ON GENERALIZATION OF WEIERSTRASS

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continuous function F(x, t) and prove that on a rectangle [a, b] × (−1, 1), a ≤ x ≤ ... In other words, assume that the function f is continuous on bounded interval [a ...
TWMS J. App. Eng. Math. V.6, N.1, 2016, pp. ...

ON GENERALIZATION OF WEIERSTRASS APPROXIMATION THEOREM FOR A GENERAL CLASS OF POLYNOMIALS AND GENERATING FUNCTIONS HEMANT KUMAR1 , M. A. PATHAN2 , § Abstract. Here, in this work we present a generalization of the Weierstrass Approximation Theorem for a general class of polynomials. Then we generalize it for two variable continuous function F (x, t) and prove that on a rectangle [a, b] × (−1, 1), a ≤ x ≤ b, |t| 0, there is a polynomial Pn (n ≥ 0) with sufficiently high degree n such that |f (x) − Pn (x)| 0, and g (x) is a probability density function defined by Z ∞ g(x)dx = 1, otherwise g(x) = 0. (2) −∞

Again, in this order we prove that Theorem 1.2. Let f : R → R be a bounded uniformly continuous function. Then Hk f converges uniformly to f as k > 0. 1

Department of Mathematics D. A-V. (P. G.) College Kanpur-208001 India e-mail: [email protected]; 2 Centre for Mathematical and Statistical Sciences, Peechi Campus, KFRI, Peechi-680653,Kerala, India e-mail: [email protected]; § Manuscript received: Month( December) Day (27), 2011. c I¸sık University, Department TWMS Journal of Applied and Engineering Mathematics Vol.5 No.1 of Mathematics 2015; all rights reserved. 1

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TWMS J. APP. ENG. MATH. V.5, N.2, 2015

 Proof. Let  > 0, then there exists δ > 0 such that |f (x) − f (y)| < , for all x, y ∈ R with 2 |x − y| < δ. Assume |f (x)| ≤ M, ∀x ∈ R. Again, g(x) is a probability density function so that with the help of Eqn. (2) we may write   Z 1 ∞ u−x f (x) = f (x)g du, k > 0. (3) k −∞ k   R ∞ δ  δ Now let k0 > 0 such that 0 < k ≤ k 0 < , where F k0 = δ (v) g (v) dv < ∞, δ 4M F

k0

k0

then on using Eqns. (2) and (3), we get   Z ∞ u−x 1 |f (u) − f (x)|g du|when|u−x| 0; ; Z Z   2M k0 |Hk0 f (x) − f (x)| ≤ + 2M g (v) dv ≤ + |v| g (v) dv 2 2 δ |v|≥ δ |v|≥ δ k0

 2M k0 = + 2 δ =

 2M k0 − 2 δ

− kδ

Z

0

−∞

Z

δ k0

k0

2M k0 (−v) g (v) dv + δ (v) g (−v) dv +



2M k0 δ

Z



(v) g (v) dv δ k0

Z



(v) g (v) dv

(5)

δ k0

Further, if in the Eqn. (5), the probability density function g (v) is even R ∞ function then g (−v) = g (v) and for odd g (v) it is g (−v) = −g (v), therefore for δ (v) g (v) dv = k0   F kδ0 < ∞, we get         + 4M k0 F δ , g (v) is even function, 2 δ k0 |Hk0 f (x) − f (x)| ≤    , g (v) is odd function. 2 δ   to get Now, in Eqn. (6) we set k0 = δ 8M F

(6)

δ 0

 |Hk0 f (x) − f (x)| ≤

, g (v) is even function,  2 , g (v) is odd function.

Finally, in Eqn. (7) as  → 0, Hk0 f (x) converges uniformly to f (x) .

(7) 

HEMANT KUMAR, M. A. PATHAN : ON GENERALIZATION OF WEIERSTRASS APPROXIMATION ...

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2. The Weierstrass Approximation Theorem for a General Class of Polynomials Here, we extend the function f to a bounded uniformly continuous function on R. This is accomplished here with the add of the following extended function. f (x) − f (a) f (x) − f (b) Let = f (a) on open interval (a − 1, a), and = −f (b) on open x−a x−b interval (b, b + 1), and f (x) = 0 for all x ∈ R\[a − 1, b + 1]. Particularly, we consider R > 0 such that R ∈ R and f (x) = 0 for |x| > R. Let  > 0 and M such that |f (x)| ≤ M, ∀ x ∈ R. Then by Theorem 2 there exists k0 > 0 such that for all x ∈ R, we have  |Hk0 f (x) − f (x)| < . (8) 2 Again, f (u) = 0 for |u| > R, R > 0, then we may write 1 Hk0 f (x) = k0

Z

R

 f (u)g

−R

u−x k0

 du, where k0 > 0.

(9)

n 1 X (−n)m An,m (x)2m , where, An,m (∀n ≥ 0, ∀m ≥ 0) is n→∞ C

Further letting g(x) = lim

m=0

(−1)m n! a bounded sequence, (−n)m = , 0 ≤ m ≤ n, and for m > n, there is (−n)m = 0, n − m! and C is any constant which may be found on satisfying the (2). Also for all equation u − x ≤ 2R, hence then on |x| ≤ R, and all |u| ≤ R, we have |u − x| ≤ 2R, so that k0 k0   2R 2R closed interval − , k0 , we get k0   n 1 u − x 1 X u − x 2m  (10) − (−n)m An,m < g 4RM k0 k0 Ck 0 k0 m=0

Thus with the help of Eqns. (9) and (10), we have  2m Z R n X u − x 1  f (u) (−n)m An,m du < Hk0 f (x) − 2 Ck 0 −R k0

(11)

m=0

Now in Eqn. (11), we define a general class of polynomials in the form Pn (x) =  2m RR Pn u−x du to get f (u) (−n) A n,m m=0 m k0 −R

1 Ck0

|Hk0 f (x) − Pn (x)|