DOI : 10.17577/IJERTV2IS110928
- Open Access
- Total Downloads : 230
- Authors : Aditya Kumar Raghuvanshi, B. K. Singh, Ripendra Kumar
- Paper ID : IJERTV2IS110928
- Volume & Issue : Volume 02, Issue 11 (November 2013)
- Published (First Online): 27-11-2013
- ISSN (Online) : 2278-0181
- Publisher Name : IJERT
- License:
This work is licensed under a Creative Commons Attribution 4.0 International License
on the degree of Approximation of function Belonging to the Lip(alpha, r) Class by (C,2) (E,1)
On the degree of Approximation of function Belonging to the Lip(, r) Class by (C,2) (E,1)
Aditya Kumar Raghuvanshi, B.K. Singh & Ripendra Kumar,
Department of Mathematics
IFTM University, Moradabad, (U.P.) India, 244001
Abstract
In this paper we obtained a theorem on the degree of approximation of functions belonging to Lip(, r) class by (C,2)(E,1) product mean of its Fourier series.
Keywords: Cesaro matrices, Euler matrices, degree of approximation.
-
Let f (t) be a periodic function with period 2 and integrable in the sense of Lebesgue. The Fourier series of f (t) is given by
.
.
1
f (t) 2 a0
A function f Lip if
+ (an
n=1
cos nt + bn sin nt) (1.1)
|f (x + t) f (x)| = O(|t| ) for 0 < 1 (1.2) And a function f Lip (, r) if
r
r
(r 2
0
0
|f (x + t) f (x)| dx
\1/r
= 0(|t|
) for 0 < |f r | (1.3)
0
0
The degree of approximation of a function f : R R by a trigonometric polynomial tn of order n is defined by (Zygmund [3])
||tn f || = sup{|tn(x) f (x)| : x R} (1.4)
if
n
n
E1 = 2n
.k=0
(n\
k
k
n
n
sk s as n (1.5)
,
,
n
then an infinite series ak with the partial sums sn is said to be summable
k=0
n
n
(E,1) to the S, (Hardy [2]). The (C,2) transform of the (E,1) transform E1
,
,
defines the (C,2) (E,1) transform of the partial sums sn of the series ak .
k=0
Thus if
n
n
2
2
n
n
(n + 1)(n + 2)
(n + 1)(n + 2)
k
k
(C , E)1 (x) = , 2(n k + 1) E1 s as n
k=0
k=0
k=0
k=0
n
n
where E1
,
,
denotes the (E,1) transform of sn, then the series ak is said to
k=0
be summable by (C,2) (E,1) means to s. We shall use the following notations:
(t) = f (x + t) + f (x t) 2f (x)
-
We shall generalize the theorem of Albayrak [1].
Theorem 2.1. If f : R R is 2 periodic, Lebesgue integrabble on [ , ] and belonging to Lip (, r) class then the degree of approximation of f by the (C,2) (E,1) product means of its Fourier series satisfies for n = 0, 1, 2, . . .
1
1
2
2
n
n
||(C E) f ||
= O / 1 \ for 0 < 1 & r 1
2
2r
2
2r
2
2r
2
2r
1
(n + 1)
1
(n + 1)
The nth partial sum Sn(x) of the series (1.1) at t = x is defined as
{
{
1
0
0
Sn(x) = f (x) + 2
(t)
sin(n + 1/2)t
dt
sin t/2
So the (E,1) means of the series (1.1) are
E = 2
E = 2
1 n
n
k=0
(n\ s
(x) for n = 0, 1, 2, . . .
k
k
= f (x) +
2n+1
= f (x) +
2n+1
k=0
k=0
k
k
r
r
1 r (t) / n
0
sin t/2
0
sin t/2
(n\ (
k
k
1\ \
sin t/2
k
sin
sin
t
t
dt
dt
sin t/2
k
k +
2
k +
2
= f (x) +
1 (t)
Im
(e 2 (1 + eit)n
2n+1
0 sin t/2
= f (x) +
1 r (t)
it
it
n
n
it
Im (e 2 )(1 + cos t + sin t) dt
2n+1
0 sin t/2
= f (x) +
= f (x) +
1 r (t)
2n+1
0
sin t/2
2n+1
0
sin t/2
( it
t ( t
2
2
2
2
Im
Im
e 2 2n cosn
e 2 2n cosn
cos
cos
+ i sin
+ i sin
dt
dt
t\n1
2
2
1 r (t)
2n+1
0
sin t/2
2n+1
0
sin t/2
t
2
2
(( t
2
2
t\ (
2
2
nt
2
2
nt\1
2
2
= f (x) +
0
0
2n cosn Im
cos
+ i sin
cos
+ i sin dt
= f (x) +
1
2 0
0
0
r
r
(t) sin t/2
cosn
t t
sin(n + 1) dt
2 2
therefore (C,2) (E,1) product means of the series (1.1) are
n
n
(C , E)1 (x) = 2(n k + 1) E1(x)
2 n
n=0
(n + 1)(n + 2) k
n n
2(n + 1) 2
(n + 1)(n + 2)
(n + 1)(n + 2)
k
k
(n + 1)(n + 2)
(n + 1)(n + 2)
k
k
= E1(x) k · E1(x)
k=0
k=0
k=0
k=0
k=0
k=0
r /
r /
1 (t) n
= f (x) +
2
2
k · cos
( t\
sin(k + 1)
t\ dt
(n + 2) 0
sin t
2 2
k
k
k=0
1 r (t) / n
0
0
k t
1\
(n + 1)(n + 2)
sin t/2
k=0
k · cos
sin(k + 1) dt
2 2
Now
= f (x) + I1 I2 (3.1)
I2 =
1 r (t) / n
2
2
k · cos
( t\
sin(k + 1)
t\ dt
(n + 1)(n + 2) 0
sin t
2 2
k
k
k=0
=
=
n+1
+ t
n+1
+ t
1 /r 1
(n + 1)(n + 2)
0
(n + 1)(n + 2)
0
r \ (t) / n
1
n+1
sin 2
1
n+1
sin 2
( t\
2
2
sin(k + 1)
sin(k + 1)
dt
dt
t\
2
2
(n + 1)(n + 2)
0
1
n+1
sin 2
k=0
2
2
(n + 1)(n + 2)
0
1
n+1
sin 2
k=0
2
2
k
k · cos
k
k · cos
I2 = I21 + I22 (3.2)
t t
t t
Applying the fact that f Lip (, r) and sin . We have
2
n r 1 O|t|
n+1
2
2
|I21| 2(n + 2)
n r
0
1
n+1
sin t dt
|t|
0
0
2(n + 2)
Using H¨older inequality.
t dt
r 1
l1 (r 1 1
n n+1 (t) r
n+1 s
r
r
|I21| dt
-
sdt
(
(
·
·
2(n + 2) 0 t 0
O 1
(n + 1)/2
1
(n + 1)1/2s
(n + 1)/2
(n + 1)/2
1 1
(n + 1)
1 1
(n + 1)
/ \
/ \
= O ( 1 / 1 \
2
2r
2
2r
2
2r
2
2r
1 1
1 1
O 1 (3.3)
2
2
2
2
2r
2r
(n + 1) +
2
2
By using k · cosk ( t ) =
2(cosk ( t ))
2
2
2
2
sin t
-
cos
t , we estimate I22.
1 r
(n + 1)(n + 2)
(n + 1)(n + 2)
(t) n
sin 2
sin 2
2
2
p/>
2(cosk ( t ))
2
2
sin 2
sin 2
t
2
2
t
2
2
|I22|
2
2
1
n+1
1
n+1
k=0
k=0
t t
2
2
· cos
sin(k + 1)
dt
t2
t2
dt
dt
2 r (cosn+1 t 1
(n + 1)(n + 2)
1
n+1
cos 2 1
(n + 1)(n + 2)
1
n+1
cos 2 1
(n + 1)(n + 2)
t
(n + 1)(n + 2)
t
(cos 1
(cos 1
2 r
t2
n+1 t
2
2
t dt
=
=
(n + 1)(n + 2)
1
r
r
n+1
cos 2 1
2
=
t2 sin t
(cosn+1 t 1)
dt
(n + 1)(n + 2)
1
n+1
(cos t 1)2
= A B. (3.4)
Now
2 r cosn t · sin t
t2
t2
(n + 2)
(n + 2)
1
n+1
1
n+1
cos t 1
cos t 1
|A| = dt
t3dt =
t3dt =
dt
dt
c1 r
(n + 2)
(n + 2)
C1 r
(n + 2)
(n + 2)
(t)
t3
t3
(n + 2)
1
n+1
(n + 2)
1
n+1
t3
(n + 2)
1
n+1
(n + 2)
1
n+1
t3
Using h older inequality.
C1
1 (t) \r
l1/r
1/s
l
l
s
|A| 3 dt · (1) dt
(n + 2)
1 t
/ \
/ \
n+1
1
n+1
C1 1 1
3
3
1
1
(n + 2)
(n + 2)
(n + 1) 2 2
(n + 1) 2 2
(n + 1) 2s
(n + 1) 2s
/ \
/ \
O ·
C1 1 1
3
3
1
1
1
1
(n + 2)
(n + 2)
(n + 1) 2 2
(n + 1) 2 2
(n + 1) 2 2r
(n + 1) 2 2r
O ·
1
1
O / 1 \ (3.5)
(n + 1) 2 2r
2
2
where C1 is positive constant.
t2
t2
(n + 1)(n + 2)
(n + 1)(n + 2)
|B| =
1
n+1
1
n+1
sin t(cosn+1 t 1)
(cos t 1)2
(cos t 1)2
dt
1
n+1
1
n+1
3 dt
3 dt
C2 (t)
(n + 1)(n + 2)
Using H¨older inequality
1 t
n+1
C2 (t) r
(n + 1)(n + 2)
1
n+1
t3
(n + 1)(n + 2)
1
n+1
t3
l1/r
l1/s
|B|
1
n+1
1
n+1
dt · (1)sdt
1
n+1
1
n+1
(t3)rdt ·
1
(t3)rdt ·
1
1
n+1
1
n+1
/ \ / \
/ \ / \
C2
(n + 1)(n + 2)
(n + 1)(n + 2)
/ 1 \
(n + 1) 2s
(n + 1) 2s
(n + 1)(n + 2)
(n + 1)(n + 2)
C2 1 1
3
3
1
1
1
1
(n + 1)(n + 2)
(n + 1)(n + 2)
(n + 1) 2 2
(n + 1) 2 2
(n + 1) 2 2r
(n + 1) 2 2r
| | / \
| | / \
|B| O ·
1
1
B O 1 (3.6)
2
2
2r
2r
(n + 1) +1
In similar way, we can obtain
1
1
1
1
I = O / 1 \ (3.7)
(n + 1) 2 2r
1
(n + 1)
1
(n + 1)
from (3.1), (3.2), (3.3), (3.4), (3.5), (3.6) and (3.7) we have
1
1
2
2
n
n
||(C , E) f (x)||
= O / 1 \
2
2r
2
2r
2
2r
2
2r
This completes the proof of the theorem.
If r = then above theorem reduces to Inci (Albayrak [1]).
1
1
2
2
n
n
(n + 1) 2
(n + 1) 2
||(C , E) f (x)|| = O ( 1 \ .
-
Albayrak Inci etc; On the degree of approximation of function belonging to the Lip by (C,2) (E,1) Int. J. of Math. Analysis, Vol. 4, 2010.
-
Hardy, G.H.; Divergent Series, First Ed. Oxford Uni. Press, 1949.
-
Zygmund, A.; Trigonometric series, 2nd rev. Ed. Cambridge Uni. Press, Cambridge. 1968.