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A new integral transform.

1. Introduction

A new integral transform is derived from the classical Fourier integral. A new integral transform was introduced by Artion Kashuri and Associate Professor Akli Fundo to facilitate the process of solving ordinary and partial differential equations in the time domain. Typically, Fourier, Laplace, Sumudu and Elzaki transforms are the convenient mathematical tools for solving differential equations. Also a new integral transform and some of its fundamental properties are used to solve differential equations.

A new integral transform is defined for functions of exponential order.

We consider functions in the set F defined by:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1)

For a given function in the set F, the constant M must be finite number, [k.sub.1], [k.sub.2] may be finite or infinite. A new integral transform denoted by the operator K(.) is defined by the integral equation:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (2)

This transform has deeper connection with the Laplace transform. We also present many different of properties of this new integral transform and few properties extent.

The purpose of this study is to show the applicability of this interesting new transform and its efficiency in solving the linear differential equations.

Definition 1.1. A function f(t) is said to be piecewise continuous on a finite interval [a, b] if f(t) is continuous at every point in [a, b], except possibly for a finite number of points at which f(t) has a jump discontinuity. A function f(t) is said to be piecewise continuous on [0, [infinity])if f(t) is piecewise continuous on [0, N) for all N > 0.

Definition 1.2. A function f(t) is said to be of exponential order 1/[k.sup.2], if there exist positive constants T and M such that, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], for all t [greater than or equal to] T.

For any function f(t), we assume that a integral equation (2) exist. In this section we find a new integral transform of some special functions.

2. A New Integral Transform Of Some Special Functions

(a) K [1] = v Take, f(t) = 1 and evaluate the integral.

(b) K [[t.sup.n]] = n![v.sup.2n+1] This is the general case.

(c) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] This result will be useful, to find a new integral transform of:

(d) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

(e) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

(f) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

(g) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

(h) If f(t) = [n.summation over (k=0)] [a.sub.k][t.sup.k] then K[f(t)] = [n.summation over (k=0)] k![a.sub.k][v.sup.2k+1]

Theorem 2.1. [Sufficient conditions for existence of a new integral transform] If f(t) is piecewise continuous on [0, [infinity]) and of exponential order 1/[k.sup.2] then K[f(t)](v) exists for [absolute value of v] < k.

Proof. We need to show that the integral [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] converges for [absolute value of v] < k.

We begin by breaking up this integral into two separate integrals:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (3)

where T is chosen so that inequality (*) holds. The first integral in (3) exists because f(t) and hence [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] f(t) are piecewise continuous on the interval [0, T) for any fixed v. To see that the second integral in (3) converges, we use the comparison test for improper integrals. Since f(t) is of exponential order 1/[k.sup.2], we have for [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and hence:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (4)

Now for [absolute value of v] < k,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (5)

Since,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

for t [greater than or equal to] T and the improper integral of the larger function converges for [absolute value of v] < k, then by the comparison test, the integral [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] converges for [absolute value of v] < k. Finally, because the two integrals in (3) exists, a new integral transform K[f(t)](v) exists for [absolute value of v] < k.

Theorem 2.2. [Duality relation] Let f(t) [member of] F with Laplace transform F(s). Then a new integral transform A(v) of f(t) is given by:

A(v) =1/v F (1/[v.sup.2]) (6)

Proof. Let f(t) [member of] F, then for -[k.sup.1] < v < [k.sup.2],

A(v) = v [?? [e.sup.-t] f([v.sup.2]t)dt (7)

Let w = [v.sup.2]t , then we have:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (8)

Theorem 2.3. [Uniqueness theorem] If F(v) and G(v) are a new integral transforms of f(t) and g(t) respectively, then:

F(v) = G(v) [??] f (t) = g(t)

Proof. Consider

F(v) = G(v) (9)

K[f(t); v] = K[g(t);v] (10)

1/v L [f(t);1/[v.sup.2]] = 1/v L [g(t);1/[v.sup.2]] (11)

L[f(t); 1/[v.sup.2]] = L [g(t);1/[v.sup.2]] (12)

By uniqueness of Laplace transform we obtain f(t) = g(t).

Theorem 2.4. [Linearity of a new integral transform] Let f(t), [f.sub.1](t) and [f.sub.2](t) be functions whose a new integral transforms exists for [absolute value of v] < k and let c be a constant. Then for [absolute value of v] < k,

1. K[([f.sub.1] + [f.sub.2])(t)] = K[[f.sub.1](t)] + K[[f.sub.2](t)]

2. K[(cf)(t)] = cK[f(t)]

Proof. Using the linearity properties of integration, we have for [absolute value of v] < k,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (13)

Hence equation (1) is satisfied.

In a similar fashion, we see that:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (14)

Theorem 2.5. [First translation theorem] Let f(t) [member of] F with a new integral transform A(v). Then:

K[[e.sup.at] f(t)] = (1/[square root of 1 - a[v.sup.2]) A [v/[square root of - a[v.sup.2]] (15)

Proof. From definition of a new integral transform we have:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (16)

that is an equivalent form of,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (17)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (18)

Let

w = t(1 - a[v.sup.2])/[v.sup.2] [??] dw = [(1 - a[v.sup.2]/[v.sup.2]] dt

Then,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (19)

So we have proof the theorem.

Theorem 2.6. [Fundamental properties of a new integral transform] Let A(v) be a new integral transform of f(t). Then:

(1) K[f'(t) = A(v)/[v.sup.2] - f(0)/v

(2) K[f'(t)] = A(v)/[v.sup.4] - f(0)/[v.sup.3] - f'(0)/v

(3) K[[f.sup.(n)](t)] = A(v)/[v.sup.2n] - [n-1.summation over (k=0)] [f.sup.(k)](0)/[v.sup.2(n-k)-1]

(4) K[tf(t)] = [v.sup.3]/2 A'(v) + [v.sup.2]/2 A(v)

(5) K[[t.sup.2]f(t)] = [v.sup.4]/4 [[v.sup.2] A"(v) + 5vA'(v) + 3A(v)]

(6) K[tf'(t)] = [v.sup.3]/2 d/dv [A(v)/[v.sup.2] - f(0)/v] + 1/2 (A(v) - vf(0))

(7) K[tf"(t)] = [v.sup.3]/2 d/dv [A(v)/[v.sup.4] - f(0)/[v.sup.3] - f'(0)/v] + [v.sup.2]/2 [A(v)/[v.sup.4] - f(0)/[v.sup.3] - f'(0)/v]

(8) K[[t.sup.2]f'(t)] = [v.sup.4]/4 [A"(v) + 1/v A'(v) - 1/[v.sup.2] A(v)]

(9) K[[t.sup.2]f"(t)] = [v.sup.2]/4 [A"(v) - 3/v A'(v) + 3/[v.sup.2] A(v)]

Proof.

(1) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] Integrating by parts to find that:

K[f'(t)](v) = A(v)/[v.sup.2] - f(0)/v (20)

(2) Let g(t) = f'(t), then: K[g'(t)](v) = K[g(t)](v)/[v.sup.2] - g(0)/v We find that, by using (1):

K[f"(t)](v) = A(v)/[v.sup.4] - f(0)/[v.sup.3] - f'(0)/v (21)

(3) We prove the general case by mathematical induction.

First way

First step it's true by (1). So, K[f'(t)](v) = A(v)/[v.sup.2] - f(0)/v.

Second step (hypothesis step): We suppose true for 1 [less than or equal to] i [less than or equal to] n - 1. So,

K[[f.sup.(i)](t)](v) = A(v)/[v.sup.2i] - [i-1.summation over (k=0) [f.sup.(k)](0)/[v.sup.2(i-k)-1] (22)

We know the fact [[f.sup.(n-1)]]' = f(n).

Third step is to prove for i = n. So,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Second way

By duality between a new integral transform and Laplace transform:

A(v) = 1/v F (1/[v.sup.2]) (23)

and the fact,

[F.sup.(n)](s) = [s.sup.n]F(s) - -[n-1.summation over (k=0)] [s.sup.n-(k+1)][f.sup.(k)](0) (24)

if we put s = 1/[v.sup.2] and multiplying both sides of equation (24) by 1/v, we have:

1/v[[F.sup.(n)] (1/[v.sup.2])] = 1/v [1/[v.sup.2n] F (1/[v.sup.2]) - 1/v [n-1.summation over (k=0)[(1/[v.sup.2]).sup.n-(k+1)] [f.sup.(k)](0)] (25)

Moreover we know the fact for [n.sup.-th] derivatives between a new integral transform and Laplace transform:

[A.sup.(n)](v) = 1/v [F.sup.(n)] (1/[v.sup.2]) (26)

If we use this result in equation (25) the relation is proved.

[A.sup.(n)](v) = A(v)/[v.sup.2n] - [n-1.summation over (k=0)] [f.sup.(k)](0)/[v.sup.2(n-k)-1] (27)

(4) Let A(v) be a new integral transform of the function f(t) [member of] F. The function tf(t) [member of] F, since f(t) is so and by applying Leibniz rule, we find that:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (28)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (29)

Then we have:

K[tf(t)](v) = [v.sup.3]/2 dA(v)/dv + [v.sup.2]/2 A(v) (30)

The (5) can be proved by using (4).

(6) Let g(t) = f'(t)then:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (31)

The proof of the other relations are similar if we use the relations above.

Theorem 2.7. [Convolution theorem] Let f(t) and g(t) be defined in F having Laplace transforms F(s) and G(s) and a new integral transforms M(v) and N(v). Then a new integral transform of the convolution of f and g:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (32)

is given

K[(f * g)(t)] = vM(v)N(v) (33)

Proof. The Laplace transform of (f * g) is given by:

L [(f * g)(t)](s) = F(s)G(s) (34)

By the duality relation (6) we have:

K[(f * g)(t)](v) = 1/v L[(f * g)(t)] (s) (35)

and since

M(v) = 1/v F (1/[v.sup.2]), N(v) = 1/v G (1/[v.sup.2]) (36)

Then,

K[(f * g)(t)](v) = 1/v [F (1/[v.sup.2]) x G (1/[v.sup.2])] = 1/v [vM(v) x vN(v)] = vM(v)N(v) (37)

So we have proof the theorem.

Let see some other relations

(a) K[(f * g')(t)] = 1/v M(v)N(v), g(0) = 0

(b) K[f * [g.sup.(n)])(t) = M(v)N(v)/[v.sup.2n-1], for all n [greater than or equal to] 1, g(0) = g'(0) = ... = [g.sup.(n-1)](0) = 0

Proof. The Laplace transform of (f * g') is given by:

L[(f * g')(t)](s) = sF(s)G(s) (38)

By the duality relation (6) we have:

K[(f * g')(t)](v) = 1/v L[(f * g')(t)](s) (39)

and since

M(v) = 1/v F (1/[v.sup.2]), N(v) = 1/v G (1/[v.sup.2]) (40)

Then,

K[(f * g')(t)](v) = 1/v[1/[v.sup.2] F (1/[v.sup.2]) x G (1/[v.sup.2])] = 1/[v.sup.3] [vM(v) x vN(v)] = 1/v M(v)N(v) (41)

So we have prove the first relation.

We can show (b) by mathematical induction.

Theorem 2.8. Let A'(v) and F'(s) denote a new integral transform and the Laplace transform of the definite integral of f(t).

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (42)

Then,

A'(v) = K[h(t)] = [v.sup.2] A(v) (43)

Proof. By the definition of Laplace transform:

F'(s) = L[h(t)](s) = F(s)/s (44)

Hence,

A'(v) = 1/v F'(1/[v.sup.2]) = 1/v [F(1/[v.sup.2])/1/[v.sup.2]] = vF (1/[v.sup.2]) = [v.sup.2]A(v) (45)

So we have proof that:

A'(v) = [v.sup.2] A(v) (46)

Theorem 2.9. [Second translation theorem] Let f(t) [member of] F with a new integral transform K [f(t)]. Then the function,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (47)

where, K[f(t)] = A(v) has a new integral transform given by the formula:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (48)

Proof. By definition of a new integral transform:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (49)

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (50)

So we have proof the theorem.

Theorem 2.10. [Initial and Final value theorem] Let f(t) [member of] F and suppose that either [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] or [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] exists. Then,

(a) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

(b) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Proof. Let w = [v.sup.2]t. The first limit is obtained as follows:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (51)

In the same manner,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (52)

3. Applications

Application of a new integral transform of Ordinary Differential Equations.

As stated in the introduction of this paper, a new integral transform can be used as an effective tool for analyzing the basic characteristics of a linear system governed by the differential equation in response to initial data. The following examples illustrate the use of a new integral transform in solving certain initial value problems described by ordinary differential equations.

Example 3.1. Consider the first-order ordinary differential equation:

y'(t) + by(t) = h(t), t > 0, y(0) = a (1)

where a and b are constants and h(t) is an external input function so that its a new integral transform exists.

First solution by Laplace transform

By apply Laplace transform we have:

sF(s) - y(0) + bF(s) = H(s) [??] F(s) = a/s + b + H(s)/s + b (2)

where that F(s) and H(s) are Laplace transforms of y(t) and h(t). Then,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (3)

Thus the solution naturally splits into two terms the first one corresponds to the response of the initial condition and the second one is entirely due to the external input function h(t). In particular if h(t) = c where c is the constant there the solution of (3) becomes:

y(t) = c/b + (a - c/b) [e.sup.-bt] (4)

Second solution by a new integral transform

Using a new integral transform of equation (1) we have:

A(v)/[v.sup.2] - y(0)/v + bA(v) = H(v) (5)

where that A(v) and H(v) are a new integral transforms of y(t) and h(t).

By applying the initial condition we have:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (6)

By the inverse of a new integral transform and convolution theorem (33) we find that:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (7)

The first term of this solution in (7) is independent of time t and is usually called the steady-state solution, and the second term depends on time t and is called the transient solution. In the limit as t [right arrow] [infinity] the transient solution decays to zero if b > 0 the steady-state solution is attained, on the other hand, when b< 0, the transient solution grows exponentially as t [right arrow] [infinity] and the solution becomes unstable.

Equation (7) describes the Law of natural growth or decay Process with an external forcing function h(t) according as b > 0 or b < 0. In particular, if h(t) = 0 and b > 0 the resulting equation (7) occurs very frequently in chemical kinetics. Such an equation describes the rate of chemical reactions.

Example 3.2. Consider the second-order ordinary differential equation:

y"(t) + [[lambda].sup.2]y(t) = g(t), y(0) = 0, y'(0) = 1 (8)

where [lambda] [not equal to] 0 is an integer and g(t) is an external input function so that its a new integral transform exists.

First solution by Laplace transform

By apply Laplace transform we have:

[s.sup.2]F(s) - sy(0) - y'(0) + [[lambda].sup.2]F(s) = G(s) (9)

where that F(s) and G(s) are Laplace transforms of y(t) and g(t).

By applying the initial conditions we have:

[s.sup.2]F(s) - 1 + [[lambda].sup.2]F(s) = G(s) (10)

Then by the inverse of Laplace transform and convolution theorem (34) we find that:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (11)

Second solution by a new integral transform

Using a new integral transform of equation (8) we have:

A(v)/[v.sup.4] - y(0)/[v.sup.3] - y'(0)/v + [[lambda].sup.2]A(v) = G(v) (12)

where that A(v) and G(v) are a new integral transforms of y (t) and g(t).

By applying the initial conditions we have:

A(v)/[v.sup.4] - 1/v + [[lambda].sup.2]A(v) = G(v) (13)

Then by the inverse of a new integral transform and convolution theorem (33) we find that:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (14)

Example 3.3. Consider the third-order ordinary differential equation:

y'"(t) + 2y'(t) = cos t, y(0) = 0, y'(0) = 1, y"(0) = 0 (15)

First solution by Laplace transform

By apply Laplace transform we have:

[s.sup.3]F(s) - [s.sup.2]y(0) - sy'(0) - y"(0) + 2sF(s) - 2y(0) = s/[s.sup.2] + 1 (16)

where that F(s) is Laplace transform of y(t). By applying the initial conditions we have:

[s.sup.3]F(s) - s + 2s F(s) = s/[s.sup.2] + 1 (17)

Then by the inverse of Laplace transform we find that:

y(t) = sin t (18)

Second solution by a new integral transform

Using a new integral transform of equation (15) we have:

A(v)/[v.sup.6] - y(0)/[v.sup.5] - y'(0)/[v.sup.3] - y"(0)/v + 2A(v)/[v.sup.2] - 2y(0)/v = v/1 + [v.sup.4] (19)

By applying the initial conditions we have:

A(v)/[v.sup.6] - 1/[v.sup.3] + 2A(v)/[v.sup.2] = v/1 + [v.sup.4] (20)

Then by the inverse of a new integral transform and convolution theorem (33) we find that:

y(t) = sin t (21)

Example 3.4. Consider the ordinary differential equation with variable coefficients:

ty"(t) + ty'(t) - 2y(t) = 2t, y(0) = 0, y'(0) = 0 (22)

First solution by Laplace transform

By apply Laplace transform we have:

-[s.sup.2]F'(s) - 2sF(s) + y(0) - sF'(s) - F(s) - 2F(s) = 2/[s.sup.2] (23)

where that F(s) is Laplace transform of y(t). By applying the initial conditions we have:

-[s.sup.2]F'(s) - 2sF(s) - sF'(s) - F(s) - 2F(s) = 2/[s.sup.2] (24)

F'(s)[-[s.sup.2] - s] + F(s) [-2s - 3] = 2/[s.sup.2] (25)

This is a linear differential equations.

First we solve this and then by the inverse of Laplace transform we find that:

y(t) = [t.sup.2] (26)

Second solution by a new integral transform

Using a new integral transform of equation (22) we have:

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (27)

By applying the initial conditions we have:

A'(v) - A(v)[3 + 5[v.sup.2]/v(1 + [v.sup.2])] = 4[v.sup.4]/(1 + [v.sup.2]) (28)

This is a linear differential equations.

First we solve this and then by the inverse of a new integral transform we find that:

y(t) = [t.sup.2] (29)

4. Conclusions

The origin of a new integral transform is traced back to the classical Fourier integral.

A new integral transform is a convenient tool for solving differential equations in the time domain. The connection of a new integral transform with the Laplace transform goes much deeper and we can find other relations of a new integral transform by this connection.

References

[1] Tarig M. Elzaki, The New Integral Transform Elzaki Transform, z Global Journal of Pure and Applied Mathematics, ISSN 0973-1768, Number 1(2011), pp. 57-64.

[2] Tarig M. Elzaki & Salih M. Elzaki, Application of New Transform Elzaki Transform z to Partial Differential Equations, Global Journal of Pure and Applied Mathematics, ISSN 0973-1768, Number 1(2011), pp. 65-70.

[3] Tarig M. Elzaki & Salih M. Elzaki, On the Connections Between Laplace and Elzaki transforms, Advances in Theoretical and Applied Mathematics, ISSN 0973-4554 Volume 6, Number 1(2011), pp. 1-11.

[4] On the ELzaki Transform and Higher Order Ordinary Differential Equations, Advances in Theoretical and Applied Mathematics, ISSN 0973-4554 Volume 6, Number 1 (2011), pp. 107-113.

[5] Tarig M. Elzaki & Salih M. Elzaki, On the Elzaki Transform and Ordinary Differential Equation With Variable Coefficients, Advances in Theoretical and Applied Mathematics, ISSN 0973-4554 Volume 6, Number 1(2011), pp. 13-18.

[6] Analytical investigations of the Sumudu Transform and applications to integral production equations FETHI BIN MUHAMMED BELGACEM, AHMED ABDULLATIF KARABALLI, AND SHYAM L. KALLA

[7] On a new integral transform and solution of some integral equations, Volume 73, No. 3 2011,299-308.

[8] Elzaki and Sumudu Transforms for Solving Some Differential Equations, Global Journal of Pure and Applied Mathematics, ISSN 0973-1768 Volume 8, Number 2 (2012), pp. 167-173.

[9] Fundamentals of differential equations.--8th ed. / R. Kent Nagle, Edward B. Saff, David Snider. (2012), pp. 358-359.

[10] Tarig M. Elzaki, Existence and Uniqueness of Solutions for Composite Type Equation, Journal of Science and Technology, (2009). pp. 214-219.

[11] Lokenath Debnath and D. Bhatta. Integral Transform and Their Application Second Edition, Chapman & Hall /CRC (2006).

[12] Hassan Eltayeb and Adem kilicman, A Note on the Sumudu Transforms and Differential Equations, Applied Mathematical Sciences, VOL, 4,2010, no. 22, 1089-1098

[13] Kilicman A. & H. ELtayeb. A Note on Integral Transform and Partial Differential Equation, Applied Mathematical Sciences, 4(3) (2010), PP.109-118.

Artion Kashuri

Department of Mathematics, University ofVlora, Albania

E-mail: artion.kashuri@univlora.edu.al

E-mail: artionkashuri@gmail.com

Akli Fundo

Department of Mathematics, Polytechnic University of Tirana, Albania

E-mail: aklifundo@yahoo.com
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