Approximate solution to linear complex differential equation by a new approximate approach

Applied Mathematics and Computation 185 (2007) 636–645 www.elsevier.com/locate/amc

Approximate solution to linear complex differential equation by a new approximate approach Mustafa Gu¨lsu *, Mehmet Sezer Department of Mathematics, Faculty of Science, Mugla University, Mugla, Turkey

Abstract The purpose of this study is to give a Taylor polynomial approximation for the solution of high-order linear complex differential equations with variable coefficients and mixed conditions about any point. For this purpose, a matrix method is introduced. This method is based on first taking the truncated Taylor expansions of the functions in the complex differential equation and then substituting their matrix forms into the given equation. Hence the result matrix equation can be solved and the unknown Taylor coefficients can be found approximately. Also, examples that illustrate the pertinent features of the method are presented and the results are discussed. Ó 2006 Elsevier Inc. All rights reserved. Keywords: Taylor polynomials; Polynomial approximations; Complex differential equations

1. Introduction Taylor and Chebyshev expansion approaches to solve linear differential, integral and integro-differential equations have been presented in many paper [1–5]. On the other hand, a Taylor expansion approach to solve high-order linear difference and integro difference equations has been given by Gu¨lsu and Sezer [6,7]. In this paper, these methods are modified and developed for solving the linear complex differential equation with variable coefficients m X P k ðzÞf ðkÞ ðzÞ ¼ gðzÞ ð1Þ k¼0

which is extended of the linear complex differential equations [8–10], under the mixed conditions m1 X 

 aik f ðkÞ ðaÞ þ bik f ðkÞ ðbÞ þ cik f ðkÞ ðz0 Þ ¼ ki ;

i ¼ 0; 1; . . . ; m  1:

k¼0

*

Corresponding author. E-mail address: [email protected] (M. Gu¨lsu).

0096-3003/$ - see front matter Ó 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2006.07.050

ð2Þ

M. Gu¨lsu, M. Sezer / Applied Mathematics and Computation 185 (2007) 636–645

637

Here the coefficients Pk(z) and the right-hand member g(z) are analytic functions in the disc D = {z 2 C : jz  z0j 6 q}, 0 < q < 1; aik, bik, cik and ki are appropriate complex constants; a, b, z0 2 D, q 2 R. We assume that the solution of Eq. (1) under the conditions (2) is expressed in the form f ðzÞ ¼

N X

n

an ðz  z0 Þ ;

ð3Þ

n¼0

so that the Taylor coefficients to be determined are an ¼

f ðnÞ ðz0 Þ ; n!

n ¼ 0; 1; . . . ; N :

2. Fundamental matrix relations Let us convert the expression defined in (1)–(3) to matrix forms. Now, let us assume that the function f(z) and its kth derivative with respect to z, respectively, can be expanded to Taylor series about the points z = z0 in the forms 1 X f ðnÞ ðz0 Þ n ð4Þ an ðz  z0 Þ ; an ¼ f ðzÞ ¼ n! n¼0 and f ðkÞ ðzÞ ¼

1 X

n aðkÞ n ðz  z0 Þ ;

ð5Þ

n¼0

where for k = 0, f (0) (z) = f(z) and an ¼ að0Þ n . First, let us derivative the expression (5) with respect to z and then put (n + 1) instead of n: 1 1 X X ðkÞ n1 n naðkÞ ðz  z Þ ¼ ðn þ 1Þanþ1 ðz  z0 Þ : f ðkþ1Þ ðzÞ ¼ 0 n n¼1

ð6Þ

n¼0

From (5), it is clear that 1 X n aðkþ1Þ ðz  z0 Þ : f ðkþ1Þ ðzÞ ¼ n

ð7Þ

n¼0 ðkþ1Þ of and By using the relations (6) and (7), we have the recurrence relation between the coefficients aðkÞ n and an (k) (k+1) f (z) and f (z) ðkÞ

anðkþ1Þ ¼ ðn þ 1Þanþ1 ;

n; k ¼ 0; 1; 2; . . .

ð8Þ

Now let us take n = 0, 1, . . . , N and assume aðkÞ n ¼ 0 for n > N. Then the system (8) can be transformed into the matrix form Aðkþ1Þ ¼ MAðkÞ ;

k ¼ 0; 1; . . . ;

ð9Þ

where 2

A

ðkÞ

ðkÞ

a0

3

6 ðkÞ 7 6a 7 6 1 7 7 6 6  7 ¼6 7; 6  7 7 6 7 6 4  5 ðkÞ

aN

2

0

6 6 0 6 6... 6 6 M ¼ 6... 6 6... 6 6 4 0 0

1

0

...

0

0 2 ...

...

0

 

... ...



0

0

...

0

0

...

0

0

3

7 0 7 7 ...7 7 7 . . . 7: 7 ...7 7 7 N 5 0

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M. Gu¨lsu, M. Sezer / Applied Mathematics and Computation 185 (2007) 636–645

For k = 0, 1, 2, . . . it follows from relation (9) that Að1Þ ¼ MAð0Þ ¼ MA; Að2Þ ¼ MAð1Þ ¼ M2 A;

ð10Þ

 AðkÞ ¼ MAðkþ1Þ ¼ Mk A; where clearly Að0Þ ¼ A ¼ ½ a0

   aN T :

a1

Consequently, the matrix equation (10) gives a relation between the Taylor coefficients matrix A of f(z) and the Taylor coefficients matrix A(k) of the kth derivative of f(z). In addition, the solution (3) and its derivatives can be written in the matrix forms ½f ðzÞ ¼ ZA

and

½f ðkÞ ðzÞ ¼ ZAðkÞ

ð11Þ

or using the relation (10) ½f ðkÞ ðzÞ ¼ ZMk A;

ð12Þ

where ðz  z0 Þ ðz  z0 Þ2

Z ¼ ½1



ðz  z0 ÞN :

On the other hand, the matrix representation of (z  z0)jf (k)(z) becomes   ðz  z0 Þj f ðkÞ ðzÞ ¼ ZIj MAðkÞ ¼ ZIj Mk A; j ¼ 0; 1; . . . ; N ;

ð13Þ

where 2

3

2

1

0 

 0

0

60 6 6 6 I0 ¼ 6 6 6 6 4 0

1   

 0 





 0 

1  0

07 7 7 7 7; . . . ; 7 7 7 5



1

3

0 0





0

0

60 0 6 6 61  Ij ¼ 6 6 1 6 6 4  0 0





0 

07 7 7 7 7; . . . ; 7 7 7 : 5

:  

 1 0

0

2

3

0

0

  0

0

60 6 6 6 IN ¼ 6 6 6 6 4 1

0 

  0 





 0

   0

07 7 7 7 7: 7 7 7 5 0

We assume that the function g(z) can be expanded as gðzÞ ¼

N X

n

gn ðz  z0 Þ ;

gn ¼

n¼0

gðnÞ ðz0 Þ : n!

Then the matrix representation of g(z) becomes ½gðzÞ ¼ ZG;

ð14Þ

where G ¼ ½ g0

g1

T

   gN  :

We now ready to construct the fundamental matrix equation corresponding to Eq. (1). For this purpose, we first reduce Eq. (1) to the form m X N X k¼0

j¼0

pkj ðz  z0 Þj f ðkÞ ðzÞ ¼ gðzÞ;

ð15Þ

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so that P k ðzÞ ¼

N X

ðjÞ

pkj ðz  z0 Þj ;

pkj ¼

j¼0

P k ðz0 Þ : j!

Substituting the matrix relations (13) and (14) into Eq. (15) and simplifying the result equation, we obtain the matrix equation m X N X k¼0

pkj Ij Mk A ¼ G:

ð16Þ

j¼0

This matrix equation is a fundamental relation for the mth-order linear complex differential equation with variable coefficients (1). Next we can obtain the corresponding matrix for the mixed conditions (2) as follows: using the relation (12), we find the matrix representations of the functions in (2), for the points a, b and z0, in the forms ½f ðkÞ ðaÞ ¼ Za Mk A; ½f ðkÞ ðbÞ ¼ Zb Mk A;

ð17Þ

½f ðkÞ ðz0 Þ ¼ Zz0 Mk A; where  Za ¼ 1 ða  z0 Þ ða  z0 Þ2  Zb ¼ 1 ðb  z0 Þ ðb  z0 Þ2 Zz0 ¼ ½ 1 0 0    0 :

3

ða  z0 Þ ðb  z0 Þ3

 N    ða  z0 Þ ;     ðb  z0 ÞN ;

Substituting the matrix representations in (17) into Eq. (2), we obtain the matrices system m1 X faik Za þ bik Zb þ cik Zzo gMk A ¼ ½ki ;

i ¼ 0; 1; . . . ; m  1

ð18Þ

k¼0

which is the fundamental matrix relation for the conditions (2). 3. Method of solution Let us consider the fundamental matrix equation (16) corresponding to the mth-order linear complex differential equation with variable coefficients (1). We can write Eq. (16) in the form WA ¼ G;

ð19Þ

where W ¼ ½wpq  ¼

m X N X k¼0

pkj Ij Mk ;

p; q ¼ 0; 1; . . . ; N

j¼0

and G is defined in (14). The augmented matrix of Eq. (19) becomes ½W; G ¼ ½wpq ; gp ;

p; q ¼ 0; 1; . . . ; N :

ð20Þ

We now consider the matrix equation (18) corresponding to the conditions (2) and can write in the matrix forms Ui A ¼ ½ki ;

i ¼ 0; 1; . . . ; m  1

or the augmented matrix forms ½Ui ; ki  ¼ ½ ui0

ui0



uiN ; ki ;

ð21Þ

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M. Gu¨lsu, M. Sezer / Applied Mathematics and Computation 185 (2007) 636–645

where Ui ¼

m1 X

faik Za þ bik Zb þ cik Zzo gMk :

k¼0

Consequently, to find the unknown Taylor coefficients an, n = 0, 1, . . . , N, related with the approximate solution of the problem consisting of Eq. (1) and conditions (2), by replacing the m row matrices (21) by the last m rows of the augmented matrix (20), we have new augmented matrix 3 2 w00 w01    w0N ; g0 6 w10 w11    w1N ; g1 7 7 6 6    ;  7 7 6 7 6 7 6 w w    w ; g N m;0 N m;1 N m;N N m 7   6 ð22Þ ½W ; G  ¼ 6 u01    u0N ; k0 7 7 6 u00 7 6 u11    u1N ; k1 7 6 u10 7 6 4    ;  5 um1;0 um1;1    um1;N ; km1 or the corresponding matrix equation W A ¼ G  : If det W* 5 0, we can write Eq. (22) as A ¼ ðW Þ1 G

ð23Þ

and the matrix A is uniquely determined. Thus the mth-order linear complex differential equation with variable coefficients (1) under the conditions (2) has a unique solution. This solution is given by the truncated Taylor series (3). Also we can easily check the accuracy of the obtained solutions as follows [1,2]. Since the Taylor polynomial (3) is an approximate solution of Eq. (1), when the solutions f(z) and its derivatives are substituted in Eq. (1), the resulting equation must be satisfied approximately; that is, for z = zi 2 D, i = 0, 1, 2, . . .,    X m   Eðzi Þ ¼  P k ðzi Þf ðkÞ ðzi Þ  gðzi Þ ffi 0   k¼0 or Eðzi Þ 6 10ki

ðk i is any positive integerÞ:

If max ð10ki Þ ¼ 10k (k is any positive integer) is prescribed, then the truncation limit N is increased until the values E(zi) at each of the points zi becomes smaller than the prescribed 10k. 4. Examples Example 1 [8, pp. 299]. Let us illustrate the method by means of the second-order complex differential equation with variable coefficients f 00 ðzÞ þ zf ðzÞ ¼ expðzÞ þ z expðzÞ; with the conditions f ð0Þ ¼ 1;

f 0 ð0Þ ¼ 1

and approximate the solution f(z) by the polynomial f ðzÞ ¼

5 X n¼0

an zn ;

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where N = 5, c = 0, P0(z) = z, P1(z) = 0, P2(z) = 1, g(z) = ez + zez. We first reduce this equation, from Eq. (16), to the matrix form 2 X 6 X k¼0

pkj Ij Mk A ¼ G:

j¼0

The quantities in the equation are computed as 3 2 3 2 1 0 0 2 0 0 0 0 7 6 60 0 0 6 0 2 0 0 7 7 6 7 6 7 6 7 6 7 6 7 6 0 0 0 0 12 0 1:5 0 7 6 7 6 7 6 7 6 2 M ¼ 6 0 0 0 0 0 20 0 7; G ¼ 6 0:6666666667 7 7 6 7 6 6 0:2083333333 7 60 0 0 0 0 0 30 7 7 6 7 6 7 6 7 6 0:05 0 0 5 5 4 40 0 0 0 0 0:0097222222 0 0 0 0 0 0 0 and the augmented 2 0 6 61 6 60 6 6 ½W; G ¼ 6 0 6 60 6 6 40

matrix is 0

2

0

0

0

0

;

0

0

6

0

0

0

;

1 0

0 1

0 0

12 0 0 20

0 0

; ;

0 0

0 0

1 0

0 1

0 0

30 0

; ;

0 0

0

0

0

1

0

;

The augmented matrix based on the 2 0 0 2 0 0 6 1 0 0 6 0 6 6 6 0 1 0 0 12 6 6   ½W ; G  ¼ 6 0 0 1 0 0 6 60 0 0 1 0 6 6 41 0 0 0 0 0 1 0 0 0

3

1

7 7 7 7 1:5 7 7 0:6666666667 7: 7 0:2083333333 7 7 7 0:05 5 0:0097222222 2

conditions f(0) = 1, f 0 (0) = 1 is obtained as 3 0 0 ; 1 7 0 0 ; 2 7 7 7 0 0 ; 1:5 7 7 20 0 ; 0:6666666667 7: 7 0 30 ; 0:2083333333 7 7 7 0 0 ; 1 5 0 0 ; 1

Solving this system, the Taylor coefficients are obtained as a0 ¼ 1; a1 ¼ 1; a2 ¼ 0:5; a3 ¼ 0:1666666667; a5 ¼ 0:0083333335; a6 ¼ 0:001388888887

a4 ¼ 0:0416666666;

and thereby the polynomial solution becomes f ðzÞ ¼ 1 þ z þ 0:5z2 þ 0:1666666667z3 þ 0:0416666666z4 þ 0:0083333335z5 þ 0:001388888887z6 : Taking N = 6, 7 the obtained solutions are compared with the exact solution f(z) = ez in Tables 1 and 2. In Table 3 we gave the absolute error for N = 7, 8, 9. Example 2. f 00 ðzÞ þ zf 0 ðzÞ þ 2zf ðzÞ ¼ 2z sin z þ z cos z  sin z;

f ð0Þ ¼ 0; f 0 ð0Þ ¼ 1:

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M. Gu¨lsu, M. Sezer / Applied Mathematics and Computation 185 (2007) 636–645

Table 1 Error analysis of Example 1 for the Re(z) value z

Exact solution (Real)

1.0  1.0I 0.8  0.8I 0.6  0.6I 0.4  0.4I 0.2  0.2I 0.0  0.0I 0.2 + 0.2I 0.4 + 0.2I 0.6 + 0.6I 0.8 + 0.8I 1.0 + 1.0I

0.1987661103 0.3130505040 0.4529537891 0.6174056479 0.8024106473 0.0000000000 1.1970560210 1.3740615390 1.5038595410 1.5505492970 1.4686939400

Present Met(z0 = 0) N = 6 (Real)

Ne = 6

N = 7 (Real)

Ne = 7

0.2000000003 0.3133226668 0.4529920000 0.6174079999 0.8024106667 0.0000000000 1.1970559990 1.3740586670 1.5038080000 1.5501439990 1.4666666680

0.1233E2 0.2721E3 0.3821E4 0.2352E5 0.1940E7 0.0000E0 0.2200E7 0.2872E5 0.5154E4 0.4052E3 0.2027E2

0.1984126987 0.3129897855 0.4529475657 0.6174053993 0.8024106464 0.0000000000 1.1970560190 1.3740612680 1.5038524340 1.5504768800 1.4682539700

0.3534E3 0.6071E4 0.6223E5 0.2486E6 0.9000E9 0.0000E0 0.2000E8 0.2710E6 0.7107E5 0.7241E4 0.4399E3

Table 2 Error analysis of Example 1 for the Im(z) value z

Exact solution (Im)

Present Met(z0 = 0) N = 6 (Im)

Ne = 6

N = 7 (Im)

Ne = 7

1.0  1.0I 0.8  0.8I 0.6  0.6I 0.4  0.4I 0.2  0.2I 0.0  0.0I 0.2 + 0.2I 0.4 + 0.2I 0.6 + 0.6I 0.8 + 0.8I 1.0 + 1.0I

0.3095598757 0.3223288692 0.3098823596 0.2610349211 0.1626566908 0.0000000000 0.2426552686 0.5809439008 1.028845666 1.596505341 2.287355287

0.3111111112 0.3226567111 0.3099264000 0.2610375111 0.1626567111 0.0000000000 0.2426552889 0.5809464889 1.0288896000 1.5968312890 2.2888888890

0.1551E2 0.3278E3 0.4404E4 0.2590E5 0.2030E7 0.0000E0 0.2030E7 0.2588E5 0.4393E4 0.3259E3 0.1533E2

0.3095238096 0.3223238298 0.3098819657 0.2610349105 0.1626566908 0.0000000000 0.2426552686 0.5809438883 1.0288451660 1.5964984080 2.2873015870

0.3606E4 0.5039E5 0.3939E6 0.1060E7 0.0000E0 0.0000E0 0.0000E0 0.1250E7 0.5000E6 0.6933E5 0.5370E4

Table 3 The absolute error for the considered method at various points for Example 1 pffiffiffiffiffiffiffi z Exact solution ðI ¼ 1Þ Ne = 7

Ne = 8

Ne = 9

1.0  1.0I 0.8  0.8I 0.6  0.6I 0.4  0.4I 0.2  0.2I 0.0  0.0I 0.2 + 0.2I 0.4 + 0.2I 0.6 + 0.6I 0.8 + 0.8I 1.0 + 1.0I

0.5644E4 0.7724E5 0.5918E6 0.1564E7 0.1000E9 0.0000E0 0.1000E8 0.1665E7 0.6673E6 0.9065E5 0.6888E4

0.8054E5 0.8805E6 0.5046E7 0.1004E8 0.1000E9 0.0000E0 0.1000E8 0.1345E8 0.5603E7 0.1017E5 0.9654E5

0.1987661103  0.3095598757I 0.3130505040  0.3223288692I 0.4529537891  0.3098823596I 0.6174056479  0.2610349211I 0.8024106473  0.1626566908I 0.0000000000  0.0000000000I 1.1970560210 + 0.2426552686I 1.3740615390 + 0.5809439008I 1.5038595410 + 1.0288456660I 1.5505492970 + 1.5965053410I 1.4686939400 + 2.2873552870I

0.3552E3 0.6092E4 0.6235E5 0.2488E6 0.9000E9 0.0000E0 0.2000E8 0.2712E6 0.7124E5 0.7274E4 0.4432E3

The exact solution is f(z) = sin z. Numerical results are given in Tables 4 and 5. In Table 6 we gave the absolute error for N = 7, 9, 11. From these tables, it can be seen that the absolute errors in the given problem are very small. Example 3 [9, Example 1]. Let us consider the equation f 00 ðzÞ þ ð25e10z  40e9z  46e8z  44e7z  25e6z  37e5z  20e4z  9e3z  4e2z Þf ðzÞ ¼ 0:

M. Gu¨lsu, M. Sezer / Applied Mathematics and Computation 185 (2007) 636–645

643

Table 4 Error analysis of Example 2 for the Re(z) value z

Exact solution (Real)

1.0  1.0I 0.8  0.8I 0.6  0.6I 0.4  0.4I 0.2  0.2I 0.0  0.0I 0.2 + 0.2I 0.4 + 0.2I 0.6 + 0.6I 0.8 + 0.8I 1.0 + 1.0I

1.2984575810 0.9594171049 0.6693640130 0.4209894110 0.2026559797 0.0000000000 0.2026559797 0.4209894110 0.6693640130 0.9594171049 1.2984575810

Present Met(z0 = 0) N = 8 (Real)

Ne = 8

N = 9 (Real)

Ne = 9

1.2984126980 0.9594111187 0.6693635657 0.4209893994 0.2026559797 0.0000000000 0.2026559797 0.4209893994 0.6693635657 0.9594111187 1.2984126980

0.4488E4 0.5986E5 0.4473E6 0.1160E7 0.0000E0 0.0000E0 0.0000E0 0.1160E7 0.4473E6 0.5986E5 0.4488E4

1.2984567900 0.9594170366 0.6693640100 0.4209894110 0.2026559797 0.0000000000 0.2026559797 0.4209894110 0.6693640100 0.9594170366 1.2984567900

0.7910E6 0.6830E7 0.3000E8 0.0000E0 0.0000E0 0.0000E0 0.0000E0 0.0000E0 0.3000E8 0.6830E7 0.7910E6

Table 5 Error analysis of Example 2 for the Im(z) value z

Exact solution (Im)

1.0  1.0I 0.8  0.8I 0.6  0.6I 0.4  0.4I 0.2  0.2I 0.0  0.0I 0.2 + 0.2I 0.4 + 0.2I 0.6 + 0.6I 0.8 + 0.8I 1.0 + 1.0I

0.6349639148 0.6187493964 0.5254528757 0.3787279455 0.1973226870 0.0000000000 0.1973226870 0.3783279455 0.5254528757 0.6187493964 0.6349639148

Present Met(z0 = 0) N = 8 (Im)

Ne = 8

N = 9 (Im)

Ne=9

0.6349206348 0.6187435479 0.5254524343 0.3783279340 0.1973226869 0.0000000000 0.1973226869 0.3783279340 0.5254524343 0.6187435479 0.6349206348

0.4328E4 0.5848E5 0.4414E6 0.1150E7 0.1000E9 0.0000E0 0.1000E9 0.1150E7 0.4414E6 0.5848E5 0.4328E4

0.6349647265 0.6187494658 0.5254528786 0.3783279456 0.1973226869 0.0000000000 0.1973226869 0.3783279456 0.5254528786 0.6187494658 0.6349647265

0.8117E6 0.6940E7 0.2900E8 0.1000E9 0.1000E9 0.0000E0 0.1000E9 0.1000E9 0.2900E8 0.6940E7 0.8117E6

Table 6 The absolute error for the considered method at various points for Example 2 pffiffiffiffiffiffiffi z Exact solution ðI ¼ 1Þ Ne = 7

Ne = 9

Ne = 11

1.0  1.0I 0.8  0.8I 0.6  0.6I 0.4  0.4I 0.2  0.2I 0.0  0.0I 0.2 + 0.2I 0.4 + 0.2I 0.6 + 0.6I 0.8 + 0.8I 1.0 + 1.0I

0.1133E5 0.9737E7 0.4172E8 0.1000E9 0.1000E9 0.0000E0 0.1000E9 0.1000E9 0.4172E8 0.9737E7 0.1133E5

0.1486E7 0.7810E9 0.1000E9 0.1000E9 0.1000E9 0.0000E0 0.1000E9 0.1000E9 0.1000E9 0.7810E9 0.1486E7

1.2984575810  0.6349639148I 0.9594171049  0.6187493964I 0.6693640130  0.5254528757I 0.4209894110  0.3783279455I 0.2026559797  0.1973226870I 0.0000000000 + 0.0000000000I 0.2026559797 + 0.1973226870I 0.4209894110 + 0.3783279455I 0.6693640130 + 0.5254528757I 0.9594171049 + 0.6187493964I 1.2984575810 + 0.6349639148I

0.6235E4 0.8368E5 0.6284E6 0.1633E7 0.1000E9 0.0000E0 0.1000E9 0.1633E7 0.6284E6 0.8368E5 0.6235E4

The exact solution of this equation is f(z) = exp(e5z + e4z + e3z + e2z) and the Taylor Expansion of f(z) about z = 0 gives f ðzÞ ¼ 54:59815003 þ 764:3741004z þ 6824:768754z2 þ 47645:98559z3 þ 282522:6772z4 þ 0:1485446408  107 z5 þ   

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M. Gu¨lsu, M. Sezer / Applied Mathematics and Computation 185 (2007) 636–645

The solution obtained using the matrix method for N = 6 is fapprox ðzÞ ¼ 54:59815003 þ 764:3741004z þ 6824:768754z2 þ 47645:98559z3 þ 282522:6772z4 þ 0:1485446408  107 z5 which is the first six terms of the Taylor expansion of the exact solution at z = 0. Example 4. Our last example is the second-order linear complex differential equation f 00 ðzÞ þ zf ðzÞ ¼

1 5 1 4 z  z; 12 6

with the condition f ð0Þ ¼ 1;

f 0 ð0Þ ¼ 1:

For N = 5, the matrix form of the problem is defined by 2 X 5 X k¼0

pkj Ij Mk A ¼ G:

j¼0

After the augmented matrices of the system and conditions are computed, we obtain the Taylor coefficients matrix A ¼ ½1

1

1=6

1=12

T

0 :

Therefore, we find the exact solution 1 1 f ðzÞ ¼ 1  z  z3 þ z4 : 6 12 5. Conclusions High-order linear complex differential equations are usually difficult to solve analytically. Then it is required to obtain the approximate solutions. For this reason, the present method has been proposed for approximate solution and also analytical solution. The method presented in this study is a method for computing the coefficients in the Taylor expansion of the solution of a linear complex differential equations, and is valid when the functions Pk(z) and g(z) are defined in the disc D = {z 2 C : jz  z0j 6 q}. The Taylor method is an effective method for the cases that the known functions have the Taylor series expansion at z = z0. In this case, the Taylor polynomial solution f(z) and the values f(zj), zj 2 D can be easily evaluated at low-computation effort. In addition, an interesting feature of this method is to find the analytical solutions if the equation has an exact solution that is a polynomial of degree N or less than N. The method can also be extended to the system of linear complex equations with variable coefficients, but some modifications are required. References [1] R.P. Kanwal, K.C. Liu, A Taylor expansion approach for solving integral equations, Int. J. Math. Educ. Sci. Technol. 20 (3) (1989) 411–414. [2] S. Nas, S. Yalc¸inbasß, M. Sezer, A Taylor polynomial approach for solving high-order linear Fredholm integro-differential equations, Int. J. Math. Educ. Sci. Technol. 31 (2) (2000) 213–225. [3] M. Sezer, A method for approximate solution of the second-order linear differential equations in terms of Taylor polynomials, Int. J. Math. Educ. Sci. Technol. 27 (6) (1996) 821–834. [4] S. Yalc¸inbasß, M. Sezer, The approximate solution of high-order linear Volterra–Fredholm integro-differential equations in terms of Taylor polynomials, Appl. Math. Comput. 112 (2002) 291–308. [5] A. Akyu¨z, M. Sezer, Chebyshev polynomial solutions of systems of high-order linear differential equations with variable coefficients, Appl. Math. Comput. 144 (2003) 237–247.

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[6] M. Gulsu, M. Sezer, The approximate solution of high-order linear difference equation with variable coefficients in terms of Taylor polynomials, Appl. Math. Comput. 168 (1) (2005) 76–88. [7] M. Sezer, M. Gulsu, A new polynomial approach for solving difference and Fredholm integro-difference equations with mixed argument, Appl. Math. Comput. 171 (1) (2005) 332–344. [8] M.R. Spiegel, Theory and Problems of Complex Variables, McGraw-Hill Inc., NewYork, 1972. [9] Y.M. Chiang, S. Wang, Oscillation results of certain higher-order linear differential equations with periodic coefficients in the complex plane, J. Math. Anal. Appl. 215 (1997) 560–576. [10] J. Heittokangas, R. Korhonen, J. Rattya, Growth estimates for solutions of linear complex differential equations, Ann. Acad. Sci. Fenn. Math. 29 (2004) 233–246.