fm60_fcaa_gololu_4

8/11/2019 fm60_fcaa_gololu_4

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COMPUTATION OF THE MITTAG-LEFFLER FUNCTION

E , (z) AND ITS DERIVATIVE

Rudolf Goreno 1, Joulia Loutchko 1 & Yuri Luchko 2

Dedicated to Francesco Mainardi,Professor at the University of Bologna,

on the occasion of his 60-th birthday

Abstract

In this paper algorithms for numerical evaluation of the Mittag-Leffler function

E , (z) =

k=0

zk

( + k ), > 0, R , z C

and its derivative for all values of the parameters > 0, R and all values of theargument z C are presented. For different parts of the complex plane differentnumerical techniques are used. In every case we provide estimates for accuracyof the computation; numerous pictures showing the behaviour of the Mittag-Leffler function for different values of the parameters and on different lines in thecomplex plane are included. The ideas und techniques employed in the paper canbe used for numerical evaluation of other functions of the hypergeometric type.In particular, the same method with some small modications can be applied forthe Wright function which plays a very important role in the theory of partialdifferential equations of fractional order.

Mathematics Subject Classication : 33E12, 65D20, 33F05, 30E15Key Words and Phrases : Mittag-Leffler function, integral representations of

special functions, numerical computation of special functions, Mathematica pro-grams


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2 R. Goreno, J. Loutchko and Yu. Luchko

1. Introduction

The function

E (z) =

k=0

zk

(1 + k ), > 0, z C , (1)

was introduced by Mittag-Leffler in 1902 in connection with his method of sum-ming divergent series. It is suited to serve as a very simple example of an entirefunction of a specied order 1 / and of normal type. Further investigations of properties of this function and its applications to various questions of mathemati-cal analysis have been carried out by Wiman [33]-[34] and Buhl [7]. An importantgeneralization,

E , (z) =

k=0

zk

( + k ), > 0, C , z C , (2)

was introduced by Humbert [22], Agarwal [1], Humbert and Agarwal [23], Cleotaand Hughes [9].

In recent years the interest in functions of Mittag-Leffler type among scientists,engineers and applications-oriented mathematicians has deepened. This interestis caused by the close connection of these functions to differential equations of fractional (meaning: non-integer) order and integral equations of Abel type, suchequations becoming more and more popular in modelling natural and technicalprocesses and situations (Babenko [2], Bagley and Torvik [3], Blair [4]-[5], Caputoand Mainardi [8], El-Sayed [12], Goreno et al. [16], Goreno and Mainardi [17]-[19], Goreno and Rutman [20], Goreno and Vessella [21], Luchko [24], Luchkoand Srivastava [27], Mainardi [28], Schneider and Wyss [30], Slonimski [31], West-erlund and Ekstam [32]).

Although there already exists a rich literature on analytical methods for solv-ing differential equations of fractional order, it is to be remarked that solutionsin closed form have been found only for such equations with constant coefficientsand for a rather small class of equations with particular variable coefficients. Ingeneral, numerical solution techniques are required.

In connection with the basic role of Mittag-Leffler type functions for the so-lution of fractional differential equations and integral equations of Abel type itseems important as a rst step to develop their theory and stable methods fortheir numerical computation. The most simple function of Mittag-Leffler type,E (z), depends on two variables: the complex argument z and the real parameter . The generalizations need at least one more argument, and may be allowed tobe complex. Experience in the computation of special functions of mathematicalphysics teach us that in distinct parts of the complex plane different numericaltechniques should be used. The aim of this paper is the development of some


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COMPUTATION OF THE MITTAG-LEFFLER FUNCTION . . . 3

methods for computing the Mittag-Leffler function (2) and its derivative as wellas the assessment of the range of their applicability and accuracy. We proposeTaylor series in the case 0 < 1 for small |z |, asymptotic representations for |z |of large magnitude, and special integral representations for intermediate values of the argument z. By aid of a recursion formula we reduce the case 1 < to the case0 < 1. We devise the needed algorithms in the system MATHEMATICA.

2. Integral representations of the Mittag-Leffler function

Integral representations play a prominent role in the analysis of entire func-tions. For the Mittag-Leffler function (2) such representations in form of animproper integral along the Hankel loop have been treated in the case = 1 andin the general case with arbitrary by Erdelyi et al. [13] and Dzherbashyan [10],[11]. They considered

E , (z) = 12i ( ;) e

1/ (1 )/ z

d, z G() ( ; ), (3)

E , (z) = 1

z(1 )/ ez1/ + 12i ( ;) e

1/ (1 )/ z

d, z G(+) ( ; ), (4)

under the conditions

0 < < 2,

2 < min{, }. (5)

The contour ( ; ) can be seen in Figure 1. It consists of two rays S andS (arg = , | | and arg = , | | ) and a circular arc C (0; )(| | = , arg ). On its left side there is a region G () ( , ), on its rightside a region G(+) ( ; ).

Using the integral representations in (3), (4) it is not difficult to get asymptoticexpansions for the Mittag-Leffler function in the complex plane. Let 0 < < 2, be an arbitrary number, and be chosen to satisfy the condition (5). Then wehave, for any p N and |z | :Whenever | arg z | ,

E , (z) = z

(1 ) ez

1/

p

k=1

zk( k )

+ O(|z |1 p). (6)

Whenever | arg z | ,

E , (z) = p

k=1

zk( k )

+ O(|z|1 p). (7)


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These formulas are also used in our numerical algorithm.In what follows we restrict our attention to the case R , the most important

in the applications. For the purpose of numerical computation we look for integralrepresentations better suited than (3) und (4). With

(, z ) = e

1/ (1 )/

z

we then have

I = 12i

( ;)(, z ) d =

12i

S

(, z ) d (8)

+ 12i C (0; ) (, z ) d + 12i S (, z ) d = I 1 + I 2 + I 3.

Now we transform the integrals I 1, I 2 and I 3. For I 1 we take = rei , r < and get

I 1 = 12i S (, z) d = 12i

+ e(re

i )1/ (re i )(1 )/re i z

ei dr. (9)

Analogously for (with = re i , r < )

I 3 = 12i S (, z ) d = 12i +

e(re i )1/

(rei

)(1

)/

re i z ei dr. (10)

S

S

C (0; )

( ; )

G(+) ( ; )

G() ( ; )

Figure 1: The contour ( ; )


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For I 2 we have = ei , and

I 2 = 12i C (0; ) (, z ) d = 12i

e( e

i )1/ ( ei )(1 )/ei z

iei d (11)

= 1+(1 )/

2

e

1/ ei/ ei ((1 )/ +1)ei z

d =

P [,, , ,z ]d,

with

P [,, , ,z ] = 1+(1 )/

2e

1/ cos(/ ) (cos() + i sin())ei z

, (12)

= 1/ sin(/ ) + (1 + (1 )/ ).

We rewrite the sum I 1 + I 3 as

I 1 + I 3 = 12i

+ e(re i )1/ (re i )(1 )/re i z

ei

e(re

i )1/ (re i )(1 )/re i z

ei dr

= 1

2i + r (1 )/ er 1/ cos( / ) e

i(r 1/ sin( / )+ (1+(1 )/ ))

rei

z

ei(r 1/ sin( / )+ (1+(1 )/ ))re i z

dr = + K [,,,r,z ]dr,

withK [,,,r,z ] =

1

r (1 )/ er 1/ cos( / ) r sin( ) z sin()r 2 2rz cos( ) + z2

, (13)

= r 1/ sin(/ ) + (1 + (1 )/ ).

By aid of (8)-(13) we can rewrite the formulas (3) and (4) as

E ,

(z) =

+ K [,,,r,z ]dr +

P [,, , ,z ]d, z G() ( ; ), (14)

E , (z) = + K [,,,r,z ]dr +

P [,, , ,z ]d (15)

+1

z(1 )/ ez1/ , z G(+) ( ; ).

Let us now consider the case 0 < 1, z = 0. By condition (5) we can choose = min {, } = . Then the kernel function (13) looks simpler:

K [,,,r,z ] = K [,,r,z ] (16)


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= 1

r (1 )/ er 1/ r sin((1 )) z sin((1 + ))r 2 2rz cos( ) + z2

.

In the formulas (14)-(16) for computation of the function E , (z) at the arbitrarypoint z C , z = 0 we will distinguish three possibilities for arg z, namelyA) | arg z | > ,B) | arg z| = ,C) | arg z| < .

First we conside the case A): | arg z| > .

z

G() ( ; ) G(+) ( ; )

( ; )

Figure 2: The case | arg z| >

In this case z always (for arbitrary ) lies in the region G() ( ; ) (see Figure2) and we arrive at

Theorem 2.1. Under the conditions

0 < 1, R , | arg z | > , z = 0

the function E , (z) has the representations

E , (z) = K [,,r,z ]dr +

P [,, , ,z ]d, > 0, R , (17)

E , (z) = 0 K [,,r,z ]dr, if < 1 + , (18)


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E , (z) = sin( )

0 er 1/

r 2 2rz cos( ) + z2 dr

1z

, if = 1 + (19)

with

K [,,r,z ] = 1


,

P [,, , ,z ] = 1+(1 )/

2e


,

= 1/ sin(/ ) + (1 + (1 )/ ).

P r o o f. The representation (17) follows immediately from (14) with = and (16) and holds for arbitrary R . Let us now consider (17) for < 1 + .In this case we have

|P [,, , ,z ]| C 1+(1 )/ , , 0 < < 1with a constant C not depending on . Consequently,

P [,, , ,z ]d 0 if 0. (20)

Further we have

K [,,r,z ] = O(r(1

)/

), r 0.This means: If < 1 + the integral

0 K [,,r,z ]dris a convergent improper integral with singularity at r = 0, but it is there not sin-gular if 1. Using (20), we can calculate a limit for 0 in the representation(17) and so arrive at the representation (18).

Finally, in the case = 1 + we have the formulas

P [,, , ,z ] = 1

2

e1/ cos(/ ) (cos( 1/ sin(/ )) + i sin( 1/ sin(/ )))

ei

z ,

lim0

P [,, , ,z ]d =

lim0

P [,, , ,z ]d (21)

=

12z

d = 1z

,

the corresponding integral converging uniformly with respect to . Hence for = 1 + there holds the formula

K [,,r,z ] = sin( )

er 1/

r 2 2rz cos( ) + z2 (22)


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and the integral

0 K [,,r,z ]dris non-singular in the point r = 0. The representation (19) now follows from (17),(21) und (22).

Example : Let = 1, 0 < < 1, z = t < 0.Theorem 2.1 then yields the representation

E , 1( t ) =

0

1

er 1/ t sin( )

r 2 + 2 rt cos( ) + t2 dr,

which by insertion of r = x t can be transformed to

E , 1( t ) = 0 1 ext x1 sin( )

x2 + 2 x cos( ) + 1 dx. (23)

For the representation (23) which can be obtained by the Laplace transformmethod see Goreno and Mainardi [17].

The next case we consider is the case B): | arg z| = .

G() ( ; ) G(+) ( ; )

( ; )

z

Figure 3: The case | arg z| =

In this case we are not allowed to make arbitrarily small in our contour ( ; ), because z is directly lying on this contour if < |z | (see Figure 3).


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0 < 1, R , | arg z| = , z = 0

the function E , (z) has the representation

E , (z) = K [,,r,z ]dr +

P [,, , ,z ]d, > |z |, (24)

with

K [,,r,z ] = 1 r (1 )/ er1/

r sin((1 )) z sin((1 + ))r 2 2rz cos( ) + z2 ,

P [,, , ,z ] = 1+(1 )/

2e


,

= 1/ sin(/ ) + (1 + (1 )/ ).

P r o o f. If > |z | then z is lying in the region G() ( ; ) and the representation(24) follows from (14) with = and (16).

Finally let us discuss the case C): | arg z| < .

G() ( ; ) G(+) ( ; )

( ; )

z

Figure 4: The case | arg z| <

In this case, if 0 < < |z |, then z is in the region G (+) ( ; ), and we have


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0 < 1, R , | arg z | < , z = 0

the function E , (z) possesses the representations

E , (z) = K [,,r,z ]dr +

P [,, , ,z ]d (25)

+1

z(1 )/ ez1/ , 0 < < |z |, R ,

E , (z) = 0

K [,,r,z ]dr +

1 z

(1

)/

ez1/

, if < 1 + , (26)

E , (z) = sin( )

0 er 1/

r 2 2rz cos( ) + z2 dr (27)

1z

+ 1z

ez1/

, if = 1 +

with

K [,,r,z ] = 1


,

P [,, , ,z ] = 1+(1 )/

2e

1/ cos(/ ) (cos() + i sin())ei z ,

= 1/ sin(/ ) + (1 + (1 )/ ).

P r o o f. Similar to the proof of Theorem 2.1; use (15) instead of (14).

3. Computation of the Mittag-Leffler function E , (z)

We have proved that for arbitrary z = 0 and 0 < 1 the Mittag- Lefflerfunction E , (z) can be represented by one of the formulas (17)- (19), (24)-(26).We intend to use these formulas for numerical computation if q < |z|, 0 < q < 1and 0 < 1. In the case |z| q, 0 < q < 1 we compute E , (z) for arbitrary > 0 by aid of the power series (2). The case 1 < can be reduced to the case0 < 1 by aid of a recursion formula. For computation of the function E , (z)for arbitrary z C with arbitrary indices > 0, R , we will distinguish threepossibilities:

A) |z | q, 0 < q < 1 (q is a xed number), 0 < ,B) |z| > q, 0 < 1 andC) |z | > q, 1 < .In each case we compute the Mittag-Leffler function with the prescribed ac-

curacy > 0.First we consider the case A): |z | q, 0 < q < 1.


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Theorem 3.1. In the case |z | q, 0 < q < 1, 0 < the Mittag-Leffler function can be computed with the prescribed accuracy > 0 by use of the formula

E , (z) =k0

k=0

zk

( + k ) + (z), |(z)| , (28)

k0 = max {[(1 )/ ] + 1 , [ln((1 | z |)) / ln( |z |)]}.

P r o o f. We transcribe the denition (2) into the form

E , (z) =

m

k=0

zk

( + k) + (z, m ), (z, m ) =

k= m +1

zk

( + k) .

For all k k0 [(1 )/ ] + 1 we have the inequality

( + k ) 1

and thus the estimate ( m + 1 k0)

|(z, m )| =

k= m +1

zk

( + k )

k= m +1

|z |k = |z |m +1 1

1 | z |.

The estimate |(z)| := |(z, k0)| < follows from the inequality k0 [ln((1 |z|)) / ln( |z |)].

Remark 3.1. For computation of the function E , (z) in case A) it isrecommendable to choose in Theorem 3.1 a number q not very close to 1 (in orderto have not a large number of terms in the sum of formula (28)). In our computerprograms we have taken q = 0 .9.

We proceed with the case B): q < |z |, 0 < 1.In this case we use the integral representations (17)-(19), (24)-(26). We then

must compute numerically either the improper integral

I =

a

K [,,r,z ]dr, a {0, },

K [,,r,z ] = 1


,

or even the integral

J =

P [,, , ,z ]d, > 0,

P [,, , ,z ] = 1+(1 )/

2e


,


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= 1/ sin(/ ) + (1 + (1 )/ ).

The second integral J (the integrand P [,, , ,z ] being bounded and the limitsof integration being nite) can be calculated with prescribed accuracy > 0 byone of many product quadrature methods.

For calculating the rst (improper) integral I over the bounded functionK [,,r,z ] we use

Theorem 3.2. The representation

I =

a

K [,,r,z ]dr =

r0

aK [,,r,z ]dr + (z), |(z)| , a {0, } (29)

is valid under the conditions

0 < 1, 0 < q < |z |,

r 0 = max{1, 2|z|, ( ln( / 6)) }, if 0,max {(| | + 1) , 2|z |, ( 2ln(/ (6( | | + 2)(2 | |)| |)) }, if < 0.

P r o o f. We try to estimate the absolute value of

(z, R ) = R K [,,r,z ]drin the expression

I = R

aK [,,r,z ]dr + (z, R ), a {0, }.

We rst consider an estimate of the function K [,,r,z ]. Let z0 = ei andr 2|z |. We then have

1|r 2 2rz cos( ) + z2 |

= 1

r 2 zr z0zr z0

1

r 2 zr | z0 |zr | z0 |

= 1

r 2 1 zr

2 4r 2

and hence also

| K [,,r,z ]| 1

r (1 )/ er 1/ |r | + |z ||r 2 2rz cos( ) + z2 |

1

r (1 )/ er 1/ 6rr 2

= 6

r (1 )/ 1er 1/ .Hence we have, for R 2|z |, the estimate

|(z, R )| R | K [,,r,z ]| dr (30)


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R 6 r (1 )/ 1er 1/ dr = 6 R 1/ t et dt = 6 (1 , R 1/ ),with the incomplete gamma function (see Erdelyi et al. [13])

(, x ) = x et t1 dt, x > 0More estimates can be obtained by aid of

Lemma 3.1. For the incomplete gamma function (1 , x) the following estimates hold:

|(1 , x )| e

x , x 1, 0, (31)

|(1 ), x) (| | + 2) x ex , x | | + 1 , < 0. (32)

P r o o f. For t x 1 and 0 we have the inequality t et et andhence also (31) from

|(1 , x )| x et dt = ex .If < 0 we determine n N such that (n + 1) < n and treat the integral(1 , x) by n -fold partial integration:

(1 , x) = x ex x 1ex + ( + 1) x 2ex (33)

+ . . . + ( 1)nn1

j =0( + j )x n ex + ( 1)n +1 x

n

j =0( + j )t n1et dt.

Now + n + 1 0 and x | | + 1 1 in the last integral, and we can use theinequality (31):

x t n1et dt ex .For x | | + 1 and (n + 1) < n we have

x > |x 1 | > | ( + 1) x 2 | > . . . > | ( + 1) . . . ( + n)|.

The latter inequalities together with the representation(33) yield the estimate

|(1 , x)| (n + 2) x ex (| | + 2) x ex , |x | | | + 1 ,

and thus Lemma 3.1 is proved.

Let us return to the proof of Theorem 3.2. (30) and Lemma 3.1 yield theestimate

|(z, R )| 6 eR

1/, 0, R 1

6 (| | + 2) R/ eR

1/, < 0, R1/ | | + 1 .

(34)


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For 0 we have |(z)| := |(z, r 0)| < , if r0 = max {1, 2|z |, ( ln(/ 6)) },and for < 0 we should nd an r 0 such that the inequality

6

(| | + 2) r /0 er1/0 (35)

is fullled. We solve this exercise by aid of

Lemma 3.2. For arbitrary x, y,q > 0 there holds the inequality

xy (qy)yex/q . (36)

P r o o f. With x = ay, a > 0, the inequality (36) can be rewritten in theform (ay)y (qy)yeay/q which is equivalent to ay q yeay/q and hence also to

aq

y ea/q

y.

The latter inequality is true because of

ea/q max {a/q, 1}, a,q > 0.

Hence Lemma 3.2 is proved.Denoting r 1/0 by x, by y, and taking q = 2 we rewrite the inequality (35)

by aid of (36) in the form

6

(| | + 2) r /0 er1/0 =

6

(| | + 2) xyex

6

(| | + 2)(2 y)yex/ 2ex = 6

(| | + 2)(2 | |)| |er1/0 / 2 < .

Solving for r0 we nd |(z)| := |(z, r 0)| for < 0 and r0 = max {(| | +

1)

, 2|z |, ( 2ln(/ (6( | | + 2)(2 | |)|

|))

}.The last case we have to discuss is the case C): q < |z |, 1 < . Here we usethe recursion formula (see Dzherbashyan [11])

E , (z) = 1m

m1

h=0E /m, (z1/m ei2h/m ) (m 1). (37)

In order to reduce case C) to the cases B) and A) we take m = + 1 in formula(37). Then 0 < /m 1, and we calculate the functions E /m, (z1/m e2ih/m ) asin case A) if |z |1/m q < 1, and as in case B) if |z |1/m > q .


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Remark 3.2. The ideas und techniques employed for the Mittag-Lefflerfunction can be used for numerical calculation of other functions of the hyperge-ometric type. In particular, the same method with some small modications canbe applied for the Wright function playing a very important role in the theoryof partial differential equations of fractional order (see for example Buckwar andLuchko [6], Goreno et al. [15], Luchko [25], Luchko and Goreno [26], Mainardiet al. [29]). To this end, the following representations (see Goreno et al. [14])can be used instead of the representations (2), (3), and (4):

(, ; z) =

k=0

zk

k!(k + ), > 1, C ,

(, ; z) = 12i Ha e + z d, > 1, C ,

where Ha denotes the Hankel path in the -plane with a cut along the negativereal semi-axis arg = .

4. Computation of the derivative of the function E , (z)

In many questions of analysis the derivative of a function plays an importantrole. The derivative of the Mittag-Leffler function can be used for example initerative methods for determination of its zeros in the complex plane. The functionE , (z) being an entire function, we nd by term-wise differentiation of its powerseries (2) the representation

E , (z) =

k=1

kzk1( + k )

=

k=0

(k + 1) zk

( + + k ). (38)

For numerical calculation of the function E , (z) with prescribed accuracy wedistinguish two cases: A) |z| q < 1 (q is a xed number) and B) |z | > q .

We start with the case A): |z| q < 1.

Theorem 4.1. In the case |z | q < 1 the derivative (38) of the Mittag-Leffler function can be calculated by aid of the formula

E , (z) =k0

k=0

(k + 1) zk

( + + k) + (z), |(z)| , (39)

k0 = max {k1, [ln((1 | z |)) / ln( |z |)]},

k1 =[(2 )/ ( 1)] + 1, > 1,[(3 )/ ] + 1, 0 < 1, D 0,max { 3 + 1 , 12 +

D2 2 + 1 }, 0 < 1, D > 0,

(40)


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5. Numerical Algorithm

The numerical scheme for computation of the Mittag-Leffler function givenin pseudocode notation by the algorithm below is based on the results presentedin Section 3. The algorithm uses the dening series (2) for arguments z of smallmagnitude, its asymptotic representations (6), (7) for arguments z of large mag-nitude, and special integral representations for intermediate values of the argu-ment that include a monotonic part K (,,,z ) d and an oscillatory part P (,, ,,z ) d, which can be evaluated using standard techniques.GIVEN > 0, R , z C , > 0 THENIF 1 < THEN

k0 = + 1E , (z) = 1k0

k01k=0 E /k 0 , (z1

k 0 exp( 2ikk0 ))ELSIF z = 0 THEN

E , (z) = 1( )ELSIF |z | < 1 THEN

k0 = max { (1 ) , ln[(1 | z|)]/ ln( |z |) }E , (z) =

k0k=0

zk( + k )

ELSIF |z | > 10 + 5 THENk0 = ln()/ ln( |z |)IF | arg z | < 4 + 12 min{, } THEN

E , (z) = 1 z(1 )

ez1/

k0k=1z k

( k )ELSEE , (z) = k0k=1

z k( k )ELSE

0 = max{1, 2|z |, ( ln( 6 ))

}, 0max {(| | + 1) , 2|z|, ( 2ln( [6(| |+2)(2 | |) | | ]

)) }, < 0

K (,,,z ) = 1 (1 )

exp( 1 ) sin[ (1 )]z sin[ (1 + )] 22z cos( )+ z2

P (,, ,,z ) = 121+ (1 ) exp(

1 cos( ))

cos( )+ i sin( )exp( i )z

= (1 + (1

)

) + 1

sin( )

IF | arg z | > THENIF 1 THEN

E , (z) = 0

0 K (,,,z ) dELSE

E , (z) = 0

1 K (,,,z ) d +

P (,, 1, ,z ) dELSIF | arg z | < THENIF 1 THEN

E , (z) = 0

0 K (,,,z ) d + 1 z

(1 ) ez

1/

ELSE


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E , (z) = 0| z |2

K (,,,z ) d +

P (,, |z|2 , ,z) d + 1 z

(1 ) ez

1/

ELSEE , (z) =

0( | z | +1)

2K (,,,z ) d +

P (,, (|z|+1)2 , ,z) d

END

Remark 5.1. The formulas for E , (z) in this algorithm are in error at mostby . It is advisable to take = m = machine precision.

6. Figures

In this section, some gures generated by aid of the methods described in thepaper are presented. We have produced them by using the programming systemMATHEMATICA.

2 4 6 8 10

-0.5

-0.25

0.25

0.5

0.75

1

Figure 5: The function E , ( t) for = 0 .25, = 1 and its derivative.Black line: E , ( t), grey line: ddt (E , ( t))


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10 20 30 40 50

-0.5

-0.25

0.25

0.5

0.75

1


20 40 60 80 100

-1

1

2



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1 2 3 4 5

50

100

150

200

250

300

350

Figure 8: |E , (z)| for = 0 .75, = 1 , arg( z) = 4

10 20 30 40 50

1.32

1.34

1.36

Figure 9: |E , (z)| 1 for = 0 .75, = 1 , arg(z) =

2


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10 20 30 40 50

0.025

0.05

0.075

0.1

0.125

0.15

Figure 10: |E , (z)| 0 for = 0 .75, = 1 , arg( z) = 34

5 1 0 15 20

0.2

0.4

0.6

0.8

1

Figure 11: E , (z) for = 0 .75, = 1 , arg( z) =


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2 4 6 8 10

10

20

30

40

50

60

70

Figure 12: |E , (z)| for = 1 .25, = 1 , arg( z) = 4

10 20 30 40 50

0.72

0.74

0.76

0.78

0.82

0.84

Figure 13: |E , (z)| 1 for = 1 .25, = 1 , arg( z) =

2


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10 20 30 40 50

0.05

0.1

0.15

0.2

0.25

Figure 14: |E , (z)| 0 for = 1 .25, = 1 , arg( z) = 34

20 40 60 80 100

-0.1

-0.05

0.05

0.1

Figure 15: E , (z) for = 1 .25, = 1 , arg( z) =


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1) Department of Mathematics and Computer Science Received:Free University of Berlin Arnimallee 3 D-14195 Berlin GERMANY e-mail: [email protected], [email protected]

2) Department of Business Informatics European University Frankfurt(Oder)POB 1786 D-15207 Frankfurt (Oder) GERMANY e-mail: [email protected]

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