Analysis of Complex Continued Fractions of The Solution of The Equation kx2-nx+n=0 Where n is An Odd Integer And k=[n/2]

IJSRR, 7(4) Oct. – Dec., 2018 Page 829

Research article

Available online www.ijsrr.org

ISSN: 2279–0543

International Journal of Scientific Research and Reviews

Analysis of Complex Continued Fractions of The Solution of The

Equation

kx

2



nx



n



0

Where

n

Is An Odd Integer And

_

    



2

n

k

Krithika S.

*1

and Gnanam A.

2

*1

Assistant professor, Department of Mathematics, Seethalakshmi Ramaswami College,Trichy-2.

2

Assistant professor, Department of Mathematics, Government Arts College, Trichy-22. Email: [email protected],[email protected]

Mobile no. 9443019706, 9994563563

ABSTRACT

In this paper we identify the complex continued fractions of the solution of the equation

0 2_nxn

kx where n is an odd integer and

    

2

n

k . The analysis of the complex continued

fractions of the above equations when n3,n3 &kn1 are found and their comparisons are made.

KEYWORDS:

Continued fraction, Finite complex continued fraction, Infinite complex continued

fraction, Periodic complex continued fraction, Gaussian integer, Euclidean algorithm, Quadratic irrational.

SUBJECT CLASSIFICATION:

MSC 11A05, 11A55, 11J70, 11Y65, 30B70, 40A15.

S. KRITHIKA

Assistant Professor, Department of Mathematics,

Seethalakshmi Ramaswami College,

Thiruchirappalli, 620002, Tamil Nadu, India.

IJSRR, 7(4) Oct. – Dec., 2018 Page 830

I.

BASIC DEFINITIONS

1, 2

Let C



zxiy/x,yR



be the set of all complex numbers. Where xand yare called the

real and imaginary parts of z. These are denoted by Re(z)and Im(z)respectively. The conjugate

of the complex number zis denoted by zxiyand the modulus of zis denoted by z x2y2.

LetZ(i)



xiy/x,yZ



be the set of all Gaussian integers. Let

   

 

 

 / , ( ) 0

)

( p q Z i andq q

p i Q

be the set of all complex rational numbers. So that Z(i)Q(i)C.Let zCthen zis rational if )

(i Q

z , zis irrational if zC\Q(i)and zis quadratic irrational if zis irrational and there exist

) ( ,

,BC Z i

A  such that Az2 BzC0and A0.

Greatest common divisor of any two complex numbers:

3, 4, 5

Let z₁,z₂Z(i),z₂ 0 . Then the greatest common divisor of z₁and z₂ is denoted by

) , (

gcd z1 z2 and it is defined to be a Gaussian integer gsuch that g|z1and g|z2and there is no

Gaussian integer hsuch that h gthat divides both z₁and z₂. In this case we writegcd(z1,z2)g

If x,yR then the floor function of xand y are denoted by

 

x and

 

y respectively. So that

 

1

 x

x _and y

 

y 1. Let zC. Then z is a quadratic irrational if and only is

s r q p

z  for

some p,q,r,sZ(i),q0,s0and ris not a perfect square.

II. INTRODUCTION

An expression of the form



      

n n

a e e a

e a

e a

e a

3 3

2 2

1 1

0 0

Where a₀,a₁,a₂,a₃, are in Z(i) and e_i's are units of complex numbers. Therefore



1,1, ,



, 1,2,3,...

 i i k

e_k is known as a complex continued fraction.6,7,8

The complex continued fraction is commonly expressed as 

   

3 3

2 2

1 1 0

a e a

e a

e

a .

The quantities , , , , 3 3

2 2

1 1 0

a e a e a e

IJSRR, 7(4) Oct. – Dec., 2018 Page 831

In a finite complex continued fraction the number of elements are finite. Where as an infinite continued fraction have infinite number of quantities.

Therefore n n a e a e a e a e a      3 3 2 2 1 1

0 is known as finite complex continued fraction and an

expression  

        1 1 3 3 2 2 1 1 0 n n n n a e a e a e a e a e

a is known as an infinite complex continued fraction. In

a finite or infinite complex continued fraction a₀,a₁,a₂,a₃,are called the partial quotients where as e₁,e₂,e₃,are known as the partial numerators.

The length of a finite complex continued fraction

n n a e a e a e a e a      3 3 2 2 1 1

0 is n1 and the

length of a infinite continued fraction  

        1 1 3 3 2 2 1 1 0 n n n n a e a e a e a e a e

a is .The value of the

finite complex continued fraction is denoted as _

     n n a e a e a e a e

a , , , ,

3 3 2 2 1 1 0 .

2.1 Convergent of complex continued fraction:

6,7,8

The successive convergent of the complex continued fractions are

 

a₀ , _

     1 1 0, a e

a , _

     2 2 1 1 0, ,

a e a e a

and so on and they are denoted by

0 0 0

q p c  ,

1 1 1

q p c  ,

2 2 2

q p

c  ,… In general the nth convergent is

denoted by _

       n n n n n a e a e a e a e a q p

c , , , ,

3 3 2 2 1 1

0 . Where pi'sand qi'sare called the numerators and

denominators of the complex continued fraction.

2.2 Properties of complex continued fraction :

6,7,8

Let _

       , , , , , 3 3 2 2 1 1 0 n n a e a e a e a e

a be an infinite complex continued fraction. We inductively

define two infinite sequences p_kand q_kwhere k1by

p0a0 p11 q01 q10

pkak pk1 ek pk2,k 0 qkakqk1 ekqk2,k 0

III. ALGORITHMS AND THEOREMS

3.1 Algorithm of complex continued fraction:

9

IJSRR, 7(4) Oct. – Dec., 2018 Page 832

Let aRe(x)



Re(x)



and bIm(x)



Im(x)



so that a0 and b0 .Now

  

x₀  Re(x)

 

iIm(x)



, a floor function. Where

    

 



 



  

b a with b

ifa i

b a with b

a if

b a if

1 1 1

0 0



Set a₀

 

x₀ . Then

_{ }

0 0 1

1

x x x



 and

  

x1  Re(x1)

 

i Im(x1)



.

Again set a₁

 

x₁ . Then

 

1 1 2

1

x x x



 and

  

x2  Re(x2)

 

iIm(x2)



_,a2

 

x2 .Then

 

2 2 3

1

x x x



 and

  

x₃  Re(x₃)

 

iIm(x₃)



_, a₃

 

x₃ ….a_k1

 

x_k1 .

Then

 

1 1

1

  



k k k

x x

x and

  

x_k  Re(x_k)

 

iIm(x_k)



.and a_k

 

x_k .

The algorithm terminates if the complex continued fraction is finite otherwise it is no terminating.

3.2 Euclid’s algorithm for complex continued fraction:

[4]

Let u₀,u₁Z(i). Then we try to find the gcd(u₀,u₁)by using Euclid’s algorithm.

Divide u₀ by u₁if N(u₀)N(u₁), where Nis the norm of the complex number.

Therefore

1 2 0 1 0

u u a u

u _{ }

where a₀ Z(i)is the quotient and u2Z(i)is the remainder.

Again

2 3 1 2 1

u u a u

u _{ }

where a1Z(i)is the quotient and u3Z(i)is the remainder.

And

3 4 2 3 2

u u a u u



 where a2Z(i)is the quotient and u4Z(i)is the remainder….

n n n n n

u u a u

u ₁

1

1 



 _{  where}

) (

1 Z i

a_n  is the quotient and u_n1Z(i)is the remainder.

And _n n

n

a u

u



1

where a_n Z(i)is the quotient and 0 is the remainder.

Here C j n

u u

j j

... , 2 , 1 , 0 , 1

 



Where ai'sZ(i),i0,1,2,..n are the quotients and

1 ,.. 3 , 2 ), (

'sZ i j  n

u_j are the remainders. These quotients are treated as the partial quotients of

the 1 0 u u

, the ratio of the complex numbers.

Therefore the complex continued fraction of



a a a_n



u

u

,..., , 1 0 1

IJSRR, 7(4) Oct. – Dec., 2018 Page 833

Note:

1. _

             1 0 1 0

0 Re Im

u u i u u

a where  is a floor function and

              b a with b ifa i b a with b a if b a if 1 1 1 0 0  ,               1 0 1 0 Re u u u u

a and _

             1 0 1 0 Im u u u u b .

The successive quotients are found by the same procedure.

2. Remainders are found by using the relationsu_j₁u_j₁a_j₁u_j,j1,2,...n.

Example 3.2.1 :

gcd(313i, 43i) _Since N(313i) 17825N(43i) , divide (313i)by )

3 4 (  i .

Therefore i i i i i 3 4 1 ) 2 2 ( 3 4 13 3        i i i i       1 1 ) 3 ( 1 3 4

(1 ).

1 1

i i_{ }

 

Hence gcd(313i, 43i)1.

Example 3.2.2 :

gcd(761i,2636i) Since N(761i) 37701972N(2636i) , divide )

61 7

(  i by (2636i).

Therefore i i i i i 36 26 17 ) 1 ( 36 26 61 7        i i i i i         17 3 7 ) 2 1 ( 17 36 26

(2 )

3 7 17

i i

i_{ }

  

Hence gcd(761i,2636i) 73i.

Definition:[3]

The function F_f :CZ(i)is said to be a floor function if F(z)Z 1for every ZC.

The function F_s:C 



1,1,i,i



is said to be a sign function.

3.3 Complex continued fraction algorithm (J.O. Shalit,1979 ):

10

If Ff :CZ(i)is a floor function and Fs:C



1,1,i,i



is a sign function together with

the following sequence is known as a complex continued fraction algorithm. (i) Z₀Z

(ii) a_n F_f(Z_n)

(iii) e_n₁F_s(Z_n)

(iv) n n n n a Z e Z   

IJSRR, 7(4) Oct. – Dec., 2018 Page 834

3.4 Theorem

If _

              n n j j b f b f b f b f b a e a e a e a e

a , , , , , , , ,

3 3 2 2 1 1 0 3 3 2 2 1 1

0 where these finite complex continued fractions

are simple if 1 j j

a e

and 1 n n

b f

then

j



n

and a_ib_i,e_i f_ifor i0,1,2,...n.

Proof:

Let i i i n n i i i i i i i i q p y b f b f b f b f b

y _ 

            , , , , , 3 3 2 2 1 1 _               n n i i i i i i i b f b f b f b f b  , , , 3 3 2 2 1 1 1 1     i i i y f b .

Therefore we have y_i b_i and yi 1for i0,1,2,...n1and yn bn1.

Also b_i

 

y_i for all i,0in.

Since _

              n n j j b f b f b f b f b a e a e a e a e

a , , , , , , , ,

3 3 2 2 1 1 0 3 3 2 2 1 1

0 , we write y0 x0 .

For that take  ₁ 1

 

 i

i i i x

q p

x for all i 0. So that a_i 

 

x_i 0i j.

Therefore b₀

   

y₀  x₀ a₀ .

Also i j

a e a e a e a e a x j j i i i i i i i

i  

               0 , , , , , 3 3 2 2 1 1 _                 j j i i i i i i i a e a e a e a e a  , , , 3 3 2 2 1 1 1 1 1 1 1 1       _{     }   i i i i i i i i i i i y f b y a x x e x e a .

Therefore , 0,1,2,... 1 1

1

1

1   

    n i y f x e i i i

i _{. So that}

1

1 

  i

i f

e and x_i1 y_i1 and

   

x_i1  y_i1 a_i1 b_i1

for i 0,1,2,...n1.

Next to prove that

j



n

.

Case (i):

j



n

.

IJSRR, 7(4) Oct. – Dec., 2018 Page 835

Using Euclid’s algorithm we get _{j j} j

j

a x q p

   

. Which is a contradiction to y_j b_j.

Case (ii): If

j



n

then we get the same contradiction.

So that

j



n

.

3.5 Theorem

If zQ(i) then the continued fraction of z is finite.

Proof:

If

n n n

q p x

   then

1 1 1

1 1

1

  

 

 _

    

 

     

n n

n n n

n n

n n n n

n n

n n

q p q a p

q f

a q p f a

x e

x .

Therefore

1 1

1 1 1

 

 

 _

    

n n n

n n n

q q f q p

x for every nN.

As x_n1 1

and fn11, qn1 qn for all nN.

As q_nZ

 

i there exist NNsuch that q_N1 0.

If 1 0 0 _  0  0

       

_{ N N N}

N N N

N N

N a x a

q p a

q p

q .

Therefore Euclidean algorithm terminates after N transformations. This completes the proof of the theorem.

3.6 Theorem

If zC\Q(i) then the continued fraction of z is infinite.

Proof:

Every rational complex number represents a finite complex continued fraction. Also in a finite continued fraction x_n a_n0where a_n 

 

x_n .

If x₀ is irrational then x₀a₀ 0or x₀

 

x₀ 0 orx₀F_l(x₀)0.

Suppose x_k is irrational then x_kF_l(x_k)0.

Consequently for every nN, x_na_n0. So that continued fraction is infinite.

This completes the proof.

In this paper we try to find the solution of the quadratic equationkx2 nx n0 where

    

2

n

k and nis odd. Since n2 4kn0, the roots of the equation are complex. Also complex roots

IJSRR, 7(4) Oct. – Dec., 2018 Page 836

If ₁and ₂are the complex roots of the equation kx2 nxn0 then their continued fractions are



c₀,c₁,c₂,...



and



d₀,d₁,d₂,...



respectively. Where c_i'sand d_i'sZ(i).

Here the continued fractions of ₁and ₂ are calculated by using either using Euclidean algorithm or

complex continued fraction algorithm and comparison of their continued fractions are made.

IV. ILLUSTRATIONS

4.1 Illustration

Find the complex continued fraction of the roots of the quadratic equation 4x2 9x90.

Solution

Roots of the above quadratic equation are ₁1.12500.9922iand ₂1.12500.9922i

Case (i): Continued fraction of the root11.12500.9922i

Take x₀1.12500.9922iThen a1.1250(1)0.1250and b0.9922(0)0.9922

Therefore

 

x₀ 10i 1ia₀1i.

Now x i

i i

x 7.9690 0.4973

)] 1 ( ) 9922 . 0 1250 . 1 [(

1

1

1 _{  }   

Round off the imaginary part to 0.5 , x17.96900.5i

Then a7.9690(7) 0.9690and b0.5(0)0.5

Therefore

 

x1 7018a18.

Again x i

i

x 0.1235 1.9923

)] 8 ( ) 5 . 0 9690 . 7 [(

1

2

2   

  

Then a0.1235(1) 0.8765and b1.9923(2)0.0077

Therefore

 

x2 12i012ia212i.

i x

i i

x 1.1408 0.0100

3 )] 2 1 ( ) 9923 . 1 1235 . 0 [(

1

3 _{    }   

Then a1.1408(1)0.1408and b0.0100(1)0.9900

Therefore

 

x₃ 1ii1a₃1.

i x

i

x 7.0666 0.5

)] 1 ( ) 0100 . 0 1408 . 1 [(

1

4

4   

 



Then a7.0666(7)0.0666 _and b0.5(0)0.5

IJSRR, 7(4) Oct. – Dec., 2018 Page 837

i x

i

x 0.2618 1.9651

)] 7 ( ) 5 . 0 0666 . 7 [(

1

5  5 

  

Then a0.2618(0)0.2618 _and b1.9651(2)0.0349

Therefore

 

x₅ 02i0 2ia₅2i.

i x

i i

x 3.7530 0.50

)] 2 ( ) 9651 . 1 2618 . 0 [(

1

6

6   

  



Then a3.7530(3) 0.7530and b0.5(1)0.5 Therefore

 

x₆ 3i14ia₆ 4i.

i x

i i

x 0.7942 1.6077

)] 4 ( ) 50 . 0 7530 . 3 [(

1

7

7   

  



Then a0.7942(1) 0.2058and b1.6077(2)0.3923

Therefore

 

x₇ 12i012ia₇12i.

Case (ii): Continued fraction of the root11.12500.9922i

Take x₀1.12500.9922i

Then a1.1250(1) 0.1250 _and b0.9922(1)0.0078

Therefore

 

x₀ 1i 01ia₀1i.

Now x i

i i

x 7.9690 0.4973

)] 1 ( ) 9922 . 0 1250 . 1 [(

1

1

1   

  



Round off the imaginary part to 0.5 , i

x17.96900.5

Then a7.9690(7) 0.9690 _and b0.5(1)0.5

Therefore

 

x₁ 7i18ia₁8i.

Again x i

i i

x 0.1235 1.9923

)] 8 ( ) 5 . 0 9690 . 7 [(

1

2

2   

   

Then a0.1235(1) 0.8765and b1.9923(2)0.0077

Therefore

 

x₂ 12i012ia₂12i.

i x

i i

x 1.1408 0.0100

)] 2 1 ( ) 9923 . 1 1235 . 0 [(

1

3

3   

   

 

IJSRR, 7(4) Oct. – Dec., 2018 Page 838

i x

i

x 7.0666 0.5

)] 1 ( ) 0100 . 0 1408 . 1 [(

1

4

4   

 



Then a7.0666(7) 0.0666and b0.5(0)0.5

Therefore

 

x₄ 7007a₄7.

i x

i

x 0.2618 1.9651

)] 7 ( ) 5 . 0 0666 . 7 [(

1

5

5   

  

Then a0.2618(0)0.2618 _and b1.9651(2) 0.0349

Therefore

 

x₅ 02i02ia₅2i.

i x

i i

x 3.7530 0.50

)] 2 ( ) 9651 . 1 2618 . 0 [(

1

6

6   

  



Then a3.7530(3) 0.7530 _and b0.5(1)0.5 Therefore

 

x₆ 3i14ia₆ 4i.

i x

i i

x 0.7942 1.6077

)] 4 ( ) 50 . 0 7530 . 3 [(

1

7

7   

  



Then a0.7942(1) 0.2058and b1.6077(2)0.3923

Therefore

 

x₇ 12i012ia₇12i.

From case(i) and case (ii) we get the complex continued fraction of the given equation are



1 ,8, 1 2, 1, 7, 2,4 , 1 2,



1 i   i  i i   i

 and



1 ,8 , 1 2, 1, 7, 2,4 , 1 2,



2 i i   i  i i   i



4.2 Illustration

Find the complex continued fraction of the roots of the quadratic equation

0 23 23

11x2 x   .

Solution

Roots of the above quadratic equation are ₁1.04550.9990iand ₂1.04550.9990i Similar to case (i) and case (ii) of previous illustration we find that the complex continued fraction of the given equation are 1



1i,22, 12i, 1, 7,12i,2i,,



and



1 ,22 , 1 2, 1, 7, 1 2 , 2,



2 i i   i   i  i



4.3 Illustration

Find the complex continued fraction of the roots of the quadratic equation x2 3x 30.

Solution

IJSRR, 7(4) Oct. – Dec., 2018 Page 839

The complex continued fraction of the roots of the given equation are



i   i i



1 ,2, 1 2, 2

1

 and ₂



1i, 2i, 12i



4.4 Illustration

Find the complex continued fraction of the roots of the quadratic equation x2 x 10.

Roots of the above quadratic equation are ₁0.500.8660iand ₂0.500.8660i

Solution

The complex continued fraction of the roots of the given equation are₁



i,2,12i,2i



and ₂



i, 2i, 12i



V. CONCLUSION

From the above illustrations we conclude that the complex continued fraction the quadratic

equation of the form kx2nxn0 where

    

2

n

k and n is odd and 3 are



₀, ₁ 2 , ₂, ₃, ₄,...



1 c c  k c c c

 and



₂



c₀,c₁ 2ki,c₂,c₃,c₄,...



_.If n3 then



i   i i



1 ,2, 1 2, 2

1

 and ₂



1i, 2i, 12i



_.If n1 and k n then



i   i i



 ,2, 1 2,2

1

 and ₂



i, 2i, 12i



. Hence If n3the complex continued fraction

of the roots of the above quadratic equation are infinite where as if n3and when nk1the complex continued fractions are periodic.

REFERENCES

1. Arumugam S., Thangapandi Isaac A., Somasundaram A., “Complex Analysis”, 9th Reprint, SciTech Publications (India) Pvt. Ltd. , December 2007.

2. Ponnusamy S., “ Foundations of Complex Analysis”, Second edition, Narosa Publishing House, New Delhi.

3. Andrews G.E., “Number Theory”, Hindustan Publishing Corporation, Delhi, India, 1989.

4. Silverman J.H., “A friendly introduction to Number Theory”, fourth edition.

5. Neville Robbins, “Beginning Number Theory”, Second edition, Narosa Publishing House, New Delhi.

IJSRR, 7(4) Oct. – Dec., 2018 Page 840

7. Jonathan Browein, Alfvander Poorten, Jeffrey Shallit, Wadim Zudilin, “Never-ending Fractions An Introduction to continued fractions”, Cambridge University Press.

8. Olds C.D., “Continued Fractions”, Random House: New York, 1963.

9. Bosma W., Bastian cijsouw, “Complex Continued Fraction Algorithms”, Thesis, Redbud University, November 2015.