Cmjv01i06p0758

Page 1

J. Comp. & Math. Sci. Vol. 1 (6), 758-761 (2010)

On

N, p n

Summability Factors of Fourier Series

k

MAHENDRA MISRA1, U. K. MISRA2 and CHITARANJAN KHADANGA3 Principal, Government Science College, Mlkangiri, Orissa1, Department of Mathematics2, Berhampur University Berhampur – 760007,Orissa, India e-mail:umakanta_misra@yahoo.com M.P.Christian College of Engineering and Technology3, P.O Box No.18, Kailash Nagar Bhilai-490026,(C.G) e-mail: chita_khadanga@redifmail.com ABSTRACT A theorem on index summabilty factors of Fourier Series has been established Key Words: Summability factors, Fourier series AMS Classification No: 40D25

1. INTRODUCTION

∑a

Let

In

be a given infinite series

n

with the sequence of partial sums {s n } . Let

{p n } be a sequence of positive numbers such

that (1.1)

n

Pn =

∑ pv → ∞ , as n → ∞

( P−i

= p− i = 0, i ≥ 1)

r =0

The sequence to sequence transformation

tn =

(1.2)

1 Pn

n

∑ v =0

p n −v s v

defines the sequence of the ( N , p n ) -mean of

the sequence {s n } generated by the sequence of coefficients {p n }. The series

∑a

n

is said to be

summable N , p n k , k ≥ 1 , if

 Pn    ∑ n =1  p n  ∞

(1.3)

k −1

t n − t n −1

k

<∞

the

case

n and x = 1, N , p n C ,1

pn = 1 ,

when

summability.

k

for

all

summability is same as For

k = 1, N , p n

k

summability is same an N , p n -summability.

Let f (t ) be a periodic function with period 2π and integrable in the sense of Lebesgue over (− π , π ) . Without loss of generality, we may assume that the constant term in the Fourier series of f (t ) is zero, so that (1.4) ∞

n=1

n=1

f (t) N ∑(an cos nt + bn sin nt) = ∑ An (t) 2. Know Theorem Dealing with

N , pn k , k ≥ 1 ,

summability factors of Fourier Series, Bor1 proved the following theorem: Theorem-A If {λ n } is a non-negative and non-increasing sequence such that p n λ n < ∞ , where {p n } is a sequence of

Journal of Computer and Mathematical Sciences Vol. 1, Issue 6, 31 October, 2010 Pages (636-768)


Mahendra Misra et al., J. Comp. & Math. Sci. Vol. 1 (6), 758-761 (2010)

759

Pn → ∞ as n → ∞ then

positive numbers such that Pn → ∞ as n → ∞ and n

∑P v =1

v

Pn λn = 0(1) as n → ∞ and

Av (t ) = 0( Pn ) . Then the factored

Fourier series

∑ A (t ) P λ n

n

n

is summable

∑P

n

5. PROOF OF THE THEOREM Let t n ( x) be the n-th ( N , p n ) mean

N , p n k , k ≥1 . We prove an analogue theorem on N, p n k -summability, k ≥ 1 , in the following form:

n =1

t n ( x) =

=

and {λ n } is a non-negative, non-increasing

sequence such that

∑p

n

(3.1)

(i).

(3.2)

(ii).

∑P

v

v =1

n = v +1

n

n

n

,

then by

definition we have

Theorem: Let {p n } is a sequence of positive numbers such that Pn = p1 + p 2 + .... + p n → ∞ as n → ∞

∑ A ( x) P λ

of the series

3. MAIN THEOREM

m +1

∆λ n < ∞ .

 Pn     pn 

k −1

n

λn < ∞ . If

Av (t ) = 0( Pn )

 p n −v −1  p    = 0 v , as m → ∞  Pn −1   Pv 

and (3.3) (iii).

Pn −v −1 ∆λv = 0( p n − v λv ) , then

the

∑ A (t ) P

series

n

=

n

λn

is

summable

N , pn k , k ≥ 1.

∑ ∑

positive numbers such that

1 Pn

v

∑ p ∑ A ( x) P λ v =0

n−v

n

∑ Ar ( x) Pr λr r =0 n

∑ A ( x) P r =0

r

r

r

r =0

r

r

n

∑p v =r

n −v

Pn − r λ r .

Then tn ( x) − tn−1 ( x) = 1 Pn

n

∑ Pn−r Pr λr Ar ( x) − r =1

1 Pn−1

n−1

∑P r =1

n−r −1

Pr λr Ar ( x)

 Pn− r  P − n − r −1  Pr λ r Ar ( x) Pn−1  r =1  Pn 1 = (Pn−v Pn−1 − Pn− r −1 Pn ) Pr λr Ar ( x) Pn Pn−1

=

4. REQUIRED LEMMA We need the following for the proof of our theorem. n Lemma [1]: P If {λ n }Pis a non-negative Pn r and Pn n r n r  sequence such that r =1 non-increasing p n λ n < ∞ , where {p n } is a sequence of

1 Pn

n

1 Pn

n

∑ 

1  n −1 ∑ ∆ {( Pn − r Pn −1 − Pn − r −1 Pn ) λr } Pn Pn−1  r =1  r   ∑ Pv A v ( x )    v =1 

=

}

with p o = 0

Journal of Computer and Mathematical Sciences Vol. 1, Issue 6, 31 October, 2010 Pages (636-768)

r

∑P v =1

v

 Av x  


2

Mahendra Misra et al., J. Comp. & Math. Sci. Vol. 1 (6), 758-761 (2010)

1  n−1 = ∑ ( pn−r Pn−1 − pn−r −1 Pn ) λr Pr Pn Pn −1  r =1 +

n −1

∑(p r =1

n − r −1

n =2

 Pn −1 − Pn − r − 2 p n ) Pr ∆λ r  

k

 Pn   n=2  n 

k −1

∑  p

Tn,1

k

P  = ∑  n  n=2  pn 

k −1

m

= 0(1)

∑ r =1

P ≤ ∑  n n =2  p n m +1

  

 Pn    ∑ n = 2  pn  m +1

n −1

1    ∑ p n −v Pv λv  k Pn  v =1 

1  n −1   ∑ p n −v Pvk λkv  Pn  v =1 

m+1 P  = O(1) ∑  n  n=2  pn 

k −1

1 Pn

n −1

∑p r =1

n −v

m

∑ v =1

k

m +1

∑ n=2

k −1

∑ n=2

m+1

∑ n =2

  

  

 Pn     pn 

 Pn   pn

  

n = v +1

Pv λv ( Pv λv ) k −1

k −1

 p n − v −1   Pn −1

  , 

=

Pv λv (Pv λv )

k −1

 Pn     pn 

m +1

k −1

1  n −1  ∑ pn−v−1 Pv ∆λv  Pnk  v=1

1  n −1 k   1  ∑ p n − v −1 (Pv ∆λv )   Pn  v =1   Pn

 Pn   pn

  

k −1

P  = O(1)∑  n  n=2  p n 

1 Pn

k −1

n −1

∑p v =1

1 Pn

n − v −1

v =1

v =1

n − v −1

  

Pv ∆λv (Pv ∆λ v )

n −1

∑p

n −1

∑p

k −1

n − v −1

k −1

k −1

Pv ∆λv , b

k −1

P  = O (1)∑ Pv ∆λv ∑  n  v =1 v =1  p n  n −1

m

p n −v Pn

= 0(1)

m

∑p v =1

v

n −1

 p n− v −1    P  n 

∆λv , using (3.2)

= 0(1) , m → ∞ .

k −1 k

 Pn     pn 

m +1

n=2

k −1

n=2

p v λv (using 3.2)

Tn.2

k

Tn,3

= O(1) ∑

k −1

= O (1), as m → ∞ . m +1

v =1

n −v −1

k −1

y the lemma

P  = O (1)∑ Pv λv ∑  n  v =1 n = v +1  p n  = O (1)

 p n −v  ∑ v =1  n −1

∑p

  

p v λv , using (3.2)

m +1

m +1

m

 1   Pn

n −1

1 Pn −1

v =1

n − v −1

Now,

m +1

k −1

  

k −n

n −1

∑p

= 0(1) , as m → ∞ .

< ∞, for i = 1,2,3,4.

m +1

1  n−1  1  ∑ p n −v −1 Pvk λkv   k Pn −1  v =1   Pn −1

v =1

Now, we have m +1

k −1

= O (1) ∑ Pv λ v

k −1

Tn,i

  

m

In order to complete the proof of the theorem, using Minkowski’s inequality, it is sufficient to show that

 Pn    ∑ n =1  p n 

 Pn   pn

m +1 P = O(1)∑  n n=  pn

= Tn ,1 + Tn , 2 + Tn ,3 + Tn , 4 , say.

m +1

760

=

m +1

∑ n=2

1  n−1  p Pk λv  k  ∑ n −v −1 v Pn  v=1 

k

 Pn     pn 

k −1

n −1

∑p v =1

P

n v Finally, 1 v v

 Pn  ∑ n=2  pn m +1

  

k −1

Tn, 4

k

=

m +1

∑ n=2

 Pn   pn

  

k −1

p nk  n −1   ∑ Pv Pn −v − 2 ∆λv  k Pn Pnk−1  v =1 

Journal of Computer and Mathematical Sciences Vol. 1, Issue 6, 31 October, 2010 Pages (636-768)

k


Mahendra Misra et al., J. Comp. & Math. Sci. Vol. 1 (6), 758-761 (2010)

761

≤ 0(1)

m+1

∑ n=2

 Pn     pn 

k −1

1  n−1 ∑ Pnk−1  v=1

k

pn−v−1

 Pv λv  , 

= 0(1), as m → ∞ , as above This completes the proof of the theorem.

using (3.3)

REFERENCE ≤ 0(1)

m+1

 Pn    n

∑  p n=2

k −1

1 Pn−1

P  = 0(1) ∑  n  n=2  p n  n +1

n−1

∑ v=1

k −1

k

 1 pn−v−1 (Pv λv )   Pn−1

1 Pn −1

n −1

∑p v =1

n − v −1

n−1

∑p v=1

n−v−1

Pv λv

  

k −1

1. Bor Huseyin: On the absolute summability factors of Fourier Series, Journal of Computational Analysis and Applications, Vol.8, No.3, 223-227, (2006).

Journal of Computer and Mathematical Sciences Vol. 1, Issue 6, 31 October, 2010 Pages (636-768)


Turn static files into dynamic content formats.

Create a flipbook
Issuu converts static files into: digital portfolios, online yearbooks, online catalogs, digital photo albums and more. Sign up and create your flipbook.