Available online through
ISSN 2229 – 5046International Journal of Mathematical Archive- 4(11), Nov. – 2013 134
USING TAYLOR EXPANSION TO PARTIAL FRACTIONS DECOMPOSITION
Mircea Ion Cîrnu*
Polytechnic University of Bucharest, Romania.
(Received on: 26-09-13; Revised & Accepted on: 11-11-13)
ABSTRACT
T
he automatic Taylor development, based on discrete convolution and deconvolution, is used to partial fractions decomposition.Keywords and phrases: Differential Taylor transform, discrete linear convolution and deconvolution, partial fractions
decomposition.
Mathematics Subject Classification: 26C15, 41A58, 30B10.
1. INTRODUCTION
Using the automatic Taylor expansion, we give some new numerical methods to partial fractions decomposition of rational functions. Firstly, we consider the Taylor differential transform that associates to a function the sequence of coefficients of its Taylor series or only a finite number of them. The Taylor transform of a product, respective quotient of functions is computed by discrete convolution, respective deconvolution, of the corresponding sequences of coefficients. This eliminates the algebraic calculation and the Taylor development becomes automatic.
As is well known, partial fractions decomposition of rational functions applied to the integration of these functions. Also it is used to determine the inverse Laplace transforms of such functions and thus to physical system theory. So far, there are numerous attempts to find a simple numerical method for this decomposition. In the paper [7] are presented nine such different methods. Using the automatic Taylor expansion, we give a new method for partial fractions decomposition of rational functions. It extend so called “particular values method”. These particular values, used to obtain the numerators of the decomposition, are even the poles of the rational function. When the poles are complex numbers, must work in complex, but there are cases in which it is possible to work only with real numbers. See Example 8.3 below. Our method of partial fractions decomposition is simpler than the already known methods, covers all cases and can be applied even to difficult examples. We give two theorems, one for a single pole of arbitrary multiplicity and the second, for a pair of poles of the same multiplicity. The second theorem can be used in the case of two complex or irrational conjugate poles of the same multiplicity. In the latest example 8.6, we do the partial fractions decomposition by both theorems. Based on the convenient decomposition, we compute the indefinite integral and the inverse Laplace transform of the considered function in that example.
Other applications of the discrete convolution and deconvolution were given by the Author in the works [1]-[5].
2.TAYLOR DIFFERENTIAL TRANSFORM
We call Taylor differential transform of an indefinite differentiable function
f
( )
x
, the sequence of functions( )
(
)
( )( )
=
=
:
0
,
1
,
2
!
1
n
x
f
n
x
f
T
n and Taylor transform in a pointx
=
x
0, the sequence of numbers( )
(
) ( )
nx
f
x
c
T
=
0 , where
( )
( )
0!
1
x
f
n
c
n=
n ,n
=
0
,
1
,
. For a natural numberm
, we also will denote( )
(
)
( )( )
=
=
f
x
n
m
n
x
f
T
m n:
0
,
1
,
,
!
1
,T
(
f
( )
x
) (
c
nn
m
)
m
x0
=
:
=
0
,
1
,
,
and( )
(
)
( )
∑
(
)
=
−
=
=
mn
n n
n m x m
x
f
x
P
c
c
x
x
P
0
0
0
0 , the Taylor polynomial of degree m.
If we denote
ab
=
(
a
nb
n)
the usual product of two sequencesa
=
( )
a
n andb
=
( )
b
n , the derivatives of the functionf
( )
x
are given by the formulaf
( )n( ) ( )
x
=
n
!
T
(
f
( )
x
)
and there values in the pointx
=
x
0 by the formulaf
( )n( ) ( )
x
n
T
x(
f
( )
x
)
0
!
0
=
.3. EXAMPLES OF TAYLOR TRANSFORMS
Some examples of Taylor transforms in an arbitrary point and in origin, are:
T
( ) (
a
=
a
,
0
,
0
,
)
, wherea
is a constant function.( )
=
−
−
,
0
,
0
,
1
,
,
,
,
,
,
1m n
n m m
m m
mx
x
n
m
mx
x
x
T
, form
=
1
,
2
,
.Particularly,
T
( ) (
x
=
x
,
1
,
0
,
0
,
)
, andT
( ) (
x
2=
x
2,
2
x
,
1
,
0
,
0
,
)
.( )
−
−
−
=
+
,
1
,
,
1
,
1
,
1
,
1
1
1 4
3
2 n
n
x
x
x
x
x
x
T
, forx
≠
0
,( )
=
,
!
1
,
,
!
3
1
,
2
1
,
1
,
1
n
e
e
T
x x( )
( )
−
−
−
=
−
,
1
,
,
4
1
,
3
1
,
2
1
,
1
,
ln
ln
1
4 3
2 n
n
nx
x
x
x
x
x
x
T
, forx
>
0
.(
)
−
−
=
sin
,
!
4
1
,
cos
!
3
1
,
sin
2
1
,
cos
,
sin
sin
x
x
x
x
x
x
T
,(
)
−
−
=
cos
,
!
4
1
,
sin
!
3
1
,
cos
2
1
,
sin
,
cos
cos
x
x
x
x
x
x
T
.Particularly,
T
0( ) (
a
=
a
,
0
,
0
,
)
,T
0( ) (
x
=
0
,
1
,
0
,
0
,
)
,T
0( )
x
2=
(
0
,
0
,
1
,
0
,
0
,
)
,( )
(
0
,
,
0
,
1
,
0
,
0
,
)
0
m m
x
T
=
,( )
=
,
!
1
,
,
!
3
1
,
2
1
,
1
,
1
0
n
e
T
x ,(
)
−
=
,
!
5
1
,
0
,
!
3
1
,
0
,
1
,
0
sin
0
x
T
,(
)
−
=
,
!
4
1
,
0
,
2
1
,
0
,
1
cos
0
x
T
.Remark: These formulas was given in the author book [1], pg. 84. However, unfortunately, in [1] the two formulas for
x
T
1
andT
( )
ln
x
contain mistakes.4. OPERATIONS WITH SEQUENCES
We consider (see [1] and [6]) the Cauchy product or (truncated or short, linear discrete) convolution
( )
=
∗
=
(
∑
= −)
=
nk k n k
n
a
b
a
b
c
c
0 of the sequences
a
=
( )
a
n andb
=
( )
b
n .The convolution can be computed by the multiplication algorithm
a
0a
1a
2
b
0b
1b
2
---
a
0b
0a
1b
0a
2b
0
a
0b
1a
1b
1
a
0b
2
---© 2013, IJMA. All Rights Reserved 136 For
m
natural numbers one denotes
m m
a
a
a
a
∗=
∗
∗
∗
.If the sequences
a
=
( )
a
n witha
0≠
0
andc
=
( )
c
n are given, then the sequenceb
=
( )
b
n such thatc
=
a
∗
b
can be determined by the deconvolution formula (see [1] and [6])
/
1
(
0,
1:
1
,
2
,
)
0
=
−
=
=
c
c
∑
=a
b
−n
a
a
c
b
n nk k n k .The deconvolution can be computed by the division algorithm
c
0c
1
c
n
a
0a
1
a
n
c
0a
1b
0
a
nb
0
b
n
a
b
a
c
b
a
c
b
0 0 1 1 1 0 0 0
−
=
=
/
c
1−
a
1b
0
c
n−
a
nb
0
c
1−
a
1b
0
a
nb
1
/
On the base of the usual (long) convolution and deconvolution, which are used to compute the product and the quotient (with rest) of two polynomials, introduced in MATLAB by the instructions
conv
anddeconv
, it is possible also to consider the truncated convolution and deconvolution by the instructionstconv
andtdeconv
, inthe following manner:
end
l
temp
res
b
a
conv
temp
a
size
l
b
a
tconv
res
function
));
2
(
:
1
,
1
(
);
,
(
);
(
)
,
(
=
=
=
=
[ ]
end
end
end
i
c
r
a
c
i
if
q
i
res
i
c
b
a
deconv
r
q
c
i
for
c
zeros
res
a
size
c
l
b
a
tdeconv
res
function
);
1
:
2
,
1
(
)
(
;
)
(
));
1
:
1
,
1
(
,
(
:
1
);
,
1
(
);
(
]
,
[
)
,
(
+
−
=
<
=
+
−
=
=
=
=
=
In the examples of partial fractions decomposition given below, we present several algorithms of convolution and deconvolution, the others being omitted.
5. TAYLOR TRANSFORMS OF PRODUCTS AND QUOTIENS OF FUNCTIONS
If a function
f
( )
x
contains products and quotients of other functions, then to determine its Taylor transform we can use the discrete convolution and deconvolution, as can see from the following theorem and its consequence.Theorem 1: If
f
( )
x
andg
( )
x
are indefinite differentiable functions, then( ) ( )
(
f
x
g
x
)
T
(
f
( )
x
)
T
(
g
( )
x
)
T
=
∗
. (1)Proof: Using Leibniz formula for derivatives of a product of functions,
( ) ( )
(
)
(
( ) ( )
)
( ) ( )( )
( )( )
0
1
1
!
!
n n k n k
k
n
T f x g x
f x g x
f
x g
x
k
n
n
− =
=
=
∑
( )
( )( )
( )( )
( )( )
g
( )( )
x
T
(
f
( )
x
)
T
(
g
( )
x
)
n
x
f
n
x
g
k
n
x
f
k
n n
n k
k n
k
=
∗
∗
=
−
=
∑
= −!
1
!
1
!
1
!
1
0
Corollary: If
f
( )
x
andg
( )
x
≠
0
are indefinite differentiable functions, then( ) ( )
(
f
x
g
x
)
T
(
f
( )
x
)
T
(
g
( )
x
)
T
=
/
. (2)Proof: From theorem,
T
(
f
( )
x
)
=
T
(
(
f
( ) ( )
x
g
x
) ( )
g
x
)
=
T
(
f
( ) ( )
x
g
x
)
∗
T
(
g
( )
x
)
, and therefore it results the above formula.Remarks: 1)Obviously, the formulas from aboveTheorem and its Corollary, are also true for Taylor transformation
0
x
T
Concentrated in a pointx
=
x
0.2) If we want to calculate only derivatives up to order
m
, then are used the Taylor transformsT
m(
f
( )
x
)
and( )
(
g
x
)
T
m, sequences of the same finite length
m
+
1
.6. PARTIAL FRACTIONS DEVELOPMENT
We will use the automatic Taylor development, based on discrete convolution and deconvolution, to get a numerical automatic method for partial fractions development of the rational functions. It is based on the following two theorems.
Theorem: 2 If
f
( )
x
is a function of the form( )
(
)
( )
∑
(
)
(
)
( )
∑
−= −
= −
+
−
−
=
+
−
=
10 1
0
1
mk
m k
k m
k
k m k
x
g
a
x
a
x
A
x
g
a
x
A
x
f
, (3)where
m
≠
0
is a natural number,a
andA
k, fork
=
0
,
1
,
,
m
−
1
, are complex numbers andg
( )
x
is a function with derivative ofm
−
1
order ina
, then the coefficientsA
k can be computed by the relation(
) ( )
[
]
km k
k
a x
k
x
a
f
x
c
dx
d
k
A
=
−
=
→
lim
!
1
,
k
=
0
,
1
,
,
m
−
1
, (4)hence the numbers
A
k are the coefficientsc
k of the Taylor development of the function(
x
−
a
) ( )
mf
x
in the pointa
x
=
, namelyA
(
x
a
)
P
am(
(
x
a
) ( )
mf
x
)
m
k
k
k
−
=
−
− −
=
∑
11
0
.
Proof: In conformity with (3), we have
=
n
c
[
(
) ( )
]
(
) (
) ( )
−
+
−
=
−
∑
−= →
→
1
0
lim
!
1
lim
!
1
mk
m k
k n
n
a x m
n n
a
x
dx
A
x
a
x
a
g
x
d
n
x
f
a
x
dx
d
n
.For
n
=
0
resultsc
0=
A
0. For1
≤
n
≤
m
−
1
, it results(
) (
)(
)
∑
−=
−
→
−
−
+
−
+
=
10
1
1
lim
!
1
mk
n k k
a x
n
A
k
k
k
n
x
a
n
c
(
) (
)(
)
( )
1
0
1
lim
1
1
!
n j m
m j
n n j
x a j
n
d
m m
m j
x a
g x
A
j
n
dx
− −
− − → =
−
− +
−
=
∑
.Remark: As is well known, if
f
( )
x
is a rational function, its nominator has degree less than the denominator anda
is a pole of multiplicitym
off
( )
x
, then the function has the form (3), therefore the Theorem 2 work.Corollary: If
f
( )
x
is a rational function with degree of the numerator smaller than the denominator, having the distinct polesx
1,
,
x
p of multiplicitiesm
1,
,
m
p, thenf
( )
x
has the partial fractions development( )
[
(
)
( )
]
(
)
j j
j
j m
j p
j
m j m
x
x
x
x
f
x
x
P
x
f
−
−
=
∑
=
−
1
1 1
, (5)
where
P
xmj[
(
x
x
j)
f
( )
x
]
j
−
−1
denotes, for
j
=
1
,
,
p
, the Taylor polynomial of degreem
j−
1
in the pointx
=
x
j© 2013, IJMA. All Rights Reserved 138
Theorem: 3 If
f
( )
x
is a function of the form( )
(
x
a
) (
x
b
)
g
( )
x
B
x
A
x
f
m
k
k m k
m k
k
+
−
−
+
=
∑
−= − −
1
0
, (6)
where
m
≠
0
is a natural number,a
,b
,A
k andB
k, fork
=
0
,
1
,
,
m
−
1
, are complex numbers andg
( )
x
is a function with derivative ofm
−
1
order ina
, then the coefficientsA
k andB
k can be computed recursively by the relations0 0
0
A
a
B
c
=
+
, (7)(
1 1)(
)
01
A
a
B
a
b
A
c
=
+
−
+
, (8)(
A
a
B
)(
a
b
)
A
a
B
A
(
a
b
)
c
=
+
−
+
1+
1+
1−
2 2
2
2 , (9)
(
)(
)
(
)(
)
(
)
12 2
2 2 3
4 4
3
A
a
B
a
b
2
A
a
B
a
b
A
a
b
A
c
=
+
−
+
+
−
+
−
+
, (10)(
+
)(
−
) (
+
−
)(
+
)(
−
)
+
(
−
)
+
=
−− − −
−
1 1
2 1
1
1
n nn n
n n
n n
n
A
a
B
a
b
n
A
a
B
a
b
A
a
b
c
(
) (
)
(
)
(
)(
)
2
2
0
2
1
2
1
!
nk n
k k
k n k
k k
k
n
A a
B
a b
n k
− −
= ≥
−
− +
+
+
−
−
∑
∑
−(
(
) (
)
)
(
)
− ≥ =
+ −
−
−
−
+
−
−
+
22 1 0
1 2
!
1
2
2
1
nn k k
n k k
a
b
A
k
n
n
k
k
k
,
1
4
≤
n
≤
m
−
, ifm
≥
5
(11)where
(
c
k:
k
=
0
,
1
,
,
m
−
1
)
=
T
am−1(
(
x
−
a
) (
mx
−
b
) ( )
mf
x
)
are the coefficients of the Taylor polynomial of degreem
−
1
of the function(
x
−
a
) (
mx
−
b
) ( )
mf
x
in the pointx
=
a
.Proof: In conformity with (6), we have
=
n
c
1
lim
(
) (
) ( )
!
nm m
n x a
d
x a
x b
f x
n
→dx
−
−
(
) (
) (
) (
) ( )
−
−
+
−
−
=
∑
−= →
1
0
lim
!
1
mk
m m
k k
k n
n
a
x
dx
A
x
a
x
b
x
a
x
b
g
x
d
n
.For
n
=
0
, it results (7). For1
≤ ≤ −
n
m
1
, it results(
) (
)(
)
(
)(
)
1
0
1
lim
1
1
!
n j m n
k j k
n x a n j k k
k j k
n
d
c
k k
k
j
x a
A a
B
x b
j
n
dx
− −
− − → = =
=
−
− +
−
+
−
∑∑
(
) (
)(
)
(
) ( )
0
1
lim
1
1
!
n j n
m j m
n j x a
j
n
d
m m
m
j
x a
x b
g x
j
n
dx
− −
− → =
+
−
− +
−
−
∑
(
)(
)
(
)(
)
0
1
!lim
!
n k n
k n
k k n n
n k x a k
n
d
k
A x
B
x b
A a
B
a b
k
n
dx
−
− → =
=
+
−
=
+
−
∑
∑
(
)
(
)
(
) (
)
(
)
−
= − −
− −
− −
→
−
−
+
−
+
−
+
10
1 1
lim
!
1
nk
k k
n k n
k k
k n
k n
k k a
x
dx
x
b
d
A
k
n
b
x
dx
d
B
x
A
k
n
,from which results the formulas (8)-(11).
Particular cases: From (11), if
m
≥
5
,
it results the formulas(
+
)(
−
)
+
(
+
)(
−
)
+
(
−
)
+
+
+
=
2 23 3
2 3
3 4 4
4
4
A
a
B
a
b
3
A
a
B
a
b
A
a
b
A
a
B
c
2
A a b
2(
−
)
, (12)(
)(
)
5(
)(
)
3(
)
4(
)(
)
5 5 5
4
4 4 43
3 3c
=
A a
+
B
a b
−
+
A a
+
B
a b
−
+
A a b
−
+
A a
+
B
a b
−
(
)
2 2 33
A
a
−
b
+
A
+
Remark: As is well known, if
f
( )
x
is a rational function, its nominator has degree less than the denominator,the last being a polynomial with real coefficients anda
is a complex pole of multiplicitym
off x
( )
,
then the function has the form (6) withb
=
a
the complex conjugate ofa
, therefore the Theorem 3 work. See Example 8.5. Same situation when the denominator has rational numbers as coefficients anda
is an irrational pole of the function of multiplicitym
. Thenb
is the irrational conjugate ofa
and the Theorem 3 also work. See Example 8.6.8. EXAMPLES OF PARTIAL FRACTIONS DEVELOPMENT
8.1.
( )
(
) (
2) (
3)
5 11(
) (
3) (
5)
21
1
1
2
1
2
1
x
x
f x
P
x
x
x
−x
x
x
=
=
+
−
−
−
−
+
(
) (
) (
)
(
) (
) (
)
2 4
1 2 5 3 2 2 3 5
1
1
1
2
1
1
1
2
x
x
P
P
x
x
x
x
x
x
+
+
+
−
−
+
−
−
(
)
(
) (
) (
+
)
+
−
∗
−
−
=
−1 ∗3 ∗5 2 11
1
1
,
3
1
,
2
1
,
1
x
P
(
)
(
) (
) (
)
2
1 2 5 3
1,1, 0
1
2,1, 0
1,1, 0
1
P
x
∗ ∗
∗ −
−
(
)
(
) (
) (
)
(
(
)
) (
)
4 1
2 2 3 5 1 2
2,1, 0, 0, 0
1
1,1
1
1944, 6156
3,1, 0, 0, 0
1,1, 0, 0, 0
2
1
P
P
x
−x
∗ ∗
−
+
=
−
∗
−
+
(
)
(
) (
)
(
(
)
) (
)
2 4
1 3 2 5
1,1, 0
1
2,1, 0, 0, 0
1
4,16, 21
1
9, 33, 46, 30, 9
2
P
P
x
x
+
+
−
−
−
−
(
)
(
)
1 2
1 2 1 3
1
13
1
1
5 59
1
,
,
,
1944
11664
1
4
4 16
1
P
P
x
x
−
=
−
−
+
−
−
+
−
(
)
(
)
(
)
1
2 5 2
2
19 13
593 2689
1
1
13
1
,
,
,
,
1
9
27 9
243 729
2
1944
11664
1
P
x
x
x
+
−
−
= −
−
+
−
+
(
)
(
)
(
)
(
)
(
)
(
)
2 2 3
3
1
5
59
1
2
19
13
593
1
1
2
2
2
4
4
x
16
x
x
1
9
22
x
9
x
243
x
+ − −
− +
−
+
−
− +
−
−
−
−
(
)
(
)
(
)
(
)
(
)
(
)
4
5 2 3 2
2689
1
1
13
1
5
2
729
x
x
2
1944
x
1
11664
x
1
4
x
1
4
x
1
+
−
= −
−
−
−
+
−
+
−
−
(
) (
)
(
)
(
)
(
)
729
(
2
)
2689
2
243
593
2
9
13
2
27
19
2
9
2
1
16
59
2 3
4
5
+
−
−
−
−
+
−
−
−
+
−
+
x
x
x
x
x
x
.For the Taylor polynomial of four degree in
x
=
2
have been performed the convolutions0 1 3 3 1
__ __________
0 1 2 1
0 0 1 2 1
__ __________
9 30 46 33 9 0
0 0 1 1
_ __________ __________
0 0 1 6 9 0
0 1 2 1
3 3 1 ___
__________ __
__________
6 18 18 6 0
0 1 3 0
0 1 1
0 9 27 27 9 0
0 0 3 9 0
0 0 1 1
__________ __________
__ __________ __
__________
0 0 1 6 9 0
0 0 1 3 0
0 0 1 1
0 1 3 3 1 0
0 0 1 3 0
© 2013, IJMA. All Rights Reserved 140 From which it results
(
1
,
1
,
0
,
0
,
0
)
∗3=
(
1
,
3
,
3
,
1
,
0
)
,(
3
,
1
,
0
,
0
,
0
)
∗2=
(
9
,
6
,
1
,
0
,
0
)
and
(
3
,
1
,
0
,
0
,
0
) (
∗2∗
1
,
1
,
0
,
0
,
0
)
∗3=
(
9
,
33
,
46
,
30
,
9
)
and the deconvolution
/ ______ __________
81 2689
81 2689 /
_____ __________
81 6523 27
593
3 142 27
593 /
________ __________
9 598 3
143 13
9 172 27
694 13
/
______ __________ __________
9 190 27
874 9
209 3
19
2 3
20 9
92 3
19 /
_________ __________
__________
729 2689 243 593 9
13 27 19 9 2 2
3 20 9
92 3
22 2
9 30 46 33
9 0
0 0
1 2
− −
− −
− −
− −
− −
− −
− −
Hence
(
)
(
) (
)
−
−
=
∗
∗∗
729
2689
,
243
593
,
9
13
,
27
19
,
9
2
0
,
0
,
0
,
1
,
1
0
,
0
,
0
,
1
,
3
0
,
0
,
0
,
1
,
2
3
2 .
8.2.
( )
(
)
(
)
∑
=(
)
−∑
=(
+
)
−+
+
−
=
+
−
=
20
5
0
6 2 2
6 2 3
1
2
~
1
2
15625
j k
k k k j
j
x
B
x
A
x
c
x
x
x
f
. We have(
)
(
)
=
(
(
)
)
=
(
(
)
)
=
+
=
∗168750
,
75000
,
15625
0
,
0
,
15625
1
,
4
,
5
0
,
0
,
1
15625
1
15625
~
,
~
,
~
6 6
2 2 2 2 1 0
x
T
c
c
c
(
1
,
−
4
.
8
,
12
.
24
)
,(
)
(
)
(
(
)
)
(
(
)
)
5
0 5 3 3
1 5 6 2 51, 0, 0, 0, 0, 0
1 5 6 2 5
, 0, 0, 0, 0, 0
15625
,
,
2 11 , 9 12 , 6 3 ,1, 0, 0
2
2
,1, 0, 0, 0, 0
i
c
c
T
i
i
i
x
i
∗
=
=
=
−
− +
− +
−
− +
=
(
−
250
−
1375
i
,
525
−
1800
i
,
1140
−
1230
i
,
1170
−
440
i
,
834
+
87
i
,
442
.
68
+
282
.
24
i
)
.(
A
i
B
)
i
i
c
1=
2
1+
1−
1375
=
525
−
1800
,A
1=
−
950
,B
1=
−
900
,(
A
i
B
)
i
i
i
c
2=
−
4
2+
2−
1900
−
950
−
900
=
1140
−
1230
,A
2=
−
405
,B
2=
−
510
,(
A
i
B
)
i
(
i
)
i
i
c
3=
−
8
3+
3+
1620
+
4
−
405
−
510
−
950
=
1170
−
440
,A
3=
−
140
,3
200,
B
= −
c
4=
16
(
A
4i
+
B
4)
+
1132
i
+
12
(
140
i
+
200
)
−
405
i
−
510
−
1620
i
=
=
834
+
87
i
,43
4
=
−
A
,66
4
=
−
B
,c
5=
32
i A i
(
5+
B
5)
−
688 32
−
i
(
−
43
i
−
66
)
+
2112
i
+
840 1200
−
i
+
1680 405
−
=
442.68 282, 24
+
i
,24
.
12
5
=
−
A
,B
5=
−
19
.
68
.Therefore,
( )
(
) (
3)
2(
2)
6(
2)
51
4.8
12.24
1375
250
950
900
2
2
2
1
1
x
x
f x
x
x
x
x
x
+
+
=
−
+
−
−
−
−
−
+
+
(
)
(
)
(
)
1
68
.
19
24
.
12
1
66
43
1
200
140
1
510
405
2 2
2 3
2 4
2
+
+
−
+
+
−
+
+
−
+
+
−
x
x
x
x
x
x
x
x
.
8.3.
( )
(
2) (
5 2)
2 2 3 5 63
1
4
3
+
+
+
+
+
+
+
=
x
x
x
x
x
x
x
x
f
.Solution: 1 (complex) We will obtain
( )
(
)
(
)
∑
∑
= − =
+
−+
+
+
+
=
10
4
0
4 2 1
2
1
3
~
~
j k
k k k j
j j
x
B
x
A
x
B
x
A
x
f
. We have(
)
(
)
(
(
)
)
6 5 3 2
1
0 1 3 2 5 5
32 7
3, 37 60
3
3
4
,
1
2, 2
3
i
i
i
x
x
x
x
x
c c
T
x
i
∗
+
+
+
+ +
− +
+
=
=
+
−
(
(
)
)
(
32
7
3
,
68
100
3
)
32
1
3
5
,
1
32
3
60
37
,
3
7
32
i
i
i
i
i
−
−
+
−
−
=
−
−
+
+
−
=
,Hence
3
32
7
1
~
3
~
~
0 0
0
A
i
B
i
c
=
+
=
−
,32
7
~
0
=
−
A
,B
~
0=
1
,(
)
3
32
100
32
68
32
7
~
3
~
3
2
~
1 1
1
i
A
i
B
i
c
=
+
−
=
+
,64
25
1
=
−
A
,16
25
1
=
B
.(
)
(
)
=
+
+
+
+
+
+
=
22 2 3 5 6 4 4 0
3
4
3
,
,
x
x
x
x
x
x
T
c
c
i(
)
(
)
2, 3 12 ,18 7 , 9 20 , 15 5
2, 2 ,1, 0, 0
i
i
i
i
i
i
∗+
−
− −
− +
