Computing Methods

682 Pages·1965·24.224 MB·English

by  I. S. Berezin and N. P. Zhidkov (Auth.)

I. S. BEREZIN and N. P. ZHIDKOV
COMPUTING METHODS
VOLUME II
Translated by
0. M. BLUNN
Translation edited by
A. D. BOOTH
College of Engineering
University of Saskatchewan
PERGAMON PRESS
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Copyright © 1965
PERGAMON PRESS LTD.
First English Edition 1965
Library of Congress Catalog Card Number 61-11157
This is a translation of the original Bussian
Memodu eWiUCMHUU (Metody vychisîenii)
published by Fizmatgiz, Moscow
MADE IN GREAT BRITAIN
FOREWORD
IN THIS second volume of the book, a study is made of numerical
methods of solving sets of linear algebraic equations, high degree
equations and transcendental equations, numerical methods of
finding eigenvalues, and approximate methods of solving ordinary
differential equations, partial differential equations and integral
equations.
The book is intended as a text-book for students in mechanical-
mathematical and physics-mathematical faculties specializing in
computer mathematics and persons interested in the theory and prac-
tice of numerical methods.
xiii
PREFACE
As STATED in the preface to the first volume of the book, this
second volume contains Chapters 6-10, which correspond to Part II
of the course on "Computing Methods" which is taken by students
in the fourth year of instruction.
I. S. BEREZIN
N. P. ZHIDKOV
XV
CHAPTER 6
THE SOLUTION OF SETS OF LINEAR
ALGEBRAIC EQUATIONS
1. CLASSIFICATION OF METHODS
The second part of our book, which is devoted to the solution of
algebraic, transcendental, differential and integral equations,
begins with a chapter on numerical methods of solving sets of linear
algebraic equations. Sets of linear algebraic equations are simple
but occur in many problems of numerical analysis.
Cramer's rule, which is well known from the course on higher
algebra, is not suitable in practice for solving sets of linear algebraic
equations since it requires an excessive amount of arithmetic.
Many suitable practical methods have therefore been proposed.
A considerable portion of the published literature dealing with
computing methods is devoted to this topic. But it is still impossible
to point to one or several of these methods as the most efficient in
the sense that the solution is obtained in the shortest time with
the necessary degree of accuracy and the minimum use of storage
devices. We need to make an extensive theoretical and experimental
comparative evaluation of the various known methods from this
point of view. In this chapter, we shall only give a few time-tested
methods and some that seem to us to be promising for practical use.
The practical methods of solving sets of linear algebraic equations
can be divided into two large groups, namely, exact methods and
methods of successive approximation. Using the exact methods
it is possible in principle to find the exact values of the unknowns
after a finite number of operations. It is naturally presupposed that
the coefficients and right-hand sides of the sets are known exactly,
and that no rounding is applied in the calculations. More often than
not they are solved in two steps. In the first step the set is simpli-
fied and in the second this set is solved and the values of the un-
knowns found.
1
2 COMPUTING METHODS
In methods of successive approximation, certain approximate
values of the unknowns are given at the very beginning. New and
"improved" approximate values of the initial approximations are
then obtained. The new approximate values are treated in the same
way, and so forth. Under definite conditions we (in theory) arrive
at the exact solution after an infinite number of steps.
Monte Carlo methods are not included in our classification. Such
methods use random quantities, the mathematical expectations of
which produce the solution. Monte Carlo methods cannot as yet
compete with the stated methods, and we shall therefore leave
them alone here.
2. ELIMINATION
We begin our study of numerical methods of solving sets of linear
algebraic equations with the exact methods. The simplest of these
methods is that of elimination.
The method of elimination will already have been studied in the
ordinary school course on algebra. By combining the equations of
the set in a certain way, one of the unknowns is eliminated in all
the equations except the first. Then, another unknown is eliminated
and then a third and so on. As a result we get a set with a triangular
or diagonal matrix which can easily be solved. No theoretical
difficulties arise in the process of elimination. But the accuracy
of the result and the amount of time required depends very greatly
on the way in which the calculations are organized. We therefore
have to pay great attention to this aspect of the matter.
Let us illustrate the use of the method of elimination by a typical
set of four equations with four unknowns :
1.1161^ + 0.1254^ + 0.1397^3 + 0.1490^ 1.5471,
0.1582^+1.1675z + 0.1768x + 0.187l# 1.6471,
2 3 4
(i)
0.1968^ +0.2071#2 +1.21683-3+ 0.227l£ 1.7471
4
0.2368^ + 0.247Lr2 + 0.2568z3 + 1.2671a?4 1.8471 :
These can be tackled in several ways. Each of these methods has
a separate name, but unfortunately there is no standardized
terminology.
THE SOLUTION OF SETS OF LINEAR ALGEBRAIC EQUATIONS 3
( 1) The Gauss method with selection of the pivotal element (pivotal
condensation)
If the calculations are not to be performed on computers, it is
convenient to write our set as follows
No mt aa Ö.-2 «<3 au b{ st
1 1.11610 0.12540 0.13970 0.14900 1.54710 3.07730
2 0.15820 1.16750 0.17680 0.18710 1.64710 3.33670
3 0.19680 0.20710 1.21680 0.22710 1.74710 3 59490
4 0.23680 0.24710 0.25680 1.26710 1.84710 3.85490
The number of the equation appears in column 1. The importance
of the second column will be clear later. The third, fourth, fifth and
sixth columns contain the coefficients of the equations and the
seventh the free terms. The last column contains the sums of the
coefficients and free terms of each row.
We now have to select the element a^ with largest absolute value.
We refer to it as the pivotal element. In our case this is a = 1.26710.
44
It is underlined in the table. We have to divide all the elements
of the column containing the pivotal element (in our case a ) by the
u
pivotal element, and then the relation with the reverse sign is placed
in the column m in the same row for which the quotient is found :
i
No TTZj an aii °<3 au bi *«
1 —0.11759 1.11610 0.12540 0.13970 0.14900 1.54710 3.07730
2 —0.14766 0.15820 1.16750 0.17680 0.18710 1.64710 3.33670
3 —0.17923 0.19680 0.20710 1.21680 0.22710 1.74710 3.59490
4 0.23680 0.24710 0.25680 1.26710 1.84710 3.85490
We now add the row that contains the pivotal element multiplied
by the corresponding m to each row of the table. The row containing
t
the pivotal element is then omitted. The column containing the
pivotal element may be omitted too, since it only consists of zeros.
In our case :
4 COMPUTING METHODS
No nit «il αί2 aiZ αί4 hi *<
1 -0.09353 1.08825 0.09634 0.10950 1.32990 2.62399
2 -0.11862 0.12323 1.13101 0.13888 1.37436 2.76748
3 0.15436 0.16281 1.17077 1.41604 2.90398
We have, however, included the values of m obtained at the next
i
step. Let us check that the calculations are correct. To do this,
we have to add the columns a^ and 6 and then compare with s
i v
If they agree to within the error allowed for rounding, they are
deemed to be correct. If the discrepancy is too large, the respective
calculations are repeated.
We then tackle this table in the same way. We select the pivotal
element (in this case 1.17077) and divide it into the elements of
this column. The results with reverse signs are written in the column
m (this has already been done here). We then successively multiply
i
the row from which the pivotal element is taken by τη and add the
ϊ
corresponding rows. We check the result and pass onto the next
step. We continue in the same way until only one row is left. In our
example we have
No 1 m{ an o<2 °<3 «*4 ^ *i
1
1 -0.07296 1.07381 0.08111 1.19746 2.35238
2 0.10492 1.11170 1.20639 2.42301
1 1.06616 1.10944 2.17560
Using the equations in which the pivotal elements have been
selected we derive a new equivalent set
1.06616»!= 1.10944, I
0.10492z +1.11170z = 1.20639, I
1 2
0.15436ζ + 0.1628ΐ£ +1.17077^3 = 1.41604, ί ( )
1 2
0.23680»!+ 0.24710x +0.25680*3 + 1.26710ar = 1.84710. j
2 4
THE SOLUTION OF SETS OF LINEAR ALGEBRAIC E QUATIONS 5
The matrix of this new set is triangular. No difficulties arise in solv-
ing it. We find
1.10944
^ = Γ066Ϊ6-=104059'
1.20639-0.10492-1.04059 1.09721
x* = ΓΪΪΪ70 = ΤΤΓΪτο = °·98697'
1.41604-0.15436.1.04059-0.16281. 0.98697
x~ =
1.17077
i(3)
1.09473
0.93505,
1.17077
1.84710-0.23680.1.04059-0.247100.98697-0.25680· 0.93505
x —
A 1.26710
1.11669
0.88130.
1.26710
This method of elimination is referred to as the Gauss method of
elimination with selection of the pivotal element. The process of elim-
ination itself is known as forward elimination, but the solution
of a set with a triangular matrix is known as back-substitution.
In practice, there is of course no need to separate the individual
stages. This was done to simplify the explanation. Neither need
the final set of equations be copied out.
In back-substitution, accuracy is checked by means of column s .
t
If we substitute s for b in the final set we should obtain x -f1 in-
t t {
stead of x The results can also be checked by substituting them in
v
the original set of equations. In our case we have 1.54711, 1.64712,
1.74710 and 1.84711 on the left-hand sides of the equations. As
we can see, the left- and right-hand sides are the same within two
units of the fifth decimal place, which must be regarded as satis-
factory.
The idea behind selecting the pivot is to make m as small as possible
i
and thereby reduce the computational error.
It is well known that multiplication and division take most time
on digital machines, or manually. It is, therefore, important to
know how many operations of multiplication and division are neces-
sary to solve the set in question. If the set is of order n, then n — 1
divisions must be made after selecting the pivot to determine the
coefficients m It is then necessary to multiply the row containing
v
6 COMPUTING METHODS
the pivot by each of these factors. This requires (n+l)(n — l) or
n2 — 1 multiplications. Thus the first stage of the Gauss method
requires n2 + n — 2 multiplications and divisions. The next stage
requires (?i — l)2 + (n— 1) — 2 such operations and so on. The total
number of operations of multiplication and division before back-
substitution is
[n*+n-2]+[(n-l)* + (n-l)-2] + ...
... +[l 2 l-2] n(n l)(2n l) n{n + l) _^
+ = + + + (4)
Back-substitution requires
1 + 2 + 3+.. . η = * ψί (5)
+
operations of multiplication and division, if no cross-check with
column s is made. As many operations are required in making this
{
check. Thus, the solution of a set of n equations by the Gauss method
with selection of the pivotal element and cross-checking requires
V 'v——■ + V ' -2n + n(n + l) = - (n2-f 6?i-l) (6)
2
operations of multiplication and division.
(2) Compact Gauss method
There are numerous modifications of the Gauss method with one
or another advantage. We shall consider one such method. First
it is supposed that there is a set of four general equations
a^+a^x^a^ + a^x, = a<°>,
5
d$Xi + €$x% + a$Xi+a$Xi = a$,
(7)
a<iixi + a3°2^2 + 0g>a? + αίΙ)χ* =
3
<ι\ 4- a$x + a$x + <>s == aa%®>, j
2 3 4
The initial data and the intermediate and final results are tabu-
lated as follows
n(0) «S? «fi? a(0) «if
a(V <> ^ ^"Ί244 40s>
a(0) a32 a(0) a^3(40 ) <>
^U3411 a(0) a43 W-44 <5>
42