• TABLE OF CONTENTS
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 Title Page
 Acknowledgement
 Table of Contents
 Introduction
 Control of translation on penultimate...
 Estimates of the zeros of g(x)
 A noncontaminating iterative process...
 Appendices
 Bibliography
 Biographical sketch














Group Title: noncontaminating iterative process for determining the zeros of a polynomial
Title: A noncontaminating iterative process for determining the zeros of a polynomial
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Permanent Link: http://ufdc.ufl.edu/UF00097876/00001
 Material Information
Title: A noncontaminating iterative process for determining the zeros of a polynomial
Physical Description: iii, 79 leaves : illus. ; 28 cm.
Language: English
Creator: Santos, Alicia Hidalgo, 1937-
Publication Date: 1966
Copyright Date: 1966
 Subjects
Subject: Polynomials   ( lcsh )
Iterative methods (Mathematics)   ( lcsh )
Mathematics thesis Ph. D
Dissertations, Academic -- Mathematics -- UF
Genre: bibliography   ( marcgt )
non-fiction   ( marcgt )
 Notes
Thesis: Thesis - University of Florida.
Bibliography: Bibliography: leaf 78.
Additional Physical Form: Also available on World Wide Web
General Note: Manuscript copy.
General Note: Vita.
 Record Information
Bibliographic ID: UF00097876
Volume ID: VID00001
Source Institution: University of Florida
Holding Location: University of Florida
Rights Management: All rights reserved by the source institution and holding location.
Resource Identifier: alephbibnum - 000566122
oclc - 13639465
notis - ACZ2549

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Table of Contents
    Title Page
        Page i
    Acknowledgement
        Page ii
    Table of Contents
        Page iii
    Introduction
        Page 1
        Page 2
        Page 3
    Control of translation on penultimate remaindering
        Page 4
        Page 5
        Page 6
        Page 7
        Page 8
        Page 9
        Page 10
        Page 11
        Page 12
        Page 13
        Page 14
        Page 15
        Page 16
        Page 17
        Page 18
        Page 19
    Estimates of the zeros of g(x)
        Page 20
        Page 21
        Page 22
        Page 23
        Page 24
        Page 25
        Page 26
    A noncontaminating iterative process for determining the zeros of g(x)
        Page 27
        Page 28
        Page 29
        Page 30
        Page 31
        Page 32
        Page 33
        Page 34
        Page 35
        Page 36
        Page 37
        Page 38
        Page 39
        Page 40
        Page 41
        Page 42
        Page 43
        Page 44
        Page 45
        Page 46
        Page 47
        Page 48
        Page 49
        Page 50
        Page 51
    Appendices
        Page 52
        Page 53
        Page 54
        Page 55
        Page 56
        Page 57
        Page 58
        Page 59
        Page 60
        Page 61
        Page 62
        Page 63
        Page 64
        Page 65
        Page 66
        Page 67
        Page 68
        Page 69
        Page 70
        Page 71
        Page 72
        Page 73
        Page 74
        Page 75
        Page 76
        Page 77
    Bibliography
        Page 78
    Biographical sketch
        Page 79
        Page 80
        Page 81
Full Text









A NONCONTAMINATING ITERATIVE

PROCESS FOR DETERMINING THE

ZEROS OF A POLYNOMIAL



















By
ALICIA HIDALGO SANTOS


A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY









UNIVERSITY OF FLORIDA

April, 1966














AC MCrVLE DTIGlTS


ine writer gratefully acknowledges r.er ad''iser, Dr. Jorm

E.. Ili1'ldel., for hi; encouraging guidance, crieer)il word:, and

pa.inrtaking efforts to ihpii her mwAe thi-i piece of work a reality.

She wishes to e>..reEs also her ird.btecnes: to tre riationda Science

Development E'-.ard of the Phllippines and to the U. .S. Air Researrn

DEvelocpm,.nt Corariand for r.tnir financial support. Tn wrlter w'ash .

al-o tor ex:pre s rier appreciation to ai thre staff rrierb.-rs of the

Uriv4erzslt of Flcrrida Computing Centr for their cooperation ard

willing a iseta-nce on the machinr programs. Sp-ecal mienti.orn of

gratitude is also due to Mrs. Edna B. Lurrick for her unusU a patience

and eFxpert tTyping crf this di-ssertation.

Finally:, the writer wis-he to 3r tke this opportunity to *-.T.re

her appreciation to her husband, Carlos, for his in-pirational guid-

ance and tolerance during sor.e .;o her trying moments.


















TABLE OF CONTENTS


Page


ACKNOWLEDGMENTS . . . . . . . . . . . .

Chapter
I INTRODUCTION . . . . . . . . . . .

II CONTROL OF TRANSLATION ON PENULTIMATE REMAINDERING .


III ESTIMATES OF THE ZEROS OF g(x)


IV A NONCONTAMNATING ITERATIVE PROCESS FOR
DETERMINING THE ZEROS OF g(x) . . . . . ..


APPENDICES . . . . . . . . . . .

A MACHINE PROGRAMS IN FORTRAN II FOR IBM 709 .

B COMMON METHODS FOR DETERMINING THE ZEROS
OF A POLYNOMIAL . . . . . . . .


. $52

. . 53


. . 63-


BIBLIOGRAPHY . . . . . . . . . . . . .


BIOGRAPHICAL SKETCH . . . . . . . . . . .


. . 20















CiHAPF'?E I


1iTRC'iJ-UCTICON


Hi s-,.r,' ran S' at-ment ci th! F r-bl.1e

In [ ] a metir,-.d was j-rijved for !he rest.imati ri of the :-er,:, of
nr1
a pl),'nn.:diaal w-iltr real ciffclcients, gi., :. *:: fl::i from the

knw.T, :ero.s of I f.' a --uminp that r is a sTiall real numberr. In [il]

J E. Hi -cl el.J t l. .st ,abli srn an iTerr ti -. Fr.r c sc, which ,ojn-

verges to thei :-r: >f g' i arnd carn tL ui- to gi'v' a bct.- r tctifriate

,f the-e z.er.-s t ran the Frevio method. 'The pr-sent Faper now ;61lr,

to pr senrt an -.t.nsl(r-t:.n ,of I1i):

Let I t.i.', It. a p'ol,Tomlal iLth kTI.-.Ti zero.: rl, ... rn

n i-i
1*1
From t .i = ... .with a = 1 Ind ir, e :erc r., o t,.: we ish
1 n
to etablirnhr n it-rativ. pr.:ce-Uar- chicr. c.:onverg t. the :eror,

r r r


nn1
I *i I = r wh*. h t1t





':.i 1 n re ,* jall
ar i tudei .


In -his pFapr. the cioeffic en-'r .f 1 .', gi -rid t.x' aC

i11ll a n trt- incrermnts 'a ', need no- tt r- trir r-al numbers.

I.Moreover, as a prc.cess for deterr.lning .e -er -s, ri. ... rn,

Sti':. Lin's FPnuiltliiate R mairnderirn I teration is studied, and it










shall be shown that a translation of the variable x can control its

convergence.


Application of the Paper

In a problem for determining the zeros of f(x) with real coef-

ficients, most of the established methods that can be applied, such as

Lin's Penultimate Remaindering, are contaminated factorization schemes

in the sense that the degree of accuracy of one factor affects that of

the succeeding factors. For instance, in Penultimate Remaindering, the

depression of the polynomial after a factor is obtained causes error

due to the rounding in the division process itself, as well as due to

the inaccuracy, perhaps, of the first factor obtained. Thus most of

these methods give only the estimate r. to the actual zeros r.. If
1 1
t(x) is the polynomial having the r.s for its true zeros, then each
1
coefficient of t(x) differs from the corresponding coefficient of f(x)

by some small number cr Using these i's and r.'s as input, the iter-

ative procedure we derive here may be applied as a corrective measure

to obtain the actual zeros r.'s, of g(x) n2 which is precisely

f(x).

Aside from being merely corrective, the said iteration can

also be used as a method by itself of determining the zeros of a poly-

nomial, independent of any other method. In this case we take for the

ri's rough estimates of the zeros being sought. For, even though the

iteration shall be proved to converge only for the increments ci's

small enough in absolute value, like other iterations, experience shows

that surprisingly, even with very poor estimates of the zeros for a

start, the iteration frequently converges.










Furt.nernorc supp; e thre eri: ?of g':: = ., : f ,

itiere I is quite smsll in magnitude, -re to be deter-dinLed. WJe r'ay

trJncate it errForaril;. by, rneglecting the lesdingr tern t. -.3bain tne

pol ynonlal ft:. i of loif r degree. If I- can fc,-ctor tn-i pFolljri d -and

with trh:-.e -er.os a. r. ', ;*e ,'i then nmpiloy our iterati'.'e cche.e to

*.bta .n tr.e zerCE o glxi ie '.s a-re :-t to ,j e::cept Cn, in c I -

tri-: irer. of fl -re obt.na'.-d to artitrary prtcisjin.

Al :j, th),-re a-re occasions whe- n thre n eros ; ..i = an, xn*

* fI .-), on, nriall in magnituje, are to Le jeterriled wrererin -om:nhow,

r.re zer.: of ft.'1 ha-.' teer previou--sl. known. solved e.xctl or e5r.i-

mated. For inz*trce. Ln -serv:omecr.ini3.r,5. gained ri rsii-. iad elec-

tronic circuits, it 15 u i Jl, e.Fedaernt to Si-plifyv the differential

'systemT ty neglecting higher-o.-rdered dern .iti.ves then fre.qunD ril.y.

after stjd;,-ng tr.e truncated ''enr, it i nece-s--a ry to reinrtroduce

tre higher-ordererd denri' native one at a time arnd study the 3au1eriented

s'terr. Here we can ti-U: fl:.' to be the chLaracter;istic function of

the tr-,nca*id ZyEttm, and xJ that of trh a ugnlnted sys'Tem, tnri. i% ,

tr.e different'rial equation fr :. rith a leading term- adjoined c.-.rreUFond-

ing to tr, ne:xt' higher-ordered derival.i'.e previously nrglected. Using

th, iterative procedures we deri'vc here, w.e an ge*. thre zer of g.

fr.)o the krnon meros of fl .. .














CHAPTER II


CONTROL OF TRANSLATION ON
PENULTIMATE REMAINDERING


Before deriving the iterative formulas for the zeros of

n+l .
g(x) = yn+2 xn + E (a +i)x1 let us first study a method for

n+1 i
i-i
determining the zeros ri's of E ax which can be used as input

to the iteration.

One of the most convenient methods for determining the zeros

of a polynomial is Lin's Penultimate Remaindering [1], which is an

iterative scheme of producing one factor of a polynomial at a time.

Like most other methods though, it is a contaminating procedure as

has been explained in the introduction. With the zeros obtained from

this as input, the iteration which shall be derived in the succeeding

chapters can serve as corrective means and give a more accurate set

of zeros.

Let us now analyze the convergence of Penultimate Remaindering,

specifically whether or not it can be controlled by translation of

variable. We shall restrict our attention to an nth-degree polynomial

f(x) with real coefficients, which for the purpose of convenience

stands for either the original polynomial whose zeros are to be deter-

mined or for one of its depressed forms, and the factor being sought be

either linear or quadratic.









Our problem at th c Joit C be : .lici'.l., staatj .a: follows:

CDoe' ther- e.idt a re'al number h fo'.r which Lin's Penultimrate Te-iinde.r-

IrgF shall con'.ergr for f, :*h

nmi r,
i-1
L.Et :'. = 1 a.:.: .:- and tre llvizor d,:'i =x
1 1 I

:x- i I.-.. t-e k I: k for k = 1,2. From [1], tre .ia.nr.'lue .of

th.- natrij.: of err.orz ri tI.h iteration are

-i ,I,n c r,- '
'" = -C1 '" .n-. k "C I'_ 1: = 1.2, ttere

' .', = f..i di ) = :-.n- c ...' .. c. c .
n -'


-i n-I .
r -c i e c. Sc = .. vlere
1 .1 "1 1 ri-i1 '


:.' i : fi' :', ' : r .: = 1 .
1 1 n


O: course rithe eigen'talu;i were deri -ed ona the ,aSsu,,ti..,ri that at

each stage of itirition, ti J-i.iation-r of the coefficient. .of the

trial di-.or fr tho,,- of tin, true dlivisor aru mriall rinough to

allow tcrms in ool'.ing said .de.-iati.jns s o-rd.r higr her thai on, to t-

nWglcCt.d.

1j obt ii tlh correstonding E:.rpression : for th- eigtn.'lueS

..th I of fix-:*h 1, we u?- thE fact that the zeros of l:,.*h i are

,.-h. j = 1. ... n anrd that 3a na \ b, more convenierntly, ,rritein a
n
*'fL -c G -J
I : -' i -: I -
S. c. .
1 1 -' n "*
i'J j. '


for i y.i quajratic, ainl k = 1.2.








n (v -Yl
Similarly for d(x) linear, p =1 -
Sj
2
n n
(y~-h-P +h) n (Yj- k)
So P (h) = 1 = 1 for d(x) quadratic;
k n n
n (v -h) T (v -h)


n (vj-l)
3 13



Similarly, for d(x) linear, Pl(h)+ 1.
n (Vj-h)
2 3

Our objective then is to find a real value h for which IPk(h)J < 1

for k = 1,2 or for k = 1, according as d(x) is quadratic or linear.


Case 1. There exists 8k equal to some yv, j = 1, ... or n


This implies that for any h, except possibly h = Bk if real

Sk(h) = 1. But h cannot be Bk, for if it were, f(x+h) would have 0

for a zero and the Penultimate Remaindering cannot be applied in that

situation. Hence there is no value of h for which the iteration would

converge in this case.

To avoid such an exceptional case, however, we can first obtain

all the multiple zeros of f(x) and leave a depressed polynomial with

simple zeros. One way to do this is to apply the Euclidean Algorithm

for determining the greatest common divisor or f(x) and its first

derivative. To illustrate, if

r Pi
f(x) = ~ (x-e.) p 1,
1 1 1

r pi-1
f'(x) = n (x-e.) H(x) where H is a polynomial of degree > 0,
1 1








r P.i -1
then gl: = f,f' = n : .-e. I wr..se zeros are the multiple :erer.
1

of f :-:. He-re Ih,ki stands for g.c.d of and n If g iz readily,

facto:rabl, we can d,-t-erLin the multiple zero. of fi' :,, includ.ing

their resFpctive multiplicities. Otherwise, wiE rmSy s W .ll consirder

fi l, gi & is .our o.riginil pol.yormial instead1 of fix :, for the zeros

of the forrimr are siFmpl, and precisely the .arme -s tht zeros of t rie

latter. Tre miulti.licities of the zeror of fix': obtained by the iter-

stion on f.:: g', can be deterrrrined later.


Case '.A All the zero. of f' x', are simple

n
Let n iv-ej, A* Ei, k = 1.:, where for eacr; v

Ak and Pk are not C- simrltaineo.:uly. Let the zeros o fi :': be

' .-1 = ti, ard 1 = t di, w r t ... :

S= c for j = 1--s, ... n. Here z 1- an integer within the closed

interval (1, n I] and the c's and d's are real numbers s:ucr trhat d > '

A slight rmolification will be necessary in case di :I is linear. na Tel~ ,

that the complex zeroc be ,2t ani v2t*, t starting from 1, ... s:

and the re-a :ero.s 3lde from -.1 r 1 be ,j,j = '*, ... n, where

s [O,,~].

rTree len-iTm s are worth., oCf norE at this point:
n
Lermis 1. r. -hi'or its corresponding e..prs-Eion wien d'.:1 is

linear' is alwuas real and nonzero. If all the zer.s of f,'x. are com-

plez, the said .expression is positive.

Proof.

n ,-h, = n ce -hi d tl][ic -n i-d iJ] n t' -hi


= n[ic -h d -h
S t t t









which is clearly real, and nonzero provided h / any zero of f(x).

In the above the empty product = 1. If all zeros are complex,

n 2 2
n (V.-h) = n [(ct-h) + d ], which is positive.
3 t t t

Lemma 2. When B 's are real, T (Yj-B) (or its corresponding

expression when d(x) is linear) are real and nonzero. For two succes-

sive real k's, they are of the same sign, positive when the k 's are

the two least real zeros.

Proof. The first statement follows from the proof of (1),

replacing h by Bk. To show the other, let us for the purpose of con-
n
venience let nk stand for n (y -Bk). Note that the sign of Tk depends
3
only on the second product n (c -Bk). For Bk to be the two least
j J
real zeros, each factor in that product is positive, so nk is positive,

for each k. When the k's are not necessarily the least but still

successive, there would be as many negative factors in nl as in r2,

hence they would be of the same sign.


Lemma 3. If the B 's are complex conjugates, 'k s are complex

conjugates, too.

Proof. To see this, let B = c1+'i = 2-" Expand T1 as

s n
2 [(ct-cl)+(dt-dl)i][(ct-cl)+(-dt-)i] 2s+ [(cJ-c1)-dli] =


s 2 2 n_
S[(ct-cl) + dt d1 2d(ct-c i] 2s -[(c Jcl)-di]

On the other hand,
s
n2 = I [(ct-cl)+(dt+dl)i[(ct-cl)+(-dt+dl)i] 2 [(j -c1)+di
^2 2( ie2s+l ^
2 Sn
s 2 +d 2 ence =
n [(ct-cl) dt dl + 2d(ct-cl)i 2s [c -c)+li]. Hence 1 T2










Let us now go back to ,.Our main problem, that of snhorirng the

.-.atenrce of a real '.-alue h for L iAichr Il ,'hil < i, k = 1,'.


Theorem 1. LEt fI'l be an rnth-degree pol.-1:nomial with re l coefficients

anrd simple zero-j .an. having ast least one real :ero. Trhen there

idts a p3olsitiv. numb.rr r for which Penultimate RemnirJndering

applied to ix+**h, will converge to a limner o.r q,'sadrqtic factor

according~ a n r i- odd or ever.


Froof. If n it odd, .r icr know- trere is at lest on. real z-ro.

Let P. be the smallest Fruch. 'rThen.
I n
n Ci. -- C
-. '.1
o.ri. = 1 -
1 n,
n -hI


iwhre, as pointed cut in L'e"1. L .


A = n -. 1 .-...



Since F ,.',,,-nr mu-t be positive tc. have |01,.r,,I < I, let it be C > ,',.


i.l h'i I | i 1 IC-AI4C e= C C -A Cr c=>C>A
n
C ri to b. tbieg i en h .so that the equation h I..,-h' = C wouil hsav a

r-al s-.lutlon in h. For Fprctical pirpo-res, we car e stiate r, iro-.ided

we krnow arn estimate e' of 1, fo.r :inc: A = 011, wr ere C = fix) d,'xi,

A can be estir-ated by g w~t re gi e'. f.. i GI* I, fro.:.

vhich C can be estimated, ard corn:eque-nti, h can be etimsted b.


n-1
C pro .idrd C is large -rough.








Henceforth, let us assume n even. The rest of the real zeros,

if any, may be obtained in a similar fashion as above, either singly

or in successive pairs starting from the smallest. If in pairs,
n
n (YJ-k) Ak
Pk(h) = 1 1 -
n (yj-h)


where A1, A2 > 0, from Lemma 2. C must necessarily be positive, too,

to have Pk(h) < 1. Moreover,

|Pk(h) <1 ;=> C-Aki < C <=>- C < C-Ak < C =>C > Ak/2, k = 1,2.


<=>C > max fAk/23

n-2
So by choosing C large enough, we can estimate h by V/C .

Thus we have shown that a real zero or a pair of real zeros

can always be made "stable" by a proper choice of h. By a stable

zero is meant one wherein the iteration can converge to its corre-

sponding factor. Therefore, all the real zeros of f(x) can be

obtained using Penultimate Remaindering, with translation on x,

if necessary.


Theorem 2. Let f(x) be an nth-degree polynomial with real coefficients

and with all its zeros complex and simple. Then there exists

a positive number h for which Penultimate Remaindering, applied

to f(x+h), will converge to a factor corresponding to a pair

of complex zeros.










Proof.
n
Oi, = 1 e-- wher- n = A Pi iana .. = A Ei,
1 '
A, E real and C C 0 using L-.rias 1 mnd 3. Since 01, I rd j r, are

cor',ie:: cor.,nj gate I1, i' = I .i h i r -o that we rnEed. to c.:.n,.iter

o:nlr, on:n of trcm.


0 i = = ,1 A C. IE. C.i.


From tries, it is clear thlat A mu-t neces arily be po..itiv e irn order

t.o ha.e Ii'r. ni 1, for if A < 0.


I irhl- = '1-A C: iB. C" > 1 since AC C < r, frcm C r,.


IUsinp L'er ma be-low. we know there e .it t wo zero f : fi x fIc.r wri-cr

A '-' .; nence e choose trh:-s two :.er:: ts and %


I, ih. : 1 = Il-A C i'- (B C) < 1 = 1 A A


U A

n--
C==> AC Af -B' < G'C >


7r.en take nh i C ti.-re i- large enough.

He'-nce a pair o: corplezx zros of fixi, n a l:,', that which

corre-spored to A > C', can alwai;, be made -tabl- by trie ch.Oicre :f

a sufficiently large h. Ey -epeating tr,.- iterati':e roc.ess on eacri

depr'e:Jsd .olrnomial, we ca r tru: .:btain a-l tlhe comiple: zeros o:f fix .


Lcn.,. L. If fi zi has real c:.effi.:ient- wit all of its. zeros.

v', co:mple.-:. then there .:-dstE a zero F whose imaginary part is prsi-

tiv e- cr that Re ,r i > r.
= ,'^ 1









Proof.
n (Vy -) = n (B-yj) since the degree n is even,


f(z)
(z-B)(z-7) z=P

f(z) f(B) 1


f'(- ) -2if'(B)
2i Im() Im (B)

-2i[Re(f'(B) + ilm(f'(B))]
Im ()

2Im(f'(B))
So Re()) Im 7


Taking B such that Im() > 0, Re( n (yj-P)) > 0 if and only if


Im(f'(B)) > 0. To show this is true for some B, consider the function
1
S, and the contour integral of the function around the curve C,
composed of the real line and the upper semicircle. C, therefore,

encloses half of the zeros of f(z), namely, those whose imaginary parts

are positive. By Cauchy's Residue Theorem, f dz/f(z) = 2n iS r where
C J
rj stands for a residue of l/f(x) at its pole inside C, and the summa-

tion taken over all such poles. Since the poles of l/f(z) are pre-

cisely the zeros of f(z) whose imaginary parts are positive,

r. = lim (z-y )(l/f(z)) = [ lim f(z)] =-1 J
Yj zy^ Y j -- f 777

if j is such a zero. Note that f'(y ) / 0 since y is a simple

zero of f(z).









iTherefore dz f = ;ni 7 1 f'- ,
r Im,"". >,'


_[F i zx' i i ,if' i 5



Ii- fJ f (\' ; rei f I< . 'ii
= 2r,:l h 1' fI1

Jf L/ *,


I'n trhe otier rianLi,


I d:z fl' = lirm :L T i xi
C ) Fl 1 I .

The fir-t integral is a poiti.ve vilue .incm a - f .1: -

..-ing n eren, Tl r ile the second i1 0 :ince trte den-:ninrator of trhe inte-

grandj is !JoLinated -,y R'n whtlre n > 1. Hence J dz fl''1. > i:' Ciiomparnir
C
this ith trie rflrt re ult, we dei.uc that I-i f' if > I'. fr. m urich

we lPrtr.-r deduce that tnere i orne, -ay of those 's sucn trat
J
Tnif'i I > '-. Prhis provSc the lei'iTla.


Concluiing R:mark:

U ha9. tnus proved that Lin' Fenuli mate ERemainlering can be

made to converge t:. a quadrartic factor t.y a pr.:per choictL of sone posi-

tive number ti. It would be er-y nice ifi h canr be estimated h ich will

work for all the zer:-. H:.we.ver, th_- is ver,' .iLfficult to di-tenrmine,

if not impossible. For. even if we have all tr e e-timiates .-f tre zero.

ure are seerinr, still e1 have no wa, cf finding out if a certain r, is

good enougr, to ma.ke a particular pair of zeros most stable, in the









sense that the said zeros would be the first ones to come out in the

iterative process, because if they are not, then further translation

might still be necessary on the depressed polynomial.


EXAMPLES


Example 1. f(x) =x x3 -x2 -x-2=0

For h = 0, the following machine output is given for the

coefficients of the divisor x2 + b(2)x + b(l):

b(l) b(2)


2.0
2.0
-0.6666667
-2. 71L287
-1.4626864
-2.4890507
-1.53336714
-2.4357176
-1.5806082


It is not converging.

With h = 1.5, the

f(x+h) = x + $x3 + 8x2 +

-0 5078125
-0.5881521
-0.7475877
-0.8380733
-0.9434391
-1.0292210
-1.1101841
-1.1704483
-1.2114062
-1.2328593
-1.2431159
-1.2472219
-1.2489542
-1.2495781
-1.2498472


1.0
-3.0
-1.6666667
-0.7142858
-1.2686565
-0.7554764
-1.2333116
-0.7821669
-1.2096235
.... .. ... .


iteration converges although

2.75x 4.0625


0.3437500
0.7404567
0.9670812
1.1892924
1.3803033
1.5618733
1.7185219
1.8419024
1.9222638
1.9660164
1.9860708
1.9945306
1.9978617
1.9991815
1.9996808


not rapidly:









Fromr tr zerom of x'" .i.. ~t8C 1.2hL%9L2 ani

depresized Fpoly.n-iTdal, tie obtain, afttr atding 1.5 to' ech

the zer.s of f :.'):

'2.0 s compared to j .C
-C.''" tre 3ctujl :-ro -1.C
-0.C'Cil. 0.9998hi i i.
Eraiplc fix L


tre
o trt
o. f Tr h


Exiple .\ = x ;.07 x .t8.. .C9L x .28

For h = C, tri, following output -srlow diivergence:


:.L12179_. -0.1 .-".7.
-0.7 57.L 2 -2.1c87 1.


o.2.C .Oi 1 C:.LCL95.L
1. 37077 .C;02LC'
LS.72hl16 '1.92L66?57
O.CCC'819 0 70L68 '-
LC.811 i. -0.,2.-'OI.


Ta~dri, n 2, so that fi- .-h' = x .91c:- 12.218' C1:*:

'.SL.rx l.1 tre iteration convcrges:

C'. 'Ll'997 0.79; "
J0.:'?L31 1.0157895
c.62L50 1.05 C 7 692
'.6 1719 1.0889571

.. .2c, C.9810' L
,-,.61c8 .:8 C' 7.0,CL 8
0.61:.3958 o;78.261
0..:.1677 71 0 O':. 5
,.c 1?r5 0.982."61
u0. .i 07 C'.98 2... l

Thus, from the *eros 01 :,: .98226.* .618072 and tne

Jepressed. Fc
:-roz of the original polm'nrdial, nael;,y.

i. ,O ..19 : .1 .' 6 i.
-O.L711? .79782 i.








Check: n(x-r = x 2.075x + .6679878x2 0.090261x + 2.2776718.

Remark: Since the coefficients of n(x-r.) differ from those of the

original by significant amount, we can improve the zeros

obtained here by applying the iteration in the next chapters.

(See Example 3 of Chapter IV.)


Example 3. f(x) = x3 8x2 + lLx 12

For h = 0, trying for a linear divisor, the iteration fails

to converge:

b(1)

-0.8571429
-1.5233160
-2.9027832
15.0731962
-0.0331687
-1.5L53726
-2.9746884
12.648L070


Taking h = 5, however, a linear divisor is obtained:

-1.8888889
-0.6591671
-1.2100789
-0.8978157
-1.0565o54
-0.9707772
-1.0156622
-0.9917623
-1.0043762
-0.9976874
-1.0012255
-0.9993515
-1.00034h3

The zeros of f(x) are then found to be 6.00034 and .99983

t 1.00086i (as compared to the actual zeros 6.0 and 1 + i).






1;


Ex T.ple L. fl':. = .&x- 1*1. c .e. .; ' '" .
= ..Z L. f-',.c 1<. -A V 15.x 8 86..cz

..-':'. + .''5

Applying Ferultimate R.naindcering without trarislatiorn, two

cerus, -.Cl9c and -.CL.., are obt i.nei after only three iterati.-e

stages Howev.-r, for ther depres-ed Flmc.miani
. 25.6.7LL09":" 1-.6..8..; '?.. ; 1":.?095:11? L:.:'019.

the iteraticn diverges.

Appl..ing then translation of .vari.able witr h = ', the p.-ly-

nori l is trarnzfl'orTmc to x 2i. .SLLiC 'W1'. 5.00 x"

* ,111.L.?.z?.:: 9- c'.;l9' :.1. and the cut.put for the coefficiernts of

the quadratic divisor o-ught conrv-rge-, though l.:.'wl,':

b' I l: )

..A2hl2 3.2KA ,,0,
5. t',,?,u r.'l.65 h..t:. .r7'77'
t..7TrC0lL L. ?L92-.

8.c'i FTc% C.. .M 1 ,0
9. ."- 100 2 '.11?.'8.?.2pl




11.9807lcr *.602672
."' .. .r:. 9


11. ,"i o ,". r.160 1'


12'.0l3.2hM7 6..8206958
12.O0779C c .'.15 73c
13 .3PT'.;'h 2 9,-715 :,


The .-ros of :x 1..97. "0 1i.'.762'h6 are -3.L9L62c'

. 1.09ci.L from which two aros of the original pol;,TiodIi.a, namely.

-.Lc89:E-, l. 1.;.,i are ottaineid.








2
The final depressed polynomial is x + 261.7624805x

+ 661.3328808 whose zeros are -259.2111402 and -2.5513402; hence two

zeros of f(x) are -256.2111402 and .1486598.

Since we have used a factor whose coefficients have not con-

verged yet, it is possible that the zeros obtained from it and conse-

quently the succeeding zeros found may not be a close estimate of the

actual zeros. Hence the iteration process which shall be established

in the next two chapters can be used to improve those zeros. (See

Example 10of Chapter IV.)


Example 5. f(x) = x6 12x5 + 50x4 70x 41x2 + 2x + 390

As in previous examples, Penultimate Remaindering is divergent

for h = 0.


For h = 6, f(x) becomes x6

+ hoh6x + 1950.

The following output shows

b(l)

0.6469092
1.1123276
1.4510382
1.7056651
1.9032140
2.0606095
2.1887029
2.2947577
2.3837912
2.4593760
2.5241261
2.5800025
2.628509b
2.6708241
2.7078851
2.7404516


+ 2Lx + 230x + 1130x3 + 3019x2



convergence, but quite slowly:

b(2)

1.3401789
1.99524h9
2.4011633
2.683033L
2.8925371
3.0554308
3.1861302
3.293L612
3.3831297
3.4590352
3.5239491
3.5799099
3.6284607
3.6707982
3.7078715
3.7L0h4h3










At thiz pcI.ntL inste-.1 :of cortinai.ng the iteration, it. eight

tbe Wrije to c.:onid:er the above c.:.efficients a; troj~e .:f a ,i,.-i.r of

f x*-h' wi dth tIe! intEntri.n of ppli-inrg 1 tsr s co.rrEctiC rEa3ire

trn it-rative proc-ess in the ne:.-. two cti pters, zince the latter co-.n-

verges q'itLe fast witer the input r are rnt t:oo far :If from.: the

actual zeror.

Applyng again Fn'altimatE Remainiring o:n the deprscse' p:oly-
L -
nora l X 20:. c :..955c 11 .L"708 07..778 : ;0 18791 .:

L.cL6721 .. 527701
7.8111.0 7 LCt. .2' 1
10.0.186076L 5.MAS -.1

3 1.,1 .2971 C.25L
1..90.101', c.Oh .i ;
1l.L92'?257 ,:..Lh9 7..0.5
lM.8'1-977 6 SL72
1L.8710879 c.l167I1
ib.;21796c 6.220,89ti
Ih..LShLL3 P.. ;2)22"
1L.179l 3i0 ~. il ? 19
l1.697c.722 0.C098298
1U ." 0c ,t.51 .018019 '
13.9&19?17 6.012'021
13.E182898. e.00,5787,
".-.''i .'?-. 0 .: "-

Ag air, t a:ing x" c .JL.E',.fx 1.Y.5, 29u? a3 dli.-i -r. tri

d*eprer: : polyTi:omi al is 1 l ...T,.Sh.. 5'2.1:8. M 09?8 Ui.sng thi

quadratic formuil on Each of the d..-i:.or-, the following :aro f f,: x

are deter-Tint-

-1.'000222 6 = .9399-77
-".7:0021 = .:c5;577
-..0:i2269j 2.099:'91i c. = 2.;71.:.L .I:9291
-7.100'i8L2 1.2?"i 6 : -1.1C~6h:'2 1.2=ji.

As in th- pre~i.:o e::xample, tio improve' the -e zer.:, we wblU are

thenr 1- inytt to the iteration trich .shall be est.ablihe in the n~.

chapter.













CHAPTER III


ESTIMATES OF THE ZEROS OF g(x)


For the purpose of estimating the zeros of g(x), we propose to

derive a set of formulas which will determine the values of I's to be

added to the known r 's to obtain the zeros r. s. These values can

also serve as the first set of iterates in case the iterative procedure,

which shall be discussed next, is to be employed to get better esti-

mates of the desired zeros.


Case 1. t(x) has a simple zero

n+l i-1
Theorem 1. Let ri, i = 1, ... n be the n zeros of t(x) = E aix ,


and ri, i = 1, ... n and r be those of
i n+l


g(x) = xn+2

Writing ri = ri + Ii where i 0

and rn+ = -1/a + 1n+ then
n+l n+2 n+l'
n+2
-Z aqr -1

(4) q s=
k k(rk-


n+l
+ 7 (a. + C )xi-1
1 1 1

as the a 's 0, for i = 1, ... n,





for all k such that rk is a

simple zero.


n n n
B ---Sr+ a -a [ ES rir (CS r)2]
n+l -Ei + n -n+2 r< i 1w i)
i,w=l










n-l
n.1 may be set to 0 by a-surring t .




Frof. n*' i n-l q 1

nix-r -TIy -- r x : =:
n1 xq-1 n i
"6: n*a x i :- rdl
n"-'




w:here from here on. i ranges from i to rn; i. from 1 to n*-, 'Lir IE

.tre eris e -:.e 1fi Ei d.

T'o J ol 'v fo'r TIk let = r in t.l:

_-lk n 1 rk-r -. OnS, rk 1 T| i = q-. r 1
Ic *p I a -n
tk i i- n-- k 1+:2 n -1l


rq-1
rlk o K

k i i1 ni n*l
qlr
q-l

TI.
n (r -r. *1 - 1 + o 'r -TI
S k 1 r-r n-* r. n1


E.rpanding thI .enorrdnator as FpOLcr Perie: and neglectin! trri-m

in a ard r.'S :f order greater than 1,
q a

q rq-1 TI
Ti| = k n [1 __ ...] z
K 'rk-ri i k r-ri r



n rk n-l n n-i

q rq-1
assur.ing bK r -TIi an r-r arae
n i r-ri
ir, ,< rk 1










each less than 1, which are valid for a 's sufficiently small, since

.i 0 with a 's, and from (6) by setting x = 0,
2- q

(-1)nI + nr.
n+2 *n+l n(r. + 1.)
1 1

which 0 with c 's.
q

To obtain (5), set x = -1/an+2 in (6). Then

(-1) n+- n
(_-) n+l a(1l)n en+l (-1)n-1 on
+ + + .
a+2 n n en+2 n-1 1


Tl+ri a n )(-1)n n[l+(ri+i ) 2]
+ n+2 n= n(-nl)(-)nn+
n+2 n 2 n+ an+2 n

Multiplying by n+2 (-1)n and dropping the term in an1 since

a1n+ = 0 from f(x) and t(x) monic,



(8) n+1
n
-[-j + E1 (-1)' a (a )u + 0(1 + r a?)+
n+ l-u n+2 i n 2
Y n+2 n[l + n+2 (ri + Bi)]


Expanding the denominator as power series and dropping powers in

a 's and T.'s in the numerator of order greater than 2,
q i


-TIn [-1 aa + 1 + i n2 + E rrW a+2 x






+- 2 2 n n+2 i n+2 i w n+ x 2


a -r. -S Er.r a + an+r.2 ri ] -
~ n i -w w~i











Case tAx has a i multiple zero


Tiheorem -.




t =


n*1


rr




Trirn a rim ni


z'rc,? c-n t..


i I-


Let r be aj zr.j of mltillicityr, n, > 1. randI r.

nll, ... n, be th- rerairning zeros. L-t r

1 ... nr. r, = nli ... n; ani r n1 the


zero. of gli, where r = r It, r r and


= Ti arji E 'i 1 i 3: previ'.usl, d.efinred.
L n l
nr


constr.'cteid where tnr co-efflcients C _ire Jet rinrej



1 n 1 r aq:r I -

rin Ir-r j


.l


C i -m
kI-r* 1


Sn]
Here n n .:-r J no e-3 tie th ideri.v tve of -r I wit-L re.p-:t.


to :*, e-T.al3ute at x':r.


Pro.,'f.
A n 1
A p.lj-nolmial C having the Ti's a i3 zero-= mls
-r k*1


har'e coefficients C *1 :ati~fying the relation C = ,-i,


mauliplied by, tlhe im. of the proJducts of re TI tsak:n inl -K st


a time. k = ... n -l anr Cn i = i. Inc:-d:d third is the value .of
-, 1 r2 q


c


e









(1/ki) [k](r) where [k] (r) def. n (x-r )
t t dx t=l
evaluated at x=r.

n+l n n
g(x) = E a xq + (x-r) 1 (x-r ) =
q=0 q+l s=n+1 s

n+l
(10) g[k](r) = S q!rq-k/(q-k)!
q=k ql

But from g(x) = n+2 (x-rnl) n(x-ri1 i)


Sn+2(x + 1 n+l) T(x) n(x), where from
n+2 +2 n1t (

here on, t ranges from 1 to nl; s, from nl+l to n, and as in the

previous theorem, i from 1 to n, and q from 1 to n+2, unless otherwise
def. d
specified; n(x) = n(x-r-It) and n(x) = '(x-rs- )'
t 5 s s 3
we can proceed by induction to show that

k-i
(11) g[k](x) = 2 k (kl) [k--m](x) -[m(x) +
n+2 M=0 m t s
k k [kin) Em]
n+2x + 1 n+2 ) 0 [k-m](x) (x).
n+2 n+2 n+l r=O M t s

It may easily be verified for k = 0,1, so assuming the identity

true for k, we can show it true for k + 1 as follows:


[ (x)]' = k k- [k- m](x) [ m](x) + [k-m-l](x) x
n+2 m=0 m t s t


[m+l](x)] + [ x + 1 a2 ] E (k)[k+-m () x
s n nm=0

W (k)[[k- ,(x),[m ](x)
S[m ]() + m[k-m](x) n[m+l] ] + t s[k-]
s t sn+2 m= M s










n m--r ,' x .



rrj=0 rj t ?
., .], .: [ ,],
i n t



ri m C," r



usins the lermn below on the iirst two groups c f ruFrrimati~on in the d'.riva-

c--
tive, with v = I. anJ : 1, r .spctively.


vL-. -- [nl-ml,] [.] (,v-1-ml] (..l] nm
L-mma. M[ I .. I I ti i I I
m-u---- t -,


S.- n[v-m]- [m, ], xI
- ri
t F


for any. positive integer v.


Proof. Left-r.ard side =


=L' t
v-i t -
9-I-

=il0 t


,, t mr



1 1-1 t
Il~nl S


v -1 v-i. ,.-1 .- 1 ,[v ].m +,J :,i' .


n, tl 1





1 t s t s S
1 [) [


rignrt-hand -ile. 'E tD









Now, equating (10) and (11) evaluated at r,




k
n+i a qn q-k k-i k Em)



[(n+2 r+l-n2 ( iT ()[k-m](r)n (r).
( n+2 mn2 n+1(
m=0 t s

Solving for n[k](r)(m=O of the last summation),
t

) n+l k-
(12) k](r) =L e q!r-k/(ql-k) n k E m(k1)[k--m](r) x
t q=k q+ n+2 m= i t


[m]( /[+ 2(r- )- (k )[k-m](r)[)(r) n(r).
sn+ m=l t s s

Expanding

l/n(r) = I/n(r-r )[1-s /(r-r )] and I/[ l+ 2(r-n+ )]
Ss n2 n+l1

as power series and neglecting terms in qa ', T 's and of Cv+l's

T[V](r)
(i.e., 's) of order greater than i, we arrive finally at




E q+ )1 ik (x-r)=r C (k-m)'
W [k]r q=k (q-k)! m x=r k-m+
t n(r r )
S


From this, (9) follows immediately.














CHAPTER IV


A IIONCOi TIIlCNAiTIG 111 '1iETRAI'. PROCESS FOR
DECTERNIJNG THE TFROS OF gi ;:


Case 1. tic': has sriipl7 zeros


Ineorem 1. Let t.i have zeros r1, i = 1, ... I. If ri'a ire the

corr-prF'P iring zerc.O of gixi trat i- r. = ri i, 1 1, ... n

:ucl-, that C, r with the V 's, and r = -1. o. TIn then
i q n*l n. n1
r J r as j a for all k vich that rk i rrimpl&e z.ro

n l I *
of tU i and -1:. r w h.r.




7 j'r, r -r r l r i
k. r Pm.nl
n
1 i-l 0un*-u rn ir n


11 -- nI '.

a 1r, i TI i are thn e.tirat'es iL I and I .1, re pectiv'-ly, olf

the previous chapter.


Proof. C..oarirng '1 1. aind iL'i with i 7 anid ,, 1, re-p-ct.iely,

1of trie previ.:-u. chapter, we sec clearly how tie it.ration f'oriTul 3 ar.

arriv.-d at, n:aa-ly,, b:. replacing all tt-,e TI's in trh right-handr ride of









(8) and (7) by '(J)'s and those at the left-hand side by O(j+l)'s.

Let us defer the proof of the convergence of these iterative proce-

dures for sufficiently small aq s, and go to Case 2.


Case 2. t(x) has a multiple zero


Theorem 2. Let r be a zero of multiplicity nl > 1 of t(x). Let rt,

t = 1, ... n1 be the corresponding zeros of g(x), i.e.,

r = r + Tt where 1t 0 with the a 's, and rs, s = nl+l, ... n

be the remaining zeros of t(x). Then r + t) rt as j -

where I 's are the zeros of the monic polynomial C ()yk,
0 K+l

the coefficients + being determined by

n+l a q!rq-k k-l
(15) C ( j) q+ a k E (k-1 k-l-m)! C (J+l) x
k+1 j q=k (q-k)i n+2 0 m k-m

(j >0)
n[m](x-r -I( ) [ (r-l ) )] x
s s x=r n+2 n+l

kk 1
S((k-m) C) (j+)[m](x-r- (< )
m=l m Ck+l-m a s x=r

k! n(r-rs- s ()[l+n+2(r-n+ ())] where as

stated before, i ranges from 1 to n and s from nl+l to n, while

(1)
Ck+ 's are the estimates (9) obtained in the previous chapter.


Proof. As was already noted in the previous chapter, Ck+1 is

precisely [k](r)/k!; comparing (12) with (15) above, the derivation
t
of the iteration formula is clear.









ve shall now est atlish the ccnr'.-.rgencce of 111', lii and

I.1i I imult areouijrly fo.r sirfficicntl t sm1aill values o 1. Uithout los1

of generality, let u1, for the lake o..f imlicity., aEsurme that t'.i has

one ruiltiFle :--roI r to witch corresFcpoin.i r z-

We -hall pr.esent.l, aim to obtain a nitr.-. equation

-Ij*1 I *1 j 1 i I
L = E I* E wnir M a-id i sr. are in-1. by n*i'i

rritrices to t- dtcnermined and L5' is a co:lurn victor with elements

Ek k for k = 1, n. ril. K -' Ti -




Let uT. then po :back and consider ,'ase 1. From 7, and '13i,

f-r k = n. 1, ... n,

j.1
I i E Ij* r r n r -r.-TI [l..j -T 1 ] -
k q 1 k [.k 1 1 r, k n.1l



ink n* k nI n


r i r -r i nI i r -r, -TI. [, -TI
irk : 1 1 *2k k i r n'2 k n*L


1k n* "k n,1 +a1 1

= k1 n rr -7111

S: k i nr k r kn 1

--1

i rn r k-TIl E1 r -+1 r. i r *



(1 rk Til n l *1 k ir -r. .







nl-1
[E(rk-r)v Cv+1 + E (rk-r) Ev+ )] x


[lk (rk-rs-s Es )] x


(1+0n+2(rk- n+1)l][1+n+2(rk- n+1 E n+1 ( ))n'

where from now on, v ranges from 0 to n1, unless otherwise specified.

Neglecting terms in E )'s of order > 1, numerator

N -_ rq-1 i(r-r-i n+2(r )] -



[ n (r -r- ) E (n (r k-r -T ))] x
s/k k s s^ p/k P s/p,k s
nl-1
[E(rk-r)V C+1 + E (rk-r)V E j)] x


[l+~n+2(rk-Iln+) n+2 En+l )

where from here on, p ranges from nl+l to n unless otherwise specified,

and empty product is defined 1.

Expanding and neglecting further terms in E(J 's of order > 1,

the terms independent of E() 's cancel out, leaving

(17) N E rq-1 [E(rk-r) Cv+](l+n+2(rk + )] E x

( (rk-r )) TT (rk-r -r )[l+e (r -1 n )] x
s/pk ks sa s/k k s s s n+2 k n+1l

ni-1
E (r -r) E (J) + a n (r -r -T )[(r -r)VC ]E .
O k v+1 n+2 s/k k ss k v+1 n+1









Srril arly, dennrirjtoc.r

n -1


A$k rk r.-- ri Tli1[ r..'n krrrk iT iln il
L r -r I.L
I 1 ] -

r:n rk-r C.

E 1 '1 n ,rk-r.-T E ',
P n-"_, EF" -1
[1 p: F'~k ] n 1 J
rV -r -, _-o "- ir -TI "
s k n.' k. nr l


E..'nt:..ing the reciprocal .:of eacr factor within the brace as

Fpc.er series, ni noting that the nunerator ri a no term inrdependtent of

Z '. e o:ni; nreci the cornstant .:.f the resulting e:'ani:.rn, rei.ily,

1, f'or orders of E' )' > 1 ar- being igno.red. Tnhu



i I k 1Ti 'I[ n* l': r* 1

.ir),-inc fr.r.m n i- t.:, n. w-e thu: derive frrm I 17i and 1. the

n.*l ist to nth rc.i; elen-rents of oar desired matrice: IH and [i. iNote that

the Irrl*lit to nthr r-ows. of 1 are- filled iwitn i' r.

r.reo-.:r. from i and i1L ,
n
-1 - .' a nlI *r.o ,
"'1 r -~ n*: n *l n -'*
I i n-'_ 1
n* '

I"[ l1..-. r.i TI' ] n([l r *i ']


1 + 1 n -.' i IL









Numerator N within the brace


= [l+e (r +1 )+c E )l][Z(-a2)nl-V (1+e r)VC
n+2 s s n+2s n+2 n+2 v+1

n1 nl-v
1 nv (j)
+ (-+ (1+a r)VE l ] n[l+a+(ri+1 )]
O n+2 n+2 v+l n+2 1.

Neglecting terms in E )'s of order > 1, and combining terms,

we get

N.- = [S(-a )nl (l+2 r)C ] SE ( n (1+a ^(r +)) +
n+2 n+2 n+2 v+1 p s/p n+ s s

l-1 n-v
+ [l+n+2 (r +s )] I (-Cn+2) (l+C 2r)V Ev+


Similarly, denominator D within the brace


Sn[l+n+a (r.+T.)]. Hence
n+2 1 i

n
-l+E(-a )ua -+n(l+r ca ) -1I
n+2 n-u+1 i n+2 n L-v
1 (j+l) 1( (a ) (1+( r)'x
(19)E n+2 n+2 1
n1 E_ ffn+2 [l2 2(ri+ ] (- ) (+12r)Vx


Ev+l(J) + n2 n+2 n l +n+2r)VCv+l x

SEpJ(s n (l++2 (rs+s)


from which we get the last row elements of M and N. Note that (19)

does not also give any nonzero contribution to the matrix M. Note

further that each term of the numerator in the expanded form of (19)

has at least an order -2 factor of a 's, so that each coefficient of

Ev+l) and E which is an element of the matrix N goes to 0 with

the c 's.
q









L.s.ring t-ri : for trne mejnrtim e. let t: *:mt.irn for Cas in

anailogous .-presion for E j re = 1. .. r1 Fro 1 ar

it z corrspor.ning expi esp i.:n for C ,l' for k= C,l, ... r -1,


E 11 I r i_ 11rq-k rq-k': .k r 1 1k -'..- 11 x1
"[ ; k n .' 0 m [ "k ,T
_ri -ai i r-T

n "' x-r -- xr o *2 r-n
n[i] i::-P- [jl mx- T [I c-c

r r .l I,
,k-mi,' .IFI n[F]l- nq-rI -'.



[1+ kI. I r-T i )] ) r r c- r q-k :: -
nL n q=k *1


Sl-I k,- 1 -n [ i-r-r -T



n-; Ir Il I r :k-l-m n
]k 1 I xL


n r-r -l' ][J 1- 2ir-Tin j ; k'n 'r-r 'l ":



nl r- -r. TI 1 jr- nl i] '.- rin '1


Frcce -tinr irn tne sme marLner s inr tr,. prev-lous c ia trhat is.

e-T-rtan ig out ai neglecting power :f r.igrer thri 1, :hr ter.is


of trne E., r, rarely,








(. +J) (k-i k-1
(20) Ek+(+) :- (r-r-s)[l+n+2(r-fn)]cn+2k ( )x


(k-l-m) n[m](x-r ) __ Ekm(J+l)

k
n(r-rss)[l+n+2)(r-n+)2 k )(k-m) x


[m](x-r s x=r Ek+l-m(j+) + (r-rs ) x


[1+cn+2(r-)n+l n+2 k (k- )(k-l-m)! C x





[1+@n+2(r-ln+)2 k )(k-m): Ck+1_mS E ) x
n+ n+2 m k-k-1


Z E ()T Ip (x-r s- ) Xr] + n(r-r -P) x









km [m(-rs s x=r] [ 2(r-n+)]2 x
[i+Y (v-P) 2 k k ()j) C fS F






J(k )(km) [m] E
n2 n-i Ck+l is k+E1 -m ) x





nn++
E =- k 2 a q/(q-k)! a+ k S ( i)(kl/m)r x
q=k q+l) n+2 0 i
Sk-rnT n (x-r -S1 ) ___I [i+cn2(r- n+l)l 2 x


S(k)(k-m) Ck+il-m r (x-rs-x~ r ) 3E E p x

n+i
(nr(r-vrP)) + +2nc (v-r-l1)[ SctqrljEkq /
q=k










I q-~2 : k- -ml C [ i.]
n C m n :- rr.


-TI I: r] En 1 i r-r -TI '
1
s1- :c'r- n*1 j



[ -a ni r-tI ]n- .I


jar'Eing k from 0? tD n1-i in I '2' we car, get trn first n-

roiw elermnts of tre mnatrrices tM and N. Ti is complete cur deterrminra-

tion cf the matrices M and 1J.

i.-ts trat in the vector eTua3ticn L 11 = E1 1 i E-

tre n-rnzerc elerenrt of M ire found .ronly In the fir-t n1 rows to the

left of .Sa e.:ciudinF the di agon.il. As for Fi, note that each of its

element a ha; in ev'.ery te r ith'r a factor o .r Ck of power greater

than C' and since C : wi. thr the c t:.ogetr with. th fact., that

thr TI'3 finit- nurt-er, we car maaice 'recr of its element arbitraLril"

small rith the ':.
Sillutre it r n d t llon pg
To illustrate it for n = ard n1 = 3, 5ee th~ following pages.








0 0
e6 1 0
[a6(r-ll5) [ r-

2a 26 2
[r-r-I4l66sr-0l5) [+<6(Tr-1l r-r4-H

o 0
0 0



- - - - - - -





o 0
-- ------------ ---------------------------













- r (r -r 1+ -(r4-r) rq/r(r-r (r-r)2(6


2 2Ia
6,(r -5)] i)2[ +6(r-r)] _-i)2[l1+o6


-[-l 6 -+(+rix [-+E(-a 6 u u+n(l+r x -[-l+Z(-6)u

c]/n2[1+a6(r )] ]r6(1+%ar)/n2[1+a [1+6r]2

(ri ) ^i)|






















., C 0

---- ------------------- ----------
i -



rI -I
i. . .. . .. .






-i rq-1 L i q -i q- r-




[ 1-i.r-Tl ] r-rl.' r


-------------- -- ...----------------- ---
} -1 [-a r q I q- 1 3 -
l'a q-'1 q-1,-?.- -.L ..' .r,["'L .rq--qr q-lL-. '::




[ [ I.: r Tl .] ', Ir -"[ lC r r ^ r- q r -

r I r -

C.-
,-1 1
r: 1 r- r r


S 1- li '' r -r -rIi 1[ l' r -rr i -


-------------- ---------------------------------------------------------

o0 n, l.r :.: [1-+-- ,, n l. +ri
L, -u i



n J1

r.. -T





38


E(j+l) = M E(+l) + N E =>) E(j+l) = (I-M)-NE(j).


Note that I-M is a lower triangular matrix with diagonal elements

1 since M is subtriangular and hence (I-M)-1 is also a triangular

matrix with diagonal elements 1. Moreover, each element of (I-M)--N

consists of finite sum of terms each of which has for a factor an

element of N. Therefore each element of (I-M)-N can be made arbi-

trarily small in magnitude with the Vr q's. Thus for sufficiently

small magnitudes of q max [I|p w] < 1 where (I-M)-N = (p ).

Therefore theleigenvaluesjof (pwu) are all less than 1. This proves

the convergence of the iteration scheme established.










ILLUSTRATIVE EL J1PL-SU


A. As a M*.th.od for Det.urdrinrg trh Zeros of a F:l;,TinOiTi.
fr.omr the Enown .cros of Itz Lower-Degr'ed Part


Ex.a.i,.ple 1

fiupp.ose Te -lcer of hi:. = ?9..?'- l: F .iv0to :LY

* 1o'..3'j?':'8:x 1199.7'-L%.1 are to be dJet. tdr d wherein '.: J?9'2L.*:'

-6 ,.L' '--': iLL:,.' s-'? O:: : 1'p?. 'Ls21 rad beer pr.i,:.u.l,

Studi.e and its zerrjz appro:imSateJ t.:' b: r' ar~ 1 W mr;,; con-icier

inc t ea d
S rI:I x f* .ix

wnei re
fi = -.oR' 1L.00':? 9011..: 11 97965,

and ,ippiv tre itinratior on i I.



oL = ,090r:




OL

oc I.

tre first tri re .*alu.- compuItejd by t.0r n0ach .:.

r 's: 1 i
I -



lI TI "


-.':":'5;8. .:'0:,O 16 -.L'?.: ; -" -7.'~0 '001
-.,:.,: .,_: .00C '726':' -.';L9S ':.' -7.S.c- :T 75,
'5r5-. .'0'2 :'i -.L2L72L1 -7.' 6Li:76
-.0'0',65. 7 .C0iC'02.21)i -.:-7. L117";.









Therefore, the zeros of g(x), and consequently of h(x), are

.999h345 + 1.0026923i
5.5652561
-107.5641174

as compared with the actual zeros of h(x):

.99943h5 + 1.0026923i
5.5652561
-107.5641174.

Example 2

Let h(x) = .001x5 + x4 x3 + 8x2 8x + L

Suppose the zeros of x h4x + 8x2 8x + are previously

known to be 1+i, each one of multiplicity 2. To determine the zeros

of h(x):

r.'s: l+i
l+i

Chi's: 0.
0.
0.
0.
0.
0.001

CO11' C012 C021' C022 5
.001 + .001i .o00o .001i -3.9920000
.000995 + .00099651 .0039801 .00099751 -3.9919533

Hence, the iteration converges after only two iterations, giving

the zeros
r 's: -1003.9919h43
1.0123783 + 0.96489251
.9836415 + 1.03411i

as compared to the computed zeros [31,

-1003.992040
1.0123775 + .96489419i
.9836425 + 1.0342299i.










B. As Ciorrective llesur." for hi.esults Cbtained
t.' Otnse Methodsj

Cx-Jple -

f = :. _- -.C'5" .c.cY- .:'9Lx .';8

,'from E.<.-mple 2 of 'Chapter II.)

r. ':: 1..c.,r.9 c. c6c.i
-.L7119 7978.i

: 0 -I ,'81










1. 0:i 8 i: 3 ..L'- i





-.U1i24:8 .?9&11)9i.

,rerct:. n ',I -r = :.h 1.' V ;i-, 7 .c 9 .0 .. 9998x 2.'. '


f.:.:' x I., :-. lrC1 ':.: ''0 ". ?L,,'

I fron Fla kagML-Iri3-a'iel. i:..rple [ 3] I

ri : .'9'''tI^ 1 .'. 1I'j 3i
-. 71'; ,77





.. .?
L*.











-.0000096 + .0000051
-.0000096 7 .000005i


~3 ~
.0161016 .2339176
.0161192 .2339000


Therefore, the zeros are:

.9996054 + 1.0019251
5.7668892
-207.7660999

as compared with actual zeros, [3]:

.9996053 + 1.001925i
5.766889
-207.7660998.


Example 5

f(x) = x + 10.65x3 + 129x2 + 203.5x + 70

(from Lin's example in his 1941 iterative process [7])


estimates r.'s:
1


.0996209
.1109277
.1123392
.1125191
.1125419


-1.3775
-0.3775
-4.455 + 9.651


ai's are: 11.2551432
0.6052341
-0.1265793
-0.015
0.0
0.0
2 3'3 1


-.1003151
-.1116175
-.1130373
-.1132171
-.1132400


.0078471 7
.0078489
.0078490 7
.0078491 7
.0078491 7


Therefore, the zeros r, are:

-1.2649581
-0.4907400
-h.4471509 +

Check: n(x-r) = + 10.6199999x
1


9.6429441i.

+ 128.9999943x2 + 203.4999714x


+ 69.9997603.


.0070o05i
.00705481
.00705571
.00705581
.00705581













1 : r-" Li -

'i.L.:n's e.*:, ,pl.e u-irng Lin'. 1 hl1

-" timia't r.': :






1l l: ,






-.0C1CT, C '0211.-7i
-. O,',b i' ? -,:'120T i

Th,-r o r,- tr. -.; rL-I &.*e,:


mr- hoi [ ,10.1)





-.1L- ,
.:.'9L' ,
,. '. ,- , 1


T4
1i ... TT

. i-"lL' 7 S ,i, ll< i
. ,,!07E 2 .,:,i S ,-


*C1IUL V.'
*v~ ~- 41


- 13 .LJ:'j L'r '.-


fI'. : ..: . .. 15 .01:.: 2- .- l -.: . ;C :"

S l.5 .' L 'Lul. -.''pl.: [i -]

r. ." -. :'h 9
-.01 0


,'-. '






S TC 110








I
-.0006630
-.0006629

Therefore


2 Y3 4b
.0000385 -.001060 + .0001377i
.0000378 -.001061 7 .0001377i

the zeros r.'s are:
1


5
.0003143
.0003650


06
-.0006102
-.0006102


-.465629
-.0129622
-.1071061 + .94663771
-.3156550
-256.2006035

Check: I(x-r) = x6 + 256.7899933x + 152.0099888x + 258.4299774x3

+ 88.3699942x2 + 4.5199998x + .0443000.

Example 8

f(x) = x4 .94x3 + .6x + 2.99x + 10.45

(from Hitchcock's example using Lin's 1941 iteration [7,1])

r.'s are: -1.04 + 1.08i
1 1.51 7 1.55i

a.'s are: .0764844
1 .0391521
.9oooo0000


01' 2
-.0028549 7 .0087951i
-.0028488 .00833801

Therefore the zeros are:


33 4
.0028519 7 .00o9815i
.0028189 7 .0049921i


-1.0428h88 + 1.07116621
1.5128489 7 1.5450079i

Check: r(x-r.) = x .9h00002x2 + .6000012x2 + 2.989999x

+ 10.4500015.









EC'aipl-1 9

fix' = ;*: F. :' .* 1;.l61r : 15.hOS.i < 7.5 "'. 512

estimated r : l+i
T T 2









1
ier '. are: -.Lfr e ie:


-. C.E0i i





t .: z <,:,3,







-,xarpl, IC,,, ,
CU.









I Tl= 7.c"1 c,. L.:









l r e:
.X^ '72 . 19-"i. "'X2'?5.. .,C972L7'5
.L",;9851 .01997171 .C,2'L',.Lr ."-, 777L
,i,' t.T. .C, 999' ..:',C,. ",,'. ..C, ';9 9 '.9
.0l :",'L"., J .0:, 'C,0 ,',01 .00C 5,'1': .'79 9 ,











Terefore tr.e zero ae:
1 .'' '';'9'?9 .98 0C."0'C0'C,01 .as comraF red t,:. trhe 1.01 + q i

1.95u0. ':'u: .,_0"5,:,": .' i ctuiil zer.:.s: I.9'.--; 1.55.


ELxample I,

frx) = .c .$ .1 i* .1 9 1 5.;07 + .j *

i from E:.-ap,Flle ,;,f Chapter II I



-. 'i5 .OC,0






*.. C'1
-5.CsL7n.

C.,:,96817
U.
O.










1' 1 2 t3 4 5 6

.2178881 T .2009548i .000000o .0000583 -.4455926 .0000761
.3283407 7 .2092154i .0000152 .0010245 -.7035223 .0000761
.3809192 7 .1660623i .0000013 -.0002359 -.8032323 .0000761
.3872277 + .13840781 .0000008 -.0001477 -.7891522 .0000761
.3795185 7 .1413487i -.oo0000008 -.ooo000161 -.7598841 .0000761
.3755483 7 .14970398 -.0000009 -.0001628 -.7581017 .0000761
.3770083 T .1504719i -.0000009 -.0001667 -.7652827 .0000761
.3783201 1 .14852631 -.0000009 -.0001627 -.7672360 .0000761
.3782191 7 .14924401 -.0000009 -.0001609 -.7658265 .0000761
.3778706 7 .1482930i -.0000009 -.0001615 -.7650339 .0000761
.3778220 T .1485242i -.0000009 -.0001621 -.7652305 .0000761
.3779017 + .1482930i -.0000009 -.0001620 -.7654734 .0000761
.377930o 4 .1484091i -.0000009 -.0001619 -.7654734 .0000761

Therefore the zeros r. are:
i
-.1066945 + .94699091
-.0157909 ~
-.029619
-.3168139
-256.2110634.

Check: r(x-r) = x6 + 256.8000183x5 + 151.9045715x4 + 258.5060730x3

+ 88.5159674x2 + 4.5009227x + .0500106.

Example 11

f(x) = x6 12 + 5x4 70x3 4lx2 + 2x + 390

(from Example 5 of Chapter II)

r.'s are: -1.1068842 + 1.233i
1 2.9771064 4 2.09291
3.2595779
4.9999778
's are: -202.5803146
1 90.9204035
-15.9860919
1.1802006
0.0001483
0.0
0.0
0.0









fli p T lj TI1L Tj
P TI.-'


.1051 icL 7: 371i .2?17.iL .0":'b8Lt -.17LC'. .1..C'0.U
.1 .7177 .-' "TLJ1 .,2?:.i:'7 '9:.61671 .25C951- .,i,:C :'
.1 5?5 7 :'@:-91? .c: ?t76L .0:92'9:71 -. 5~ 7 .:" ?L
.1* 8.511 *- .2 .*2989'6i .,2' 1i9 -7 ,9289' i -.:'..c .,:,O : L


Tri.rf.:.re trie zero- ri' are:

-1. ':Coo071 1. CC'llLi
'-.99959r 2.9p0C' 171
c '?' 8 :; '.o .:' ,
5.i c':' "0 :

a; compared t*) t ihe actual sd ro:

--i Ni:; i : .



C. A- a Method, Ind'-pendernt of O'ther Method r .
fcr Finaring the Zero: of a Pol,-norra.


L:':E ,ple 1:

f, *.: = A" c.'' c'002:
Io)tinj that fl101 I: airn fil l < I: n d tr.at f L < 1:1 while

frli > o that trere are real '' zro- betw'ein C' and I and between

L and ., -we tA.:e for initial guesse3 .5 and L.5 for tr,.- rial :er.:..: ...

f ::', .and Eince fl :' >-.5', :.-Li. + w.ho e z r, are : .r71 ,



Here ', 's ire: -.'O''Li'LL
1. ,bLT; .

-. 0: ni'.3o
.U
,ix-r c,,
L L









711

*.065513
.0721217
.0754045
.0752518
.0752249
Therefore, the zeros


72 3 l4

-.0855367 .0194928 .2118031i
-.0886485 .0104368 + .2406905i
-.0890344 .0069015 7 .24237761
-.0889806 .0068349 + .2422557i
-.0889739 .0068747 + .24225161
are: .5752249
4.4110261
.0068747 + .6277484i


as compared with the actual zeros

.5752290
4.4110253
.0068730 + .6277460i.

Example 13

h(x) = x3 + 8.0000038x2 + 8.0799952x 15.8399957

Considering x3/8.0000038 + (x2 + 1.0099990x 1.9799985),

the zeros of the quadratic part are readily obtainable, in fact,
2
S-2 and 1 since the quadratic polynomial x + x 2. So tempor-

arily truncating the cubic polynomial, we get for our input

rl's: -2 and 1

S.'s: .0200015
.0099990
.0
.1249999


-.3333319
-.5879487
-.6142955
-.6243220
-.6254606
-.6258970
-.6259472
-.6259662


2
-.0516668
-.Ol88511
-.0466220
-.0467498
-.0466695
-.0466748
-.0466714l
-.0466716


3
1.3849987
1.5871481
1.6612947
1.6690061
1.6721429
1.6724822
1.6726190
1.6726339










Therefor- the Zero: of hi: arc:



3s c-impFre,' witr the -actual zeros:


-Jr

-t '2,rp


E.carnpie lb

L
h x' = :6' 9.999*999' .O':"0'i!i .x' 9.?,9C ,'-c'

S.,r-' 'j x L9. c999;990'

The zero: of hi x are the -amre Acs trose of gl x' = .1" fi ;





Since the zeros of : X 2 are rea.il, found, ~e use tre tecrlrique

of temporarily truncatiing gi'x' to obtain the zeros + an- i 1 of fili.

jo taYe ri ': *
1. lh21


1 '


- iii~luii
* o:: cc's0


TI1 TI;

.Ol, .i. d 6
Ci T 'c. 'C l,': ':'Li
.'Ic'FTT_ 7 .Cilc: f'i-
c TTE 772 ."i' .5611C

'Th r fore, the zeroc are:




a- comrpared -rith tr, actual


Tl Ti
9
-0. ':* .C"2' -.. '-'C9c96 -'9?
.:it'?7c; -.*..921'r 9
-. .i29'9? -.079.671
- .c.'582852' -. 07''T90c 20

,l_ 5772 .9?9L91i

-1.L09&26:


-1. 93 '2,,1,
-9. 9c;. 7c1P




1. 2"1..'h'

-'?.897971.


TI,

.099999"
.100120

.1, 20,31










Conclusions


We have thus established that the iterative formulas proposed

in this paper converge whenever theaq I's in

xn+l +
Cn+2 + (ai+ai)x-


are sufficiently small. As a means of solving for the zeros of

n+l i
E (a.+a. )x ,
1 1

that is, an+2 = 0, that condition amounts to requiring that the esti-

mates r is of the actual zeros rid's, which are input of the program, be

not too far off. However, as shown in Examples 10 and 11, the process

frequently converges even when the u 's are as big as a three-digit

figure.

The iteration is then particularly effective as a corrective

measure to other methods which obtain the zeros one at a time and thus

are "contaminating" procedures. Note from the example that as such,

this noncontaminating iterative process usually obtains the actual

zeros after only a few iterations.

As a method, by itself, of obtaining the zeros of a polynomial,

it is especially applicable to a polynomial of degree 3 or one of higher

degree, wherein ignoring the leading term gives a polynomial of lower

degree whose zeros are readily obtainable or are known. (Examples 13

and 1I illustrate this point.) Then, after the zeros of the truncated

polynomial are found, the zeros of the original polynomial can be






51



.jterrirind. Here, 5' C 0 and tr it.irati,.'e procTss OCC ee5 to

converge hrn ). n-| is quite s.all, trat is, i.hen trhe Eeconl c-'ffi-

cient of the oriiral monic' p.:l;,.rmonial is rather large in magnitude.




































APPENDICES








APPENDIX A. MACHINE PROGRAMS IN FORTRAN II FCR IBM 709
C LIN'S PENULTIMATE REMAINDERING WITH TRANSLATION FOR QU
ADRATIC OR L
1INEAR FACTOR,WI1H GCD ALGO TO DETERMINE PRESENCE OF MU
LTIPLE ZERCS
CIMENSION A(201, A1(20), B(2,2C), C(20), D(20), AD(2C)
I DIMENSION G(10), Y(2)
1 REALINPUTTAPE5, 10, N, JLIM
10 FURVAT(212)
REACINPLTTAPE5, 14, (A(I), I=1,N), H
14 FORMAT(4F15.7)
C GCO ALGURITHM TO DETERMINE PRESENCE CF MULTIPLE ZERCS
CO 49 I=1,N
49 Al(I)= A I)
N1 = N-I
AN= N
CC 51 I=1,Nl
AI=I
51 AD(I)= AI*A1(I+i )/AN
NN=N
M=N1
C PERFORM DIVISION OF A'S BY AD'S
79 N1= NN-1
00 53 I=2,M
MN= NN-M+I
53 C(MN)= 0.0
MN=NN-M+1
C(MN)= 1.0
CO 55 I=N,N1
CELTA = 0.0
CO 57 K=1,M
MN= NN+M+1-I-K
57 LELIA= C(MN)*AD(K) + DELTA
NI= NN-I
NIP= NI+M
55 C(NI)= AI(NIM)- DELTA
NM1= NN-P+1
C(NMI)= 1.0
CO 59 I=1,M
CELTA =0.0
CO 61 K=1,I
IK= I+1-K
61 CELTA= DELTA+ AU(K)*C(IK)
59 C(I) = Al(I)- DELTA
C CHECK WHETHER DIVISION NEED BE REPEATED OR GCD IS OBTA
INED
CO 63 I=1,M
PM= M-I+1
IF(ABSF(D(O M ))-i.OE-5) 63,63,65
63 CONTINUE
WRITECUTPUTTAPE6, 67, (AC(I), 1=1l, )
67 FORMAT (14H THE GCD IS /(6F12.6))
GO TC 81
65 IF(PM-1) 69,69,71








69 lRITEOUTPUTTAPE6, 73
73 FORVAT(9H GCC IS 1)
GO TC 91
71 NN=W
v= FV-1I
00 75 I=1,NN
75 AlI( )= AOll)
CO 77 I=1,M
77 AO(I)= C(1)/O(MM)
GO 10 79
81 CO 83 I=2,M
NN= N-V+I
83 C(MN)= 0.0
VN= N-M+l
C(MN)= 1.0
NI= N-1
CO 85 I=M,NI
CELIA= 0.0
00 97 K=1,M
MN= N+M+I-I-K
97 CELTA= C(MN)*AD(K) + DELTA
NI= N-I
NIM=NI+M
85 C(NI)= A(NI?)-DELIA
N=N-I
CO 89 I=1,N
89 A(I)= C(I)
91 WRITEOUTPUTTAPE6, 32, (A(I), I=1,N)
32 FORPAT(46h rHE CGEFF GF POLYNCMIAL BEING FACTORED ARE
/(4F15.1))
IF(N-2) 1,1,93
L PERFCRM TRANSLATION ON X OF F(X)
93 CU 174 =1,1N
174 C(I) = A(1)
00 175 1= 1,N
NO = N-2-1
D \C) = 1.0
CC 176 K=1I,
NK= N+1-K
176 D(NK) = C(NK) + HO* (NK+I)
AI(I)= 0(1)
IF(I-N) 2CC, 175,175
200 NI= N-I
CO 201 K=1,NI
201 C(K) = O(K+1)
175 CONTINUE
AI(N+1)= 1.0
WRITECUTPUTTAPE6, 32, (A I(1), I=1,N)
L FORM THE IST TRIAL DIVISCR
IF(XMCDF(N,2)) 22,23,22
22 M = 1
GO TC 25
23 V=2
25 IF(ABSF(A1(V+1))-1.OE-5) 27,27,29








27 WRITEOU1PUTTAPE6, 20
20 FORMAT(39H 0 DIVISOR, CHANGE STARTING VALUES OF B)
GO TO 1
29 00 31 I=1,M
31 B(I,1)= AI(I)/A1(M+1)
C PERFORM DIVISION
J=l


34 WRITEOUTPUTTAPEo, 16, J
16 FORPAT(1X, 18H ITERATION NO. IS ,12)
WRITEOUTPUTTAPE6, 14, (B(I,J), I=1,M)
CO 36 I=2,M
MN = N-M+I
36 C(tK)= 0.0
VN= N-V+1
C(MN)= 1.0
N1=N-I
CO 38 I=M,N1
CELTA = 0.0
CO 40 K=1,M
MN= N+V+1-I-K
40 DELTA = C(MN]*B(K,J]+DELTA
I = N-1
NIM = N-I+M
38 C(N[)=Ai(NIM)-OLLTA
IF(J-I) 42,42,44
44 T=ABSF(B(1,J)-B( 1,J- ))
IF(T-1.OE-3) 50,50,48
50 IF(V-1) 1,68,46
46 T= ABSF(B(2,JI-l(2,J-1))
IF(T-1.OE-3) 70,7C,48
48 IF(J-JLIM) 42,42,54
42 IF(ABSF(C(1))-1.OE-5) 27,27,56
56 B(1,J+1) = Al(1)/C(1)
IF(P-1) 1, 62, 64
64 B(2,J+1)= (A1(2)-C(2)*B(1,J))/C(I)
62 J=J+1
GO FC 34
54 WRITEOUTPUTTAPE6, 66,N
66 FORMAT(40H ITERATION NOT CONVERGING F
GO TC 1
C OBTAIN THE ZEROS
68 X= -B(1,J)+ H
WRITEOUTPUTTAPE6, 106, X
106 FORPAr (13H ZEROS ARE /(2F12.5))
GO TC 72
70 CO 74 I=1,2
I G(I) = (0.0,C.0)
74 G(I) = B(I,J)
I 82 CALL OUADRA(G,Y)
I hH = (0.0,0.0)
hH = H
CC 87 1=1,2
1 87 Y(I] = Y(I) + hH
WRITEOUTPUTTAPE6, 106, Y(l), Y(3), Y(


OR DEGREE


2), Y(4)


/12)








72 N=N-M
IF( ) ,L,84
84 CO 80 I=1,N
80 AI(1) = C(I)
IF(N-2) 1, 150, 12
150 Y=2
00 151 I=1,2
I G(I)= (0.0,0.0)
151 G(I)= AlII)
GO TC 82
12 WRITEOUTPUTTAPE6, 32, (Al(I), I=1,N)
GO TO 23
END
SUBRCUFINE CLACRA(G,Y)
I DIMENSION G(10), Y(2)
I DELTA= SQRTF(G(2)**2-(4.0,O.O)*G(1))
I YL1) = (-G(2)+DELTA)/(2.0,0.0)
I Y(2)= (-G(2)-DELTA)/(2.0,0.0)
I RETURN
END

NDFILE


For the data, n = the degree of the polynomial f(x) (; 20)

jlim = upperbound for the iteration number

h = amount of translation on x
n+l
(A(i)} = real coefficients of f(x) = 2 A(i)x1-
1


If GCD = 1, Penultimate Remaindering is applied to f(x+h).

If GCD is a polynomial of positive degree, Penultimate Remaindering

is applied to f/GCD, in order to have simple zeros. The

multiplicity of the zeros found can later be determined

either by studying the GCD or by repeatedly depressing f

by the zeros obtained.








C A NCNCONTAMINATING ITERATIVE PROCESS FOR DETERMINING T
HE ZEROS CF
IA POLYNOMIAL
CIkENSION NT(5)
I CIMENSION A(10), ALPHA(12), C(10), CC(10,10,20), 0(10)
ETA(11,20)
1,G(10), H(10, 5), R(10), RR(11), X(11), Y(2), PNUM(10)
PNUV (10O,5
2), CENUM(10), SUM(5,10), SSLM(5)
100 REACINPLITAPE5, 1, N, NS,L
1 FORMAT (313)
N1 = N+1
N2 = N+2
REACINPUTTAPE5, 2, (A(I), A(1+10), I=1,, )
READINPUT1APE5, 2, (R(I), R(I+10), I=1,N)
REACINPUTTAPES, 2, ALPHA(N2), ALPHA(N2+12)
WRITEObLPUTTAPEb, 7
7 FORVAT(46H HERE ARE THE A'S, R'S + ALPHA'S, RESPECTIVE
LY)
WRITEOUTPUTTAPE6, 2, (A(I), A(I+10), I=I,N)
,RITECU1PUTTAPEo, 2, (R(I), R(I+10), I=1,N)
2 FORMAT(4F15.7)
IF(L) 3,3,4
4 REACINPUTrAPES, 1, (NT(I), 1=1,L)
C OBTAIN ALPHA'S
I 3 C(1) = -R(1)
CO 10 1=2,N
I C(1)= -R(I)
I CALL PULTIP(C,D,G,I-1,1)
CO 10 J=1,I
I 10 C(J)= G(J)
0O 11 J=1,N
I 11 ALPHA(J)= A(J)-G(J)
I ALPIA(N1)= (0.0,0.0)
WRITECUTPUTTAPE6, 2, (ALPHA(J), ALPHA(J+I2), J=1,N2)
C FOR SIMPLE R'S AND R(N+I)
J=l
I ETA(N1,2) = G(N) ALPHA(N2)*(G(N-1)-G(N)**2) + ALPHA(
N)
IF(NS) 27,27,12
12 00 13 K=1,NS
I PNUV(K) = (C.0,C.0)
I CENCM(K)= (1.0,0.0)
CO 14 I= 1,N2
I 14 PNUV(K) = PNUM(K)- ALPHA(I)*R(K)**(I-1)
00 15 I=1,N
IF (K-I)16,15,16
I 16 CENCM(K) = CENOMIK)*IR(K)-R(I))
15 CONTINUE
I 13 ETA(K,J+1)= PNUM(K)/DENOV(KI
GO TO 17
I 18 IF(ABSF(ALPHA(N2))) 19,19,20
I 20 PNUF2 = (0.0,0.0)
I PNUV3 = (1.C,0.0)








CO 21 I= 1,N
IN = N-I+1
I PNUV2 = PNUf2 + ALPHA(IN)*(-ALPHA(N2))**I
I 21 PNUV3 = ((L.O,C.0) + R(I)*ALPHA(N2))*PNLr3
I PNUP4 = (I.C,O.L) PNUM2-PNUM3
I CENCIV = ALPHA(N2)
IF(NS) 22,22,23
23 CO 24 I=1,NS
I 24 LENOMI = DENOMI *((1.0,.0)+ALPHA(N2)*(R(I)+l TA(I,J)))
IF (L) 25,25,22
22 CO 26 I=1,L
I 26 OENCYl = UENOMl*SSUM[(I)
I 25 EIA(N1,J+1)= PNUM4/DENOM1
19 IF(I S) 27,27,28
28 CO 29 K = 1,NS
I CENCY(K) = (1.0,0.0)
CO 30 I=1,NS
IF(K-I) 31,30,31
I 31 CENOV(K)= DENOM(K)*(R(K)-R(I)-ETA(I,J))
30 CGNIINUE
IF(L) 32,32,23
33 CO 34 I=1,L
I 34 OENCM(K) = CENOM(K)*SUM(I,K)
I 32 IF(ABSF(ALPFA(N2))) 29,29,35
I 35 OENOM(K) = DENOM(K) *((1.0,0.0)+ALPHA(N2)*(R(K)-ETA(N1
,J)))
1 29 ETA(K,J+1) = PNUM(K)/DENOK(K)
17 WRI EOUTPUTIAPE6, 2, (ETA(K,J+1), ETA(K,J+2L), K=1,NS)
I 27 IF(ABSF[ALPHA(N21)) 36,36,31
37 WRITECUTPUIIAPE6, 2, ETA(N1,J+L), ETA(NI,J+21)
36 WRIIECUTPUTTAPE6, 38
38 FORMAT (32F ABOVE ARE ETAS FOR SIMPLE ZLRLS)
C FOR MULTIPLE ZEROS
IF(L) 39,39,41
39 IF(J-1) 94,94,40
41 NSS = NS+1
NSI = NS+1
CO 42 I=L,L
NTI = NT(I)
NSI = NSS
I C(1) = (1.0,0.0)
IF(NS) 43,43,44
44 CO 45 K=1,NS
IF(J-I) 220,220,47
I 220 C(1)= -R(K)
GO TC 46
I 47 C(l)= -R(K) EIA(K,J)
I 46 CALL MULTIP(C,O,G,K-1,I)
00 48 KK=l,K
I 48 C(KK)= G(KK)
45 CONTINUE
43 NCEG= NS
IF(J-1) 49,49,50
49 CO 51 K=NS1,N








IF(R(K)-R(NSI)) 52,53,52
53 IF(R(K+10)-R(NS1+10)) 52,51,52
I 52 C(1) = -R(K)
I CALL MULTIPIC,D,G,NDEG,1)
NCEG= NLEG+1
00 54 KK=1,NDEG
I 54 C(KK)= G(KK)
51 CONTINUE
CO 61 K= I,NTI
I PNUPI(K,I) = (0.0,0.0)
KI = K-1
00 61 M= K,N2
Ml = M-l
1 61 PNUF'(K,I) = PNUML(K,I)+PROD( MI,Kl)*ALPHA(M)*R(NSI)**(
M-K)
GO TL 55
50 CO 56 II=1,L
NTII = NT(II)
IF(I-II) 57,56,57
57 CO 58 K=1,NTII
I 58 D(K)= h(K,11)
I CALL MULTIP(C,D,G,NDEG,NTII]
NDEG = NCEG + NIII
CO 59 K = L,NDEG
I 59 C(K) = G(K)
56 CONTINUE
[ IF(ABSF(ALPHA(N2))J 210,210,67
I 67 CCN= (1.0,0.0) + ALPHA(N2)*(R(NSI)-ETAIN1,J))
GO TO 55
I 210 CON = (1.0,0.0)
I 55 RT = R(NSIJ
[ CALL DERIV(O,C,X,N-NTI, RT)
CO 60 K= 1,NTI
KI = K-1
I DENCP2 = X(1)*PROD(Kl,K1)
I PNUF2 = (O.C O, .C)
I PNUM3 = (0.C,0.0)
IF(KI) 9,9,63
9 IF(J-1) 60,60,8
63 UD 64 M=1,K1
K1M= Kl-M
KM= K-M
I RT= R(NSI)
I CALL DERIV(M,C,X,N-NTI, RT)
[ 64 PNUM3 = PNUM3-PROD(Kl,KIM)*X(M+1)*CO(KM,I,J+1)
IF(J-1) 60,60,65
65 CO 66 M=1,K1
KIM = KI-M
KM= K-M
I 66 PNUM2 = PNUM2 + PROD(K-2, K1M)*CO(KM,I,J+1)kX(M)
I AK= K1
I PNUP2 = -ALPHA(N2)*AK*PNLM2
I PNUP3 = CON*PNUM3
I 8 DENOV2 = CON *DENOM2








I 60 CG(K,I,J+1) = (PNUM1(K,I) +PNLM2+PNUM3)/DENOV2
RI EOUUTPUT APE6, 2, (CO(K,I,J+1),CO(K,I,J+21), K=1,NT
I)
WRITEGUIPUTTAPE6, 69, 1
69 FORYAT(35H ABUVE ARE C'S FOR MULTIPLE ZERO /13)
C CATA NEEDED FOR NEXT ITERATICN OF BOTH THE SIMPLE AND
MULTIPLE ZER
10S
IF(AS) 62,62,70
70 CO 71 K=1,NS
I SUM(I,K)= (0.0,0.0)
DO 271 KL=1,NTI
I 271 SUM(I,K)= SUM(I,K) + CO(KL,I,J+1)*(R(K)-R(NSI))**(KL-1

I 71 SUf(I,K)= SUM(I,K) + (R(K)-R(NSI))**n TI
I 62 IF(ABSF(ALPFA(N2)))68,68,72
I 72 SSUM[I)= (0.0,0.0)
00 73 K=1,NTI
I 73 SSU ((IJ= SSUM(I) + CO(K,I,J+1)*((-ALPHA(N2))**(NTI-K+1
))*((1.0,0.0
1) + ALPHA(N2)*R(NSI))**(K-1)
I SSUM(I)= SSUM(I) + ((1.0,0.0) + ALPHA(N2)*R(NSI))**NTI
68 CO 74 K=1,NTI
I 74 C(K) = CC(K,I,J+1)
CO 75 M= 1,NTI
NO = NTI+2-M
I L(NC) = (1.C,0.C)
CO 76 K= V,NTI
NTK= NTI+1-K
1 76 D(NTK)= C(NTK)-R(NSI)*D(NTK+1)
I (M, I) = C(i)
IF (Y-NTI) 2CO,75,75
200 NT" = NTI-M
DO 201 K=1,NTM
1 201 C(KJ = C(K+1)
75 CONTINUE
42 NSS= NSS+NTI
C TEST FOR CCNVERGENCc
IF(J-1) 94,94,7d
78 00 80 I=1,L
NTI = NT(I)
DO 80 K=1,NTI
I T = ABSF(CO(K,I,J+1)-CO(K,I,J))
IF(r-1.OE-4) 80,80,82
80 CONTINUE
IF(NS) 81,81,40
40 CO 83 K=1,NS
I T = ABSF(ETA(K,J+1)-ETA(K,J))
IF(T-1.OE-4) 83,83,82
83 CONTINUE
I 81 IF(AESF(ALPHA{N2))) 84,84,71
1 77 T = ABSF(ETA(N1,J+1I-ETA(N1,J))
IF(T-1.OE-4) 84,84,82
84 WRITEOUTPUTTAPE6, 85








85 FORMAT( 14H THE ROOTS ARE)
IF(NS) 86,86,87
87 00 88 K=1,NS
I RR(K)= R(K)+ ETA(K,J+11
88 WRITEOUTPUTTAPE6, 2, RR(K), RRIK+11)
1 86 IF(ABSF(ALPhA(N2)))7,97,97,99
I 99 RR(NI) = -(1.O ,0.0)/ALPHA(N2) + ETA(NI,J+1)
WRITEOUTPUTTAPE6, 2, RR(N1), RR(Ni+11)
97 IF(L) 100,100,89
89 NSS= NS+1
CO 90 1= 1,L
NSI = NSS
IF(rT(I)-2) 91,91,90
91 CO 92 K=1,2
S 92 G(K) = CC(K,I,J+1)
I CALL OUACRA(G,Y)
00 93 K= 1,2
I RR(K)= R(NSI)+Y(K)
93 WRITEOUTPUTTAPE6, 2, RR(K), RR(K+11)
90 NSS = NSS+NT(I)
GO TC 100
82 IF(J-l5) 94,94,95
94 WRI1EOUIPUTTAPE6,96,J
96 FORMAl (25H STATUE OF ITERATION IS //I3)
79 J = J+1
GO TO 18
95 WRIIEOUTPUTTAPE6, 98
98 FORMAT( 32H ThE IIERATION IS NOT CONVERGING)
GO TC 100
ENC
SUBROUTINE MULTIP(C,D,G,NP,MP)
I CIMENSION C(10), 0(10), G(10)
NP2 = NP*2
MP2 = MP+2
NM= NP+MP
CO 1C1 K=NP2,NM
I 101 C(K)= (0.0,0.0)
CO 1C2 K = MP2, NM
I 102 0(K)= (0.0,0.0)
I C(NP+1) = (1.0,0.0)
I C(MP+1) = (1.0,0.0)
CC 103 I= 1,NM
I G(I) = (0.0,0.0)
CO 103 J= 1,1
IJ= I-J+1
I 103 G(I)= G(I)+C(J)*D(IJ)
I RETURN
END
FUNCTION PROD(NU,NV)
MPRCC = I
IF(NU) 105,105,106
106 IF(KV) 105,105,107
107 CO 108 I= 1,NV
NUI = NU-I+1








108 MPRCC = VPRCODNUI
I 105 PRGC = MPROD
I RETURN
END
SUBROUTINE DERIV(M,G,X,Mi ,RI)
I DIMENSION X(11) G(10)
I X(M+1) =(O.CO.O)
IF(P-ME 110,110,111
110 lV = M+1
CC 112 I=V1, ME
I 112 X(M1) = XIMI)+ PRUD(I-1,M)*G(I)*RT**(I-V1)
I X(M1) = X(M1) + PROC(ME, M)RT** (ME-M)
I 111 RETURN
END
SUBROUTINE GUACRA(G,Y)
I DIMENSION G(10), Y(2)
I CELTA = SCRTF(G(2)**2-(4.0,O.0)*G(1))
I Y(1) = (-G(2)+DELTA)/(2.0,0.C)
I Y(2) = (-G(2)-CLLTA)/(2.0,C.C)
I RETURN
END


NDFILE


For the data, n = degree of the polynomial f(x) (< 10)

ns = number of simple zeros of t(x)

L = number of multiple zeros of t(x)

n+l
(A(i)} = coefficients of f(x) = E A(i)xi-
1
(r(i)] = zeros of t(x) with the simple zeros listed first

and the multiple zeros repeated according to

respective multiplicities

fnt(i)}= multiplicities of the multiple zeros of t(x)

alpha (n2) = coefficient of xn+l term of g(x).















AFP EIDI X B


COMMON ME'iHOD'D FOR LDEIELRINIG C-EIROS OF A FOLThCIOr~AL


I. For Re. 2eros:

1. Half Int-rval Search [6]

i a' First approyimate by -rsFrnical or any other meair ian

interval count dining a real zer- of f'ix., say I.lbl.

Assurne f1id' < C. for the purpo.:e of discussion.

d't
ib1 T 3t wrhther ffl--1 is les- than or greater than or

J'b.
equal to 0 If f'--I 0, th:n the r-quired z-ro is

obtained: 1 if it i; iess than CO. then the recpjired zero
d*t
iS in the interval i--, b, w+ere-a iif greater than 0.

it is ir, the interval id,- b .


Ic' Sc. if .till nece- ary, rperat tht- process with the inter-
-t L -- b
-il td,b' replaced by eitnrcr 1--, bi or id. d-- as

th-e cae may be.

Id'i By continuing the process, we trhu narrow down tre irter-

.al as much as we want to, and hence Ilocate a zero to anr

accuracy of ny,, reasonable number of si5mifi.a-nt digits.


-. method of Falt Fo~itio.n it

(al As in the previous method, start ilth an interval id,bl

In which f, x changes sign.









(b) Linear interpolation is then used to find an approximate

zero x, as follows:

Suppose f(d) < 0 and f(b) > 0. The line joining f(d)

add f(b) intersects the real axis in x, where x is deter-
-f(d) x-d
mined by the relation f(b) f(d) = d that is, by

x = d + (d)(b-d) (The same formula is derived when
f(d)-f(b)
f(d) > 0 and f(b) < 0.)

(c) To determine which of the two intervals (d,x) or (x,b)

contains the true zero xt, check the sign of f(x):

if f(x) >0, then xt is in (d,x); if < 0, then xt is in

(x,b); if = 0, then xt = x.

(d) If the zero is not yet determined, repeat the procedure

with (d,b) replaced by either (d,x) or (x,b), as the

case may be.

(e) Continue the process until either If(x) < eps or the

length of the interval < eps, where eps is a predetermined

small positive number.


3. Newton-Raphson Method [6]

(a) Start with an arbitrary approximate to the zero, say xl.

(b) To improve the approximation, draw the tangent to the

curve f(x), and determine its intersection with the real

axis, say x2. The relation therefore is

f(xt) f(x)
tan a = f'(x) = 22, that is, x2 = x 7)






c-



S': I L. iterative ilequnce can now be Set up a? fo.ll.s:

f, .<
n


nn



nn*
:'r. .nI .' p-. o r ei e ,.-,-n it. :r? i' ;I-r,.S of d.v r-

g -rie '-ci a .hen If' n:1. i [ or th- number ,:f

iteration, e:.eeds jlim. Here ~lirri and _Fs are predeter-

rned 'value-.



ii. For .eal an, Imari.-ar' e:r.:.

1. Fenuilimrate Femundering [i1,

'a Str.art wIrth a first trial .lvii-.)r to f ., u-ui.llv, tlhe-

i--t rn terms .f fi::. redJuced ta a ronrLic n "r..:."yTz:'pi"

dL'-i s;r, w.tEri n is th tgree c the :i"izcr being

soJgn t.


St' Divide fil: b;, th-e 4 pr'.:mt.. dj.;.i:sor, until a Fenulti-

ma te rer aintrr i :.btained treat i, a rrmaider .:.f

*lcgreCe m.


'c Ti ':ing tris- pren'ltinate rir-ind.Zr i'.-ded b' its i.eadin

c--efficient 3- tr, ne:. r tri.il divis=.r. the Fprc-,ceure i

re e ated.


Id Thr: proce i c-onitliui-d until tihe njmbeir of it-eratiCon

e.:ceed'- a c-rt.in number .ith. ut sh,': rirn sign of










convergence, or else when the coefficients of the succeed-

ing trial divisors converge.

A slight variation to this iteration is to take alternatively

for trial divisor the penultimate remainder of degree m+l and the

ultimate remainder which is degree m, each time reduced to monic poly-

nomial.


2. Lin's Ultimate Remaindering Iteration [7,8]

nil m+l
When f(x) = 2 a.x1 is divided by d(x) = E b.x'1, the
1 1 1 1

ultimate remainder has coefficients which are expressions involving

the coefficients a's, b's, and the quotient coefficients c's.

If d(x) were a factor of f(x), each of those remainder terms is

identically 0. So to obtain an iterative set of formulas for the b's

which hopefully will converge to the exact coefficients of d(x), we

set each of the terms in the expression for the remainder to 0, and

solve for the b's as functions of the a's and the c's.

To illustrate for the case when m=2, the quotient when f(x)

is divided by the quadratic d(x) has coefficients expressed as a

recursive relation cn-k- = an-k+l b2cn-k b ln-k+l,

k = 1,2,... n-2 where cn = 0 and Cn- = 1; while the remainder

coefficients are r = a2-b2cl-b c2

s = al-blcl.

Solving for b1 and b2, after setting r and s to 0, we get

al a2-b1c2
b -a and b -
1 c1 2 c1











front urhicr the iteration formTul a- are der.red, n;Inamely,


a-
LII,j .i = -









r, ni -, respectivelyy, urles:s a C= .



Lu.e an- iUfford' Lx'tenicrn of 1 [?]

InsteaJ of derivbag the form- l.---- that wll enable u- t- get

ar, estirmate- of one factor at e3ch stage of itertion, thi: ex.:te-riom

derive tih forimlas tIat will di pla, tet e'i.ilmtes of two or rore,


F...ibil. al the factors, at esac stage of the :teration. This,

trerrefore., relii dntes the, contained nation of .:ne fIjc.tor t: j b.:ther .Jiej

t1 *'i'silorn inrr te Jdere-eon cf te rf:l;,ri ml U and permiap- dJl to

the inrccuracy, of 0 fact-,r obtained. In tre -aid iteratior- if ori

-et- of coefficiicnts turn ou. to be c,,n'.erernt, ,t-ilie Othner j3.ri't,

then depr esion of F.oi,Tionr il ill .be in ijt able, the re--t 3f the

factors can- tbe obtained by situ .lar pr -ce.dure from the deprec-:jed poly-

n..mi. al.

Ho:iie'eer to a.tter.pt to wrrite a general e:..Tres-sio.n or set of

forrnnl i to sepr-ate. j ,llTorri all into a -, number of factors present.

crnsl.erable algebrjic difficulties. ,3 trhe iterative fonrmula:- are

set Iup according t:. tre inlirlldual protl-r..








7
i-1
To illustrate for a sectic Z a.x seeking three quadratic
1 i

factors, let the quadratics be x2 + b2x + bl, x2 + d2x + d, and

x + e2x + e Dividing f(x) by x b2 + b bl, we obtain as in Lin's

Method a set of formulas for the quotient coefficients. If the

5 i-1
quotient is Z c.xi with remainder r x + rl, the recursive relations
1 1 x
are:

Sa6 b2

c3 = a5 c4b2 b1

c2 =a cb2 cb1

c = a c2b2 c b1


r2 =a2 clb2 2b1

rl = a clb1


5 i-l 2
Dividing further cix by x + d2x + d, we derive a

similar set offormulas for the coefficients of the quotients
2
x + e2x + el

e2 = e d2

el =c3- e2d2 -


r =2 eld2 -

r = C el

To obtain a set of formulas for b's, d's and e's from which the

iterative formulas are to be based, the idea is to set the remainder










coefficients r' ar.d r'- to 0 arnd elirrinate tre c's using the abj,:'~

equati.:.ns. Thu- by,' first -.tting 7 anxd r. to 1i, we get


C. ed
j= and =






a. < *_ a-b. _a b- b bb a 12tb b -




L ab: ab -1 ab -2 1



Sirilarly, by setting r1 and r. to C', and ellrdnating the c's,


b i a U blel 1
b b.. = I 1ad
1 C1*- 'l dl


e. = b.. el 41 b1 b; B b !1 *12

3o tih it-rati.ve- formula' canr no'J be :t up for there b' d's

.ind' t' in tinr of their previrouF 'v;ues anr .of the original c-:ffi-

cient .

For the caC e .of linea-r factors, Ix*.Ci', 1i 1, ... n, the

formulas are

: = a 'L:.
n n l

a a .:, i ~. :.; '. :;
nL-1 -n-1 r 1 "i j n



: [a a j r .*:,' -r x*. .. 1 .. .
n- n- n-l1 1 'i I i i ,1 n-1 n









Xn-3 [an-3 a-2x. + an (x +x ) a(x+xx

+ (x.+ x3x )]/ x x
1 i hn-2 n-1 n


S= a/X2x3 ... xn

where the notation E is used in the symmetric function sense, it being

understood that in the formula for x ,

k k-P1 k k-I
E(x+ x x ) = (x + xi x ).
1 1 j ij=l 1 j
ifj

h. Bairstow's Method for Determining
a Quadratic Factor [6]

From the recursive relations among the coefficients of the

n-1 -1
quotient E b.x of f(x) by the quadratic factor x + px + q,

namely,

bn-k- = an-k+1 Pbn-k qbn-k+ k = 1, ... n-2, bn_ = bn = O,
we get the remainder terms with coefficients

r = a2 pb1 qb2 = b

s = a qb1 = bl + pb0
where b0 and bl are obtained by taking k = n-1, n, respectively.

We require r and s, which are actually functions of p and q,

say f (p,q) and f2(p,q), respectively, to go to 0. To do this,

express the functions f and f2 as Taylor Series in two variables,
neglecting all terms beyond the first three, and equate to 0:

0 = f(p',) f i(P ) + Ap flp Pkk) + Aq fiq(Pk,qk) and

0 = f2(p,q) f2(Pk^qk) + Ap f2p(Pk^ k) + Aq f2q(P k ) where










Ipk'^ka 1- th'e e'timrtr for Ip.q' a t iterat-ion k. -p = p,1 k'
i.r
-I = qk1- qk. ij k' denote: -- ;-aluat.ed t 'F.k *

-p and .q may be .col'.'vd for ;iraltan-eouly; from the above.:.' two

eqi't.ion, a_- follows-

Sub'ttitut ing tre values lor tre partial :..f f and f. with

re_ pect to p anid q. s, obtaineca from the equiati.onr f r and s.


= -p' -- q -' -

'.b 6b bbt tb..
-1. 0 -1
0 = b t -p, ' ., .* -q - Fp -r- i
-1b ^ 0 Pt. I'-;Z L Lq Iq


Adding (-pr.i ime- he firt equation to the -econd. we obtain


111 -7


--


tfaw tj obtain the

reiati:.n


partial of b- and b we may u -e trhe recursi.e


.-Lb b .7
-n- -.-1k


tb .b C.D
n- -1 n-k: n-k*i
= p -r-- q r - brk1 whdch

are obtained from the r.cur-ive formula' *5f the L's aba:,e. However,

for computational purposeC it is wortr, of note that if the quo.tient


1 t
a"









recursive relation cn-k-3 = bnk- Pcnk-2 -qcnk1 for the

coefficients of the quotient. Replacing k by k-1, and comparing with

the first recursive relation for the partial of b's (that is, with

respect to p), we see that

8b
n-k-l
p- n-k-2;

similarly, replacing k by k-2, and comparing with the second (that is,

partial with respect to q),

Sb
n-k-1
S= c n-k-l.

Thus (1) and (2) above reduce to


(1) bl + Ap(-c_2 + bo) + Aq(-c_l) = 0

(2) bO + Ap(-c_1) + Aq(-cO) = 0


from which we derive the formulas


b0c-1-cOb_1
Ap = 2- and
Cl + CO(bo0-C_-2)

b_ c-1 + b0(b0-c-2)
Aq= 2
C-l + cO(b c2)

If IApj +|Aql < e, a small preassigned number, then the iter-

ation has converged to the desired p and q; otherwise, let pk+l = Ap+qk

and qk+l = q+qk' and repeat the procedure.









I'.:. recapitulate tihe procedure:

ai laik a starting value fo.r p as well as :.r q.

ibt Cbt tain b . ,,b using
n-& z -1
b = afnk1 bp qbfr.
bIn-k1-1 n-k-1 Ft. n-k n-k-l'

k = 1. ... n urere b = 0 ard b 1.
n n-l

i. i Derive or i- ... C : firo the relatio:.n

b -pc r -qc,
n--- P n-k- n--- qn--'

R = 1, ... r-1.

'di .lie C;.,, C- c _- to .:btairn I and .q from


O'1 c- an
c" c.b c .
-1 0 -


b l_ 1 + bO b, -r.._-
aq =

-1 cbt .,--c-

'e;I Tet far p| I ql :

If < e, t a pkl i to be the dei red i p.q: nd solve

the equjati.;n .: p..: q for t Le zeros. Othrwie. rep:st tlhe

iteration using pk1 = k.p pk and -q qk = -q unl-sIs tlh

numrlbr oc iitrati:ons as -.*:ce-dc. tthe pper limit st at tihe start.


r,. 'ra;effe 's' -ot quartr"g tL eti: f.r f,.-:
av-ing Comp.iex _ero- [12'

n -1
Fron f :-, = 1 a..: derive 'i :-: = I .. I wru, e zers
1
-,re squares of the :-eros j fi :1:.. Fr instance, for n = L,

a L i transform into <* Za.:









2 2 2 2
(2al 2a a2 + a )x + (2ala a2)x +al so that if

n+l i-1
g(x) = E b.x we have the formulas
1
2
b = -a + 2a3

2
b = a 2a a2 + 2al

2
b2 = -a2 + 2a3al

2
bI = a

If the zeros of f(x) differ in absolute value, the zeros of

g(x) must differ by a greater amount. Thus by doing this transforma-

tion m times, an equation may be found with roots equal to the 2mth

power of the original and which roots differ in absolute value as

greatly as may be desired. In this case, the coefficient of xn-2

will nearly be the square of the absolute value of the zeros with

largest magnitude, since if a + bi are those zeros, the Zr.r

21 j
is dominated by a + b The greater m is, the more nearly true this

is. The root-squaring process is terminated when the coefficient of
n-2
xn in the last derived equation is nearly equal that in the previous

step.

Having found in this manner the 4mth power of the largest

absolute value among the zeros of the original, the square of such

absolute value may be obtained by logarithm. The zeros can then

be determined from this by using relation among the zeros.








To illustrate for n = L, if tne :er.: are d.l, J = 1,2

.hen i = :c ic trhat ince Ir = c is Known,

the ,qui re f '.re at-.:,lute v.-ales of tre Eecond pSdr f' zeror,

Ir., = cI .i can be- etert-in.rd. Then Liing thn relsaion

S= -'l 2 c nd c, c ~

be 317v3 for sirultan-ousl.;, and c.ronsequentl, f'ror the values of

Ir I- and Ir. I , a d. can bet .det.e rLd.r-.d.


r.. HiitcnrcociK' GCD hIetrro for fi:*:,
Hrivin All v-r-s Cncle..: [L,b

First aslrum th- co.E Ifici-nts i of' fi r'eal.

FL+l

1

"- ,[r n-- i-L ?
f ix, = [: n l


i:- ,q a.q ... E nnq x : .q ... ]

where = rn 2' nd. trhe in trie lat ter-m f each of t'..- iast two ceeifi-

ciKnt e:O'pressio:nr is detrnine-d ,uch that the signs are alternating.

If +_ . .ere a pair of :.er- f .' Ir i, tren tre laS-t two coef-

ficientr e*:prezsi.:on would te identical 0. Hence we deri'.e trh two

test equationr for a pir of pare i.aginary zeros. na-liy,

-m- -
i11 i aq a q' a q, = u

i 31 n
1 1a q -qq ".'.. u









So given f(x), if the above polynomials (1) and (2) have a common

factor, say q-ql, then we shall have determined a pair of zeros of

f(x), namely, + lql i. The GCD algorithm can be employed for this

purpose. However, in most cases the common factor is 1. So the idea

now is to use Homer's root diminishing process and move a pair of

zeros of f(x) until they are transformed to pure imaginary ones.

The process then boils down to using the GCD algorithm on the

test equations. If the last real number remainder in the algorithm

is positive (or negative), try translation of variable by a peg&tive

(or pegstive) number h, and apply the GCD algorithm again on the new

test equations. If the remainder changes sign, then we know there is

a pair of zeros whose real part lies between 0 and +h. So try a num-

ber between 0 and h for a translating factor, and repeat the process.

Continuing this way, we can thus narrow down the interval containing

the desired real part by translations till the remainder in the GCD

algorithm approximates 0. This way, we also find at the same time the

imaginary part of that particular pair of zeros from the common factor

of the two test polynomials. We may also employ interpolation to

estimate the amount of translation necessary, after we have tried a

couple of values for h. Thus if translation of variable by h gives

a common factor x-q to the test polynomials, the desired zeros are

then +h 'q i.

For a 's complex, the procedure is practically the same except

that there is a different set of test equations for a pure imaginary

zero qi. If f(x) = xn + (an + bi)xn1 + ... (al + bli), then the
n n










remainder in di'.vidinE fI' ty' x-qi 1: fqi I, *mJiclc i. what we need to

'e.:t whether C or not. B,- -eparating rertl ard imaginary parts -:' fiql

the two t st cquatiorn-s thern are


i1 b.,q aq q a,qb tbq iq ,'


2 a q t,3q aq t.q aq' b q -... =


wh.r. it -ill be noted .that th, ig: are 2 plus and r rinuc alter-

nativel. e.-.cept at 'he start of' 11. .














BIBLIOGRAPHY


[1] Aitken, A. C., "Studies in Practical Mathematics VI. On the
Factorization of Polynomials by Iterative Methods," Proc.
Roy. Soc. Edinburgh, Set A, Vol. 63, pp. 17h-191, 1951.

[2] Aitken, A. C., "Studies in Practical Mathematics VII. On the
Theory of Methods of Factoring Polynomials by Iterated
Division," Proc. Roy. Soc. Edinburgh, Set A, Vol. 63,
pp. 326-335, 1952.

[3] Flanagan, C., and Maxfield, J. E., "Estimates of the Roots of
Certain Polynomials," Journal of the Society for Industrial
and Applied Mathematics, Vol. 7, No. I, 1959.

[h] Hitchcock, F. L., "Finding Complex Roots of Algebraic Equations,"
Journal of Mathematics and Physics, Vol. 17, No. 2, 1938.

[5] Hitchcock, F. L., "Algebraic Equations with Complex Coefficients,"
Journal of Mathematics and Physics, Vol. 18, pp. 202-210, 1939.

[6] Kuo, Shan, Numerical Methods and Computers, Addison-Wesley Pub-
lishing Company, Reading, Massachusetts, 1965.

[7] Lin, Shih-Nge, "A Method of Successive Approximations of Eval-
uating the Real and Complex Roots of Cubic and Higher-Order
Equations," Journal of Mathematics and Physics, Vol. 20,
pp. 231-2L2, 19l1.

[8] Lin, Shih-Nge, "A Method for Finding the Roots of Algebraic
Equations," Vol. 22, pp. 60-77, 1943.

[9] Luke, Y., and Ufford, D., "On the Roots of Algebraic Equations,"
Journal of Mathematics and Physics, Vol. 30, pp. 94-101, 1951.

[10] Lyon, W. V., "Note on a Method of Evaluating the Complex Roots
of a Quartic Equation," Journal of Mathematics and Physics,
Vol. 3, No. 3, 1924.

[11] Maxfield, J. E., Santos, A. H., and Selfridge, R., "An Iterative
Process for Determining Zeros of Certain Polynomials."
(Submitted to S.I.A.M. Journal.)

[12] Sharp, H. S., "A Comparison of Methods of Evaluating the Complex
Roots of Quartic Equations," Journal of Mathematics and Physics,
Vol. 20, pp. 243-258, 19hl.















BIOGCAPFHICAL StICH


Alilcia Hida.Lg.:. .ant.z r taz .: to a;' :',n 'a: 2. at 3BiiL n,

Laglura, rnilippines. Inr April, i9L., r'e prad'iated a3 class *vale-

diLtorian from B ijar Elementary ch: Schol.,i fi r ye car rience, from

Late: S ore Hirn .-h.:.l ithr, tnr. Zan-, nr:.n r3. fr. receive ti

derrere .?f Bachel,:r of Science irn Eduicastion fr.n the l.lrver;i, .;.f

Prilippines in 1:6. mragna curi iaade I From 1?956 t: 19E -h: tiirent

as the i.. F. College :.f Education, after iti-icr, sre t.ranrlferr-A .: t te

U . College of Literal Arts, ,-ith, haicr. ,hre I, benli connected up

to tr., Fresent tim-. Eie :- granted a U.P. 1.:.- i fellowship fra;m

19'0 L:. 19,1. d Jring urhcl ta-i s-e obt gained a Ma.Itetr of Sc rence de'r.-e.

riaj.r i"' lathematlc: In aeptiemt.er, 12lc.:, she bec a1.7-e Frillippine

Ila'.ional.l 3 cienc-. Dev:.l-.',pient Board iellou under Fr.)Ject .. i,

urni.:n rinabtl e- h r to Fi;r-ue e:-r o:.r. t.f:i.'ir d the deCree of [..ctor

ol' Fiil:~:sopr.

Alici a Hidalo.:. S an ts is: married to C arilo 5S.rianr:. 5,.zanI-3,

an architect, and iE the n.,mothe.r .:o tuo children. nre i- ; member

.:f ihe I:atrenaticil Azf.:'cisat.in :of Armerica, the Fhi :.ap.-r Phi Inter-

nati :nal Hon:.or Soc.iet and thr Fi -.ar.'va H1 Irternartionl H,:onor

Doci t.i .











This dissertation was prepared under the direction of the

chairman of the candidate's supervisory committee and has been

approved by all members of that committee. It was submitted to the

Dean of the College of Arts and Sciences and to the Graduate Council,

and was approved as partial fulfillment of the requirements for the

degree of Doctor of Philosophy.


April 23, 1966




Dean, College of Ats and Sciences





Dean, Graduate School



Supervisory Committee:



C airman




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