An Introduction to Numerical Analysis - Solutions

190 Pages·0.484 MB·English

by  Suli E., Mayers D.

An Introduction to
Numerical Analysis
Solutions to Exercises
Endre Su(cid:127)li and David F. Mayers
University of Oxford
CAMBRIDGE UNIVERSITY PRESS
Cambridge
New York Port Chester Melbourne Sydney

1
Solution to Exercise 1.1
The (cid:12)xed points are the solutions of
2
q(x) x 2x+c=0:
(cid:17) (cid:0)
Evidently q(0) = c > 0, q(1) = c 1 < 0, q(x) + as x ,
(cid:0) ! 1 ! 1
showing that 0<(cid:24)1 <1<(cid:24)2.
Now
1 2
xn+1= 2(xn+c)
1 2
(cid:24)1= 2((cid:24)1 +c);
so by subtraction
1 2 2 1
xn+1 (cid:24)1 = 2(xn (cid:24)1)= 2(xn+(cid:24)1)(xn (cid:24)1):
(cid:0) (cid:0) (cid:0)
It follows that if xn + (cid:24)1 < 2, then xn+1 (cid:24)1 < xn (cid:24)1 . Now
j j j (cid:0) j j (cid:0) j
(cid:24)1+(cid:24)2 =2,soif0 x0 <(cid:24)2 thenx0+(cid:24)1 <2,andevidentlyx0+(cid:24)1 >0.
(cid:20)
Hence x1 is closer to (cid:24)1 then was x0, so also 0 x1 <(cid:24)2. An induction
(cid:20)
argument then shows that each xn satis(cid:12)es 0 xn <(cid:24)2, and
(cid:20)
n
x0+(cid:24)1
xn (cid:24)1 < x0 (cid:24)1 ;
j (cid:0) j 2 j (cid:0) j
(cid:18) (cid:19)
and xn (cid:24)1.
!
Now xn+1 is independent of the sign of xn, and is therefore also in-
dependent of the sign of x0, and it follows that xn (cid:24)1 for all x0 such
!
that (cid:24)2 <x0 <(cid:24)2.
(cid:0)
The same argument shows that if x0 >(cid:24)2 then x1 >x0 >(cid:24)2, and so
xn . As before this means also that xn if x0 < (cid:24)2.
!1 !1 (cid:0)
If x0 = (cid:24)2 then of course xn = (cid:24)2 for all n > 0. If x0 = (cid:24)2, then
(cid:0)
x1 =(cid:24)2, and again xn =(cid:24)2 for all n>0.
2
Solution to Exercise 1.2
0 x 00 x 0 00
Since f (x) = e 1 and f (x) = e , f (x) > 0 and f (x) > 0 for
(cid:0)
all x > 0. It therefore follows from Theorem 1.9 that if x0 > 0 then
Newton’s method convergesto the positive root.
0 00
Similarlyf (x)<0andf (x)>0in( ;0)andthesameargument
(cid:0)1
shows that the method convergesto the negative root if x0 <0.
0
If x0 =0 the method fails, as f (0)=0, and x1 does not exist.
For this function f, Newton’s method gives
exp(xn) xn 2
xn+1=xn (cid:0) (cid:0)
(cid:0) exp(xn) 1
(cid:0)
1 (xn+2)exp( xn)
=xn (cid:0) (cid:0)
(cid:0) 1 exp(xn)
(cid:0)
xn 1 n>1:
(cid:25) (cid:0)
(cid:0)100
In fact, e is very small indeed.
In the same way, when x0 is large and negative, say x0 = 100,
(cid:0)
xn 2
xn+1 xn (cid:0) (cid:0) = 2:
(cid:25) (cid:0) 1 (cid:0)
(cid:0)
Hence when x0 = 100, the (cid:12)rst few members of the sequence are
100;99;98;:::; after 98 iterations xn will get close to the positive root,
andconvergencebecomesquadraticandrapid. About100iterationsare
required to give an accurate value for the root.
However, when x0 = 100, x1 is very close to 2, and is therefore
(cid:0) (cid:0)
veryclosetothenegativeroot. Three,orpossiblyfour,iterationsshould
give the value of the root to six decimal places.
3
Solution to Exercise 1.3
Newton’s method is
f(xn)
xn+1 =xn 0 :
(cid:0) f (xn)
Toavoidcalculatingthederivativewemightconsiderapproximatingthe
derivative by
0 f(xn+Æ) f(xn)
f (xn) (cid:0) ;
(cid:25) Æ
where Æ is small. The given iteration uses this approximation, with
Æ = f(xn); if xn is close to a root then we might expect that f(xn) is
small.
If xn (cid:24) is small we can write
(cid:0)
0 1 2 00 3
f(xn)=f((cid:24))+(xn (cid:24))f ((cid:24))+ 2(xn (cid:24)) f ((cid:24))+ (xn (cid:24))
(cid:0) (cid:0) O (cid:0)
0 1 2 00 3
=(cid:17)f + 2(cid:17) f + ((cid:17))
O
0 00
where (cid:17) =xn (cid:24), and f and f are evaluated at x=(cid:24). Then
(cid:0)
0 1 2 00
f(xn+f(xn)) f(xn)=f((cid:24)+(cid:17)+(cid:17)f + 2(cid:17) f ) f((cid:24)+(cid:17))
(cid:0) (cid:0)
0 2 1 2 0 00 0 2 00 3
=(cid:17)(f ) + 2(cid:17) [3f f +(f ) f ]+ ((cid:17)) :
O
Hence
2
[f(xn)]
xn+1 (cid:24)=xn (cid:24)
(cid:0) (cid:0) (cid:0) f(x+n+f(xn)) f(xn)
(cid:0)
2 0 2 0 00
(cid:17) [(f ) +(cid:17)f f ] 3
=(cid:17) 0 2 1 0 0 00 + ((cid:17))
(cid:0) (cid:17)[(f ) + 2(cid:17)f (3+f )f ] O
00 0
2f (1+f ) 3
=(cid:17) + ((cid:17)) :
0
f O
This shows that if x0 (cid:17) is suÆciently small the iteration converges
(cid:0) 000
quadratically. The analysis here requires that f is continuous in a
3
neighbourhoodof(cid:24), to justify the terms ((cid:17) ). A morecarefulanalysis
O 00
might relax this to require only the continuity of f .
The leading term in xn+1 (cid:24) is very similar to that in Newton’s
(cid:0)
method, but with an additional term.
Theconvergenceofthis method,startingfromapointclosetoaroot,
isverysimilartoNewton’smethod. Butifx0 issomewayfromtheroot
0
f(xn) will not be small, the approximation to the derivative f (xn) is
very poor, and the behaviour may be very di(cid:11)erent. For the example
x
f(x)=e x 2
(cid:0) (cid:0)
4
starting from x0 =1; 10 and 10 we (cid:12)nd
(cid:0)
0 1.000000
1 1.205792
2 1.153859
3 1.146328
4 1.146193
5 1.146193
0 10.000000
1 10.000000
2 10.000000
3 10.000000
4 10.000000
5 10.000000
0 -10.000000
1 -1.862331
2 -1.841412
3 -1.841406
4 -1.841406
The convergence from x0 =1 is satisfactory. Starting from x0 = 10
(cid:0)
we get similar behaviour to Newton’s method, an immediate step to x1
quite close to 2, and then rapid convergenceto the negative root.
(cid:0)
However, starting from x0 = 10 gives a quite di(cid:11)erent result. This
time f(x0) is roughly 20000 (which is not small), and f(x0 +f(x0))
9500
is about 10 ; the di(cid:11)erence between x0 and x1 is excessively small.
Althoughtheiterationconverges,therateofconvergenceissoslowthat
for any practical purpose it is virtually stationary. Even starting from
x0 =3 many thousands of iterations are required for convergence.
5
Solution to Exercise 1.4
The number of correct (cid:12)gures in the approximation xn to the root (cid:24)
is
Dn =integerpartof log10 (cid:24) xn :
f(cid:0) j (cid:0) jg
From (1.24) for Newton’s method we have
00
(cid:24) xn+1 f ((cid:24))
j (cid:0) 2j j 0 j:
(cid:24) xn ! 2f ((cid:24))
j (cid:0) j j j
Hence
Dn+1 2Dn B;
(cid:25) (cid:0)
where
00
f ((cid:24))
B =log10 j 0 j:
2f ((cid:24))
j j
If B is small then Dn+1 is close to 2Dn, but if B is signi(cid:12)cantly larger
than 1 then Dn+1 may be smaller than this.
In the example,
x
f(x)=e x 1:0000000005
(cid:0) (cid:0)
0 x
f (x)=e 1
(cid:0)
00 x
f (x)=e ;
and (cid:24) =0:0001. Hence
0:0001
e
B =log10 0:0001 =3:7
2(e 1)
(cid:0)
andthenumberofsigni(cid:12)cant(cid:12)guresinthenextiterationisabout2k 4,
(cid:0)
not 2k.
Starting from x0 =0:0005 the results of Newton’s method are
0 0.000500000000000 3
1 0.000260018333542 3
2 0.000149241714302 4
3 0.000108122910746 5
4 0.000100303597745 6
5 0.000099998797906 9
6 0.000099998333362 14
where the last column shows the number of correct decimal places.
The root is (cid:24) =0:000099998333361to 15 decimal places.
The number of correct (cid:12)gures increases by a factor quite close to 4.
6
Solution to Exercise 1.5
From (1.23) we have
2 00
((cid:24) xn) f ((cid:17)n)
(cid:24) xn+1 = (cid:0) 0 :
(cid:0) (cid:0) 2f (xn)
0
Now f ((cid:24))=0, so by the Mean Value Theorem
0 0 00
f (xn) f ((cid:24))=(xn (cid:24))f ((cid:31)n)
(cid:0) (cid:0)
for some value of (cid:31)n between (cid:24) and xn. Hence
00
((cid:24) xn)f ((cid:17)n)
(cid:24) xn+1 = (cid:0) 00 :
(cid:0) 2f ((cid:31)n)
00 00
Now f ((cid:17)n) <M and f ((cid:31)n) >m, and so
j j j j
(cid:24) xn+1 <K (cid:24) xn ;
j (cid:0) j j (cid:0) j
where
M
K = <1:
2m
Henceifx0liesinthegiveninterval,allxnlieintheinterval,andxn (cid:24).
00 00 00 00 !
Then (cid:17)n (cid:24), f ((cid:17)n) f ((cid:24)) and f ((cid:31)n) f ((cid:24)). This shows that
! ! !
(cid:24) xn+1 1
(cid:0) 2
(cid:24) xn !
(cid:0)
and convergenceis linear, with asymptotic rate of convergence ln2.
x 0
Fortheexamplef(x)=e 1 x,f(0)=0,f (0)=0. Startingfrom
(cid:0) (cid:0)
x0 =1, Newton’s method gives
0 1.000
1 0.582
2 0.319
3 0.168
4 0.086
5 0.044
6 0.022
7 0.011
8 0.006
9 0.003
10 0.001
1
showing (cid:24) x0 reducing by a factor close to 2 at each step.
(cid:0)
7
Solution to Exercise 1.6
0 00
Whenf((cid:24))=f ((cid:24))=f ((cid:24))=0wegetfromthede(cid:12)nitionofNewton’s
000
method, provided that f is continuous in some neighbourhood of (cid:24),
f(xn)
(cid:24) xn+1=(cid:24) xn+ 0
(cid:0) (cid:0) f (xn)
1 3 000
6(xn (cid:24)) f ((cid:17)n)
=(cid:24) xn+ 1 (cid:0) 2 000
(cid:0) 2(xn (cid:24)) f ((cid:31)n)
(cid:0)
000
f ((cid:17)n)
=((cid:24) xn) 1 000 :
(cid:0) (cid:0) 3f ((cid:31)n)
(cid:26) (cid:27)
If we now assume that in the neighbourhood [(cid:24) k;(cid:24)+k] of the root
(cid:0)
000
0<m< f (x) <M; where M <3m;
j j
then
(cid:24) xn+1 <K (cid:24) xn ;
j (cid:0) j j (cid:0) j
where
M
K =1 <1:
(cid:0) 3m
Hence if x0 is in this neighbourhood, all the xn lie in the neighbour-
hood, and Newton’s method convergesto (cid:24). Also,
(cid:24) xn+1 2
j (cid:0) j ;
(cid:24) xn ! 3
j (cid:0) j
sothatconvergenceislinear,withasymptoticrateofconvergenceln(3=2).
8
Solution to Exercise 1.7
The proof follows closely the proof of Theorem 1.9.
From (1.23) it follows that xn+1 < (cid:24), provided that xn lies in the
interval I = [X;(cid:24)]. Since f is monotonic increasing and f((cid:24)) = 0,
f(x) < 0 in I. Hence if x0 I the sequence (xn) lies in I, and is
2
monotonic increasing. As it is bounded above by (cid:24), it converges; since
(cid:24) is the only root of f(x) = 0 in I, the sequence converges to (cid:24). Since
00
f is continuous it follows that
00 00
(cid:24) xn+1 f ((cid:17)n) f ((cid:24))
(cid:0) 2 = 0 0 ;
((cid:24) xn) (cid:0)2f (xn) ! (cid:0)2f ((cid:24))
(cid:0)
so that convergenceis quadratic.