 Open Access
 Total Downloads : 147
 Authors : Pawan Saxena
 Paper ID : IJERTV3IS20828
 Volume & Issue : Volume 03, Issue 02 (February 2014)
 Published (First Online): 14032014
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
on the Euler’s Summability of a Differentiated Fourier Series
Pawan Saxena
Department of Applied Science (Mathematics) IIMT College of Engineering, Knowledge ParkIII Plot No. A 20, Gr. Noida (G.B. Nagar) 201308 India
Abstract In this paper we have proved a theorem On the Euler Summability of a differentiated Fourier series which
1.2.7. P t = (t)
A sin 3 t
2
C , where C is constant K denotes an
generalizes known results however our theorem is as follows
Theorem: If (u) du = o(log 1) as t 0 and nq
is a
absolute constant not necessarily the same in each occurrence.

Introduction: The following theorem on Euler summability of derived Fourier series is due to Ray 7 .
t u t
monotonic convergence sequence
Such that q x and
n Theorem A: If g (t) is of bounded variation to the right of t=0 and g (t) = O (1) as t 0,then derived Fourier series at t = x,
is summable E, q for q > 0,to the value s, where
1x 2q (x)
E n,t dt = o q x
As x 1, then the series
g(t)= t 4sin t
1
s and
1x t3
1x 2q (x) 2
nBn (t), at t = x is summable E, q for q > 0 to the
t = f x + t f x t .
n=1
value s.
2000 Mathematics subjects classification:
40D25, 40E25, 40F25, 40G25.
Chandra 1 generalized above theorem for Euler summability of Fourier series in the following form:
Theorem B: If t log 1 = o 1 , t 0, then, An x
t n=0
Keywords and Phrases;
=1 (), at = is summable s E, q for q > 0.The object of this paper is to generalize
=0
1. Definition and Notations: Let be a given infinite series with partial sums . The sequence Eulers transform of
Theorem A and B by establishing the following theorem.

We assert the following main theorem
a sequence is defined by
1.1 = + 1
, where > 0, if
Theorem:

If (u) du = o(log 1) as t 0
=0
t u t
As n , we say that sn or
an are summable
and nqn is a monotonic convex sequence such that
n=0
E , q q > 0 , to s or symbolically we write sn s

q x
,
1x 2 q (x)
E , q for q > 0 , to s or symbolically we write sn
and
E n,t dt = o q x
s E, q , for q > 0, Hardy 3 ,
1x t3
1x 2q (x)
as x 1, then the series nBn (t), at t = x is summable
It is evident that that , 0 is equivalent to convergence.
1.2 Let () be a periodic function with period 2 and
E, q for q > 0
n=1
to the value S.
integrable in the Lebesgue sense over the interval , and


Proof of the theorem: note that
Br x = 1 t sin + 1 dt so that
let
1.2.1. f t ~ 1
2
a0 +
n=1
an cosnt + sinnt =
An (t) be
Sn x =
0
d
1
t
0 dt
n
n=1
1 + cos( + 1) dt 2
the Fourier series of f(t).Then the differentiated series of
1.2.1 at t = x, is
=1
3
= 1 t d sin n + 2 t
n=1
1.2.2.
n bn cosnx ansinnx =
nBn (x)
0 dt
3
2 sin 2 t
dt
n=1
Throughout we use the following notations for
0 < < 1.
1.2.3 t = f x + t f x t
3
1 2n sin 3/2 t cos n + 2 t sin n + 1 t
0

P q, t = 1 + q2 + 2q cost
= t
4sin2 3/2 t dt

Q q, t = tan1 sint
q+cos t

E n, t = n nk sin k + 1 t
k=0 k q 2
3
1.5.5.
1 t sin(n + 1)
2n t
cos n + 2 t
2 +2 = 1 +1
=
0 4sin2
3
2 t
dt
0
3 dt
4 sin 2 t
=1
=1
= 1 3
= 1 +1 – 2 + 3
Also,
2
0 3
2
0 2
2 nq
n = n + 2 qn+2
2 n + 1 q
n+1 + nqn
= 0
+ 1
2 + 3 + 1
N + 1 2qn = n + 1 qn+2 2 n + 1 qn+1 + n + 1 qn
Therefore,
0 2
This implies that
2 nqn n + 1 2 qn = qn+2 qn
Or,
Q x = 2
P t
q xn sin n + 1 t dt
n + 1 2 q
= 2 nq
+ q q
q x 0 t
2
n=1 n
n
n
Hence,
n n n +2
q(x) P(t) nqn x
cos n + 3/2 t dt + o(1)
0 n=1
Or
Q x o(1) = 2
0
q x
2
P t
t E(n, t)dt
n + 1 2qn xn+2
n=1
= 2 nqn
n=1
xn+2
q(x) P t P(q, t) dt
=I I
0
, say
+ qn xn+2 qn+2xn+2
1 2
q xn+1
int
n=1
n=1
Now, E n, t = I
n=1 n e
n it
+1
3 1 x 3
= I n=1 qn =1 e x
= 1 x q x x q x
n=1
= I
qn
xeit (1xn +1eint )
it
1xe
1.5.6. and
n=1
= I
qn
xn +2ei n +1 t
1xeit
n + 1 q
n+2
xn+2 = x2 q x 1 x2 q x
=
n=1
qn
xn +2ei n +1 t
1xeit
n=1
So that
1
= ,
+2 + 2
1.5.7.
n q
xn +1
q x
+ 1 x2 q x
=1
2 +1 +2 + 2
n x2
n=1
Now using (1.5.1), (1.5.6) and (1.5.7)
=1
I 2
1
q x
1x P t
t dt +
P t E n, t dt
t
1.5.1.=
0
2 1x P t
=
dt +
1x
P t
dt 1
1
x
Q q,t
n=1
2qn xn+2sin n + 2 t + 1
q x 0 t
1x t
Q q, t
2
n=1
We have
1.5.2.
qn+1xn+2sin n + 2 t
2 +2 + 2
=1
+ 1 3 +1 +2 + 2
qn
xn+2 = 1 x Pn xn+1 = 1 x q x
2 1x P t
=1
P t 1
n=1
n=1
q x
t dt +
t dt Q q, t
1.5.3
0 1x
1 x 2
+ 1 + 2 +2 + 1 3
2 qn xn+2 = 1 x qn xn+1 =
x q x
n=1
1.5.4.
n=1
=1
+ + 1 +1
+2
nqn xn+2 = 1 x nqn xn+1
=1
n=1
n=1
= x 1 x 2q x
2qn xn+2sin n + 2 t 1
n=1
x3 qn+1xn+2sin n + 2 t
n=1
2 1x P t
P t 1
1
1
0
q x
dt +
t
1x
t dt Q x, t
3
1 0
+ n + 1 2qn xn+2 + 1 x3
+
2 +2
n=1
+ n + 1 qn+1xn+2
1
+ 1
2
=1
5
n=1
3 +1 + 2
2 1x 1
k 1 1x
=1
= q x P(t) dt + P(t) dt Q q, t
3 P(t) dt
0 1x
q(x) 1 x 0
P(t)
+ n + 1 1 x3 n qn xn+1
+ dt 2 nqn xn+2
k 1
n=1
1x P t dt +
n=1
1x t2
n=1
q x
1x 3 0
P t
+ 1 x3 nqn
xn+1
t2 dt 1 x 3q x x + 3
1x
n=1
1x
k 1
P t dt + P t dt
k x + 3 q (x)
(1x)
3 P(t)
q x
1 x 3 0
1x t2
= q(x) P(t) dt + 1 x
t2 dt
1 3 + 3
0 1x
k x+3 q (x)
1x
3 P(t)
= 2k x+3 q (x) 1x 3
E(n,t) dt + E(n,) by
= q (x) 0
k 1x 3q (x)
P t
E n,t
dt +
1 x
E n,
1x
t2 dt
q (x)
(1x) t3 2
= q (x) 1x
t3 dt +
2 , by integration by parts
integrating by parts
= o 1 as x 1 by using (1.4.2) and (1.4.3)
Proceeding as in E n, t , we easily get, 1.5.8
By hypothesis (1.4.2) and (1.4.3). This completes the proof of the theorems.
REFERENCES
P q, t = 1
Q(q, t)
2 nqn
n=1
xn+2
5
cos n + 2 t

P. Chandra, On Euler Summability of Fourier series, Ranchi University, Mathematical journal, Publisher, Location, Date, pp. 110.

On the E, q summability of a Fourier series, University Parma,
+ 1 3 nqn xn+1cos n +
=1
Therefore, Using (1.5.4), (1.5.5) and (1.5.8),
2 1
5 t 2
Riv.Maths (4), 3(1977), 6578.

H.G.Hardy, Divergent series, Oxford University Press, Oxford (1949).

K.Knopp, and G.G. Lorentz,: B. Kwee.
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0
2 +2
1
+ 5 +
2
32(1968).

B.K. Ray. : On , summability of derived Fourier series and a derived conzugate series Indian Jour. Math., 11(1969), 4350.

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1 3 +1 5
integrals, To hoku Math. Jour. (2) 4(1952), 153156.
=1
=1
+ 2

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