 Open Access
 Total Downloads : 258
 Authors : Ripendra Kumar, B.K. Singh & Aditya Kumar Raghuvanshi
 Paper ID : IJERTV2IS90263
 Volume & Issue : Volume 02, Issue 09 (September 2013)
 Published (First Online): 14092013
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Degree of Approximation of function belonging to Lip(α, r) functions by Product Summability Method
Lip(, r) functions by Product Summability Method
Ripendra Kumar, B.K. Singh & Aditya Kumar Raghuvanshi
Department of Mathematics
IFTM University, Moradabad, (U.P.) India, 244001
Abstract
In this paper author have been determined the degree of approxi mation of certain functions belonging to Lip(, r) class by (C, 1)(E, q) means of its Fourier series.
Let f(t) be periodic functions with period 2 and integrable in the Lebesgue sense. The fourier series f(t) is given by
)
)
1
f(t) = a0
2
+
n=1
(an

cos nt + bn

sin nt) (1)
A function f Lip(, r) for 0 x 2, if
r
r
(r 2
0
0
f(x + t) f(x) dx
\1/r
= O(t
), 0 < 1, r 1, t > 0 (2)
0
0
The degree of approximation of a function f : R R by trigonometrical polynomial tn of order n is defined by Zygmund [1]
tn f = sup{tn(x) f(x) : x R} (3)
1 )
1 )
n
n
n
If (E, q) = Eq
)
)
Â·
Â·
= qnk sk
(1 + q)n
k=0
s as n . Then an infinite
series uk with the partial sums sn is said to be summable (E, q) to the
k=0
definite number s. (Hardy [4]).
Ripendra Kumar, B.K. Singh & Aditya Kumar Raghuvanshi
The series uk
k=0
1 n
k
k
is said to be (C, 1) summable to s. If (C, 1) = s
(n + 1)
k=0
s as n . The (C, 1) transform of the (E, q) transform defines the (C, 1)(E, q)
transform of the partial sums sn of the series uk.
k=0
Thus if
(CE)q
n
1
1
= Eq s as n (4)
n (n + 1)
k
k=0
n
n
where Eq denotes the (E, q) transform of sn, then the series uk is said to be
k=0
summable (C, 1)(E, q) means or simply summable (C, 1)(E, q) to s. We shall
use following notation:
(t) = f(x + t) + f(x t) 2f(x)

In this paper we have generalized the theorem of S. Lal [12].
Theorem 2.1. If f : R R is 2 periodic, Lebesgue integrable on [ , ] and belonging to the Lipschitz (, r) class then the degree of approximation of f by the (C, 1)(E, q) product means of its Fourier series satisfies for n = 0, 1, 2, 3, …
q
q
n
n
(CE) (x) f(x)
= O / 1 \ for 0 < < 1 and r > 1
1
(n + 1)r
1
(n + 1)r

For proof of our theorem, we shall use the following lemmas [12].
Lemma 1. Let
then
Mn(t) =
n
1
1
2(n + 1)
k=0
1 k
(q + 1)k
r=0
k
r
r
qkr
1
1
sin(r + )t
sin(r + )t
2
2
2
sin t
Lemma 2.
Mn(t) = O(n + 1) for 0 < t < n + 1
n
n
t
t
n + 1
n + 1
M (t) = O 1 , for 1 < t <
Degree of Approximation of function belonging to Lip(, r) functions by …

2
2
The nthpartial sum sn(x) of the series (1) at t = x is written as
r
r
1
0
0
sn(x) = f(x) + 2
(t) Â·
sin(n + 1 )t sin ( t ) dt
2
2
So that (E, q) means of the series (1) are
n
n
Eq (x) =
n
1
1
(q + 1)n
k=0
n
k
k
qnksk(x)
= f(x) +
2(q + 1)n
= f(x) +
2(q + 1)n
sin ( t )
2
sin ( t )
2
sin
sin
k +
2
k +
2
t
t
dt.
dt.
1 r (t) / n
0
0
n
k
k
1\
0
k=0
k
0
k=0
k
Therefore (C, 1)(E, q) means of the series (1) are
n
n
n
n
(n + 1)
(n + 1)
k
k
(CE)q (x) = 1 Eq (x) (n = 0, 1, 2, 3, …)
k=0
k=0
n
n
1 ( 1 r (t) / k
2
2
k
kr
1
\'l
= f(x) +
2(n + 1)
r
Â·
Â·
k=0
(q + 1)k
0 sin ( t )
r q
r=0
sin r +
2
t dt
where
= f(x) +
(t) Mn(t)dt (5)
0
Mn(t) =
so
n
1
1
2(n + 1)
k=0
1 k
(q + 1)k
r=0
k
r
r
qkr
1
sin(r + )t
sin(r + )t
2
sin(t/2)
n
n
(CE)q (x) f(x) =
r
r
Â·
Â·
(t) Mn(t)dt
/r
/r
0
1
n+1
0
0
r \
= +
1
n+1
(t) Â· Mn(t)dt
Now
\
\
I1 =
1
r
r
n+1
/r
/r
0
(t) Â· Mn(t)dt
= I1 + I2 (6)
1
n+1
1 1
r 1 s
n+1
r s
Â·
Â·
I1
[(t)] dt0
[Mn(t)] dt/r
/r
\
\
0
, using HÂ¨olders inequality
Ripendra Kumar, B.K. Singh & Aditya Kumar Raghuvanshi
( 1 \
!r
!r
\
\
1
1 s
n+1
s
I1 O
(n + 1) Â·
(n + 1) dt
0
( 1 \
r(n + 1)s 1
I1 O
(n + 1)
Â· n + 1
s
s
1
1
(n + 1)
(n + 1)
1s
(n + 1) s
1s
(n + 1) s
I  O ( 1 \ Â· ! 1 \
1
1
1
(n + 1)+ 1s
1
(n + 1)+ 1s
I  O ! 1 \
1
1
(n + 1)(1 1 )
s
(n + 1)(1 1 )
s
r
r
s
s
I  O ! 1 \ 1 + 1 = 1
1
1
1
(n + 1)r
1
(n + 1)r
I  O ! 1 \
Next
r
r
Â·
Â·
I2 = (t) Mn(t)dt
1
n+1
I2 =
I2 =
1
n+1
1
n+1
1
n+1
1
n+1
r
1
n+1
1
n+1
(t) Â· Mn(t)dt
1
n+1
1
n+1
r
s
r
s
!r
I2
I2
\1 !r \1
s(Mn(t)) dt
s(Mn(t)) dt
I2
r((t)) dt
s(Mn(t)) dt
I2
r((t)) dt
s(Mn(t)) dt
( \
( \
1
(n + 1)
1
(n + 1)
1
n+1
1
n+1
1
s dtt
1
s dtt
s
s
!r \1
I2 O
I2 O
I2 O
I2 O
s
s
( 1 \ r 1 1s
I2 O
(n + 1) n + 1
2
2
1s
(n + 1)+ s
1s
(n + 1)+ s
I  O ! 1 \
2
2
1
(n + 1)+ 1s
1
(n + 1)+ 1s
I  O ! 1 \
2
2
1
(n + 1)r
1
(n + 1)r
I  O ! 1 \
Then from (6) and the above inequalities we have
tn
f
= sup{tn
(x) f(x) : x R = O 1 , 0 < < 1, r > 1.
1
1
 } / \
 } / \
(n + 1)r
This completes the Proof of the theorem.
for 0 < < 1
for 0 < < 1
If r then degree of approximation of a function f Lip is given by
q
n
which reduces to the theorem of S. Lal [12].

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Ripendra Kumar, B.K. Singh & Aditya Kumar Raghuvanshi

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