<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article  PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd"><article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article"><front><journal-meta><journal-id journal-id-type="publisher-id">AM</journal-id><journal-title-group><journal-title>Applied Mathematics</journal-title></journal-title-group><issn pub-type="epub">2152-7385</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/am.2018.97059</article-id><article-id pub-id-type="publisher-id">AM-86296</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Physics&amp;Mathematics</subject></subj-group></article-categories><title-group><article-title>
 
 
  Newton, Halley, Pell and the Optimal Iterative High-Order Rational Approximation of √&lt;span style='margin-left:-2px;margin-right:2px; border-top:1px solid black'&gt;N&lt;/span&gt;
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Isaac</surname><given-names>Fried</given-names></name><xref ref-type="aff" rid="aff1"><sub>1</sub></xref><xref ref-type="corresp" rid="cor1"><sup>*</sup></xref></contrib></contrib-group><aff id="aff1"><label>1</label><addr-line>Department of Mathematics, Boston University, Boston, MA, USA</addr-line></aff><author-notes><corresp id="cor1">* E-mail:<email>if@math.bu.edu</email></corresp></author-notes><pub-date pub-type="epub"><day>12</day><month>07</month><year>2018</year></pub-date><volume>09</volume><issue>07</issue><fpage>861</fpage><lpage>873</lpage><history><date date-type="received"><day>29,</day>	<month>June</month>	<year>2018</year></date><date date-type="rev-recd"><day>27,</day>	<month>July</month>	<year>2018</year>	</date><date date-type="accepted"><day>30,</day>	<month>July</month>	<year>2018</year></date></history><permissions><copyright-statement>&#169; Copyright  2014 by authors and Scientific Research Publishing Inc. </copyright-statement><copyright-year>2014</copyright-year><license><license-p>This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/</license-p></license></permissions><abstract><p>
 
 
  
    In this paper we examine single-step iterative methods for the solution of the nonlinear algebraic equation 
   <em>f</em> (
   <em>x</em>) = 
   <em>x</em>
   <sup>2</sup> - 
   <em>N</em> = 0 , for some integer 
   <em>N</em>, generating rational approximations 
   <em>p</em>/
   <em>q</em> that are optimal in the sense of Pell’s equation 
   <em>p</em>
   <sup>2</sup> - 
   <em>Nq</em>
   <sup>2</sup> = 
   <em>k</em> for some integer 
   <em>k</em>, converging either alternatingly or oppositely. 
  
 
</p></abstract><kwd-group><kwd>Iterative Methods</kwd><kwd> Super-Linear and Super-Quadratic Methods</kwd><kwd> Square Roots</kwd><kwd> Pell’s Equation</kwd><kwd> Optimal Rational Iterants</kwd><kwd> Root Bounds</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>We present ever higher order single-point iterative methods for the numerical solution of the nonlinear equation f ( x ) = 0 . Then we show that for f ( x ) = x 2 − N these methods are optimal in the sense of Pell’s equation (see [<xref ref-type="bibr" rid="scirp.86296-ref1">1</xref>] [<xref ref-type="bibr" rid="scirp.86296-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.86296-ref3">3</xref>] [<xref ref-type="bibr" rid="scirp.86296-ref4">4</xref>] ), namely, that if the initial guess x 0 = p 0 / q 0 satisfies the diophantine Pell’s equation p 0 2 − N q 0 2 = k , for some integer k, then the iterated value x 1 = p 1 / q 1 , obtained by a method of order n, satisfies the Pell equation p 1 2 − N q 1 2 = k n .</p><p>Using a generalization of the recursive solution to Pell’s equation we generate super-linear and super-quadratic methods that converge alternatingly and oppositely to provide upper and lower bounds on the targeted root (see [<xref ref-type="bibr" rid="scirp.86296-ref5">5</xref>] [<xref ref-type="bibr" rid="scirp.86296-ref6">6</xref>] [<xref ref-type="bibr" rid="scirp.86296-ref7">7</xref>] ).</p></sec><sec id="s2"><title>2. Pell’s Equation</title><p>Let N be a positive integer which is not a square. The pair of natural numbers p, q satisfying the general Pell’s equation (see [<xref ref-type="bibr" rid="scirp.86296-ref1">1</xref>] [<xref ref-type="bibr" rid="scirp.86296-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.86296-ref3">3</xref>] [<xref ref-type="bibr" rid="scirp.86296-ref4">4</xref>] ):</p><p>p 2 − N q 2 = k (1)</p><p>are such that</p><p>( p q ) 2 = N + k q 2     or     p q = N ( 1 + k 2 N q 2 ) (2)</p><p>nearly, if k / ( N q 2 ) ≪ 1 . If k &gt; 0, then p/q is an overestimate of N , and if k &lt; 0, then p/q is an underestimate of N .</p><p>We verify (see [<xref ref-type="bibr" rid="scirp.86296-ref1">1</xref>] , Chapter 32) that a new solution pair ( p 1 , q 1 ) to the optimal, or minimal ( k = &#177; 1 ), Pell’s equation is obtained from a known solution pair ( p 0 , q 0 ) by the expansion of</p><p>p 1 + N q 1 = ( p 0 + N q 0 ) n , (3)</p><p>where variable n is taken odd for k = − 1 .</p><p>For example, if we take in Equation (3) n = 2, then (see [<xref ref-type="bibr" rid="scirp.86296-ref8">8</xref>] [<xref ref-type="bibr" rid="scirp.86296-ref9">9</xref>] , and also [<xref ref-type="bibr" rid="scirp.86296-ref10">10</xref>] [<xref ref-type="bibr" rid="scirp.86296-ref11">11</xref>] )</p><p>p 1 + N q 1 = ( p 0 2 + N q 0 2 ) + N ( 2 p 0 q 0 ) (4)</p><p>and</p><p>p 1 = p 0 2 + N q 0 2 ,   q 1 = 2 p 0 q 0 ,     or     x 1 = x 0 2 + N 2 x 0 ,   x 1 = p 1 q 1 , x 0 = p 0 q 0 , (5)</p><p>which is Newton’s method, preferably written as</p><p>x 1 = x 0 − x 0 2 − N 2 x 0 ,   x 1 = x 0 − f ( x 0 ) f ′ ( x 0 ) ,   f ( x ) = x 2 − N . (6)</p><p>Here</p><p>p 1 2 − N q 1 2 = ( p 0 2 − N q 0 2 ) 2 . (7)</p></sec><sec id="s3"><title>3. Super-Linear Iterative Method</title><p>We start with the general power series expansion around N</p><p>x 1 = F ( x 0 ) = F ( N ) + F ′ ( N ) ( x − N ) + 1 2 ! F ″ ( N ) ( x − N ) 2                                     + O ( ( x − N ) 3 ) (8)</p><p>and ask that N = F ( N ) , or that N is a fixed-point of iteration function F ( x ) . Now we pass from the general to the specific</p><p>x 1 = A x 0 + B C x 0 + D = A N + B C N + D + A D − B C ( C N + D ) 2 ( x 0 − N ) + O ( ( x 0 − N ) 2 ) (9)</p><p>for parameters A , B , C , D , and we ask, here specifically, that</p><p>N = A N + B C N + D , (10)</p><p>or, again, that N is a fixed-point of the rational iteration function in Equation (9). To satisfy Equation (10) we take B = N C , D = A , and are left with</p><p>x 1 = A x 0 + N C C x 0 + A ,     x 1 − N = A / C − N A / C + N ( x 0 − N ) ,   A / C &gt; 0 (11)</p><p>in which A / C + N in the second denominator is written for A / C + x 0 .</p><p>Writing x 0 = p 0 / q 0 and x 1 = p 1 / q 1 , the iterative method assumes the form</p><p>x 1 = p 1 q 1 = A p 0 + C N q 0 C p 0 + A q 0     or   p 1 = A p 0 + C N q 0 ,   q 1 = C p 0 + A q 0 ,   x 0 = p 0 q 0 ,   C ≠ 0 (12)</p><p>for parameters A and C. Referring to Equation (12) we have</p><p>Lemma 1. If p 0 , q 0 are such that p 0 2 − N q 0 2 = k , and A 2 − N C 2 = m , then p 1 , q 1 are such that p 1 2 − N q 1 2 = k m .</p><p>Proof. We verify that</p><p>p 1 2 − N q 1 2 = ( A p 0 + N C q 0 ) 2 − N ( C p 0 + A q 0 ) 2 = ( A 2 − N C 2 ) ( p 0 2 − N q 0 2 ) = k m , (13)</p><p>and the result follows.</p><p>For instance, if N = 2 , C = A = 1 , p 0 = 3 and q 0 = 2 , then</p><p>p 1 = p 0 + 2 q 0 = 7 ,   q 1 = p 0 + q 0 = 5     and     p 1 2 − 2 q 1 2 = − 1. (14)</p><p>Observe that the iterative method (11) converges linearly for any A / C &gt; 0 , since then</p><p>− 1 &lt; A / C − N A / C + N &lt; 1 (15)</p><p>and | x 1 − N | &lt; | x 0 − N | .</p></sec><sec id="s4"><title>4. Alternating Convergence</title><p>If in Equation (11), A / C &gt; N , then x 0 − N and x 1 − N are of the same sign, but if A / C &lt; N , they are of opposite signs. Also, the smaller | A / C − N | , the faster the convergence.</p><p>The method of Equation (11), as well as higher order methods, can be derived directly, in reverse, from the generalized Equation (3)</p><p>p 1 + N q 1 = ( p + N q ) n ( p 0 + N q 0 ) m (16)</p><p>with n = 1 and m = 1. Indeed, expansion of Equation (16) brings it to the form</p><p>p 1 + N q 1 = ( p p 0 + N q q 0 ) + N ( q p 0 + p q 0 ) , (17)</p><p>which elicits the pair of equations</p><p>p 1 = p p 0 + N q q 0 ,   q 1 = q p 0 + p q 0 (18)</p><p>with A = p and C = q in Equation (11).</p><p>For example, taking in Equation (11) N = 7 , p 0 / q 0 = 8 / 3 , p 0 2 − 7 q 0 2 = 1 , p / q = A / C = 5 / 2 , p 2 − 7 q 2 = − 3 , we obtain from it the alternating sequence of convergents:</p><p>x 1 = p 1 q 1 = { 8 3 , 82 31 , 844 319 , 8686 3283 , 89392 33787 , 919978 347719 , ⋯ } (19)</p><p>with</p><p>( p 1 q 1 ) 2 = { 7.1 , 6.997 , 7.00009 , 6.9999975 , 7.00000007 , 6.9999999979 , ⋯ } (20)</p></sec><sec id="s5"><title>5. The Method of Newton and Its Opposites</title><p>Taking in Equation (9) A / C = x 0 , or A = p 0 and C = q 0 , the linear method rises to become the quadratic method of Newton, otherwise directly obtainable from Equation (3) with n = 2 (see Equations (5)-(7)).</p><p>Here, for Newton’s method</p><p>x 1 − N = 1 2 N ( x 0 − N ) 2 (21)</p><p>nearly, if x 0 is close to N .</p><p>The method</p><p>p 1 q 1 = 2 N p 0 q 0 p 0 2 + N q 0 2 ,   or   x 1 = 2 N x 0 x 0 2 + N (22)</p><p>is such that</p><p>p 1 2 − N q 1 2 = − N ( p 0 2 − N q 0 2 ) 2 , (23)</p><p>or</p><p>x 1 − N = − 1 2 N ( x 0 − N ) 2 (24)</p><p>if x 0 is close to N . Here, convergence is quadratic and from below. Compare Equations ((21) and (24)).</p><p>The average of methods (5) and (22)</p><p>x 1 = 1 2 ( x 0 2 + N 2 x 0 + 2 N x 0 x 0 2 + N ) (25)</p><p>is quartic</p><p>x 1 − N = 1 8 N N ( x 0 − N ) 4 . (26)</p><p>Or</p><p>p 1 q 1 = p 0 4 + 6 N p 0 2 q 0 2 + N 2 q 0 4 4 p 0 q 0 ( p 0 2 + N q 0 2 ) ,     and   p 1 2 − N q 1 2 = ( p 0 2 − N q 0 2 ) 4 . (27)</p><p>For example, for N = 2 and x 0 = 3 / 2 we obtain from method (6) x 1 = 17 / 12 , from method (22) x 1 = 24 / 17 , and for their average x 1 = 577 / 408 , and</p><p>17 2 − 2 &#215; 12 2 = 1 ,   24 2 − 2 &#215; 17 2 = − 2 ,   577 2 − 2 &#215; 408 2 = 1. (28)</p><p>Here, ( 17 / 12 ) 2 = 2.007 , ( 24 / 17 ) 2 = 1.993 , ( 577 / 408 ) 2 = 2.000006 .</p><p>The biased average method</p><p>x 1 = x 0 2 + N 2 x 0 ( 1 2 − ϵ ) + 2 N x 0 x 0 2 + N ( 1 2 + ϵ ) ,   ϵ = 1 16 N 2 ( x 0 2 − N ) 2 (29)</p><p>produces an oppositely converging quartic method such that, asymptotically</p><p>x 1 − N = − 1 8 N N ( x 0 − N ) 4 . (30)</p><p>Compare Equations ((26) and (30)).</p><p>The biased average method</p><p>x 1 = x 0 2 + N 2 x 0   ( 1 2 − ϵ ) + 2 N x 0 x 0 2 + N ( 1 2 + ϵ ) ,   ϵ = 1 32 N 2 ( x 0 2 − N ) 2 (31)</p><p>is a quintic method and such that</p><p>x 1 − N = − 1 4 N 2 ( x 0 − N ) 5 + O ( ( x 0 − N ) 6 ) , (32)</p><p>implying that the convergence of method (31) is alternating. Indeed, starting with x 0 = 3 / 2 we obtain from method (31)</p><p>x 0 2 = 2 + 0.25 ,   x 1 2 = 2 − 7.6 &#215; 10 − 7 ,   x 1 2 = 2 + 2.5 &#215; 10 − 34 . (33)</p></sec><sec id="s6"><title>6. More Convergence from Below</title><p>The noteworthy method</p><p>x 1 = 3 N − x 0 2 2 N x 0 (34)</p><p>converges to N quadratically and from below,</p><p>x 1 − N = − 3 2 N ( x 0 − N ) 2 . (35)</p><p>We write x 0 = p 0 / q 0 and x 1 = p 1 / q 1 and have for Equation (34) that</p><p>p 1 2 − N q 1 2 = ( p 0 2 − 4 N q 0 2 ) ( p 0 2 − N q 0 2 ) 2 . (36)</p></sec><sec id="s7"><title>7. Super-Linear Alternating Methods</title><p>We put in Equation (11)</p><p>A = x 0 ( 1 + 2 ϵ ) ,   C = 1 , (37)</p><p>and obtain</p><p>x 1 − N = ϵ ( x 0 − N ) + 1 2 N ( x 0 − N ) 2 (38)</p><p>nearly, if x 0 is close to N and ϵ ≪ 1 , the super-linear method</p><p>x 1 = x 0 − x 0 2 − N 2 x 0 ( 1 + ϵ )     or     x 1 = x 0 − x 0 2 − N 2 x 0 ( 1 − ϵ ) . (39)</p><p>A small negative ϵ causes method (39) to ultimately oscillate, or alternate.</p><p>With</p><p>ϵ = − 1 4 N ( x 0 2 − N ) (40)</p><p>method (39) becomes cubic and of alternating convergence</p><p>x 1 − N = − 3 4 N ( x 0 − N ) 3 + O ( ( x 0 − N ) 4 ) . (41)</p></sec><sec id="s8"><title>8. Stacked Methods</title><p>From</p><p>p 2 + N q 2 = ( p 0 + N q 0 ) ( p 1 + N q 1 ) = ( p 0 p 1 + N q 0 q 1 ) + N ( p 0 q 1 + p 1 q 0 ) (42)</p><p>we have the stacked method</p><p>p 2 q 2 = p 0 p 1 + N q 0 q 1 p 0 q 1 + p 1 q 0 , (43)</p><p>or</p><p>x 2 = x 0 x 1 + N x 0 + x 1 . (44)</p><p>It is such that if</p><p>x 0 = N + ϵ 0 ,     and     x 1 = N + ϵ 1 , (45)</p><p>then</p><p>x 2 = N + ϵ 0 ϵ 1 2 N = N + 1 2 N ( x 0 − N ) ( x 1 − N ) (46)</p><p>nearly, if both epsilons are small compared with N .</p><p>If ϵ 0 ϵ 1 &lt; 0 , then x 2 &lt; N , and if ϵ 0 ϵ 1 &gt; 0 , then x 2 &gt; N . For example, for N = 2 we obtain from the stacked method of Equation (44) the alternatingly converging sequence</p><p>x 2 = { 1 1 , 3 2 , 7 5 , 41 29 , 577 408 , 47321 33461 , ⋯ } (47)</p><p>with</p><p>x 2 2 = { 1 ,   2.25 ,   1.96 ,   1.9988 ,   2.000006 ,   1.9999999991 , ⋯ } (48)</p></sec><sec id="s9"><title>9. Halley’s Third-Order Method</title><p>Halley’s cubic iterative method</p><p>x 1 = x 0 − det [ 1 f 0 0 2 f ′ 0 ] det [ f ′ 0 f 0 f ″ 0 2 f ′ 0 ] ⋅ f 0 (49)</p><p>becomes for f ( x ) = x 2 − N and x 0 = p 0 / q 0</p><p>x 1 = p 1 q 1 ,   p 1 = p 0 ( p 0 2 + 3 N q 0 2 ) ,   q 1 = q 0 ( 3 p 0 2 + N q 0 2 ) , (50)</p><p>and is verified to be such that</p><p>p 1 2 − N q 1 2 = ( p 0 2 − N q 0 2 ) 3 = k 3     if   p 0 2 − N q 0 2 = k , (51)</p><p>implying that if p 0 / q 0 is an underestimate (k &lt; 0), then so is p 1 / q 1 , and if p 0 / q 0 is an overestimate (k &gt; 0), then so is p 1 / q 1 .</p><p>Otherwise, here</p><p>x 1 − N = 1 4 N ( x 0 − N ) 3 (52)</p><p>nearly, if p 0 / q 0 is close to N .</p></sec><sec id="s10"><title>10. Fourth-Order Method</title><p>The quartic method (see [<xref ref-type="bibr" rid="scirp.86296-ref12">12</xref>] [<xref ref-type="bibr" rid="scirp.86296-ref13">13</xref>] for higher order methods):</p><p>x 1 = x 0 − det [ 1 f 0 0 0 2 f ′ 0 f 0 0 3 f ″ 0 3 f ′ 0 ] det [ f ′ 0 f 0 0 f ″ 0 2 f ′ 0 f 0 f ‴ 0 3 f ″ 0 3 f ′ 0 ] ⋅ f 0 (53)</p><p>becomes for f ( x ) = x 2 − N and x 0 = p 0 / q 0</p><p>x 1 = p 1 q 1 ,   p 1 = p 0 4 + 6 N p 0 2 q 0 2 + N 2 q 0 4 ,   q 1 = 4 p 0 q 0 ( p 0 2 + N q 0 2 ) , (54)</p><p>observed to be a repeated second order method and such that</p><p>p 1 2 − N q 1 2 = ( p 0 2 − N q 0 2 ) 4 . (55)</p><p>Otherwise, here</p><p>x 1 − N = 1 8 N N ( x 0 − N ) 4 (56)</p><p>if x 0 is close to N . Convergence here is from above.</p></sec><sec id="s11"><title>11. Fifth-Order Method</title><p>The quintic method</p><p>x 1 = x 0 − det [ 1 f 0 0 2 f ′ 0 f 0 0 3 f ′ ′ 0 3 f ′ 0 f 0 0 4 f ′ ′ ′ 0 6 f ′ ′ 0 4 f ′ 0 ] det [ f ′ 0 f 0 f ′ ′ 0 2 f ′ 0 f 0 f ′ ′ ′ 0 3 f ′ ′ 0 3 f ′ 0 f 0 f ′ ′ ′ ′ 0 4 f ′ ′ ′ 0 6 f ′ ′ 0 4 f ′ 0 ] ⋅ f 0 (57)</p><p>becomes for f ( x ) = x 2 − N and x 0 = p 0 / q 0</p><p>x 1 = p 1 q 1 ,   p 1 = p 0 ( p 0 4 + 10 N p 0 2 q 0 2 + 5 N 2 q 0 4 ) ,   q 1 = q 0 ( 5 p 0 4 + 10 N p 0 2 q 0 2 + N 2 q 0 4 ) , (58)</p><p>and happens to be such that</p><p>p 1 2 − N q 1 2 = ( p 0 2 − N q 0 2 ) 5 . (59)</p><p>Otherwise, here</p><p>x 1 − N = 1 16 N 2 ( x 0 − N ) 5 (60)</p><p>if x 0 is close to N .</p></sec><sec id="s12"><title>12. A Rational Quadratic Method</title><p>The method</p><p>x 1 = 2 2 + 4 x 0 + 2 x 0 2 2 + 2 2 x 0 + x 0 2 (61)</p><p>is merely x 1 = 2 in disguise. Replacement of 2 by the good rational approximation p/q turns the scheme into</p><p>x 1 = 2 p + 4 q x 0 + p x 0 2 2 q + 2 p x 0 + q x 0 2 , (62)</p><p>and for the specific p / q = 7 / 5 ,   7 2 − 2 &#215; 5 2 = − 1 , it becomes</p><p>x 1 = p 1 q 1 = 14 + 20 x 0 + 7 x 0 2 10 + 14 x 0 + 5 x 0 2 ,   x 0 = p 0 q 0 , p 1 2 − 2 q 1 2 = − ( p 0 2 − 2 q 0 2 ) 2 . (63)</p><p>Starting with x 0 = 7 / 5 we obtain x 1 = 239 / 169 , x 1 2 = 1.999965 . Starting with x 0 = 17 / 12 we obtain x 1 = 8119 / 5741 , x 1 2 = 1.99999997 . Then</p><p>x = 1855077841 / 1311738121 ,     1855077841 2 − 2 &#215; 1311738121 2 = − 1 ( 1855077841 / 131173812 ) 2 = 1.99999999999999999942. (64)</p><p>From</p><p>x 1 = p 1 q 1 = 6 + 8 x 0 + 3 x 0 2 4 + 6 x 0 + 2 x 0 2 ,   x 0 = p 0 q 0 ,   p 1 2 − 2 q 1 2 = ( p 0 2 − 2 q 0 2 ) 2 , (65)</p><p>obtained from Equation (62) with p / q = 3 / 2 ,   3 2 − 2 &#215; 2 2 = 1 , we compute</p><p>x 1 = { 3 / 2 , 99 / 70 , 114243 / 80782 , 152139002499 / 107578520350 } (66)</p><p>with</p><p>152139002499 2 − 2 &#215; 107578520350 2 = 1 ( 152139002499 / 107578520350 ) 2 = 2.000000000000000000000086. (67)</p></sec><sec id="s13"><title>13. The General Rational Super-Quadratic Method</title><p>We start by writing</p><p>x 1 = p 1 q 1 = A x 0 2 + B x 0 + C P x 0 2 + Q x 0 + R (68)</p><p>to have</p><p>x 1 − N = p 1 ( x 0 ) − q 1 ( x 0 ) N q 1 ( x 0 ) . (69)</p><p>To have a factor ( x 0 − N ) 2 in the numerator of the right-hand side of Equation (69), we ask that</p><p>p 1 ( x ) − q 1 ( x ) N = 0 ,   and   that     ( p 1 ( x ) − q 1 ( x ) N ) ′ = 0 ,   at   x = N (70)</p><p>resulting in</p><p>P = 1 , B = 2 N , R = N , C = A N , Q = 2 A , (71)</p><p>and the method</p><p>x 1 − N = A − N x 0 2 + 2 A x 0 + N ( x 0 − N ) 2 (72)</p><p>that can be raised to cubic with the choice A = x 0 .</p><p>Instead, we leave A = p / q , x 0 = p 0 / q 0 to have the method</p><p>x 1 = p 1 q 1 = A x 0 2 + 2 N x 0 + A N x 0 2 + 2 A x 0 + N ,   A = p q ,   x 0 = p 0 q 0 (73)</p><p>such that</p><p>p 1 2 − N q 1 2 = ( p 2 − N q 2 ) ( p 0 2 − N q 0 2 ) 2 . (74)</p><p>For example, for N = 7 , A = 8 / 3 , x 0 = 8 / 3 , 8 2 − 7 &#215; 3 2 = 1 , we obtain from Equation (73)</p><p>x 1 = 2024 765 ,   x 1 2 = 7.0000017 ,   2024 2 − 7 &#215; 765 2 = 1 , x 1 = 130576328 49353213 , x 1 2 = 7.0000000000000004 ,   130576328 2 − 7 &#215; 49353213 2 = 1. (75)</p><p>For N = 7 , A = 5 / 2 , x 0 = 5 / 2 , 5 2 − 7 &#215; 2 2 = − 3 , we obtain from Equation (73)</p><p>x 1 = 545 206 ,   x 1 2 = 6.99936 ,   545 2 − 7 &#215; 206 2 = − 27 ,   x 1 = 6113945 2310854 , x 1 2 = 6.9999999996 ,   6113945 2 − 7 &#215; 2310854 2 = − 2187. (76)</p><p>Equation (73), as well as higher order methods, could have been derived directly, in reverse, from</p><p>p 1 + N q 1 = ( p + N q ) n ( p 0 + N q 0 ) m (77)</p><p>with n = 1 , m = 2 .</p></sec><sec id="s14"><title>14. The Direct Construction of a Super-Quadratic Method</title><p>To locate root a of f ( x ) , f ( a ) = 0 , we start by writing the fixed-point iterative method</p><p>x 1 = F ( x 0 ) ,   F ( x ) = x + A f ( x ) + B f 2 ( x ) (78)</p><p>for constants A and B. Then we require that</p><p>F ′ ( a ) = 0 ,   F ″ ( a ) = ϵ , (79)</p><p>where ϵ is any parameter.</p><p>Differentiating F ( x ) once and twice, the previous system of two equations in the two unknowns A and B becomes</p><p>[ f ′ 2 f f ′ f ″ 2 ( f ′ 2 + 2 f f ″ ) ] [ A B ] = [ − 1 ϵ ] , (80)</p><p>which we solve to have</p><p>A = − 1 f ′ 3 ( ϵ f f ′ + f ′ 2 + f f ″ ) , B = 1 2 f ′ 3 ( ϵ f ′ + f ″ ) . (81)</p><p>Since root a of f ( x ) is unknown we replace a by x 0 to have the method</p><p>x 1 = x 0 − f 0 f ′ 0 − 1 2 f 0 2 f ′ 0 2 ( ϵ + f ″ 0 f ′ 0 ) , (82)</p><p>where f 0 = f ( x 0 ) etc. Here</p><p>x 1 − a = − 1 2 ϵ ( x 0 − a ) 2 + O ( ( x 0 − a ) 3 ) , (83)</p><p>and convergence is from below if ϵ &gt; 0 , while convergence is from above if ϵ &lt; 0 .</p><p>For</p><p>f ( x ) = x 2 − N (84)</p><p>the method becomes</p><p>x 1 = 1 8 x 0 3 ( − N 2 + 6 N x 0 2 + 3 x 0 4 − ϵ   ( x 0 2 − N ) 2 x 0 ) . (85)</p><p>For example, for N = 2 , ϵ = 1 / 25 , and x 0 = 1.5 we have x 1 = 1.414213 and x 1 2 = 1.9999983 . For ϵ = 0 we have x 1 = 1.414352 and x 1 2 = 2.00039 .</p><p>The choice</p><p>ϵ = 1 2 N N ( x 0 2 − N ) (86)</p><p>makes method (82) the quartic</p><p>x 1 − N = − 7 8 N N ( x 0 − N ) 4 + O ( ( x 0 − N ) 5 ) . (87)</p></sec><sec id="s15"><title>15. The Simplest of All Methods</title><p>A simple routine for constructing a rational approximation to an irrational number consists of starting with any good rational approximation p/q to, say, 2 , then adding one to p if ( p / q ) 2 &lt; 2 , or adding one to q if ( p / q ) 2 &gt; 2 . Starting with 3/2 we obtain this way the alternating sequence</p><p>3 2 , 3 3 , 4 3 , 5 3 , 5 4 , 6 4 , 6 5 , 7 5 , 8 5 , 8 6 , 9 6 , 9 7 , 10 7 , 10 8 , 11 8 , 12 8 , 12 9 , 13 9 , ⋯ , (88)</p><p>where ( 12 / 9 ) 2 = 1.78 ,   ( 13 / 9 ) 2 = 2.086 .</p><p>The method is sluggish, yet we can glean from this long sequence some very good Pell approximations to 2 , such as 1 / 1 , k = − 1 ; 3 / 2 , k = 1 ; 7 / 2 , k = − 1 ; 17 / 12 , k = 1 ; 41 / 29 , k = − 1 ; 99 / 70 , k = 1 ; 239 / 169 , k = − 1 ; 577 / 408 , k = 1 . Number k = p 2 − N q 2 .</p><p>Going up to 4-digit approximations we find ( 3363 / 2378 ) 2 = 2.000000180 , 3363 2 − 2 &#215; 2378 2 = 1 , and then ( 8119 / 5741 ) 2 = 1.999999970 , 8119 2 − 2 &#215; 5741 2 = − 1 . Among the 5-digit approximations we find ( 19601 / 13860 ) 2 = 2.000000005 , 19601 2 − 2 &#215; 13860 2 = 1 and ( 47321 / 33461 ) 2 = 1.999999999 , 47321 2 − 2 &#215; 33461 2 = − 1 .</p><p>Thus, the alternating sequence of rational approximations to 2</p><p>1 1 , 3 2 , 7 5 , 17 12 , 41 29 , 99 70 , 239 169 , 577 408 , 1393 985 , 3363 2378 , 8119 5741 , 19601 13860 , 47321 33461 , 114243 80782 (89)</p><p>is of excellent p/q rational approximations to 2 such that p 2 − 2 q 2 = − 1 if p / q &lt; 2 , and p 2 − 2 q 2 = 1 if p / q &gt; 2 .</p><p>For N = 7 we find this way 8 / 3 , k = 1 ; 127 / 48 , k = 1 ; 2024 / 765 , k = 1 for the upper bounds, and 2 / 1 , k = − 3 ; 5 / 2 , k = − 3 ; 37 / 14 , k = − 3 ; 82 / 31 , k = − 3 for the lower bounds.</p><p>To understand the convergence mechanism of this algorithm, let p/q be the last fraction less then 2 , namely, such that p / q &lt; 2 , but ( p + 1 ) / q &gt; 2 . Then</p><p>p q &lt; 2 &lt; p q + 1 q , (90)</p><p>and the bounds on 2 become tighter as q increases by the repeated addition of 1 to it.</p></sec><sec id="s16"><title>16. Bisection by Mediants</title><p>Mediant m of the two nonzero rationals a / b &lt; c / d is</p><p>m = a + c b + d . (91)</p><p>Lemma 2. We have</p><p>a b &lt; m &lt; c d . (92)</p><p>Proof. Since a / b &lt; c / d , b c − a d &gt; 0 , and the result follows.</p><p>Lemma 3. We have</p><p>If b c − a d = k , then m − a b = k b ( b + d ) , and c d − m = k d ( b + d ) . (93)</p><p>Proof. The result follows by some simple algebra.</p><p>For example, from Equation (89) we have that 7 / 5 &lt; 2 &lt; 3 / 2 with 3 / 2 − 7 / 5 = 1 / 10 . Here the mediant m = 10 / 7 , and 7 / 5 &lt; 2 &lt; 10 / 7 with 10 / 7 − 7 / 5 = 1 / 35 . The next m = 17 / 12 , and 7 / 5 &lt; 2 &lt; 17 / 12 with 17 / 12 − 7 / 5 = 1 / 60 ; all spreads between the upper and lower bounds having a numerator equal to one.</p><p>Unlike ordinary bisections, bisection by mediants converges to a rational number in a finite number of steps. For example, by mediants</p><p>1 1 &lt; 2 &lt; 4 1 ,   1 1 &lt; 2 &lt; 5 2 ,   6 3 ≤ 2 ≤ 6 3 , (94)</p><p>while by ordinary bisection</p><p>1 1 &lt; 2 &lt; 4 1 ,   2 2 &lt; 2 &lt; 5 2 ,   7 4 &lt; 2 &lt; 10 4 ,   14 8 &lt; 2 &lt; 17 8 ,   28 16 &lt; 2 &lt; 31 16 , ⋯ . (95)</p></sec><sec id="s17"><title>17. Root Bracketing</title><p>We start with the following result.</p><p>Lemma 4. Let the integer pair ( p 0 , q 0 ) satisfy Pell’s equation p 0 2 − N q 0 2 = 1 , and let p 1 = A p 0 + C N q 0 , q 1 = C p 0 + A q 0 . Then</p><p>p 0 q 0 − p 1 q 1 = C q 0 q 1 . (96)</p><p>Proof. The result follows by common denominator.</p><p>Numerical example. For N = 7 , A = 2 , C = 1 we have that A 2 − 7 C 2 = − 3 . Hence, in accordance with Lemma 1</p><p>p 1 = 2 p 0 + 7 q 0 ,   q 1 = p 0 + 2 q 0 ,   are   such   that   p 1 2 − 7 q 1 2 = − 3 ( p 0 2 − 7 q 0 2 ) = − 3. (97)</p><p>Choosing the Pell (k = 1) pair ( p 0 , q 0 ) = ( 127 / 48 ) , 127 2 − 7 &#215; 48 2 = 1 , we obtain the Pell ( k = − 3 ) pair ( p 1 , q 1 ) = ( 590 , 223 ) , 590 2 − 7 &#215; 223 2 = − 3 , and</p><p>590 223 &lt; 7 &lt; 127 48     of   spread   127 48 − 590 223 = 1 10704 (98)</p><p>of a numerator equal to one.</p><p>Similarly, choosing the Pell (k = 1) pair ( p 0 , q 0 ) = ( 2024 , 765 ) we obtain the Pell ( k = − 3 ) pair ( p 1 , q 1 ) = ( 9403 , 3554 ) , and</p><p>9403 3554 &lt; 7 &lt; 2024 765     of   spread   2024 765 − 9403 3554 = 1 2718810 . (99)</p><p>The mediant in Equation (99) is m = ( 9403 + 2024 ) / ( 3554 + 765 ) = 11427 / 4319 , and with it</p><p>9403 3554 &lt; 7 &lt; 11427 4319     of   spread     11427 4319 − 9403 3554 = 1 15349726 . (100)</p></sec><sec id="s18"><title>18. Conclusion</title><p>In this paper we have examined single-step iterative methods for the solution of the nonlinear algebraic equation f ( x ) = x 2 − N = 0 , for some integer N, which produce rational approximations p/q that are optimal in the sense of Pell’s equation p 2 − N q 2 = k for some integer k. We have also considered the most elementary bisection method for iteratively creating upper and lower bounds on the targeted root.</p></sec><sec id="s19"><title>Conflicts of Interest</title><p>The authors declare no conflicts of interest regarding the publication of this paper.</p></sec><sec id="s20"><title>Cite this paper</title><p>Fried, I. (2018) Newton, Halley, Pell and the Optimal Iterative High-Order Rational Approximation of . Applied Mathematics, 9, 861-873. https://doi.org/10.4236/am.2018.97059</p></sec></body><back><ref-list><title>References</title><ref id="scirp.86296-ref1"><label>1</label><mixed-citation publication-type="other" xlink:type="simple">Silverman, J.H. 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