Solving A Square Root Problem Using the Pierce Continued Fraction

Page 1

Solving A Square Root Problem Using the Pierce Continued Fraction

1Department of Mechanical Engineering, Open University of Sri Lanka, Nawala, Nugegoda, Sri Lanka.

2Department of Mathematics, Faculty of Science, University of Peradeniya, Sri Lanka.

Abstract In this paper, we have developed a new continued fractionusingthe pierceexpansionwhich is called the pierce continued fraction. There are several applications of continued fractions and one important application is the approximation to the square root problem. In our work, we have introduced new convergent sequences to solve this problem using pierce continued fraction. The convergent rate of these two continued fractions is different from each other. Furthermore, an algorithm was developed using matlab programmed to find the pierce continued fraction. As a future work we hope to construct a new approach to solve the pelle’s equation.

Indexed Terms Approximation to square root problem, continued fraction, convergent, pierce expansion, pierce continued fraction.

I. INTRODUCTION

Continued fraction expansions of numbers are contained in the Euclidian algorithm [3] and are important in giving rational approximation of real numbers. They have been used since the beginning of science in many ancient civilizations including Greek, Indian, and Chinese [3]. The famous golden ration can be represented in an attractive manner using the corresponding continued fractions.

equations [2], and solving Pelle’s equation [2]. One of themostimportantaspectsofthesecontinuedfractions is their convergent which was introduced by the Dutch mathematician /astronomer Christiaan Huygens (1629 1695).

Definition 1.1 Continued Fraction [1]

x be a any real number,

There are several applications based on continued fractions in Linear Algebra [1], Group Theory [4], Convex Analysis [2], Design Theory [2], Approximation [2], solving linear Diophantine

where

are natural numbers with ��1,��2

positive is called a finite continued fraction.

Definition 1.2 (Convergent of a Continued Fraction) [2]

The continued fraction [��

,.....��

where k is a non negative integer is called the nth convergent of the simple continued fraction [��

,��

and is denoted by ��

When solving the approximation square root problem, the convergent rate of the continued fraction is relatively slower. Therefore, a new continued fraction has been introduced using the Pierce expansion.

Definition 1.3 (Pierce Expansion)

Any real number

can be written as

natural numbers

© OCT 2022 | IRE Journals | Volume 6 Issue 4 | ISSN: 2456-8880 IRE 1703837 ICONIC RESEARCH AND ENGINEERING JOURNALS 16
�� =1+ 1 1+ 1 1+ 1 1+ 1 1+1 1
Let
�� =��0 + 1 ��1 + 1 ��2 + 1 ��3 + 1 ���� 1 + 1 ����
��1,��2, ����
, ����
0,��1
��]
0
1, ����]
��
x,0<�� ≤1,
�� = 1 ��1 1 ��1��2 1 ��1��2��3 ...+ ( 1)��+1 ��1��2��3 ���� , where ��1,��2,��3 are

Only disadvantage of this representation is x should be

than or equal to ‘.’ respectively.) Using this we can compute the corresponding partial quotients.

Using the above expansion, a new continued fraction, called the pierce continued fraction, can be obtained. pierce continued fraction are defined for real numbers

Definition 1.4 Pierce Continued Fraction

Any real number

are positive real numbers is called a finite pierce continued fraction. If the positive real numbers are integers then above finite pierce continued fraction is called a simple finite pierce continued fraction. The above simple finite pierce continued fraction is denoted by {��

Further, a MATLAB programmed has been developed to find the pierce continued fraction using above mapping as below:

For example, pierce continued fraction of 1⁄�� is, {2,3,4,5,6,7 }

Theorem 1.1

i. Every rational number x, 0 < x < 1 less than 1 can be written as a simple pierce continued fraction.

are called partial quotients of a simple finite pierce continued fraction.

,��2

When the number is greater than 1, (say) n, we take

ii. Every simple finite pierce continued fraction represent a rational number.

Proof

i. Clearly by the definition of pierce expansion.

By using the Principle of Mathematical Induction

new mapping has been proposed to construct the pierce continued fraction,

from the interval (0,1] to the set of positive

a mapping

numbers, given by

greatest

than or equal to

© OCT 2022 | IRE Journals | Volume 6 Issue 4 | ISSN: 2456-8880 IRE 1703837 ICONIC RESEARCH AND ENGINEERING JOURNALS 17
0<�� ≤1
��; 0<�� ≤1
x,0<�� ≤1, �� = 1 1 1 1 1 1 ��3 ��4 ��3 ��2 ��1 ={��1,��2,......����} where ��1,��2, ����
1,��2, ����} ��1
, ����
�� as �� =�� ⌈��⌉ Example 1 3 8 = 1 1 1 2 1 2 ={2,1,2} A
Define
real
ℎ(��)=1 ��(⌈1 ��⌉ 1) Set���� =⌊ 1 ℎ(�� 1)(��)⌋, where ℎ(��)(��)= ℎ (ℎ(�� 1)(��)) and ℎ(0)(��) =�� (Here ⌈ ⌉and⌊ ⌋ denote
integer less
‘ ’, least integer greater
ii.
Consider [��0,��1,.....����] When n=1, [��0,��1]= 1 1 ��1 ��0 = ��1 1 ��1 ��0 = ��1 1 ��0��1 ∈ℚ Assume that [��0,��1, ����] is a rational number �� ��,�� ≠0,�� ≤�� for ����ℤ+ Consider, [��0,��1, ����,����+1]= 1 1 1 1 ����+1 ���� ��2 ��1 ��0 k terms

By the principle of mathematical induction, the results is true for ����ℤ+

Convergent property can be defined as follows:

Definition 1.3

The pierce continued fraction {��0,��1, ����} where �� is a non negative integer is called the nth convergent of the simple finite pierce continued fraction {��0,��1, ����}and is denoted by ����

Two sequences {[��

},{����} have been defined to obtain a better for the approximation to square root problem

Theorem 1.2 Let

be finite or infinite sequence of real numbers. Define two sequences {[��

we know that

} respectively as follows,

Then

Proof

},{��

When we prove this theorem, general formula for {��

},{��

}have been found and then we can consider the convergent part.

Now consider {��

} sequence,

for

Now consider,

© OCT 2022 | IRE Journals | Volume 6 Issue 4 | ISSN: 2456-8880 IRE 1703837 ICONIC RESEARCH AND ENGINEERING JOURNALS 18 = 1 1 . 1 1 ����+1 ���� . ��2 ��1 ��0 = 1 �� �� ��0 = �� �� ����0 ��ℚ ��0,�� ≠0
��
��0,��1, ����
��
��
��0 =1 ��1 =��1 1 . ���� =�������� 1 +( 1)��
���� = ���� ����
��
��
��
��0 =1 ��1 =��1 1 ���� =�������� 1 +( 1)�� Now clearly ���� =��1��2 ���� +( 1)��2 ���� + ( 1)2��3.....����+.....+( 1)�� 1���� +( 1)�� We can prove this by P.M.I �� =1 ��1 =��1 +( 1)=��1 1 Assume that it is true for �� =��; Therefore ���� =��1��2......���� +( 1)��2......���� + ( 1)2��3 ����+ +( 1)�� 1���� +( 1)�� ……(1) Then by (1)×����+1, we get ��������+1 =��1��2 ��������+1 +( 1)��2 ��������+1 +( 1)2��3.....��������+1+.....+( 1)�� 1��������+1 +( 1)������+1 But
����+1 =��������+1 +( 1)��+1 ����+1 ( 1)��+1 =����+1���� ����+1 =��1��2 ��������+1 +( 1)��2 ��������+1 +( 1)2��3.....��������+1+.....+( 1)�� 1��������+1 + ( 1)������+1 +( 1)��+1 Therefore, the result is true for �� ∈ℤ+ . Therefore ���� =��1��2 ���� +( 1)��2 ���� + ( 1)2��3.....����+.....+( 1)�� 1���� +( 1)�� …..(2) Similarly,
{����} sequence, ���� =��0��1.....���� ………………………………(3)
��0 =��0 ��1 =��1��0 ���� =���� 1���� for �� ∈��+

Using above algorithm, we can construct simple finite pierce continued fraction (S.F.P.C.F) for (√5 2).(√5 2)=[4,17,19,57,77, ] Then we can construct two sequences that we have mentioned above.

(By definition).

Using above results following identities can be easily formed, Theorem 1.3

convergent of the pierce continued fraction {��

be

where n is positive integer. Then

Proof.

Proofs can be done by Principle of Mathematical Induction Corollary 1

=22352941176

=22360679775

Now to get final answers we have to add 2 for each ���� value. Our new ���� sequence is, Exact value of the √5 is, √5=2.2360679977

By definition 1.3, nth convergent is denoted by Cn, the actual value x of the infinite pierce continued fraction will lies between two strings, one string made up of odd convergent, and other made up of even convergent. Thus,

Proofs can be prepared using Theorem 1.4.

Example 2

Approximation value of √5 can be determined using pierce continued fraction as follows;

have re written √5, this way

According to this we can obtain following theorems, Theorem 1.4

convergent. Then,

where

is

© OCT 2022 | IRE Journals | Volume 6 Issue 4 | ISSN: 2456-8880 IRE 1703837 ICONIC RESEARCH AND ENGINEERING JOURNALS 19 (2) (3), ���� ���� = 1 ��1 1 ��1��2 + 1 ��1��2��3 ...+ ( 1)��+1 ��1��2��3 ���� . = 1 1 1 1 1 1 ��3 ��4 ��3 ��2 ��1 ={��1,��2, ����}=����
Let ���� = ���� ����
nth
0,��1,.....����....}
i. �������� 1 ���� 1���� =( 1)������ 1 ii. �������� 2 ���� 2���� =( 1)�� 1(���� 1)���� 2;�� ≥2
i. ���� ���� �� = ( ��)�� ���� ii. ���� ���� �� = ( ��)�� ��(���� ��) ���� .
We
�� =√5=2+(√5 2)
��0 =2.25 ��1
��2 =2.2360681115 ��3
��0 ≥��2 ≥��4 ≥�� ≥ ≥��5 ≥��3 ≥��1
Let x =[��0,��1,��2, ���� ],
����
nth
��1 �� 0 1 2 3 ��1 �� 4 17 19 77 ���� 1 16 305 1761984 ���� 4 68 1292 7463884 ���� 0.25 0.2352941176 0.2360681115 0.2360679775
© OCT 2022 | IRE Journals | Volume 6 Issue 4 | ISSN: 2456-8880 IRE 1703837 ICONIC RESEARCH AND ENGINEERING JOURNALS 20 i. Cx n n = → lim . Further �� ⋛���� according to n is odd or even. ii. |�� ����|≥|�� ����+1| iii. |�� ����|< 1 ����+1 Proof i. Since ��1 ≤��3 ≤��5 ≤......is increasing sequence. Bounded above by any even convergent. 21lim →  n C n exist. n n CCCCC 13572 lim →  Similarly, ��0 ≥��2 ≥��4........ is an decreasing sequence which is bounded below by any odd convergent. n n C2lim →  exist. 02421 lim..... →  n n CCCC Since ���� =���� 1����:{����}is an increasing sequence Therefore ���� → ∞as �� → ∞ Since ���� ���� �� = ( ��)�� ���� by corollary 1 part (iv) Therefore 221limlim →→ = n n n n CC n n n n n n CCC 2212limlimlim →→→ == i.e. �� =[��0,��1,��2, ����,����+1 ] �� = 1 ��1 1 ��1��2 + 1 ��1��2��3 + ( 1)��+1 ��1��2��3...���� + ( 1)��+2 ��1��2��3...����+2 �� = 1 ��1 1 ��1��2 + 1 ��1��2��3 ...+ ( 1)�� ��1��2��3 ���� + ( 1)��+1 ��1��2��3 ���� { 1 ����+1 + ( 1)1 ����+1����+2} This can be written as, �� =���� + ��( 1)��+1 ���� ………………………………..(4) Now consider, |�� ����|=|��( 1)��+1 ���� |= �� ���� but we know �� ���� ≥0 But we know �� <1 Then, �� ���� < 1 ���� ,0≤|�� ����|< 1 ���� ……………………(5) But �� → ∞, 1 ���� →0(∵���� >0) By sandwich lemma with (5) Cx n n = → lim Therefore, ��0 ≥��2 ≥��4 ≥�� ≥ ≥��5 ≥��3 ≥��1 i.e. �� ⋛���� according to �� is odd or even. ii. By (4), �� =���� + ��( 1)��+1 ���� |�� ����|=|�� ����|= �� ���� (��) �� =����+1 + ��( 1)��+2 ����+1 (6) |�� ����+1|=|��( 1)��+2 ����+1 | = �� ����+1 (��),�� ≥1 We know ���� is an increasing sequence ∴���� ≤����+1 ⇒ 1 ���� ≥ 1 ����+1 ⇒ �� ���� ≥ �� ����+1 Therefore |�� ����|≥|�� ����+1| iii. By Corollary 1 part (iv) ���� ���� �� = ( ��)�� ���� |���� ���� 1|=|( 1)�� ���� |= 1 ���� (7) ���� ��

above part ii.,

between

and

Last part has been given for upper bound for given approximation square root problem.

II. RESULTS AND DISCUSSION

Using our proposed algorithm, we establish couple of results.

Result 1. For any real number x (0<�� <1), suppose that the Pierce contined fraction of ��is⟨��,��,��,��,..⟩ (for example, = 11 3 ⟨3,4,3,4,...⟩).

This infinite Pierce continued fraction can be written as a finite continued fraction, �� =⟨��,��,��,��,..⟩=⟨��,��,���� 1⟩, where �� =��+1

Result 2. For any real number ,x (0<�� <1), supposethatthePiercecontinued fractionof��is⟨��,��⟩ (for example, = 117 12 ⟨9,13⟩). This finite continued fraction can be written as an infinite continued fraction, �� =⟨��,��⟩=⟨��,�� 1,�� 1,�� 1, ⟩

In previous theorem last part have been discussed an upper bound for square root problem, for an example considering the value of √5 correct to four decimal places,

then

gives the value of √5 correct to four decimals places.by theorem 1.3 part (iii),

and

According example 2,

Using continued fraction approximation value of √5in to 4 decimal places is given by the 6th convergent (

6)

But using the piercecontinued fraction same result can be obtained by 4th convergent or less.

In addition, we have illustrated above results graphically. Table 1, we get this value when n=3. But if we used standard continued fraction minimum value for n is 6.

1 2 3 4 B 2.25 2.23529412.2360681 2.236068 T 2.236068 2.236068 2.236068 2.236068

Fig. 1 exact value of √5 vs ����

1.8 1.9 2 2.1 2.2 2.3

2.225 2.23 2.235 2.24 2.245 2.25 2.255 1 2 3 4 T 2.236068 2.236068 2.236068 2.236068 D 2 2.25 2.235 2.23616

Fig. 2 exact value of √5 vs

1.8 1.9 2 2.1 2.2 2.3

2 3 4

2 2.25 2.235 2.23616

2.25 2.235294122.236068112.23606798

Fig. 3

© OCT 2022 | IRE Journals | Volume 6 Issue 4 | ISSN: 2456-8880 IRE 1703837 ICONIC RESEARCH AND ENGINEERING JOURNALS 21 By
�� lies
����
����+1 |�� ����|<|����+1 ����|= 1 ����+1 |�� ����|< 1 ����+1
If |√5 ����|<000005
���� 000005<√5< ���� +000005
����
|�� ����|< 1 ����+1 If 1 ����+1 <000005then|√5 ����|<000005 i.e., ���� >20,000
��
����
��1 �� vs ����
1
D
B

In figure 1, graph of ���� has been coincide with graph of √5 subsequently �� =3. According to the above graphs, the variance of the partial quotients of pierce continued fraction are smaller than partial quotients of usual continued fraction.

CONCLUSION

In this work we introduce a continued fraction called the Pierce continued fraction which has a better convergent rate than the usual continued fraction. Using this advantage, we got appropriate approach for approximation to square root problem. Moreover, two sequences {����} and {����} are defined to obtain the relevant convergence. In addition, we have proved theorem which are necessary to obtain the convergence of a pierce continued fraction. Most of the proofs are similar to the proofs of their continued fraction.

There are lots of applications related to continued fractions. Solving linear Diophantine equations is one of the major applications. As a future work, we expect to establish a result to solve the Pelle’s equation using Pierce continued fraction.

REFERENCES

[1] Cor. Kraaikamp and Jun Wu, On a New Continued Fraction Expansion with Non Decreasing Partial Quotients (2004).

[2] Peter J. Cameron A Course on Number Theory, Queen Mary, University of London (2009).

[3] William Stein Elementary Number Theory: Primes, Congruence’s, and Secrets. (2011).

[4] Rotman, Joseph, An introduction to the theory of groups, New York: Springer Verlag, ISBN 0 387 94285 8 A standard contemporary reference, (1994)

© OCT 2022 | IRE Journals | Volume 6 Issue 4 | ISSN: 2456-8880 IRE 1703837 ICONIC RESEARCH AND ENGINEERING JOURNALS 22

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