
+
(2.30)
n k!(l+1)Pk,k + (n + 1)!( l+1 )Pn+1,n+1
Q (x) =
k=0
x
l+1 n
(P
k=1
k
n j=k
(Pj,jk) j!(l+1)
n+1
j
k
( )
(l+1)
n+1 k=0
k,k
)k!(l+1)
n+1
(Pn+1,n k+1) (n + 1)! l+1 Ql k+1
k=0
k
xl+1 n+1 n+1 (j!(l+1)Pj,jk Qlk+1(x)
+
n+1(Pk,k)k!(l+1)
(2.31)
Ql+1(x) =
n+1 k!(1)P
, (2.32)
k=1
j=k
j
k=0
k
k,k
Thus, the formula for Qr(x) given by (2.11) holds for all r and for all m. Hence, its validation.
The method of section 2 is also applied here to obtain a general result for the derivative of the canonical polynomial. So doing, we considered the individual cases below:
Case m = 1: First Derivative:
l rxr1 rP1,0Qlr1(x)
Second Derivative:
Qr (x) =
P0,0
+ rP1,1
(3.1)
ll r(r 1)xr2 rP1,0Qlrl1(x)
Qr (x) =
Third Derivative:
P0,0
+ rP1,1
(3.2)
lll
r(r 1)(r 2)xr3 rP1,0Qlrll 1(x)
Qr (x) =
P0,0
+ rP1,1
(3.3)
If we continue with this process, we shall obtain the nth derivative for case
m = 1 as
r
Q(n)(x) =
n
P0,0 + rP1,1
r1
(3.4)
n!(r )xrn rP1,0Q(n) (x)
Following a similar procedure, the nth derivatives for cases m = 2 and m = 3 are
r
n!(r )xrn r(r 1)P2,1Q(n) (x) r(r 1)P2,0Q(n) (x)
Q(n)(x) = n
r1
P0,0 + rP1,1 + r(r 1)P2,2
r2
(3.5)
and
r
n!(r )xrn [r(r 1)(r 2)P3,2 + r(r 1)P2,1 + rP1,0] Q(n) (x)
Q(n)(x) =
n
P0,0 + rP1,1 + r(r 1)P2,2 + r(r 1)(r 2)P3,3
r1
[r(r 1)(r 2)P3,1 + r(r 1)P2,0] Q(n) (x) + r(r 1)(r 2)P3,0Q(n) (x)
r2
P0,0 + rP1,1 + r(r 1)P2,2 + r(r 1)(r 2)P3,3
r3
(3.6)
respectively.
We deduce from that for general mth order equation that the nth deriva tive of Qr(x) is
n!(r )xrn m
r
( m
j!(r)Pj,jk Q(n) (x)
Q(n)(x) = n
Theorem
k=1
m k=0
j=k
k
k!(r)P
j
k,k
rk
(3.7)
If the canonical polynomials associated with the mth order linear DE:
Ly(x) P0(x)y(0)(x) + … + Pm(x)y(m)(x) = f (x)
where Pk(x) is a kth degree polynomial, y(k)(x) is the kth derivative of y(x) and f (x) is a polynomial function, is given by
Qr(x) =
then
r m
x
k=1
m j=k
(
m
k=0
r j
( ) j! P Q (x)j,jk rk
k
k,k
k!(r)P
then
n!(r )xrn m
r
( m
j!(r)Pj,jk Q(n) (x)
Q(n)(x) = n
Proof
k=1
m k=0
j=k
k
k!(r)P
j
k,k
rk
(3.8)
We shall establish this by the principle of mathematical induction. We start by fixing r = 1 and subject m to varied values:
Let us try for m = 1,
(n)
n!(1)x1n 1!{1
1
(n)
(1\
0
(1\
1
1
0
1
1
Q1 (x) =
n
P1,0Q0 (x)0!
P0,0 + 1!
P1,1 (3.9)
Q(n)(x) =
n
P0,0 + P1,1
0
(3.10)
n!(1)x1n P1,0Q(n)(x)
which shows that it is true for m = 1, since this (i.e 3.10) is in conformity with (3.8) when r = 1.
We assume that it is true for m = q,
n!(r )xrn q
r
( q
k=0
j!(r)Pj,jk Q(n) (x)
k,k
Q(n)(x) = n
k=1
q
j=k j
k
k!(r)P
rk
(3.11)
Now we shall prove that (3.8) holds for m = q + 1.
1
From our construction of Q(n)(x) in (3.8) (with r = 1) for m = q up to
m = q + 1, we have
n!(1)x1n ( q
l
q (Pj,jk) j!(1) Q(n) (x)
k,k
q+1,q+1
Q(n)(x) =
n k=1
k=0
k
q k!(1)P
j=k j
q+1
+ (q + 1)!( 1 )P
1k
(Pq+1,qk+1) (q + 1)!( 1 ) Q(n) (x)
+
k,k
q+1,q+1
q k!(1)P
q+1 1 k
k=0
k
q+1
+ (q + 1)!( 1 )P
(3.12)
Q(n)(x) =
n
k=1
n!(1)x1n q
q (Pj,jk) j!(1) Q(n) (x)
j=k
j
1k
k=0
k
l q+1 k!(1)Pk,k
(Pq+1,qk+1) (q + 1)!( 1 ) Q(n) (x)
+
q+1
k=0
k
n!(1)x1n q+1 q+1 (Pj,jk) j!(1)Q(n) (x)
q+1 k!(1)Pk,k
1k
(3.13)
Q(n)(x) =
n
k=1
j=k
j
1k
(3.14)
k=0
Thus, it is true for m = q + 1 also.
k
l q+1 k!(1)Pk,k
If we decide to fix r = l, so that we can apply the principle of mathe matical induction on m again, we have
n!(1)xln m m (Pj,jk) j!(1)Q(n) (x)
l
k=0
k
k,k
Q(n)(x) = n
We try for m = 1,
k=1
m
j=k
( )
k! 1 P
j lk
(3.15)
We try for m = 1,
( )+ 1! P
Q(n)(x) =
n
0!(1)P
1
1
1
l1
1,1
(3.16)
n!(1)xln 1!(1)P1,0Q(n) (x)
l
0
0,0
l
0
0,0
Q(n)(x) =
n
l1
(3.17)
n!(1)xln lP1,0Q(n) (x)
l
P0,0 + lP1,1
l
P0,0 + lP1,1
Since this tallies with (3.8) with r replaced by l, thus it is true for m = 1.
Let us assume that (3.8) holds for m = q. Now to prove that (3.8) holds for m = q + 1.
l
From our construction of Q(n)(x) in (3.8) for m = q up to m = q + 1, we
have
l
n!(1)xln q
q (Pj,jk) j!(1) Q(n) (x)
k,k
q+1,q+1
Q(n)(x) =
n k=1
k=0
k
q k!(1)P
j=k j
q+1
+ (q + 1)!( 1 )P
lk
(Pq+1,qk+1) (q + 1)!( 1 ) Q(n) (x)
+
( )
q k!(1)P
q+1 l k
+ (q + 1)!( 1 )P
(3.18)
Q
(n)
l
(x) =
k=0 k
n
n! 1 xln
k=0
k
q+1 (Pk,k) k!(1)
k,k
q+1
q+1,q+1
q q
j!(1) (Pj,jk) + (q + 1)!( 1
) (Pq+1,q+1) Q(n) (x)
k=1
j=k j
q+1
k=0
k
n!(1)xln q+1 ( q+1 j!(1)Pj,jk Q(n) (x)
q+1 (Pk,k) k!(1)
lk
(3.19)
(3.19)
l
Q(n)(x) = n
k=1
k=0
q
j=k j
k
k,k
k!(1)P
lk
(3.20)
Thus it is true for m = q + 1
Finally, if we choose to fix r = l + 1, and we apply the principle of math ematical induction on m. With our r = l + 1, (3.15) becomes
n!(l+1)xln+1 m
l+1
m (Pj,jk) j!(l+1)Q(n)
k
(x)
Q(n) (x) = n
k=1
m k=0
j=k
k!(l+1)Pk,k
j lk+1
(3.21)
We try for m = 1,
Q(n)
n!(l+1)xln+1 1!(l+1)P1,0Q(n)(x)
(3.22)
0
1
(l+1)
0!(l+1)P0,0 + 1!(l+1)P1,1
Q
(x) =
n
1
l
(n)
(l+1)
(x) =
n
l
(3.23)
n!(l+1)xln+1 (l + 1)P1,0Q(n)(x)
P0,0 + (l + 1)P1,1
P0,0 + (l + 1)P1,1
and this is the same as (3.8) with r replaced l + 1, confirming that it is true for m = 1.
We assume that it is true for m = q, thus (3.21) can now be written as:
n
n!(l+1)xln+1 q
( q
j!(l+1)Pj,jk Q(n)
(x)
k
Q(n) (x) =
k=1
j=k
j
l+1
q k=0
k!(l+1)Pk,k
lk+1
(3.24)
The next stage of our work is to prove that it is true for m = q + 1.
l+1
From our construction of Q(n) (x) in (3.21) for m = q up to m = q + 1,
we have
n!(l+1)xln+1 q q (Pj,jk) j!(l+1) Q(n)
(x)
l+1
q k!(l+1)Pk,k + (q + 1)!(l+1)Pq+1,q+1
Q(n) (x) =
n
k=1
j=k
j
lk+1
k=0
k
q+1
q+1
lk+1
(Pq+1,qk+1) (q + 1)!(l+1) Q(n) (x)
+
(3.25)
( )
q k!(l+1)Pk,k + (q + 1)!(l+1)Pq+1,q+1
Q
(n)
l+1
(x) =
k=0 k
n
n! l+1 xln+1
k=0
k
q+1 (Pk,k) k!(l+1)
q+1
q q
j!(l+1) (Pj,jk) + (q + 1)!(l+1) (Pq+1,qk+1) Q(n)
(x)
k=1
j=k j
q+1
k=0
k
q+1 (Pk,k) k!(l+1)
lk+1
(3.26)
(3.26)
Q(n) (x) =
n
k=1
j=k
j
lk+1
n!(l+1)xln+1 q+1 ( q+1 j!(l+1)Pj,jk Q(n)
(x)
k=0
k
l+1 q+1 k!(l+1)Pk,k
(3.27)
(3.27)
From the above, we can conveniently say that (3.8) holds for all calues of m
and r.
We now want to fix our n at n = 1 and apply the principle of mathe matical induction on m.
With n = 1, (3.8) becomes
1!(r)xr1 m
r
( m
k=0
j!(r)Pj,jk QI
k,k
(x)
QI (x) = 1
k=1
m
j=k j
k
k!(r)P
rk
(3.28)
We try for m = 1,
( )+ 1! P
Qr (x) =
1
0!(r)P
1
r
1
r1
1,1
(3.29)
I 1!(r)xr1 1!(r)P1,0QI
0
0,0
(x)
0
0,0
l rxr1 rP 1,0Qlr1(x)
Qr (x) =
P0,0
+ rP
1,1
(3.30)
which is the same as (3.4), confirming the correctness of (3.36), that is, it is true for m = 1.
k=1
j=k
j
rk
We assume that it is true for m = q, that is (3.28) becomes
rxr1 q
q (j!(r)Pj,jk QI
k=0
k,k
(x)
I
k
Qr (x) =
q k!(r)P
(3.31)
Next is to prove that it is true for m = q + 1,
I
rxr1 q
k=0
q j!(r)Pj,jk QI
k,k
(x)
k
Qr (x) = q
k!(r)P + (q + 1)!( r )P
k=1
j=k
j
rk
q+1
q+1,q+1
(Pq+1,qk+1) (q + 1)!( r
k,k
) QI
(x)
+ q
k!(r)P
q+1 r k
k=0
k
q+1
q+1,q+1
+ (q + 1)!( r )P
(3.32)
I
k=0
Qr (x) = q
k!(r)P
rxr1
k,k
q+1
q+1,q+1
+ (q + 1)!( r )P
k
q q
j!(r)Pj,jk + (Pq+1,qk+1) (q + 1)!( r
k=0
) QI
(x)
k=1
j=k
q
j
k
q+1
k!(r)Pk,k + (q + 1)!( r
q+1
)Pq+1,q+1
rk
(3.33)
(3.33)
QI (x) =
k=1
j=k
j
rk
k=0
k
rxr1 q+1 ( q+1 Pj,jkj!(r) QI (x)
(3.34)
k,k
r q+1 k!(r)P
Thus it is true for m = q + 1
Thus it is true for m = q + 1
k=1
j=k
j
rk
We again fix n at n = l and apply the principle of mathematical induc tion on m. With n = l, (3.8) becomes
1
l!(r)xrl m
m (j!(r)Pj,jk Q(l)
k=0
k,k
(x)
(l)
k
Qr (x) =
m k!(r)P
(3.35)
We try for m = 1,
r
( )+ 1! P
Q(l)(x) =
1
0!(r)P
0,0
1
r
1
r1
1,1
(3.36)
l!(r)xrl 1!(r)P1,0Q(l) (x)
0
r
Q(l)(x) =
1
P0,0 + rP1,1
r1
(3.37)
l!(r)xrl rP1,0Q(l) (x)
k=1
j=k
j
rk
We assume that it is true for m = q, so that (3.35) becomes
1
l!(r)xrl q
q (j!(r)Pj,jk Q(l)
k=0
k,k
(x)
(l)
k
Qr (x) =
q k!(r)P
(3.38)
We now prove for m = q + 1.
r
From our construction of Q(l)(x) in (3.35) for m = q up to m = q + 1,
we have
r
l!(r)xrl q
k=0
q (Pj,jk) j!(r) Q(l)
k,k
q+1,q+1
(x)
Q(l)(x) =
1 k=1
k
q+1
q k!(r)P
j=k j
+ (q + 1)!( r )P
rk
(Pq+1,qk+1) (q + 1)!( r
) Q(l)
(x)
+ q
k!(r)P
q+1 r k
+ (q + 1)!( r )P
(3.39)
Q(l)(x) =
l!(r)xrl
k=0
k,k
k=0
k
k k,k
q+1
q+1,q+1
1
r q+1 (P
) k!(r)
q q
j!(r) (Pj,jk) + (q + 1)!( r
) (Pq+1,q+1) Q(l)
(x)
k=1
j=k j
q+1
k=0
k
q+1 (Pk,k) k!(r)
rk
(3.40)
(3.40)
1
k=1
j=k
j
rk
l!(r)xrl q+1 ( q+1 j!(r)Pj,jk Q(l)
k,k
(x)
(l)
k=0
Thus it is true for m = q + 1 also.
k
Qr (x) =
q k!(r)P
(3.41)
Finally, we shall fix n = l + 1 and apply the principle of mathematical in duction on m again.
With n = l + 1, (3.8) can now be written as
(l+1)
r+1
(l + 1)!( r
)xr(l+1) m
m (j!(r)Pj,jk Q(l+1)(x)
Qr (x) =
Let us try for m = 1,
m k=0
k!(r)P
k
k=1
j=k
j
rk
k,k
(3.42)
r
( )0! P
( )+ 1! P
Q(l+1)(x) =
r+1
r
0
0,0
1
r
1
1,1
r1
(3.43)
(l + 1)!( r )r(l+1) 1!(r)P1,0Q(l+1)(x)
r
Q(l+1)(x) =
l+1
P0,0 + rP1,1
r1
(3.44)
(l + 1)!( r )xr(l+1) rP1,0Q(l+1)(x)
And (3.44) is the same as (3.4) with n replaced by l + 1, thus it is true for
m = 1.
Next, we assume that it is true for m = q, in which case, (3.42) can now be written as
r+1
(l + 1)!( r
)xr(l+1) q
q (j!(r)Pj,jk Q(l+1)(x)
(l+1)
Qr (x) =
q k=0
k!(r)P
k
k=1
j=k
j
rk
k,k
(3.45)
Now we want to prove that (3.42) holds for m = q + 1.
r
From our construction of Q(l+1)(x) in (3.42) for m = q up to m = q + 1,
we have
(l+1)
Qr (x) =
(l + 1)! r xr(l+1) q q (Pj,j k) j! r Q(l+1)(x)
l+1
k=1
j=k
j
rk
k=0
k
k,k
q+1
q+1,q+1
( ) ( )
q k!(r)P + (q + 1)!( r )P
(Pq+1,qk+1) (q + 1)!( r
k,k
) Q(l+1)(x)
q+1,q+1
+ q
k!(r)P
q+1 r k
k=0
k
q+1
+ (q + 1)!( r )P
(3.46)
(l + 1)!( r
r
k=0
)xr(l+1)
k
Q(l+1)(x) =
r+1
k,k
q+1 (P
) k!(r)
q q
j!(r) (Pj,jk) + (q + 1)!( r
) (Pq+1,q+1) Q(l)
(x)
k=1
j=k j
q+1
k=0
k
(l + 1)!( r )xr(l+1) q+1 ( q+1 j!(r)Pj,jk Q(l+1)(x)
q+1 (Pk,k) k!(r)
rk
(3.47)
(3.47)
(l+1)
Qr (x) =
l+1
q k=0
k!(r)P
j
k,k
rk
(3.48)
k
k=1
j=k
From the foregoing, it can be concluded that (3.8) holds for all values of m, r and n. Thus, the validity of our formula.
The derivation of a general formula for the canonical polynomials as sociated with mth order nonoverdetermined linear ODE together with its associated nth order derivative has been presented.
The formula are recursive and hence makes for easy determination of particular cases for which m will be specified. The use of canonical poly nomial in the Tau approximation problem to the solution of ODEs is very attractive as they do not depend on the boundary conditions and when Tau approximations of higher degrees are needed, the process of their (i.e canonical polynomial) determination does not begin from the scratch. These are some of the shortcomings of the two other variants of the Tau method namely, the differential form and the integrated form.
It is intended that the polynomial reported above will be incorporated into the Tau method for purpose of generalizing the recursive formulation of the Tau method itself.

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