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MTH 309

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Compi
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edby…………………….
.
Dept
.Chemi
cal
Engi
neer
i
ng 1
UNI
VERSI
TYOFMAI
DUGURI
Facul
t
yofSci
ence
DEPARTMENTOFMATHEMATI
CSANDSTATI
STI
CS
MATH309
Numer
i
cal
Anal
y
si
sI
(
2Uni
t
s)
1
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
I
nt
r
oduct
i
on
Manyt
i
mes,
i
ti
snotpossi
bl
et
of
i
ndt
heexactsol
ut
i
onoft
hepr
act
i
calpr
obl
em whi
chhasbeen
t
r
ansf
or
medi
nt
omat
hemat
i
calpr
obl
em usi
ngst
andar
dt
echni
ques.Forexampl
e,t
her
ear
e
f
or
mul
asf
orsol
v
i
ngquadr
at
i
candcubi
cpol
y
nomi
alequat
i
onsbutnomuchf
or
mul
aexi
st
sf
or
pol
y
nomi
al
equat
i
onsofdegr
eegr
eat
ert
hanf
ourorev
enf
orasi
mpl
eequat
i
onsuchas
x=cosx
Numer
i
cal
anal
y
si
si
nv
ol
v
est
hedev
el
opmentandev
al
uat
i
onofmet
hodsf
orcomput
i
ngr
equi
r
ed
numer
i
cal
r
esul
t
sf
r
om gi
v
ennumer
i
cal
dat
a.
Pr
esenceofEr
r
or
Anexactsol
ut
i
onandnumer
i
calsol
ut
i
onsar
enotsame.Thedev
i
at
i
onofnumer
i
calsol
ut
i
on
f
r
om exactsol
ut
i
oni
sknownaser
r
or
.Thus
Er
r
or=Exactv
al
ueAppr
oxi
mat
ev
al
ue
I
ft
heer
r
ori
ssmal
l
t
hent
heappr
opr
i
at
ev
al
ueorcomput
edv
al
uei
sadmi
ssi
bl
e,
ot
her
wi
sei
ti
st
o
ber
ej
ect
ed.Er
r
ori
ni
t
ssi
mpl
estf
or
m,
i
st
hedi
f
f
er
encebet
weent
heexactanswerA,
sayandt
he
̅
comput
edr
esul
t
,A.Hence,
wecanwr
i
t
e,
̅
Er
r
or=A A
Si
ncewear
eusual
l
yi
nt
er
est
edwi
t
ht
hemagni
t
udeorabsol
ut
ev
al
ueoft
heer
r
orwecanal
so
def
i
ne
| |
Absol
ut
eEr
r
or=̅
AA
I
npr
act
i
cewear
eof
t
enmor
ei
nt
er
est
edi
nsocal
l
edr
el
at
i
v
eer
r
ort
hant
heabsol
ut
eer
r
orand
wedef
i
ne
̅
|
AA|
Rel
at
i
v
eEr
r
or=
|
A|
Sour
cesofer
r
ori
nanycomput
at
i
onar
e
1.Humaner
r
or
2.Tr
uncat
i
oner
r
or
3.Roundi
nger
r
or
Ty
pi
cal
humaner
r
ori
s
2
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
 Ar
i
t
hmet
i
cer
r
or
 Pr
ogr
ammi
nger
r
or(
Bl
under
)
Tr
uncat
i
onEr
r
or
At
r
uncat
i
oner
r
ori
spr
esentwhensomei
nf
i
ni
t
epr
ocessi
sappr
oxi
mat
edbyaf
i
ni
t
e
pr
ocess.Forexampl
e,
consi
dert
heTay
l
orser
i
esexpansi
on
x2
xn
ex =1 +x+ +… + +…
n!
2!
I
ft
hi
sf
or
mul
ai
susedt
ocal
cul
at
ef=e0 +1weget
n
2
3
(
(
(
0.
1)
0.
1)
0.
1)
f=1 +0.
1+
+
+… +
+…
n!
2!
3!
Thepr
obl
em her
ei
st
hat
,wher
edowest
opt
hecal
cul
at
i
on?Howmanyt
er
msdowei
ncl
ude?
Theor
et
i
cal
l
yt
hecal
cual
t
i
onwi
l
lnev
erst
op.I
fwedost
opaf
t
erf
i
ni
t
enumberoft
er
ms,wewi
l
l
notgett
heex
actanswer
.Forexampl
e,i
fwet
aket
hef
i
r
stf
i
v
et
er
msast
heappr
oxi
amt
i
onwe
get
,
4
2
3
(
(
(
0.
1)
0.
1)
0.
1)
̅
f=e0.1 ≅1 +0.
1 + 2! + 3! + 4! =f ≈1.
105
Fort
hi
scal
cul
at
i
on,
t
het
r
uncat
i
oner
r
orTE(
t
hati
st
het
er
mst
hathav
ebeenchoppedof
f
)i
s
̅
TE =ff
Roundi
ngEr
r
or
̅
Consi
dert
hecal
cul
at
i
onoffabov
e;
1 =1
(
0.
1)
=0.
1
1!
2
(
0.
1)
2!
=0.
005
3
(
0.
1)
3!
=0.
0001666
4
(
0.
1)
=0.
000004166
4!
3
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
̅
Summi
ngabov
e =1.
105170833 =f
̅
Theexactanswert
ot
het
r
uncat
edpr
obl
em,f,
i
sani
nf
i
ni
t
est
r
i
ngofdi
gi
t
sand,
assuch,
i
snot
v
er
yusef
ul
.Wecanr
oundt
hi
st
osi
xorsev
endeci
mal
pl
aces.Nowt
osi
xdeci
mal
pl
acewehav
e
̃
̅
f ≈1.
105171 =f
̃
̅
Thedi
f
f
er
encebet
ween fand f
̅ ̃
f-f =0.
000000166 =RE
i
st
her
oundi
nger
r
orRE.
⌂1⌂
METHODOFSOLVI
NGANONLI
NEAREQUATI
ON
I
nt
hi
ssect
i
on,
wewi
l
l
st
udymet
hodsf
orf
i
ndi
ngsol
ut
i
onsoff
unct
i
onsofsi
ngl
ev
ar
i
abl
e,
x
x
()=0.Tof
()=0,west
t
hati
s,v
al
uesofx sucht
hatf
i
ndt
her
ootoff
ar
twi
t
haknown
appr
oxi
mat
esol
ut
i
onandappl
yanyoft
hef
ol
l
owi
ngmet
hods:
 BI
SECTI
ONMETHOD
x)=0bet
(
Thi
smet
hodconsi
st
si
nl
ocat
i
ngt
her
ealr
ootoft
heequat
i
onf
weenaandb.I
f
x
a
()i
()andf
(
b)ar
f
scont
i
nuousbet
weenaandb,andf
eofopposi
t
esi
gnst
hent
her
ei
sa
a)beposi
(
r
ootbet
weenaandb.Fordef
i
ni
t
eness,l
etf
t
i
v
e.Thent
hef
i
r
stappr
oxi
mat
i
ont
o
t
her
ooti
s
x1 =a +b
2
x1)=0,t
x)=0.Ot
(
(
I
ff
henx1 i
st
her
ootoff
her
wi
se,t
her
ootl
i
esbet
weena andx1 or
x
(1)i
bet
weenx1 andbaccor
di
ngl
yasf
sposi
t
i
v
eornegat
i
v
e.Thenwebi
sectt
hei
nt
er
v
alas
bef
or
eandcont
i
nuet
hepr
ocessunt
i
l
t
her
ooti
sf
oundt
odesi
r
edaccur
acy
.
4
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
x1)= +Ve,sot
(
I
nt
hedi
agr
am abov
e,f
hatt
her
ootl
i
esbet
weenaandx1.Thent
hesecond
appr
oxi
mat
i
ont
ot
her
ooti
s
a +x1
x2 =
2
x
(2)i
I
ff
s–Ve,
t
her
ootl
i
esbet
weenx1andx2.Thent
het
hi
r
dappr
oxi
mat
i
ont
ot
her
ooti
s
x +x
x3 = 1 2
2
andsoon.
I
l
l
ust
r
at
i
v
eEx
ampl
e
1.Fi
ndar
ootoft
heequat
i
onx34x9 =0usi
ngt
hebi
sect
i
onmet
hodi
nf
ourst
ages
Sol
ut
i
on
x)=x3(
Letf
4x9
a)=f
(
(
2)=f
9<0
⇒ negat
i
v
e
(
(
3)=6>0
b)=f
f
t
i
v
e
⇒ posi
Thef
i
r
stappr
oxi
mat
i
ont
ot
her
ooti
s
a +b 2 +3 5
x1 =
=
= =2.
5
2
2
2
x1)=f
(
(
2.
5)=f
3.
375
x1)i
(
Si
ncef
snegat
i
v
e,
t
her
ooti
sbet
weenx1andb
Thesecondappr
oxi
mat
i
ont
ot
her
ooti
s
5
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
x1 +b 2.
5 +3
x2 =
=
=2.
75
2
2
x2)=f
(
(
2.
75)=0.
f
7969
x2)i
(
Nowf
s +Ve
Thet
hi
r
dappr
oxi
mat
i
ont
ot
her
ooti
s
a +x2 2 +2.
75
=
=2.
375
x
3 =
2
2
x3)=f
(
(
2.
375)=f
5.
1035
x3)=(
Si
ncef
Vet
hef
our
t
happr
oxi
mat
i
ont
ot
her
ooti
s
x3 +b 2.
375 +3
x4 =
=
=2.
6875
2
2
Hence,
t
her
ooti
s2.
6875appr
oxi
mat
el
y
x11 =0,
2.Fi
ndt
her
ootoft
heequat
i
onx3cor
r
ectt
o4deci
mal
usi
ngbi
sect
i
onmet
hod.
Sol
ut
i
on
x)=x3(
Gi
v
enf
x11 =0
Leta=2andb =3
a)=f
(
(
2)=f
5<0 (
negat
i
v
e)
(
(
3)=13<0
b)=f
f
(
posi
t
i
v
e)
The1stappr
oxi
mat
i
ont
ot
her
ooti
s
a +b 2 +3
x1 =
=
=2.
5
2
2
x1)=f
(
(
2.
5)=2.
f
125
x
(1)i
Si
ncef
s +Ve,
t
her
ootl
i
esbet
weenaandx1
The2ndappr
oxi
mat
i
ont
ot
her
ooti
s
a +x1 2 +2.
125
=
=2.
0625
x2 =
2
2
x2)=(
4.
2888
∴ f
x2)i
(
Si
ncef
s–Ve,
t
her
ootl
i
esbet
weenx2andb
The3r
dappr
oxi
mat
i
ont
ot
her
oot….
x2 +b 2.
0625 +3
x3 =
=
=2.
5313
2
2
6
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
x3)=2.
(
Now f
6880
Ther
ef
or
e,
t
her
ootl
i
esbet
weenx2andx3
The4t
happr
oxi
mat
i
oni
s
x2 +x3 2.
0625 +2.
5313
x4 =
=
=2.
2969
2
2
x4)=f
(
(
2.
2969)=Al
so, f
1.
1790
The5t
happr
oxi
mat
i
ont
ot
her
ooti
s
x3 +x4 2.
5313 +2.
2969
x5 =
=
=2.
4141
2
2
x5)=0.
(
6550
f
x4 +x5 2.
2969 +2.
4141
x6 =
=
=2.
3555
2
2
x6)= (
f
0.
2863
x
x6 2.
4141 +2.
3555
5 +
x7 =
=
=2.
3848
2
2
x7)=0.
(
f
1782
x6 +x7 2.
3555 +2.
3848
x8 =
=
=2.
3702
2
2
x8)= (
0.
0548
f
x
x8 2.
3848 +2.
3702
7 +
x9 =
=
=2.
3775
2
2
x9)=0.
(
f
0613
x8 +x9 2.
3702 +2.
3775
x10 =
=
=2.
3739
2
2
x10)=0.
(
0040
f
x8 +x10 2.
3702 +2.
3739
x11 =
=
=2.
3721
2
2
x11)= (
0.
0246
f
x10 +x11 2.
3739 +2.
3721
x12 =
=
=2.
3730
2
2
x12)= (
0.
0103
f
x10 +x12 2.
3739 +2.
3730
=
=2.
37345
x13 =
2
2
x13)= (
0.
00318
f
x10 +x13 2.
3739 +2.
37345
x14 =
=
=2.
37368
2
2
x14)=0.
(
00048
f
x13 +x14 2.
37345 +2.
37368
x15 =
=
=2.
37357
2
2
7
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
x15)= (
0.
00127
f
x14 +x15 2.
37368 +2.
37357
x16 =
=
=2.
37363
2
2
3736
x16 ≈2.
Not
ex
15 ≈
Hencet
her
ooti
s2.
3736
3.Fi
ndt
heposi
t
i
v
er
ootofx3x=1cor
r
ectt
of
ourdeci
mal
pl
acesbybi
sect
i
onmet
hod.
Sol
ut
i
on:
x)=x3(
Letf
x1
(
0)=03f
01 =1 =v
e
(
1)=13f
11 =1 =v
e
(
2)=23f
21 =5 = +v
e
1 +2)
Sor
ootl
i
esbet
ween1and2,
wecant
ake(
/
2asi
ni
t
i
al
r
ootandpr
oceed.
(
1.
5)=0.
i
.
e.
,f
8750 = +v
e
(
1)=f
1 =v
e
Sor
ootl
i
esbet
ween1and1.
5,
Letx0 =(
ni
t
i
al
r
ootandpr
oceed.
1 +1.
5)
/
2asi
(
1.
25)=f
0.
2969
Sor
ootl
i
esbet
weenx1bet
ween1.
25and1.
5
/
2 =1.
3750
Nowx1 =(
1.
25 +1.
5)
(
1.
375)=0.
f
2246 = +v
e
Sor
ootl
i
esbet
weenx2bet
ween1.
25and1.
375
/
2 =1.
3125
Nowx2 =(
1.
25 +1.
375)
(
1.
3125)=f
0.
051514 =v
e
Ther
ef
or
e,
r
ootl
i
esbet
ween1.
375and1.
3125
Nowx3 =(
1.
375 +1.
3125)
/
2 =1.
3438
(
1.
3438)=0.
f
082832 = +v
e
Sor
ootl
i
esbet
ween1.
3125and1.
3438
8
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
Nowx4 =(
1.
3125 +1.
3438)
/
2 =1.
3282
(
1.
3282)=0.
f
014898 = +v
e
Sor
ootl
i
esbet
ween1.
3125and1.
3282
Nowx5 =(
1.
3125 +1.
3282)
/
2 =1.
3204
(
1.
3204)=f
0.
018340 =v
e
Sor
ootl
i
esbet
ween1.
3204and1.
3282
Nowx6 =(
1.
3204 +1.
3282)
/
2 =1.
3243
(
1.
3243)=f
v
e
Sor
ootl
i
esbet
ween1.
3243and1.
3282
Nowx7 =(
1.
3243 +1.
3282)
/
2 =1.
3263
(
1.
3263)= +v
f
e
Sor
ootl
i
esbet
ween1.
3243and1.
3263
Nowx8 =(
1.
3243 +1.
3263)
/
2 =1.
3253
(
1.
3253)= +v
f
e
Sor
ootl
i
esbet
ween1.
3243and1.
3253
Nowx9 =(
1.
3243 +1.
3253)
/
2 =1.
3248
(
1.
3248)= +v
f
e
Sor
ootl
i
esbet
ween1.
3243and1.
3248
Nowx10 =(
1.
3243 +1.
3248)
/
2 =1.
3246
(
1.
3246)=f
v
e
Sor
ootl
i
esbet
ween1.
3248and1.
3246
Nowx11 =(
1.
3248 +1.
3246)
/
2 =1.
3247
(
1.
3247)=f
v
e
Sor
ootl
i
esbet
ween1.
3247and1.
3248
Nowx12 =(
1.
3247 +1.
3248)
/
2 =1.
32475
Ther
ef
or
e,
t
heappr
oxi
mat
er
ooti
s1.
32475
9
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
4.Fi
ndt
heposi
t
i
v
er
ootofxcosx=0bybi
sect
i
onmet
hod.
Sol
ut
i
on
x)=xLetf
cosx
(
f
cos0 =01 =1 =v
e
(
0)=0f
5cos(
0.
37758 =v
e
(
0.
5)=0.
0.
5)=f
cos(
42970 = +v
e
(
1)=11)=0.
Sor
ootl
i
esbet
ween0.
5and1
Letx0 =(
ni
t
i
al
r
ootandpr
oceed
0.
5 +1)
/
2asi
f
75cos(
018311 = +v
e
(
0.
75)=0.
0.
75)=0.
Sor
ootl
i
esbet
ween0.
5and0.
75
x1 =(
0.
5 +0.
75)
/
2 =0.
625
f
625cos(
0.
18596
(
0.
625)=0.
0.
625)=Sor
ootl
i
esbet
ween0.
625and0.
750
x2 =(
0.
625 +0.
750)
/
2 =0.
6875
(
0.
6875)=f
0.
085335
Sor
ootl
i
esbet
ween0.
6875and0.
750
x3 =(
0.
6875 +0.
750)
/
2 =0.
71875
f
71875cos(
0.
033879
(
0.
71875)=0.
0.
71865)=Sor
ootl
i
esbet
ween0.
71875and0.
750
/
2 =0.
73438
x4 =(
0.
71875 +0.
750)
(
0.
73438)=f
0.
0078664 =v
e
Sor
ootl
i
esbet
ween0.
73438and0.
750
742190
x5 =0.
(
0.
742190)=0.
f
0051999 = +v
e
x6 =(
0.
73438 +0.
742190)
/
2 =0.
73829
10
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
(
0.
73829)=f
0.
0013305
Sor
ootl
i
esbet
ween0.
73829and0.
74219
x7 =(
0.
73829 +0.
74219)
/
2 =0.
7402
f
7402cos(
0018663
(
0.
7402)=0.
0.
7402)=0.
Sor
ootl
i
esbet
ween0.
73829and0.
7402
73925
x8 =0.
(
0.
73925)=0.
f
00027593
7388
x9 =0.
Ther
ooti
s0.
7388
x1)= +Ve,
(
I
nsummar
y
:I
ff
x1)=(
Ve,
I
ff
x2)= +Ve,
(
I
ff
x2)=(
Ve,
I
ff
a +x1
x2 =
2
x +b
x2 = 1
2
a +x2
x3 =
2
x +b
x3 = 2
2
andt
hepr
ocesscont
i
nuesunt
i
l
anyt
woconsecut
i
v
ev
al
uesar
eequal
andhencet
her
oot
.
 METHODOFFALSEPOSI
TI
ONORREGULARFALSIMETHOD
x0)andf
x1)ar
(
(
Her
e,
wechooset
wopoi
nt
sx0 andx
ucht
hatf
eofopposi
t
esi
gns,
t
hati
s,
t
he
1s
gr
aphofy=f
ossest
hexaxi
sbet
weent
hesepoi
nt
s.
(
x)cr
11
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
Fi
gur
e1
x0)
x1)<0.
(
(
Thi
si
ndi
cat
est
hatar
ootl
i
esbet
weenx0andx1consequent
l
yf
f
x0)
x1)
x0,f
f
x
(
(
Theequat
i
onoft
hechor
dj
oi
ni
ngt
hepoi
ntA[
sobt
ai
nedasf
ol
l
ows:
]i
]andB[
1,
Sl
ope,
m ofAB
x0)
x1)y
y (
(
f
2 1 f
=
=
x2x1
x1x0
Usi
ngf
or
mul
af
orequat
i
onofast
r
ai
ghtl
i
ne
y
y=m(
xx1) weget
1
x0)
x1)(
(
f
f
xx0)
(
x1x0
x0)=
(
yf
Theabsci
ssaoft
hepoi
ntwher
et
hechor
dcut
st
hexaxi
s(
sgi
v
enby
y=0)i
x0)
x1)(
(
f
f
x2x0)
(
x1x0
x0)
x1x0)
f
(
(
=x
x0
2x
x
(1)(0)
f
f
xx
x0)
x2 =x0- 1 0 f
(
x
x
(
)
(
)
f1 f0
x0)=
(
0f
whi
chi
st
heappr
oxi
mat
i
ont
ot
her
oot
.
Al
t
er
nat
i
v
epr
oof
:
I
nt
hebi
sect
i
onmet
hod,wei
dent
i
f
ypr
operv
al
uesx0 (
l
owerboundv
al
ue)andx1 (
upperbound
x0)
x1)<0.
(
(
v
al
ue)sucht
hatf
f
Thenextpr
edi
ct
ed/
i
mpr
ov
edr
ootx2canbecomput
edast
hemi
dpoi
ntbet
weenx0andx1as
x0 +x1
x2 =
2
Thepr
ocedur
ei
sr
epeat
edunt
i
l
t
heconv
er
gencei
sachi
ev
ed(
sucht
hatt
henewl
owerandupper
boundsar
esuf
f
i
ci
ent
l
ycl
oset
oeachot
her
)
.
12
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
Basedont
wosi
mi
l
art
r
i
angl
es,
showni
nFi
gur
e2,
oneget
s
x0) 0x1)
(
(
0f
f
=
x2x0
x2x1
x
x
x
x
x
x
(2 0)
(1)=(2 1)
(0)
f
f
x0)x0)
x1)x1)=x2f
x2f
x0f
x1f
(
(
(
(
x
x0)
x1)x1)x2f
x2f
x0f
x1f
(
(
(
(
0)=
x0)
x0)
x1)x1)x2{
x1f
(
(
(
(
f
f
}=x0f
x0)
x1)xf
x1f
(
(
x2 = 0
x0)
x1)(
(
f
f
i
st
heappr
oxi
mat
i
ont
ot
her
oot
Thi
scanal
sobewr
i
t
t
enas
x0)xx
xx
x1)
xf
x0f
(
(
or x =x - 1 0 f
or x =x - 0 1 f
x
x1)
(
)
(
0
1
2
0
2
x
x
x
x
x
x
(1)(0)
(0)(1)
f
f
f
f
(0)(1)
f
f
x2 = 1
Fal
sePosi
t
i
onAl
gor
i
t
hm
x)=0ar
(
Thest
epst
oappl
yt
ot
hef
al
seposi
t
i
onmet
hodt
of
i
ndt
her
ootoft
heequat
i
onf
eas
f
ol
l
ows.
x0)
x1)<0,ori
(
(
f
1.Choosex0 a
ndx1 ast
woguessesf
ort
her
ootsucht
hatf
not
herwor
ds,
x)changessi
(
f
gnbet
weenx0a
ndx1.
x)=0as
(
2.Est
i
mat
et
her
oot
,
heequat
i
onf
x2oft
x0)
x1)xf
x1f
(
(
x0)
x1)(
(
f
f
x2 = 0
3.Nowcheckt
hef
ol
l
owi
ng
x
x
(0)
(2)<0,
f
I
ff
t
hent
her
ootl
i
esbet
weenx0a
ndx2;t
henx0 =x0andx1 =x2.
x
x
(0)
(2)>0,
f
I
ff
t
hent
her
ootl
i
esbet
weenx2a
ndx1;t
henx0 =x2andx1 =x1.
x0)
x2)=0,
(
(
I
ff
f
t
hent
her
ooti
sx2.St
opt
heal
gor
i
t
hm
4.Fi
ndt
henew est
i
mat
eoft
her
oot
13
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
x0)
x1)xf
x1f
(
(
x0)
x1)(
(
f
f
x2 = 0
Fi
ndt
heabsol
ut
er
el
at
i
v
eer
r
oras
|
|
xnew xold
εa|= 2 new2 ×100
|
x2
n
e
w
wher
e x2 = est
i
mat
edr
ootf
r
om pr
esenti
t
er
at
i
on
x2old =est
i
mat
edr
ootf
r
om pr
ev
i
ousi
t
er
at
i
on
5.Compar
et
heabsol
ut
er
el
at
i
v
eappr
oxi
mat
eer
r
or|
t
ht
hepr
especi
f
i
edr
el
at
i
v
eer
r
or
εa|wi
t
ol
er
anceεs.I
f|
t
hengot
ost
ep3,
el
sest
opt
heal
gor
i
t
hm.
εa|>εs,
2x5byt
Exampl
e1:Fi
ndar
ealr
ootoft
heequat
i
onx3hemet
hodoff
al
seposi
t
i
on,cor
r
ectt
o
t
hr
eedeci
mal
pl
aces.
Sol
ut
i
on
x)=x3(
Letf
2x5
Wechoosex0 =2andx1 =3
x0)=f
(
(
2)=f
1<0
x1)=f
(
(
3)=16>0
f
Theappr
oxi
mat
i
ont
ot
her
ooti
s
x1x0
x0)=2 +1 =2.
x2 =x0f
0588
(
x0)
x1)17
(
(
f
f
x2)=f
(
(
2.
0588)=Nowf
0.
3911
x
x
(0)
(2)=((1)
0.
3911)>0,
Andf
f
t
hati
s,
t
her
ootl
i
esbet
weenx2andx1
0588
∴New x0 =x2 =2.
x1 =x1 =3
x0)=(
f
0.
3911
x
(
)
f 1 =16
32.
0588
0.
3681
05880588 +
=2.
0813
x3 =2.
(0.
3911)=2.
1
6
.
3911
(160.
3911)
x2)=(
0813;
Taki
ngx2 =2.
f
0.
1468
x
x
(0)
(2)=((0.
3911)
0.
1468)>0,
Si
ncef
f
t
her
ootl
i
esbet
weenx2andx1
0813
∴New x0 =x2 =2.
x1 =x1 =3
x0)=(
f
0.
1468
x1)=16
(
f
32.
0813
0.
1349
x4 =2.
08130813 +
=2.
0897
(0.
1468)=2.
16.
1468
16(0.
1468)
0928 , x6 =2.
0939 ,
Repeat
i
ng t
hi
s pr
ocess,t
he successi
v
e appr
oxi
mat
i
ons ar
e x5 =2.
0943,x8 =2.
0945.Forx9t
x7 =2.
her
ootr
epeati
t
sel
fmaki
ngx8 =x9.
Hencet
her
ooti
s2.
095cor
r
ectt
o3deci
mal
pl
aces.
No.of
i
t
er
at
i
on
x0
x1
Tabl
eSummar
y
x0)
(
f
14
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
x1)
(
f
x2
x2)
(
f
1
2
3
1
16
2.
0588
0.
3911
2
2.
0588
3
0.
3911
16
2.
0813
0.
1468
3
2.
0813
3
0.
1468
16
2.
0897
0.
0540
4
2.
0897
3
0.
0540
16
2.
0928
0.
0195
5
2.
0928
3
0.
0195
16
2.
0939
0.
0073
6
2.
0939
3
0.
0073
16
2.
0943
0.
0028
7
2.
0943
3
0.
0028
16
2.
0945
0.
0006
8
2.
0945
3
0.
0006
16
2.
0945
Theansweri
s2.
095appr
oxi
mat
el
y
2
x)=(
(
(
x4)
x+2)=0,usi
2.
5 and
Exampl
e2:Fi
ndt
her
ootoff
ngt
hei
ni
t
i
alguessesofx0 =1.
0,
x1 =andapr
especi
f
i
edt
ol
er
anceofεs =0.
1%
Sol
ut
i
on
Thei
ndi
v
i
duali
t
er
at
i
onsar
enotshownf
ort
hi
sexampl
e,butt
her
esul
t
sar
esummar
i
zedi
nt
he
Tabl
ebel
ow.I
tt
akesf
i
v
ei
t
er
at
i
onst
omeett
hepr
especi
f
i
edt
ol
er
ance.
I
t
er
at
i
on
x0
x1
x0)
(
f
x1)
(
f
x2
εa|
|
%
1
2.
5
1
21.
13
25.
00
1.
813
N/
A
2
2.
5
1.
813
21.
13
6.
319
1.
971
8.
024
3
2.
5
1.
971
21.
13
1.
028
1.
996
1.
229
4
2.
5
1.
996
21.
13
0.
1542
1.
999
0.
1828
5
2.
5
1.
999
21.
13
0.
02286
2.
000
0.
02706
|
|
|
|
xnew xold
2.
000(1.
999)
εa|= 2 new2 ×100 =
|
×102 =0.
05
x2
2.
000
Now,|
ooti
sεa|>εs
∴Ther
2
2m
ε
|
|
Al
so, a ≤0.
5×10
0.
02706≤0.
5×102-m
m ≤3.
2666
[
mi
st
heno.ofsi
gni
f
i
cantdi
gi
t
st
hatar
eatl
eastcor
r
ecti
nt
hel
asti
t
er
at
i
v
enumber
]
Hence,
3si
gni
f
i
cantdi
gi
t
scanbet
r
ust
edt
obeaccur
at
e.
00i
So x=2.
scor
r
ect
Exampl
e3:
Fi
ndar
ootoft
hef
ol
l
owi
ngbyFal
seposi
t
i
onmet
hod;
(
i
)x=cosx
(
i
i
)xl
2
og10x =1.
Sol
ut
i
on:
15
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
(
i
)
Gi
v
en x=cosx
[
Not
e:
angl
e‘
x’
i
st
akeni
nr
adi
ans]
x)=xLetf
cosx
(
5andx1 =1
Choosi
ngx0 =0.
x
(0)=f
(
0.
5)=f
0.
377583
x1)=f
(
(
1)=0.
f
459698
Theappr
oxi
mat
i
ont
ot
her
ooti
s
10.
5
5725482
x2 =0.
(0.
377583)=0.
0.
377583)
0.
459698(-
x2)=f
(
(
0.
725482)=So,
f
0.
022698
x
x
(0)
(2)=((0.
377583)
0.
022698)>0;
Si
ncef
f
t
her
ootl
i
esbet
weenx2andx1;
725482
x0 =x2 =0.
x1 =x1 =1
x0)=(
f
0.
022698
x
(1)=0.
f
459698
10.
725482
x3 =0.
725482738399
(0.
022698)=0.
0.
022698)
0.
459698(x2)=f
(
(
0.
738399)=So,
f
0.
001148
x
x
(0)
(2)=((0.
022698)
0.
001148)>0;
Si
ncef
f
t
her
ootl
i
esbet
weenx2andx1;
738399
x0 =x2 =0.
x1 =x1 =1
x0)=(
f
0.
001148
x1)=0.
(
f
459698
10.
738399
x4 =0.
738399739051
(0.
001148)=0.
0.
001148)
0.
459698(x2)=f
(
(
0.
739051)=So,
f
0.
000057
x0)
x2)=((
(
(0.
001148)
0.
000057)>0;
Si
ncef
f
t
her
ootl
i
esbet
weenx2andx1;
739051
x0 =x2 =0.
x1 =x1 =1
x0)=(
f
0.
000057
x1)=0.
(
f
459698
10.
739051
x5 =0.
739051739083
(0.
000057)=0.
0.
000057)
0.
459698(x2)=f
(
(
0.
739083)=So,
f
0.
000004
x
x
(0)
(2)=((0.
000057)
0.
000004)>0;
Si
ncef
f
t
her
ootl
i
esbet
weenx2andx1;
x0 =x2 =0.
739083
x1 =x1 =1
x0)=(
f
0.
000004
x1)=0.
(
f
459698
10.
739083
x6 =0.
739083739085
(0.
000004)=0.
0.
000004)
0.
459698(7391
Hencet
her
ooti
s ≈0.
16
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
(
i
i
)
Gi
v
en xl
og10x =1.
2
x)=xl
f
1.
2
og10x(
Choosi
ngx0 =2andx1 =3
x0)=f
(
(
2)=f
0.
597940
x
(
)
(
)
f 1 =f3 =0.
231364
Theappr
oxi
mat
i
ont
ot
her
ooti
s
x0)
x1)xf
x1f
(
(
x2 = 0
x0)
x1)(
(
f
f
0.
462728 +1.
793820
x2 =
=2.
721014
0.
829304
x2)=(
So,
f
0.
017091
x0)
x2)=((
(
(0.
597940)
0.
017091)>0;
Si
ncef
f
t
her
ootl
i
esbet
weenx2andx1;
721014
x0 =x2 =2.
x1 =x1 =3
x0)=(
f
0.
017091
x1)=0.
(
f
462728
(
(
2.
721014)
0.
462728)0.
017091)
3(x3 =
=2.
730951
0.
462728(0.
017091)
x2)=f
(
(
2.
730951)=So,
f
0.
008448
x0)
x2)=((
(
(0.
017091)
0.
008448)>0;
Si
ncef
f
t
her
ootl
i
esbet
weenx2andx1;
730951
x0 =x2 =2.
x1 =x1 =3
x0)=(
f
0.
008448
x1)=0.
(
f
462728
(
(
2.
730951)
0.
462728)0.
008448)
3(x4 =
=2.
735775
0.
462728(0.
008448)
x2)=f
(
(
2.
735775)=So,
f
0.
004246
x
x
(
)
(
)
(
)
(
0.
008448 0.
004246)>0;
Si
ncef 0 f 2 = t
her
ootl
i
esbet
weenx2andx1;
735775
x0 =x2 =2.
x1 =x1 =3
x0)=(
f
0.
004246
x1)=0.
(
f
462728
(
(
2.
735775)
0.
462728)0.
004246)
3(x5 =
=2.
738177
0.
462728(0.
004246)
x2)=f
(
(
2.
738177)=So,
f
0.
002153
x
x
(0)
(2)=((0.
004246)
0.
002153)>0;
Si
ncef
f
t
her
ootl
i
esbet
weenx2andx1;
x0 =x2 =2.
738177
x1 =x1 =3
x0)=(
f
0.
002153
x
(1)=0.
f
462728
(
(
2.
738177)
0.
462728)0.
002153)
3(x6 =
=2.
739390
0.
462728(0.
002153)
17
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
x2)=f
(
(
2.
739390)=So,
f
0.
001095
x0)
x2)=((
(
(0.
002153)
0.
001095)>0;
Si
ncef
f
t
her
ootl
i
esbet
weenx2andx1;
739390
x0 =x2 =2.
x1 =x1 =3
x0)=(
f
0.
001095
x1)=0.
(
f
462728
(
(
2.
739390)
0.
462728)0.
001095)
3(x7 =
=2.
740005
0.
462728(0.
001095)
x2)=f
(
(
2.
740005)=So,
f
0.
000559
x
x
(0)
(2)=((0.
001095)
0.
000559)>0;
Si
ncef
f
t
her
ootl
i
esbet
weenx2andx1;
740005
x0 =x2 =2.
x1 =x1 =3
x0)=(
f
0.
000559
x1)=0.
(
f
462728
(
(
2.
740005)
0.
462728)0.
000559)
3(x8 =
=2.
740319
0.
462728(0.
000559)
x2)=f
(
(
2.
740319)=So,
f
0.
000285
x
x
(0)
(2)=((0.
000559)
0.
000285)>0;
Si
ncef
f
t
her
ootl
i
esbet
weenx2andx1;
740319
x0 =x2 =2.
x1 =x1 =3
x0)=(
f
0.
000285
x
(1)=0.
f
462728
(
(
2.
740319)
0.
462728)0.
000285)
3(x9 =
=2.
740479
0.
462728(0.
000285)
x2)=f
(
(
2.
740479)=So,
f
0.
000146
x
x
(0)
(2)=((0.
000285)
0.
000146)>0;
Si
ncef
f
t
her
ootl
i
esbet
weenx2andx1;
740479
x0 =x2 =2.
x1 =x1 =3
x0)=(
f
0.
000146
x1)=0.
(
f
462728
(
(
2.
740479)
0.
462728)0.
000146)
3(x10 =
=2.
740561
0.
462728(0.
000146)
x2)=f
(
(
2.
740561)=So,
f
0.
000074
x0)
x2)=((
(
(0.
000146)
0.
000074)>0;
Si
ncef
f
t
her
ootl
i
esbet
weenx2andx1;
740561
x0 =x2 =2.
x1 =x1 =3
x0)=(
f
0.
000074
x1)=0.
(
f
462728
(
(
2.
740561)
0.
462728)0.
000074)
3(x11 =
=2.
740602
0.
462728(0.
000074)
x2)=f
(
(
2.
740602)=So,
f
0.
000038
18
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
x0)
x2)=((
(
(0.
000074)
0.
000038)>0;
Si
ncef
f
t
her
ootl
i
esbet
weenx2andx1;
x0 =x2 =2.
740602
x1 =x1 =3
x0)=(
f
0.
000038
x1)=0.
(
f
462728
(
(
2.
740602)
0.
462728)0.
000038)
3(x12 =
=2.
740623
0.
462728(0.
000038)
The11thand12thi
t
er
at
i
onar
eal
mostt
hesame.Hencet
her
ooti
s ≈2.
7406
 NEWTONRAPHSONMETHOD
I
ti
sal
soknownasNewt
on’
sMet
hodorSuccessi
v
eSubst
i
t
ut
i
onMet
hod.
Letx0beagoodest
i
mat
eofrandl
etr=x0 +h.Si
ncet
het
r
uer
ooti
sr
,
andh =rx0,
t
henumber
eshowf
art
heest
i
mat
ex0i
sf
r
om t
het
r
ut
h.
hmeasur
Si
ncehi
s‘
smal
l
’
,
wecanuset
hel
i
near(
t
angentl
i
ne)appr
oxi
mat
i
ont
oconcl
udei
t
sv
al
ue.
x0 +h)byTay
Expandi
ngf
l
orser
i
esweobt
ai
n
(
h2 ''x
'x
x0)+hf
x0 +h)=f
(
(0)+ f
(0)+… + =0
(
f
2!
Negl
ect
i
ngt
hesecondandhi
gheror
derder
i
v
at
i
v
esoft
heequat
i
onweobt
ai
n
'x
x0)+hf
(
(0)=0
f
x0)
(
f
whi
chgi
v
es
h ≈-'
x0)
(
f
x0)
(
f
I
tf
ol
l
owst
hat r=x0 +h ≈x0-'
x0)
(
f
Ournewi
mpr
ov
edest
i
mat
ex1ofri
st
her
ef
or
egi
v
enby
x0)
(
f
x1 =x0'x
(0)
f
Thenextest
i
mat
ex2i
sobt
ai
nedf
r
om x1i
nexact
l
yt
hesamewayasx1wasobt
ai
nedf
r
om x0:
x1)
(
f
x2 =x1'x
(1)
f
Cont
i
nuei
nt
hi
sway
.I
fxni
st
hecur
r
entest
i
mat
e,
t
hent
henextest
i
mat
exn +1i
sgi
v
enby
x
(n)
f
xn +1 =xn'x
(n)
f
whi
chi
st
heNewt
onRaphsonFor
mul
a
AGeomet
r
i
cI
nt
er
pr
et
at
i
onoft
heNewt
onRaphsonI
t
er
at
i
on
I
nt
hi
spi
ct
ur
ebel
ow,t
hecur
v
ey=f
st
hexaxi
sat.
hecur
r
entest
i
mat
eof.
rLetabet
r
(
x)meet
Thet
angentl
i
net
oy=f
hepoi
nt(
a,f
(
a)
)hasequat
i
on
(
x)att
'
a)+(
xa)
a)
(
(
y=f
f
Letbbet
hexi
nt
er
ceptoft
het
angentl
i
ne.Then
a)
(
f
b =a-'
a)
(
f
19
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
sj
ustt
he‘
next
’
Newt
onRaphsonest
i
mat
eof.
bi
r
2x5 =0
Exampl
e1:
UseNewt
onRaphsonmet
hodt
of
i
ndar
ootoft
heequat
i
onx3Sol
ut
i
on
'
x)=3x2x)=x3(
(
f
2x5 =0 and f
2
xn)
(
f
Usi
ng xn +1 =xn'x
(n)
f
'x
x
(0)=10
(
Choosi
ngx0 =2,
weobt
ai
nf
1andf
0)=1
∴ x1 =2-- =2.
1
10
2
3
'x
x1)=(
(1)=3(
2.
1)
(
2.
1)
2.
1)Nowf
2 =11.
23
2(
5 =0.
061and f
0.
061
Hence x2 =2.
1=2.
094568
11.
23
I
tshowst
hat
,compar
et
opr
ev
i
ousexampl
esi
nt
heot
heri
t
er
at
i
onmet
hod,Newt
onRaphson
met
hodconv
er
gesmor
er
api
dl
yt
ot
hedesi
r
edr
oot
.
( )
x1 =0cor
Exampl
e2:UseNewt
on’
sMet
hodt
of
i
ndt
heonl
yr
ealr
ootoft
heequat
i
onx3r
ectt
o
9deci
mal
pl
aces.
'
x
x)=x3(
)=3x2(
(
(
1)=2)=5,t
x1andf
1andf
Wehav
ef
1.Si
ncef
hef
unct
i
onhasa
r
ooti
nt
hei
nt
er
v
al[
1,2]si
ncet
hef
unct
i
onchangessi
gnbet
ween[
1,2]
.Letusmakeani
ni
t
i
al
guessx0 =1.
5.
Newt
on’
sf
or
mul
aher
ei
s
1 2x3 +1
x3xxn +1 =xn-n n = n
3xn21 3xn21
5,
So,
wi
t
hourv
al
ueofx0 =1.
ourappr
oxi
mat
i
onf
orx1i
sgi
v
enby
3
1.
5)
2(
+1
x1 =
≈1.
347826087
2
1.
5)
3(
1
3
2(
+1
1.
347826087)
x2 =
=1.
325200399
2
3(
1
1.
347826087) 3
2(
+1
1.
325200399)
x3 =
=1.
324718174
2
3(
1
1.
325200399) 3
2(
+1
1.
324718174)
x4 =
=1.
324717957
2
3(
1
1.
324718174) 3
2(
+1
1.
324717957)
x5 =
=1.
324717957
2
3(
1
1.
324717957)
20
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
324717957cor
Ther
ef
or
e r=1.
r
ect
l
yr
oundedt
o9deci
mal
pl
aces
Exampl
e3:Usi
ngNewt
onRaphsonmet
hod,ev
al
uat
et
ot
wodeci
malf
i
gur
es,t
her
ootoft
he
equat
i
onex =3xl
y
i
ngbet
ween0and1.
x)=ex(
Sol
ut
i
on:f
3x
(
)
f0 =1
(
1)=e1f
3 =0.
2817
Themi
ddl
epoi
ntoft
hei
nt
er
v
al
(
0,
1)i
s0.
5
'x
x
x
(
)
n
Letx0 =0.
=ex 3
5;f
3xn;f
(n)=e n
n
xn)
ex -3xn xnex -ex
(
f
=xn= x
∴x
xnn +1 =
x
'x
e3
(n)
e -3
f
x
x
x0e -e 0.
5e0.5 -e0.5
=
=0.
6101
x1 = x
e3
e0.5 -3
x
x
x1e -e 0.
6101e0.6101 -e0.6101
=
=0.
6190
x2 = x
e3
e0.6101 -3
x
x
x2e -e 0.
6190e0.6190 -e0.6190
x3 = x
=
=0.
6191
0.
6190
e3
-3
e
n
n
n
0
n
n
0
0
1
1
1
2
2
2
3
Exampl
e4:Wr
i
t
et
heNewt
onRaphsonpr
ocedur
ef
orf
i
ndi
ng N,
wher
eNi
sar
ealnumber
.Use
3
i
tt
of
i
nd 18cor
r
ectt
o2deci
mal
s,
assumi
ng2.
5ast
hei
ni
t
i
al
appr
oxi
mat
i
on.
3
Sol
ut
i
on:
Letx= N x3 =Norx3N =0
'
x)=3x2
x)=x3(
(
Letf
N =0 f
ByNewt
onRaphsonMet
hod,
xn)
(
f
xn +1 =xn'x
(n)
f
3
xn -N 2xn3 +N
=xn=
, n =0,
1,
2,
…
3xn2
3xn2
5
LetN =18,x=appr
oxi
mat
ecuber
ootof18 =2.
3
2(
+18
2.
5)
=2.
62667
x1 =
2
2.
5)
3(
3
2(
+18
2.
62667)
x2 =
=2.
62075
2
2.
62667)
3(
3
2(
+18
2.
62075)
x3 =
=2.
62074
2
3(
2.
62075)
3
62074
∴ 18 =2.
Exampl
e5:
Comput
et
her
eci
pr
ocal
ofAi
.
e.1usi
ngNewt
onRaphsonmet
hodi
fA =3
A
Sol
ut
i
on
1
Letx=
x-1 =A
x-1A =0
A
1
'
x)=-2
x)=x-1(
(
A =0
f
Letf
x
21
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
xn)
(
f
x-1A
∴ xn +1 =xn=xn=2xnAxn2
'x
1
(
)
fn
x2
25
ForA =3 & Choosi
ngx0 =0.
2
3(
=0.
3125
x1 =2(
0.
25)0.
25)
2
x2 =2(
3(
=0.
3320
0.
3125)0.
3125)
2
x3 =2(
3(
3333
0.
3320)0.
3320) =0.
1
Not
et
hat0.
3333i
st
he4di
gi
tappr
ox
i
mat
i
onof (
cor
r
ectupt
o4deci
mal
f
i
gur
es)
.
3
AModi
f
i
edNewt
onMet
hodf
orMul
t
i
pl
eRoot
s
Thi
si
sal
socal
l
edGener
al
i
sedNewt
on’
sMet
hod.
x)=0wi
(
I
fx* i
sar
ootoff
t
hmul
t
i
pl
i
ci
t
yP,t
hent
hei
t
er
at
i
onf
or
mul
acor
r
espondi
ngt
o
t
heequat
i
oni
st
akenas
xn)
(
f
xn +1 =xnP
'x
(n)
f
1'x
(0)i
whi
chmeanst
hat f
st
hesl
opeoft
hest
r
ai
ghtl
i
nepassi
ngt
hr
ough(
andi
nt
er
sect
i
ng
xn y
n)
P
t
hexaxi
satt
hepoi
nt(
.
xn +1,0)
'
*
x)=0wi
x)=0wi
(
(
Si
ncex i
st
her
ootoff
t
hmul
t
i
pl
i
ci
t
yP,
i
tf
ol
l
owst
hatx*i
sal
soar
ootoff
t
h
'
'
x
(
)
mul
t
i
pl
i
ci
t
y(
nar
ootoff =0 wi
t
hmul
t
i
pl
i
ci
t
y(
he
P1),agai
P2) andsoon.Hencet
expr
essi
on
'x
'x
x0)
(
f
(0)
(0)
f
f
P
, x0,
x0x0(
)
(
)
P1
P2
'x
'
'
'
'
x0)
x0)
(0)
f
f(
f(
musthav
et
hesamev
al
uei
far
ootwi
t
hmul
t
i
pl
i
ci
t
yP,pr
ov
i
dedt
hatt
hei
ni
t
i
alappr
oxi
mat
i
onx0
i
schosensuf
f
i
ci
ent
l
ycl
oset
ot
her
oot
.
x)=x3(
Exampl
e:
Fi
ndt
hedoubl
er
ootoft
heequat
i
onf
x+1 =0
x28,
Sol
ut
i
on:
Choosi
ngx0 =0.
wehav
e,
'
'
'
x)=3x2x)=6x(
(
2x1 and f
2
f
Mul
t
i
pl
i
ci
t
yP =2;
x
(
f
0)
0.
072
x02
=0.
82
=1.
012
'x
(0)
(0.
68)
f
'x
'x
'x
(0)
(0)
(0)
f
f
f
x0and
=0.
8=0.
8(
(
P1)''
21)''
'
'
x0)
x0)
x0)
f(
f(
f(
(0.
68)
=0.
8=1.
043
2.
8
Thecl
osenessoft
hesev
al
uesi
ndi
cat
est
hatt
her
ei
sadoubl
er
ootnear
eruni
t
y
.Fort
henext
appr
oxi
mat
i
onwechoosex1 =1.
01andobt
ai
n
x1)
(
f
x12
=1.
010.
0099 =1.
0001
'x
(1)
f
'x
(1)
f
x1and
=1.
010.
0099 =1.
0001
'
'x
f(1)
0001whi
Weconcl
udet
her
ef
or
et
hatt
her
ei
sadoubl
er
ootatx=1.
chi
ssuf
f
i
ci
ent
l
ycl
oset
ot
he
act
ual
r
ootuni
t
y
.
22
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
 SECANTMETHOD
Themaj
ordi
sadv
ant
ageoft
heNewt
onMet
hodi
st
her
equi
r
ementoff
i
ndi
ngt
hev
al
ueoft
he
der
i
v
at
i
v
eof f
oxi
mat
i
on.Ther
ear
ef
unct
i
onsf
orwhi
cht
hi
sj
obi
sei
t
her
(
x) ateachappr
ext
r
emel
ydi
f
f
i
cul
t(
i
fnoti
mpossi
bl
e)ort
i
meconsumi
ng.A wayouti
st
oappr
oxi
mat
et
he
der
i
v
at
i
v
ebyknowi
ng t
hev
al
uesoft
hef
unct
i
onatt
hatand t
hepr
eci
ousappr
oxi
mat
i
on.
'
xn)andf
xn-1)
xn)as:
Knowi
ngf
,
wecanappr
oxi
mat
ef
(
(
(
xn)xn-1)
f
f
(
(
'
xn)≈ x f
(
n x
n1
Then,
t
heNewt
oni
t
er
at
i
onsi
nt
hi
scasebecome:
xn)
(
f
xn +1 =xn'
(
xn)
f
xnxn-1)
xn)
(
(
f
≈xnxn)xn-1)
f
f
(
(
Thi
sconsi
der
at
i
onl
eadst
ot
heSecantmet
hod
Al
gor
i
t
hm oft
heSecantMet
hod:
x)-Thegi
(
I
nput
s:
f
v
enf
unct
i
on
x0,
x1-Thet
woi
ni
t
i
al
appr
oxi
mat
i
onsoft
her
oot
r
ort
ol
er
ance
∈-Theer
N-Themaxi
mum numberofi
t
er
at
i
ons
Out
put
s:Anappr
oxi
mat
i
ont
ot
heexactsol
ut
i
onunt
i
l
t
hest
eppi
ngcr
i
t
er
i
ai
smet
.
x
x
(n)andf
(n-1)
 Comput
ef
 Comput
et
henextappr
oxi
mat
i
on:
xnxn-1)
xn)
(
(
f
x
xnn +1 =
xn)xn-1)
f
f
(
(
xn +1xn|< ∈,
 Testf
orconv
er
genceormaxi
mum numberofi
t
er
at
i
ons:
I
f|
St
op.
Not
e:
Thesecantmet
hodneedst
woappr
oxi
mat
i
onsx0andx1t
ost
ar
twi
t
h,
wher
east
heNewt
on
x0.
met
hodj
ustneedsone,
namel
y
,
Exampl
e:Appr
oxi
mat
et
hePosi
t
i
v
eSquar
eRootof2,choosi
ngx0 =1.
Dof
our
5 andx1 =1 (
i
t
er
at
i
ons)
.
x)=x2
(
Sol
ut
i
on: f
I
ni
t
i
al
appr
oxi
mat
i
ons:
x0 =1.
5andx1 =1
xnxn-1)
xn)
(
(
f
For
mul
at
obeused:
xn +1 =xnxn)xn-1)
f
f
(
(
.
x1x
x1)
(
(
((1)
0.
5)
f
0)
0.
5
=1=1 +
=1.
4
x2 =x1x1)10.
25
1.
25
x0)
f
f
(
(
x2)
x2x1)
(
(
((
0.
04)
0.
4)
f
=1.
4=1.
4167
x3 =x2-x2)f
f
(
(x1)
1)
0.
04)(
(
23
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
x3)
x3x2)
(
(
f
=1.
4142
x4 =x3x
x
(
)
(
)
f3 f2
x
x3)
x
(4)
(4f
x5 =x4=1.
4142
x
x4)(
(
)
f
f
3
Not
e:Bycompar
i
ngt
her
esul
t
soft
hi
sexampl
ewi
t
ht
hoseobt
ai
nedbyt
heNewt
onmet
hod,we
seet
hati
tt
ook4i
t
er
at
i
onsbyt
hesecantmet
hodt
oobt
ai
na4di
gi
taccur
acyof 2 =1.
4142;
wher
easf
ort
heNewt
onmet
hod,
t
hi
saccur
acywasobt
ai
nedj
ustaf
t
er2i
t
er
at
i
ons.
I
ngener
al
,
t
heNewt
onmet
hodconv
er
gesf
ast
ert
hant
hesecantmet
hod.
AGeomet
r
i
cI
nt
er
pol
at
i
onoft
heSecantMet
hod
 x2i
st
hexi
nt
er
ceptoft
hesecantl
i
nepassi
ngt
hr
ough(
.
x0)
x1)
x,f
x,f
(
(
)
)and(
0
1
 x3i
st
hexi
nt
er
ceptoft
hesecantl
i
nepassi
ngt
hr
ough(
.Andso
x2)
x1)
x2,f
x1,f
(
(
)and(
)
on.
Not
e:
Thati
swhyt
hemet
hodi
scal
l
edt
heSecantMet
hod.
 FI
XEDPOI
NTI
TERATI
ON
Def
i
ni
t
i
on:
Anumberξi
saf
i
xedpoi
ntofaf
unct
i
ong(
fg(
.
x)i
ξ
)=ξ
x)=0i
(
Supposet
hatt
heequat
i
onf
swr
i
t
t
eni
nt
hef
or
m x=g(
;
t
hati
s,
x)
x
x
f
g()=0
-(
1)
()=xx)=0because
(
Thenanyf
i
xedpoi
ntξofg(
sar
ootoff
x)i
ξ
ξ
f
g(
ξ=0
(
)=ξ)=ξSt
ar
twi
t
hani
ni
t
i
al
guessx0oft
her
ootandf
or
m asequence{
i
nedby
xn}def
xn +1 =g(
xn)
, n =0,
1,
2,
…
-(
2)
x)=0
(
I
ft
hesequence{
er
ges,
t
henl
l
l
bear
ootoff
i
m xn =ξwi
xn}conv
n→
Ther
ear
emanyway
sofdoi
ngt
hi
s.Forexampl
e,
t
heequat
i
onx3 +x21 =0canbeexpr
essedas
1
-
(
a)x=(
1 +x)2
1
2
(
b)x=(
x3)
1-
24
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
1
3
(
c)x=(
et
c
x2)
1Letx0 beanappr
oxi
mat
ev
al
ueoft
hedesi
r
edr
ootx*.Put
t
i
ngi
tf
orxont
her
i
ghtsi
deof
equat
i
on(
2)
,
weobt
ai
nt
hef
i
r
stappr
oxi
mat
i
on
x1 =g(
x0)
Thesuccessi
v
eappr
oxi
mat
i
onsar
et
hengi
v
enby
x2 =g(
x1)
x3 =g(
x2)
………….
xn-1)
xn =g(
………….
xn +1 =g(
xn)
xndoesnotal
Thesequencex0 ,x1,x2 ,
…,
way
sconv
er
get
ot
her
ootx*.Thef
ol
l
owi
ngt
heor
em
gi
v
esasuf
f
i
ci
entcondi
t
i
onong(
chensur
est
heconv
er
genceoft
hesequence{
w
i
t
h
x)whi
x
n}
anyi
ni
t
i
al
appr
oxi
mat
i
onx0i
n[
.
a,b]
x)=0bewr
(
Theor
em:Letf
i
t
t
eni
nt
hef
or
m x=g(
.Assumet
hatg(
i
sf
i
est
hef
ol
l
owi
ng
x)
x)sat
pr
oper
t
i
es:
a,b]
xi
(
i
)
Foral
l
n[
,
g(
x)∈[
;
t
hati
sg(
akesev
er
yv
al
uebet
weenaandb.
x)t
a,b]
(
i
i
)
st
son(
t
ht
hepr
oper
t
yt
hatt
her
eexi
st
saposi
t
i
v
econst
ant0<r<1
g'(
x)exi
a,b)wi
sucht
hat
xi
f
oral
l
n(
.
a,b)
Then
(
i
)
(
i
i
)
,
g'(
x)
|
|≤r
t
her
ei
sauni
quef
i
xedpoi
ntx=ξofg(
n[
,
x)i
a,b]
f
oranyx0i
n[
,
t
hesuccessi
v
e{
i
nedby
xn}def
a,b]
xn +1 =g(
, k=0,
1,
…
xn)
x)=0.
(
er
gest
ot
hef
i
xedpoi
ntx=ξ
;
t
hati
s,
t
ot
her
ootξoff
conv
Toest
i
mat
et
heer
r
oroft
heappr
oxi
mat
i
onr
ootobt
ai
ned,
wehav
e
k x-(
3)
x*xn|≤ |
n x
|
n1|
1k
I
ngener
al
,t
hespeedoft
hei
t
er
at
i
ondependsont
hev
al
ueofk;t
hesmal
l
ert
hev
al
ueofk,t
he
f
ast
erwoul
dbet
heconv
er
gence.I
fεi
st
hespeci
f
i
edaccur
acy
,
t
hati
s,
i
f
x*xn|≤ε
|
t
henf
or
mul
a(
3)gi
v
es
k
xnxn-1|≤1.
ε
|
k
whi
chcanbeusedt
of
i
ndt
hedi
f
f
er
encebet
weent
wosuccessi
v
ei
t
er
at
esnecessar
yt
oachi
ev
e
aspeci
f
i
caccur
acy
.
Pr
oofoft
heFi
xedPoi
ntTheor
em:
Thepr
oofcomesi
nt
hr
eepar
t
s:
Exi
st
ence,
Uni
queness,
andConv
er
gence.
Pr
oofofExi
st
ence
Fi
r
st
,
consi
dert
hecasewher
eei
t
herg(
b)=b.
a)=aorg(
25
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
x)=0
(
 I
fg(
t
henx=ai
saf
i
xedpoi
ntofg(
.Thus,
x=ai
sar
ootoff
x)
a)=a,
x)=0
(
 I
fg(
t
henx=bi
saf
i
xedpoi
ntofg(
.Thus,
sar
ootoff
x=bi
x)
b)=b,
Next
,
consi
dert
hegener
alcasewher
et
heabov
eassumpt
i
onsar
enott
r
ue.Thati
s,
g(
a)≠aand
g(
b)≠b
I
nsuchacase,
si
sbecause,
byAssumpt
i
on1,
bot
hg(
ei
n
g(
b)<b.Thi
a)>aandg(
a)andg(
b)ar
,
andsi
nceg(
eat
ert
hanaandg(
eat
ert
han
g(
a)≠aandg(
b)≠b,
a)mustbegr
b)mustbegr
a,b]
[
b.
x)sat
x)=g(
x)Now,
def
i
net
hef
unct
i
onh(
l
lnowshowt
hath(
i
sf
i
esbot
ht
hehy
pot
hesi
s
x.Wewi
oft
heI
nt
er
medi
at
eVal
ueTheor
em (
I
VT)
.
I
nt
hi
scont
ext
,
wenot
e
(
i
)
si
nceg(
scont
i
nuousbyourassumpt
i
ons,h(
sal
socont
i
nuouson[
.
x) i
x) i
a,b]
(
Hy
pot
hesi
s(
i
)i
ssat
i
sf
i
ed)
.
a)=g(
a)(
i
i
) h(
Hy
pot
hesi
s(
i
i
)i
ssat
i
sf
i
ed)
a>0andh(
=
b<0.(
)
b g(
b)Si
ncebot
hhy
pot
hesi
soft
heI
VTar
esat
i
sf
i
ed,
byt
hi
st
heor
em,
t
her
eexi
st
sanumberci
n[
a,b]
sucht
hat
c)=0
h(
t
hi
smeansg(
c)=c.
Thati
s,
saf
i
xedpoi
ntofg(
.Thi
spr
ov
est
heExi
st
encepar
toft
heTheor
em.
x=ci
x)
Pr
oofofUni
queness(
byCont
r
adi
ct
i
on)
Ourl
i
neofpr
oofwi
l
l
beasf
ol
l
ows:
Wewi
l
lf
i
r
stassumet
hatt
her
ear
et
wof
i
xedpoi
nt
sofg(
n[
henshow t
hatt
hi
s
x)i
a,b]andt
assumpt
i
onl
eadst
oacont
r
adi
ct
i
on.
Supposeξ
ndξ
r
et
wof
i
xedpoi
nt
si
n[
,andξ
Thepr
oofi
sbasedont
heMean
1a
1 ≠ξ
2a
2.
a,b]
Val
ueTheor
em appl
i
edt
og(
n[
.
x)i
ξ,ξ]
1
2
I
nt
hi
scont
ext
,
not
et
hatg(
sdi
f
f
er
ent
i
abl
eandhence,
cont
i
nuouson[
.
x)i
ξ,ξ]
1
2
So,
byMVT,
wecanf
i
ndanumberci
n(
sucht
hat
ξ
1,ξ
2)
g(
g(
ξ
ξ
1)
2)
c)
=g'(
ξ
ξ
(
1)
2ξ
ξ
ξ
ξ
Si
nceg(
andg(
becauset
heyar
ef
i
x
edpoi
nt
sofg(
,
weget
x)
1)=
1,
2)=
2,
ξ
ξ
2 1
c)
=g'(
ξ
ξ
1
2c)=1,
ξ
Thati
s,
whi
chi
sacont
r
adi
ct
i
ont
oourAssumpt
i
on2.Thus,
annotbedi
f
f
er
entf
r
om
g'(
1c
ξ
,
t
h
i
s
p
r
o
v
e
s
t
h
e
u
n
i
q
u
e
n
e
s
s
.
2
Pr
oofofConv
er
gence
x)=0.Thent
(
Letξbet
her
ootoff
heabsol
ut
eer
r
oratst
epk+1i
sgi
v
enby
ek+1 =|
ξxk+1|
Topr
ov
et
hatt
hesequenceconv
er
ges,
weneedt
oshowt
hatl
i
m ek+1 =0.
k→
26
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
Toshowt
hi
s,
weappl
yt
heMeanVal
ueTheor
em t
og(
x)i
n[
.Si
nceg(
sosat
i
sf
i
esbot
h
x)al
x,ξ
]
k
t
hehy
pot
hesi
soft
heMVTi
n[
,
weget
x,ξ
]
k
xk)
g(
g(
ξ
)c)
=g'(
ξxk
-(
*
)
wher
exk <c<ξ
.
x=ξi
Now,
saf
i
xedpoi
ntofg(
,
sowehav
e(
bydef
i
ni
t
i
on)
:
x)
(
i
)
g(
=
ξ
ξ
)
Al
so,
f
r
om t
hei
t
er
at
i
onx=g(
,
wehav
e
x)
xk)
(
i
i
) xk+1 =g(
So,
f
r
om (
*
)
,
weget
xk+1
ξξxk
c)
=g'(
Taki
ngabsol
ut
ev
al
uesonbot
hsi
deswehav
e
Or
xk+1|
ξ|
=|
c)
g'(
|
ξ
x
| k|
ek+1
=g'
c)
(
e | |
k
Si
nce|
,
wehav
eek+1 ≤r
ek.
c)
g'(
|≤r
2
Si
ncet
heabov
er
el
at
i
oni
st
r
uef
orev
er
yk,
wehav
eek ≤r
ek-1.Thus,
ek+1 ≤r
ek-1.
k+1
Cont
i
nui
ngi
nt
hi
sway
,wef
i
nal
l
yhav
eek+1 ≤r
ee0 i
st
hei
ni
t
i
aler
r
or
.(
Thati
s,
e0,wher
)
.
e0 =|
x0ξ
|
k+1
Si
ncer<1,
wehav
er
→0ask→ .
Thus,
l
i
m ek+1 =l
i
m|
xk+1ξ
|
k→
k→
k+1
≤l
i
mr
e0 =0
k→
Thi
spr
ov
est
hatt
hesequence{
er
gest
ox=ξandt
hatx=ξi
st
heonl
yf
i
xedpoi
ntof
xk}conv
g(
.
x)
Exampl
e1:Fi
ndar
ealr
ootoft
heequat
i
onx3 +x21 =0ont
hei
nt
er
v
al[
t
hanaccur
acy
0,1]wi
of10-4.
Sol
ut
i
on
Tof
i
ndt
hi
sr
oot
,
wer
ewr
i
t
et
hegi
v
enequat
i
oni
nt
hef
or
m
1
1
x=
=(
x+1)2
x+1
1 1
1
x)=g(
Thus,
, g'(
x)=
3
2
x+1
(
x+1)2
Max ' = 1
and
= 1 =k=0.
2
x)
g(
|
|
3
0,1]
[
28
2(
x+1)2
k
xnxn-1|≤1Now,
usi
ng |
.
ε
k
(
10.
2)×10-4
<
=0.
0004
0.
2
27
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
5,
St
ar
t
i
ngwi
t
hx0 =0.
weobt
ai
n
1
1
x0)=
=
=0.
8165
x1 =g(
x0 +1 0.
5 +1
1
x1)= 1 =
x2 =g(
=0.
7420
x1 +1 0.
8165 +1
1
x2)= 1 =
x3 =g(
=0.
7577
x2 +1 0.
7420 +1
1
x3)= 1 =
x4 =g(
=0.
7543
x3 +1 0.
7577 +1
1
x4)= 1 =
x5 =g(
=0.
7550
x4 +1 0.
7543 +1
1
x5)= 1 =
x6 =g(
=0.
7549
x5 +1 0.
7550 +1
Att
hi
sst
age,
wef
i
ndt
hat
x
x
n|=
|
0.
75490.
7550 =0.
0001
n +10004
whi
chi
s <0.
Hencewest
opher
et
aki
ng0.
7549(
4d.
p)ast
her
oott
ot
her
equi
r
edaccur
acy
.
Exampl
e2:
Fi
ndt
her
ootoft
heequat
i
on2x=cosx+3,
cor
r
ectt
ot
hr
eedeci
mal
.
Sol
ut
i
on
1
Rewr
i
t
i
ngi
nt
hi
sf
or
m x=g(
.
e. x= (
x) i
cosx+3)
2
1
g(
x)= (
cosx+3)
2
St
ar
t
i
ngwi
t
hx0 =1.
5
xn
cosxn +3
cosxn +3
n
2
0
1.
5
3.
0707
1.
5354
1
1.
5354
3.
0354
1.
5177
2
1.
5177
3.
0531
1.
5265
3
1.
5265
3.
0443
1.
5221
4
1.
5221
3.
0487
1.
5243
5
1.
5243
3.
0465
1.
5232
6
1.
5232
3.
0476
1.
5238
7
1.
5238
3.
0470
1.
5235
8
1.
5235
3.
0473
1.
5236
9
1.
5236
3.
0472
1.
5236
28
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
Hence,
wet
aket
hesol
ut
i
onas1.
524cor
r
ectt
o3d.
p.
Exampl
e3:Fi
ndaf
i
xedpoi
nti
t
er
at
i
v
eschemef
ordet
er
mi
ni
ng a whena>0,andusei
tt
o
cal
cul
at
e 2t
oanaccur
acyofsi
xdeci
mal
pl
aces
Sol
ut
i
on
Let x=a
x2 =a
2x2 =x2 +a
1
a
x= x+
x
2
1
a
So,
g(
x)= x+
x
2
Nowr
epl
aci
ngxont
hel
ef
tbyxn +1andxont
her
i
ghtbyxnt
oobt
ai
n
a
1
xn +1 = xn +
xn
2
Thei
t
er
at
i
oni
sst
ar
t
edbyset
t
i
ngn=0andx0 =k,
wher
eki
sanappr
oxi
mat
i
ont
oa
Toi
l
l
ust
r
at
et
heschemewewi
l
l
cal
cul
at
e 2,
sot
hata=2
xn +1 =1xn +2
xn
2
andf
orsi
mpl
i
ci
t
ywest
ar
tbyset
t
i
ngx0 =1.Ther
esul
t
soft
hecal
cul
at
i
onar
e
x0 =1
x1 =1.
5
x2 =1.
41666667
41421569
x3 =1.
41421356
x4 =1.
41421356
x5 =1.
Ast
hex4andx5i
t
er
at
esar
ei
dent
i
cal
,
r
educi
ngt
her
esul
tofx5t
osi
xdeci
mal
pl
acesgi
v
es
2 =1.
414214
( )
( )
(
)
(
)
I
mpor
t
antAddi
t
i
on
Thef
i
xedpoi
nti
t
er
at
i
v
eschemesconv
er
gedr
api
dl
y
,andi
ti
st
hi
sschemet
hati
susedi
n
comput
er
st
odet
er
mi
net
hesquar
eofr
ootofanyposi
t
i
v
enumbert
oanaccur
acyt
hati
swi
t
hi
n
t
hecapabi
l
i
t
yoft
hecomput
i
ngsy
st
em andsof
t
war
ebei
ngused.
I
nt
heBi
sect
i
onmet
hod,ev
er
yi
nt
er
v
alunderconsi
der
at
i
oni
sguar
ant
eedt
ohav
et
hedesi
r
ed
r
ootx=ξ.Thati
swhyt
heBi
sect
i
onmet
hodi
sof
t
encal
l
edabr
acketmet
hod,becauseev
er
y
i
nt
er
v
albr
acket
st
her
oot
.Howev
er
,
t
heNewt
onmet
hodandt
heSecantmet
hodar
enotbr
acket
met
hodsi
nt
hatsense,becauset
her
ei
snoguar
ant
eet
hatt
het
wosuccessi
v
eappr
oxi
mat
i
on
wi
l
l
br
ackett
her
oot
.
CONVERGENCEANALYSI
SOFTHEI
TERATI
VEMETHODS
Whenamet
hodconv
er
ges,how f
astdoesi
tconv
er
ge?I
not
herwor
ds,whati
st
her
at
eof
conv
er
gence?Tot
hi
send,
wef
i
r
stdef
i
ne:
Def
i
ni
t
i
on:(
Rat
eofConv
er
genceofanI
t
er
at
i
v
eMet
hod)
.Supposet
hatt
hesequence {
xn}
ξ
conv
er
gest
oξ.Thent
hesequence{
i
s
s
a
i
dt
oc
o
n
v
e
r
g
et
o
w
i
t
ht
h
eo
r
d
e
r
o
f
c
o
n
v
e
r
g
e
n
c
e
α
xn}
i
ft
her
eexi
st
saposi
t
i
v
econst
antpsucht
hat
29
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
l
i
m
n→
ξ
xn +1e
|
|=l
i
m n +1 =p
α
ξ
xn|
|
n→
enα
 I
fα =1,
t
heconv
er
gencei
sl
i
near
 I
fα =2,
t
heconv
er
gencei
squadr
at
i
c
 I
f1<α<2,
t
heconv
er
gencei
ssuper
l
i
near
Usi
ngt
heabov
edef
i
ni
t
i
on,
wewi
l
l
nowshow:
 Ther
at
eofconv
er
genceoft
hef
i
xedpoi
nti
t
er
at
i
oni
susual
l
yl
i
near
,
 Ther
at
eofconv
er
genceoft
heNewt
onmet
hodi
squadr
at
i
c,
 Ther
at
eofconv
er
genceoft
heSecantmet
hodi
ssuper
l
i
near
.
Compar
i
sonofDi
f
f
er
entMet
hodsf
ort
heRoot
f
i
ndi
ngPr
obl
em
Met
hod
Bi
sect
i
on
Wor
kRequi
r
ed
Conv
er
gence
Two
f
unct
i
on Al
way
s
ev
al
uat
i
ons per conv
er
ges(
but
i
t
er
at
i
on
sl
ow)
Remar
ks
(
i
)
The i
ni
t
i
ali
nt
er
v
alcont
ai
ni
ng
t
her
ootmustbef
oundf
i
r
st
(
i
i
)
A br
acket
i
ng met
hod – ev
er
y
i
t
er
at
i
onbr
acket
st
her
oot
.
Fi
xedPoi
nt One
f
unct
i
on Rat
e
of Fi
ndi
ngt
her
i
ghti
t
er
at
i
onf
unct
i
onx=g(
x)
ev
al
uat
i
on
per conv
er
gencei
s i
sachal
l
engi
ngt
ask.
i
t
er
at
i
on
l
i
near
Newt
on
Secant
Two
f
unct
i
on
ev
al
uat
i
ons per
i
t
er
at
i
on
–
f
unct
i
on
v
al
ue
and v
al
ueoft
he
der
i
v
at
i
v
e att
he
i
t
er
at
e
Quadr
at
i
c
conv
er
gence
at a si
mpl
e
r
oot
.
Conv
er
gence
i
sl
i
nearf
ora
mul
t
i
pl
er
oot
Two
f
unct
i
on Super
l
i
near
ev
al
uat
i
ons per
i
t
er
at
i
on
–
howev
er
,
onl
yone
oft
he t
wo i
sa
newev
al
uat
i
on
(
i
)
I
ni
t
i
alappr
oxi
mat
i
on mustbe
chosen car
ef
ul
l
y– i
fi
ti
sf
ar
f
r
om t
her
oot
,t
hemet
hodwi
l
l
di
v
er
ge.
(
i
i
)
For some f
unct
i
on, t
he
der
i
v
at
i
v
e i
s di
f
f
i
cul
t t
o
comput
e,
i
fnoti
mpossi
bl
e
Needst
woi
ni
t
i
al
appr
oxi
mat
i
ons
LessonEndAct
i
v
i
t
i
es
1.Fi
ndaposi
t
i
v
er
ootoft
hef
ol
l
owi
ngequat
i
onbybi
sect
i
onmet
hod:
4x9
x3(
i
)
Ans:
2.
7065
(
i
i
) 3x=cosx+1
Ans:
0.
66664
(
i
i
i
) x3 +3x1
Ans:
0.
322
30
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
3x
(
i
v
) exAns:
0.
6190
(
v
) cosx2x+3
Ans:
0.
3604
2.(
I
)Sol
v
et
hef
ol
l
owi
ngbymet
hodoff
al
seposi
t
i
on(
Regul
af
al
si
Met
hod)
:
3
2
(
i
)
Ans:
1.
732
x +x 3x3
20
(
i
i
) x3 +2x2 +10xAns:
1.
3688
(
i
i
i
) 2x3si
nx=5
Ans:
2.
8832
(
i
v
) e-x =si
nx
Ans:
0.
5885
x
3x
(
v
) eAns:
6.
0890
(
v
i
) cosx2x+3
Ans:
1.
5236
x
(
I
I
)Uset
hef
al
seposi
t
i
onmet
hodt
of
i
ndt
her
ootofxsi
n(
1 =0t
hati
sl
ocat
edi
nt
he
)i
nt
er
v
al
(
t
h
e
f
u
n
c
t
i
o
n
s
i
n
(
x
)
i
s
e
v
a
l
u
a
t
e
di
n
r
a
d
i
a
n
s
)
.
A
n
s
:
1
.
1
1
4
1
5
7
1
4
0,2]
[
x)=x2(
01,
01andst
(
I
I
I
)Consi
derf
i
ndi
ngt
her
ootoff
3.Letεstep =0.
andεabs =0.
ar
twi
t
h
t
hei
nt
er
v
al
A
n
s
:
1
.
7
3
4
4
1,2]
[
3.I
.(
i
)Comput
et
her
eal
r
ootofxl
1.
2 =0usi
ngNewt
on’
smet
hod Ans:
2.
7407
og10x(
i
i
)x2si
nx=0
Ans:
1.
8955
(
i
i
i
)cosxx
Ans:
0.
5177
ex
2
+7 i
I
I
.(
i
)Fi
ndt
her
ootoft
hef
unct
i
ony=x3 +4x
nt
hev
i
ci
ni
t
yofx=r
ectt
o5
4 cor
deci
mal
pl
aces
x)=2x3(
(
i
i
)Fi
ndt
heposi
t
i
v
er
ootoff
3x6 =0byNewt
onRaphsonmet
hodcor
r
ectt
o
f
i
v
edeci
mal
pl
aces.
Ans:
1.
783769
I
I
I
.Appr
oxi
mat
et
heposi
t
i
v
esquar
er
ootof2usi
ngNewt
on’
smet
hodwi
t
hx0 =1.
5(
Do
t
hr
eei
t
er
at
i
ons)
.
Ans:
1.
4142
 GRAEFFEROOTSQUARI
NGMETHOD
Thi
smet
hodhasagr
eatadv
ant
ageov
ert
heot
hermet
hodst
hati
tdoesnotr
equi
r
epr
i
or
i
nf
or
mat
i
onaboutt
heappr
oxi
mat
ev
al
uesoft
her
oot
s.I
ti
sappl
i
cabl
et
opol
y
nomi
alequat
i
on
hav
i
ngr
eal
anddi
st
i
nctr
oot
s.
Exampl
e:
Appl
yGr
aef
f
e’
sr
ootsquar
i
ngmet
hodt
osol
v
et
heequat
i
onx34x2 +5x2 =0
Sol
ut
i
on
x)=x3(
Gi
v
en f
4x2 +5x2 =0
-(
1)
x)hast
(
Cl
ear
l
yf
hr
eechangesi
.
e.
,
f
r
om +t
o,-t
o +and +t
o.Hencef
r
om Descar
t
esr
ul
e
ofsi
gnsf
et
hr
eeposi
t
i
v
er
oot
s.
(
x)hav
xweget
Separ
at
i
ngt
heev
enandoddpower
sof,
x3 +5x=4x2 +2
2
Rewr
i
t
e,
squar
ebot
hsi
deandputx =y
x(
x2 +5)=4x2 +2
2
2
y
=(
y+5)
4y+2)
(
3
2
+9y=6y
+4
y
2
2
y
y +9)=6y +4
(
2
Squar
i
ngagai
nandput
t
i
ngy
,
weget
=z
2
2
z(
=(
6z+4)
z+9)
z3 +18z2 +81z=36z2 +48z+16
z3 +33z=18z2 +16
z(
z2 +33)=18z2 +16
31
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
Squar
i
ngagai
nandput
t
i
ngz2 =u
2
2
u(
=(
u +33)
18u +16)
Wr
i
t
et
hi
si
nt
hef
or
m:u3 +λ1u2 +λ2u +λ3 =0
u3258u2 +513u256 =0
-(
2)
I
ft
her
ootof(
1)ar
e P1,P2,P3 andt
hoseof(
2)ar
eq1,q2,q3,t
hen
1
1
1
8
8
8
q1)
P1 =(
=(=(
=2.
001946482 ≅2
λ1)
258)
1
1
1
5138
λ
8
q2)
P2 =(
= -2 8 =
=1.
089713189 ≅1
λ1
258
1
1
1
2568
λ
8
q3)
P3 =(
= -3 8 =
=0.
9167804109 ≅1
λ2
513
(
(
(
1)=f
1)=0
2)=f
∴ f
Hencet
her
oot
sar
e2,
1and1
( ) ( )
( ) ( )
8x2 +17x10 =0
Tr
yt
hi
s→ Appl
yGr
aef
f
e’
sr
ootsquar
i
ngmet
hodt
osol
v
et
heequat
i
onx3Ans:
5,
2,
1
⌂2⌂
SOLUTI
ONOFASETOFLI
NEAREQUATI
ONS
x2,
Asy
st
em ofnl
i
nearequat
i
ons(
setofnsi
mul
t
aneousl
i
nearequat
i
ons)i
nnunknownsx
…
1,
xni
sasetofnequat
i
onsoft
hef
or
m
a11x1 +a12x2 +… +a1nxn =b1
a21x1 +a22x2 +… +a2nxn =b2
….
. …… …. ……. ….
an1x1 +an2x2 +… +annxn =bn
-(
1)
wher
et
hecoef
f
i
ci
ent
sajandbjar
egi
v
ennumber
s.Thesy
st
em i
ssai
dt
obehomogeneousi
fal
l
t
he bj ar
e zer
o;ot
her
wi
se t
he sy
st
em i
s sai
dt
o be nonhomogeneous.Usi
ng mat
r
i
x
mul
t
i
pl
i
cat
i
on,
wecanwr
i
t
esy
st
em (
1)asasi
ngl
ev
ect
orequat
i
on
AX =B
-(
2)
aij]i
wher
et
hecoef
f
i
ci
entmat
r
i
xA =[
st
hen×nmat
r
i
x
a11 a12
a21 a22
A=
⋮ ⋮
an1 an2
[
… a1n
x1
… a2n
x2
,
X
=
… ⋮
⋮
… ann
xn
] [] [ ]
b1
b
andB = 2
⋮
bn
andcol
umnv
ect
or
s.Theaugment
edmat
r
i
xAoft
hesy
st
em (
1)i
s
a11 a12
a21 a22
̅ [
A|
b]
A=
=⋮ ⋮
an1 an2
[
32
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
|]
… a1nb1
… a2nb2
… ⋮⋮
… annbn
Asol
ut
i
onofEquat
i
on(
1)i
sasetofnumber
sx1,x2,…xn t
hatsat
i
sf
yal
lt
henequat
i
onsanda
sol
ut
i
onv
ect
orof(
1)i
sav
ect
orxwhosecomponent
sconst
i
t
ut
easol
ut
i
onofEquat
i
on(
1)
Thesol
ut
i
onofal
i
nearequat
i
oncanbeaccompl
i
shedbynumer
i
calmet
hodwhi
chf
al
l
si
none
oft
wocat
egor
i
es:
di
r
ectori
t
er
at
i
v
emet
hods.
 DI
RECTMETHOD
 Adj
oi
ntMet
hod
LetAbeanonsi
ngul
armat
r
i
xsot
hatA-1exi
st
s.Then,
pr
emul
t
i
pl
y
i
ngbot
hsi
desofEquat
i
on(
2)
1
byA ,
weobt
ai
n
A-1AX =A-1B
[
]
A-1A =I
I
X =A-1B
X =A-1B
1
I
fA i
sknown,
t
hent
hesol
ut
i
onv
ect
orxcanbef
oundoutf
r
om t
heabov
emat
r
i
xr
el
at
i
on.
t
hati
s
x1 +2x2 +x3 =4
Exampl
e1:
Sol
v
et
hesetofequat
i
ons
3x14x22x3 =2
5x1 +3x2 +5x3 =1
Fi
r
stwr
i
t
et
hesetofequat
i
onsi
nmat
r
i
xf
or
m,
whi
chgi
v
es
1 2 1 x1
4
4 3 2 x2 = 2
1
5 3 5 x3
i
.
e. Ax=b
∴ x=A-1b
1 2 1
|
A|=3 4 2 =1450 +29 =35
5 3 5
7 0
14 14 25 29
T
25 0 5
7 0
7
Cof
act
orC = adj
A =C = 29 7 10
5 0
10
7
1
4
0
adj
A
1
25 0 5
A-1 =
=- |
A| 35
29 7 10
7 0 4
14 1
25 0 5 2
∴ x=A-1b =- 35
1
29 7 10 70
2
1
=- 105 = 3
35
4
140
x1
2
2; x2 =3;x3 =4
∴ x
Sof
i
nal
l
yx=x2 = 3
1 =
x3
4
(
|
(
|
)( ) ( )
(
)
(
)
(
)( )
( )()
() ( )
Exer
ci
se
1.Sol
v
et
heequat
i
ons3x+y+2z=3
33
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
)
2x3yz=3
x+2y+z=4
x2 +3x3 =2
2.I
f 2x1x1 +3x2x3 =11 ,f
i
ndx1,x2 andx3.
2x12x2 +5x3 =3
Ans:
x =1,y=2,z=1
x1 =Ans:
1,x2 =5,x3 =3
 Gaussi
anEl
i
mi
nat
i
onMet
hod
Thi
si
st
heel
ement
ar
yel
i
mi
nat
i
onmet
hodandi
tr
educest
hesy
st
em ofequat
i
ont
oan
equi
v
al
entuppert
r
i
angul
arsy
st
em whi
chcanbesol
v
edbybacksubst
i
t
ut
i
on.Letusconsi
der
t
hesy
st
em ofequat
i
on.
a11x1 +a12x2 +a13x3 =b1
a21x1 +a22x2 +a23x3 =b2
a31x1 +a32x2 +a33x3 =b3
1.Toel
i
mi
nat
ex1f
r
om t
hesecondequat
i
on,
mul
t
i
pl
yt
hef
i
r
str
owoft
heequat
i
onmat
r
i
xby
a21
- a andaddi
tt
osecondequat
i
on.Si
mi
l
ar
l
yel
i
mi
nat
ex1 f
r
om t
het
hi
r
dequat
i
onand
11
subsequent
l
yal
l
ot
herequat
i
ons.Wegetnewequat
i
onoft
hef
or
m
a11x1 +a12x2 +a13x3 =b1
b22x2 +b23x3 =c2
b32x2 +b33x3 =c3
a
wher
e b22 =a22- 21a ×a12
11
a21
b23 =a23- a ×a13
11
a21
c2 =b2- a ×b1
11
a31
b32 =a32- a ×a12
11
a31
b33 =a33- a ×a13
11
a
c3 =b3- 31a ×b1
11
2.Toel
i
mi
nat
ex2 f
r
om t
het
hi
r
dequat
i
on,mul
t
i
pl
yt
hesecondr
owoft
heequat
i
onmat
r
i
x
b
by -32 andaddi
tt
ot
hi
r
dequat
i
on.Si
mi
l
ar
l
yel
i
mi
nat
ex2 f
r
om t
het
hi
r
dequat
i
onand
b22
subsequent
l
yal
l
ot
herequat
i
ons.
a11x1 +a12x2 +a13x3 =b1
b22x2 +b23x3 =c2
b33x3 =d3
b
wher
e c33 =b33- 32
×b23
b22
×c2
d3 =c3-b32
b22
3.Fr
om t
heabov
er
educedsy
st
em ofequat
i
onsubst
i
t
ut
et
hev
al
uesx3,x2 and x1byback
subst
i
t
ut
i
onwegett
hesol
ut
i
onoft
hegi
v
enequat
i
ons.
(
(
(
(
(
(
)
)
)
)
)
)
( )
( )
Exampl
e2:
Sol
v
et
hesy
st
em ofequat
i
onbyGaussel
i
mi
nat
i
onmet
hod
34
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
Sol
ut
i
on
x+2y+z=3
2x+3y+3z=10
3xy+2z=13
1 2 1x
3
2 3 3 y =10
1 2 z 13
3 1 2 13
Theaugment
edmat
r
i
xbecomes 2 3 310
1 213
3 Theabov
est
epspr
ov
i
dedar
esummar
i
zedasf
ol
l
ows:
1 2 13
a21
R2→R2- R1 0 1 14
a11
1 213
3 1 2 13
a31
R3→R3- R1 0 1 14
a11
7 14
0 Fr
om t
heabov
e:
1 2 1 3
a32
R3→R3- R2 0 1 1 4
a22
24
0 0 8Not
et
hatasar
esul
toft
hesest
eps,t
hemat
r
i
xofcoef
f
i
ci
ent
sofx hasbeenr
educedt
oa
t
r
i
angul
armat
r
i
x.
Fi
nal
l
y
,
wedet
acht
her
i
ght
handcol
umnbackt
oi
t
sor
i
gi
nal
posi
t
i
on:
1 2 1 x
3
1 1 y= 4
0 24
0 0 8z
Then,
by‘
backsubst
i
t
ut
i
on’
,
st
ar
t
i
ngf
r
om t
hebot
t
om r
owweget
:
8z=24
∴ z=3
4 =34 =1
y+z=4
∴ y=zx+2y+z=3
2yz=3 +23 =2
∴ x=3Ther
ef
or
e x=2; y=1;z=3
(
Wr
i
t
ei
nt
hef
or
m Ax=b;
)() ( )
( |)
|)
|)
( |)
(
(
(
)() ( )
Not
e:
Asexampl
e(
2)i
ssol
v
ed,
y
oucanal
sosuccessf
ul
l
yuset
hi
sst
epf
ory
ourel
i
mi
nat
i
on:
a23
R1→R2- R3
a33
a13
R2→R1- R3
a33
a12
R2→R1- R2
a22
Exampl
e3:
UseGaussi
anel
i
mi
nat
i
ont
osol
v
et
hesy
st
em ofl
i
nearequat
i
ons
2x2 +x3 =8
2x23x3 =0
x1x1 +x2 +2x3 =3
Sol
ut
i
on
Let
’
swr
i
t
eAx=b
35
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
I
naugment
edf
or
m wehav
e
0 2 1 x1
8
x
=
1 2 3 2
0
1 1 2 x3
3
0 2 18
1 2 30
1 1 23
(
(
|)
)( ) ( )
|)
1 2 30
SwapRow1andRow2 0 2 1 8
1 1 23
(
|)
( |)
1 2 30
0R1
R2→R20 2 18
R3→R3 +R1
1 13
0 1 2 3
1 0 2 10
R3→R3 + R2
8
2
1
0 0
1
2
1 2 3x
0
1
0 2 1 x
Now,
8
2 =1
x
0 0
1
3
2
1
Bybacksubst
i
t
ut
i
on
-x3 =1
∴ x3 =2
2
1
1
2x2 +x3 =8
∴ x2 =-(
5
x3 +8)=-(
2 +8)=2
2
2x23x3 =0
10 =4
x1∴ x1 =3x3 +2x2 =64; ∴ x2 =5 and ∴ x3 =2
∴ x1 =-
(
( )( ) ( )
Exampl
e4:
UseGaussi
anel
i
mi
nat
i
ont
osol
v
et
hesy
st
em ofl
i
nearequat
i
ons
x1 +5x2 =7
2x17x
5
2 =Sol
ut
i
on
Wr
i
t
ei
nt
hef
or
m Ax=b
1 5 x1
7
=
x
5
7 2
2 Wr
i
t
et
heabov
ei
naugment
edf
or
m
1 57
A |
b)=
(
5
72 -
(
(
)( ) ( )
|)
a21
R2→R2- R1 1 57
a11
0 39
x
7
1 5 1
Ther
ef
or
e
=
9
0 3 x2
Bybacksubst
i
t
ut
i
on
( )( ) ()
( |)
3x2 =9
x1 +5x2 =7
36
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
9
∴ x2 = =3
3
5(
8
∴ x1 =73)=-
8 and x2 =3
Fi
nal
l
y x1 =Not
et
hatwhendeal
i
ngwi
t
ht
heaugment
edmat
r
i
x,
wemay
,
i
fwewi
sh:
(
a)i
nt
er
changet
wor
ows
(
b)mul
t
i
pl
yanyr
owbyanonz
er
of
act
or
(
c)add(
orsubt
r
act
)aconst
antmul
t
i
pl
eofanyoner
owt
o(
orf
r
om)anot
her
.
Exer
ci
se
UseGaussi
anel
i
mi
nat
i
ont
osol
v
et
hesy
st
em ofl
i
nearequat
i
ons
2x+y+z=10
I
.
3x+2y+3z=18
9;z=5
Ans:
x=7;y=x+4y+9z=16
I
I
.
20x+y+4z=25
8x+13y+2z=23
4x11y+21z=14
Ans:
x=y=z=1
I
I
I
.
2x26x3 =12
x12x1 +4x2 +12x3 =17
4x212x3 =22
x1-
Ans:
nosol
ut
i
ons
 GaussJor
danMet
hod
Thi
smet
hodi
sasl
i
ght
l
ymodi
f
i
cat
i
onoft
heabov
eGaussEl
i
mi
nat
i
onmet
hod.Her
eel
i
mi
nat
i
on
i
sper
f
or
mednotonl
yi
nt
hel
owert
r
i
angul
arbutal
souppert
r
i
angul
ar
.Thi
sl
eadst
ouni
tmat
r
i
x
andhencesol
ut
i
oni
sobt
ai
ned.Thi
si
sJor
dan’
smodi
f
i
cat
i
onoft
heGaussel
i
mi
nat
i
onand
hencet
henamei
sGaussJor
danMet
hod.
Exampl
e5:
Sol
v
et
hesy
st
em ofequat
i
onbyGaussJor
danmet
hod.
2x+y+4z=12
8x3y+2z=20
4x+11yz=33
Sol
ut
i
on
2 1 4 x 12
LetAx=b i
.
e.
8 3 2 y =20
4 11 1 z 33
2 1 4 12
Wr
i
t
i
ngi
ti
naugment
edf
or
m: 8 3 2 20
4 11 1 33
2 1 4 12
a21
Fr
om (
i
)
: R2→R2- R1 0 28
7 14 a11
4 11 1 33
(
(
a31
R3→R3- R1
a11
)() ( )
(
(
|)
|)
|)
2 1 4 12
28
7 14 0 0 9 9 9
37
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
-(
i
)
-(
i
i
)
Fr
om (
i
i
)
:
a32
R3→R3- R2
a22
(
1
R1→ ×R1
2
1
Fr
om (
i
i
i
)
:
R2→ - ×R2
7
1
R3→ - ×R3
27
Cont
i
nuef
r
om (
i
v
)successi
v
el
y
:
a13
R1→R1- R3
a33
a23
R2→R2- R3
a33
|)
2 1 4 12
28
7 14 0 0 0 27 27
( |)
1 12 26
0 1 24
0 0 11
( |)
( |)
( |)
1 12 04
0 1 24
0 0 11
-(
i
v
)
-(
v
)
1 12 04
0 1 02
0 0 11
1 0 03
0 1 02
0 0 11
Rewr
i
t
i
ngt
hef
or
m ofAx=b wehav
e
1 0 0x 3
x 3
y =2
0 1 0 y =2
z
z
0 0 1
1
1
a12
R1→R1- R2
a22
-(
i
i
i
)
( )() ( ) () ( )
∴ x=3;y=2;and z=1
Exampl
e6:
Sol
v
et
hef
ol
l
owi
ngsy
st
em ofequat
i
onsbyGaussJor
danmet
hod:
2x12x2 +3x3 +4x4 =18
4x1 +x211
x3 +2x4 =26
x1x2x3 +5x4 =2x13x2 +2x33
x4 =Sol
ut
i
on:
Let Ax=b sot
hat
18
2 3 4 x1
2 x
11
4 1 1 2 2
=
x
1 1 1 5 3
26
1 x4
2 3 2 3
Wr
i
t
et
heabov
ei
naugment
edf
or
m:
18
2 3 42 11
4 1 1 21 1 1 526
12 3 2 3
(
(
38
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
)( ) ( )
|)
a21
R2→R2- R1
a11
a31
R3→R3- R1
a11
a41
R4→R4- R1
a11
a32
R3→R3- R2
a22
a42
R4→R4- R2
a22
a43
R4→R4- R3
a33
1
R1→ R1
2
1
R2→ R2
5
2
R3→ -R3
5
25
R4→ - R4
227
a14
R1→R1- R4
a44
|)
|)
|)
|)
|)
(
2
5
1
3
2
0
0
2
(
(
(
2 3
7
5 5
0 2
3 2
418
6 25
317
13
2
0
0
0
2 3
7
5 5
0 2
1 1
418
6 25
317
5 15
2
0
0
0
2 3
7
5 5
0 -2
1 1
418
6 25
317
5 15
2
0
1
2
(
2
0
0
0
(
2
0
0
0
(
(
1
0
0
0
1
0
0
0
3
7
1
2
18
46 25
526
13
Nochangebecausea32i
sal
r
eady0.
3
4 18
2 7
6
5
25
5
3 0
2
17
3
1
0 12
5 20
5
4
18
2 3
6
7
25
5 5
3
17
0 -2
227 908
0 0 25
25
|)
|)
|)
3
2 9
1
2 6 - 5
1 7
5
5
0
6 345
5
0 1
4
1 0
3
0 1
1
2 6 - 5
1 7
5
5
0
6 345
5
0 1
4
1 0
39
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
a24
R2→R2- R4
a44
a34
R3→R3- R4
a44
a13
R1→R1- R3
a33
Ther
ef
or
e
(
(
3
1
1
2 0 1
0 5
1 7
6
5 - 34
0
5 5
1
0
1 4
0
1
0
0
0
3
1
2
1 7
5
0
1
0
0
1
0015
02
14
(
1 0
7
1 5
0 1
0 0
4
0015
02
14
1
0
0
0
(
(
( )( ) ( )
() ( )
a23
R2→R2- R3
a33
1
0
0
0
1
1
0
0
a12
R1→R1- R2
a22
1
0
0
0
0
1
0
0
1
0
0
0
0
1
0
0
0
0
1
0
|)
1
0
0
0
1
0 x1
x
3
0 2
=
2
0 x3
x
4
1 4
|)
|)
4
003
02
4
1-
0
0
1
0
0
0
1
0
|)
|)
1
003
02
4
1-
x1
1
x2
3
=
x3
2
x4
4
4
whi
chmeans x1 =1;x2 =3;x3 =2;x4 =Exer
ci
se
Sol
v
et
hesy
st
em ofequat
i
onsusi
ngGaussJor
danmet
hod
3x+yz=3
I
.
x=1,y=1,z=1
Ans:
2x8y+z=5
x2y+9z=8
I
I
.
x+2y+zw =2;
2x+3yz+2w =7;
40
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
x+y+3z2w =6;
x+y+z+w =2
Ans:
x=1,y=0,z=1,w =2
I
I
I
.
10x+y+z=12
2x+10y+z=10
x+y+5z=13
x=y=z=1
Ans:
I
V.
10x+y+z=13
2x+10y+z=14
x+y+15z=32
x=y=1,z=2
Ans:
 Met
hodofFact
or
i
sat
i
onorTr
i
angul
ar
i
sat
i
on
Consi
dert
hesy
st
em ofequat
i
ons
a11x1 +a12x2 +a13x3 =b1
a21x1 +a22x2 +a23x3 =b2
a31x1 +a32x2 +a33x3 =b3
Theseequat
i
onscanbewr
i
t
t
eni
nmat
r
i
xf
or
m asAX =B
-(
1)
x1
a11 a12 a13
b1
x
a
a
a
wher
e A = 21 22 23 ,X = 2 and B =b2
x3
a31 a32 a33
b3
I
nt
hi
smet
hodweuset
hef
actt
hatt
hesquar
emat
r
i
xAcanbef
act
or
i
sedi
nt
ot
hef
or
m
wher
e
LU,
tl
owert
r
i
angul
armat
r
i
x
L=auni
r
i
angul
armat
r
i
x
U =uppert
i
fal
l
t
hemi
nor
sofAar
enonsi
ngul
ar
.
(
) ()
()
Let A =LU
-(
2)
u11 u12 u13
1 0 0
wher
e L=l
nd U = 0 u22 u23
21 1 0 a
l
0 0 u33
31 l
32 1
(
)
(
Ther
ef
or
eequat
i
on(
1)becomesLUX =B
)
y
1
Set
t
i
ng
e Y =y
UX =Y wher
2
y
3
∴ LY =B
St
eps:
(
i
)Fi
ndt
heunknownel
ement
si
nLandUbycomput
i
ngA =LU
(
i
i
)UseLY =B t
ogetYsi
nceLi
snowknown
(
i
i
i
)Fi
nal
l
yuseUX =Yt
of
i
ndXwhi
chi
st
her
equi
r
edunknownby‘
backsubst
i
t
ut
i
on’
.
()
Exampl
e7:
Sol
v
et
hef
ol
l
owi
ngsy
st
em,
bymet
hodoft
r
i
angul
ar
i
sat
i
on
2x1 +x2 +2x3 =30
x14x23x3 =54
5x1 +2x2x
18
3 =
Sol
ut
i
on
41
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
x1
2 1 2
4 Our A = 1 3 , X =x2
x3
5 2 1
) ()
(
30
54
and B = 18
()
u11 u12 u13
1 0 0
Al
so,L=l
nd U = 0 u22 u23
21 1 0 a
l
0 0 u33
31 l
32 1
(
Now,LU =A
u11 =2
u12 =1
u13 =2
)
(
)
1 0 0 u11 u12 u13
2 1 2
u
u
i
.
e. l
4 1 3
21 1 0 0
22
23 = u
5 2 1
l
0 33
31 l
32 1 0
(
)(
)(
)
1
1
∴l
=21 =
u11
2
1
7
∴ u22 =41 - =2
2
1
32 - =2
∴ u23 =2
5 5
∴ l
=
31 =
u11 2
5 2 1
- =
∴ l
1
32 =22 7 7
5
40
1
12 ∴ u33 =(2) =7
7
2
u11l
1
21 =-
()
()
u12l
u22 =4
21 +
u13l
u23 =3
21 +
u11l
5
31 =
[ ()]( )
u12l
u22l
2
31 +
32 =
() ()
u13l
u23l
u33 =1
31 +
32 +
( ) ( )
( )( ) ( )
1 0 0
1
So, L= 2 1 0
5 1 1
2 7
Then,LY =B
1
7
and U =0
2
2
2
40
0 0 7
1 0 0
y
30
1 1 0 1
y
54
2
2 =y
5 1 1 3
18
2 7
y
=30
1
1
-y
+y
=54
1
2
2
5
1
y
+y
+y
=18
1
2
3
7
2
Last
l
y
,UX =Y
2
() (
=39
∴y
2
39 360
=1875 + =∴y
3
7
7
)
30
y
1
39
Y =y
2 =
360
y
3
7
42
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
(
1
7
i
.
e. 0
2
2
0
Bybacksubst
i
t
ut
i
on
0
)( ) ( )
2
30
x1
39
2 x = 2
360
40 x3
7
7
40x =360
3
7
7
x3 =9
7
-x22x
39
3 =2
7
-x22(
9)=39
2
x2 =6
2x1 +x2 +2x3 =30
2x1 +6 +2(
9)=30
x1 =3
∴ x1 =3;x2 =6; x3 =9
 Cr
out
’
sMet
hod
Thi
smet
hodi
ssuper
i
ort
ot
heGaussel
i
mi
nat
i
onmet
hodbecausei
tr
equi
r
esl
esscal
cul
at
i
on.I
t
i
sbasedont
hef
actt
hatev
er
ysquar
emat
r
i
xAcanbeexpr
essedast
hepr
oductofal
ower
t
r
i
angul
armat
r
i
xandauni
tuppert
r
i
angul
armat
r
i
x
.LetAX =B bet
hegi
v
ensy
st
em andl
et
e
A =LUwher
l
11 0 0
1 u12 u13
L=l
nd U =0 1 u23 ,t
hen LUX =B
21 l
22 0 a
0 0 1
l
31 l
32 l
33
∴ UX =Y and LY =B
NowLandUcanbef
oundf
r
om LU =A.
a11 a12 a13
l
11 0 0 1 u
12 u
13
u
i
.
e. l
21 a
22 a
23
23 =a
21 l
22 0 0 1
a
a
a
1
l
31
32
33
31 l
32 l
33 0 0
(
)
(
(
l
11
l
11u
12
(
)(
l
11u
13
)
)(
)
)(
)
a11 a12 a13
or l
=a21 a22 a23
l
l
l
21u
22
21u
22u
21 l
12 +
13 +
23
a31 a32 a33
l
l
l
l
31u
32 l
31u
32u
33
31 l
12 +
13 +
23 +
Equat
i
ngt
hecor
r
espondi
ngel
ement
sonbot
hsi
des,
weget
,
,
l
=
l
=
l
=
a
a
a
11
31
11 21
21
31
a
l
∴ u12 = 12
a12
11u
12 =
a11
a
∴ u13 = 13
l
a13
11u
13 =
a11
l
+
l
=
l
=
l
u
∴
a
a
21 12
22
22
21u
22
2212
l
l
l
∴l
a32
a3231u
32 =
32 =
31u
12 +
12
1
l
l
∴ u23 = (
a23
l
a2321u
22u
13 +
23 =
21u
13)
l
22
43
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
l
l
l
a33
31u
32u
33 =
13 +
23 +
l
l
∴l
a3333 =
31u
32u
1323
Thusal
l
t
heunknownsar
edet
er
mi
nedf
orLandU
Cr
outdur
i
ngt
heabov
edecomposi
t
i
onoft
hecoef
f
i
ci
entmat
r
i
xA,dev
i
sedat
echni
que
whi
chi
sgi
v
enher
eundert
odet
er
mi
ne12unknownssy
st
emat
i
cal
l
y
.
Comput
at
i
onSchemebyCr
out
’
sMet
hod
Theaugment
edmat
r
i
xoft
hesy
st
em AX =Bi
s
a11 a12 a13 b1
A |
B)=a21 a22 a23 b2
(
a31 a32 a33 b3
Themat
r
i
xof12unknownssocal
l
edder
i
v
edmat
r
i
xorauxi
l
i
ar
ymat
r
i
xi
s
u12 u13 y
l
11
1
u23 y
l
2
21 l
22
y
l
l
3
31 l
32
33
andi
st
obecal
cul
at
edasf
ol
l
ows:
(
)
(
)
St
ep1:
Thef
i
r
stcol
umnoft
heder
i
v
edmat
r
i
x(
D.
M)i
si
dent
i
cal
wi
t
ht
hef
i
r
stcol
umnof(
.
A|
B)
l
a11;l
a21; l
a31
11 =
21 =
31 =
St
ep 2:Thef
i
r
str
ow t
ot
her
i
ghtoft
hef
i
r
stcol
umn oft
heD.
M i
sgotbydi
v
i
di
ng t
he
cor
r
espondi
ngel
ementi
n(
hel
eadi
ngdi
agonal
el
ementoft
hatr
ow.
A |
B)byt
a12
a13
b
u12 = ; u13 = ; y
=1
1
a11
a11
l
11
St
ep3:
Remai
ni
ngsecondcol
umnofD.
M
l
l
l
l
a22a3222 =
21u
32 =
31u
12;
12
Cor
r
espondi
ngel
ementi
n(
A|
The
B)Eachel
ementonorbel
ow
=pr
oductoft
hef
i
r
stel
ementi
nt
hatr
ow
t
hedi
agonal
andi
nt
hatcol
umn
St
ep4:
Remai
ni
ngel
ement
sofsecondr
owofD.
M.
A|
[
Cor
r
espondngel
ementi
n(
The
B)-
}{
{
Eachel
ement=pr
oductoft
hef
i
r
stel
ementi
nt
hatr
owandi
nt
hat
col
umn]+[
l
eadi
ngdi
agonal
el
ementi
nt
hatr
ow]
i
.
e.u
23
b2 l
a l
21y
21u
13
1
; y
=
= 23
2
l
l
22
22
St
ep5:
Remai
ni
ngel
ement
soft
hi
r
dcol
umnofD.
M
Cor
r
espondngel
ementi
n(
A|
Sum oft
he
B)i
n
n
e
r
pr
oduct
soft
hepr
ev
i
ousl
ycal
cul
at
edel
ement
s
Eachel
ement=
{
i
nt
hesamer
owandcol
umn
i
.
e.l
l
l
u
a3333 =
31u
3
2
13
23
St
ep6:
Remai
ni
ngel
ement
soft
hi
r
dr
owofD.
M
44
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
{
Eachel
ement=
i
.
e.y
3
A|
[
Cor
r
espondngel
ementi
n(
Sum oft
hei
nner
B)pr
oduct
soft
hepr
ev
i
ousl
ycal
cul
at
edel
ement
s
i
nt
hesamer
owandcol
umn]+[
t
hel
eadi
ngdi
agonal
i
nt
hatr
ow]
l
+l
b (
31y
32y
1
2)
=3
l
33
Nowt
hemat
r
i
cesL,
UandYcanbewr
i
t
t
enandhenceXi
scal
cul
at
edf
r
om UX =Y
Exampl
e8:
Sol
v
et
hesy
st
em ofequat
i
ons
2x+y+4z=12,8x3y+2z=20,4x+11yz=33 byCr
out
’
smet
hod.
Sol
ut
i
on:
2 1 4 12
Augment
edmat
r
i
x =(
=
A|
)
B 8 3 2 20
1 33
4 11 -
(
)
u12 u13 y
l
11
1
u
y
Lett
heder
i
v
edmat
r
i
x(
D.
M) =l
l
23
2
21 22
y
l
l
l
33
3
31 32
(
)
(
1)El
ement
soft
hef
i
r
stcol
umnofD.
M ar
el
2,l
8, l
4.
11 =
21 =
31 =
(
2)El
ement
soft
hef
i
r
str
owt
ot
her
i
ghtoft
hef
i
r
stcol
umn
a13 4
a
b1 12
u12 = 12 =1; u13 = = =2; y
= = =6
1
a11 2
a11 2
2
l
11
(
3)El
ement
soft
her
emai
ni
ngsecondcol
umn
1
l
l
a227
(3)- (
8)=22 =
21u
12 =
2
1
l
l
11- (
a324)=9
32 =
31u
12 =
2
(
4)El
ement
soft
her
emai
ni
ngsecondr
ow
()
()
l
a2321u
1
13
=-[
u23 =
2(
(
8)
2)
]=2
7
l
22
b2 l
0(
(
6)
21y
8)
1 2
=
=
=4
y
2
7
l
22
(
5)Remai
ni
nngt
hi
r
dcol
umn
l
l
l
a33 +(
1-9)
27
(
(
(
4)
2)(
2)=33 =
31u
32u
1323)=(
6)Remai
ni
ngt
hi
r
dr
ow
l
+l
b3 (
3(
(
-9)
(
4)
31y
32y
4)
6)(
1
2) 3
=
=
=1
y
3
2
7
l
33
(
)
1
2 2 6
∴
7 2 4
27 1
9 Nowt
hesol
ut
i
oni
sgotf
r
om t
hesy
st
em UX =Y
2
D.
M =8
4
45
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
( )() ()
1 2
x 6
2
y =4
1 2
z 1
0 1
1
whi
chi
sequi
v
al
entt
o x+ y+2z=6; y+z=4; z=1
2
Bybacksubst
i
t
ut
i
on x=3,y=2 and z=1
Exer
ci
se
1.Sol
v
et
hef
ol
l
owi
ngsy
st
em byt
hemet
hodoft
r
i
angul
ar
i
sat
i
on:
2x3y+10z=3, x+4y+2z=20, 5x+2y+z=12
Ans:x=4,y=3 and z=2
2.Sol
v
et
hesy
st
em ofequat
i
ons
6x2y3z=2
x+y+z=6
5x3y3z=2
UseCr
out
’
smet
hod.
2
Ans:x=y=z=1
i
.
e. 0
0
 I
NDI
RECTMETHODORI
TERATI
VEMETHOD
Thi
smet
hodi
sappl
i
edi
ft
heabsol
ut
ev
al
ueoft
hel
ar
gestcoef
f
i
ci
enti
sgr
eat
ert
hant
he
absol
ut
ev
al
uesofal
l
r
emai
ni
ngcoef
f
i
ci
ent
si
neachequat
i
on(
condi
t
i
onf
orconv
er
gence)
.
Consi
dert
hesy
st
em ofequat
i
ons
a11x1 +a12x2 +a13x3 =b1
a21x1 +a22x2 +a23x3 =b2
a31x1 +a32x2 +a33x3 =b3
Thi
smet
hodi
sappl
i
edonl
ywhendi
agonalel
ement
sar
eexceedi
ngal
lot
herel
ement
si
nt
he
r
espect
i
v
eequat
i
onsi
.
e.
,
a12|+|
a13|+… +|
a11|>|
a1n|
|
a22|>|
a21|+|
a23|+… +|
a2n|
|
⋮
an,n-1|
an2|+… +|
an1|+|
a
|nn|>|
Def
i
ni
t
i
on ofSt
r
i
ct
l
ydi
agonal
l
ydomi
nantmat
r
i
x:An n×n mat
r
i
xA i
sst
r
i
ct
l
ydi
agonal
l
y
domi
nanti
ft
heabsol
ut
ev
al
ueofeachent
r
yont
hemai
ndi
agonali
sgr
eat
ert
hant
hesum oft
he
absol
ut
ev
al
uesoft
heot
herent
r
i
esi
nt
hesamer
ow.
Asseenabov
e;
Whi
choft
hef
ol
l
owi
ngsy
st
emsofl
i
nearequat
i
onshasast
r
i
ct
l
ydi
agonal
l
ydomi
nantcoef
f
i
ci
ent
mat
r
i
x?
x2 =4
(
a)3x12x1 +5x2 =2
1
x3 =(
b)4x1 +2x2x1 +2x3 =4
46
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
3x15x2 +x3 =3
Sol
ut
i
on
1
3 (
a)Thecoef
f
i
ci
entmat
r
i
xA =
i
sst
r
i
ct
l
ydi
agonal
l
ydomi
nantbecause|
1|and
3|>|2 5
|
5|>|
2|
4 2 1
(
b)Thecoef
f
i
ci
entmat
r
i
xA =1 0 2 i
snotst
r
i
ct
l
ydi
agonal
l
ydomi
nantbecauset
he
5 1
3 ent
r
i
esi
nt
hesecondandt
hi
r
dr
owsdonotconf
or
mt
ot
hedef
i
ni
t
i
on.
Howev
er
,i
nt
er
changi
ngt
he2ndand3r
dr
owsi
nt
heor
i
gi
nalsy
st
em ofl
i
near
equat
i
onsmakesi
tst
r
i
ct
l
ydi
agonal
l
ydomi
nant
.
4 2 1
'
5 1
A =3 1 0 2
Cor
ol
l
ar
y
:I
famat
r
i
xAi
sst
r
i
ct
l
ydi
agonal
l
ydomi
nantori
r
r
educi
bl
ydi
agonal
l
ydomi
nant
,t
heni
t
i
snonsi
ngul
ar
.
[ ]
[
]
[
]
Mor
eNot
es:
Di
agonal
l
yDomi
nantMat
r
i
ces
Def
i
ni
t
i
on:
Amat
r
i
xAi
s
 (
weakl
y
)di
agonal
l
ydomi
nanti
f
i=n
∑
ajj|≥ |
aij|
,
|
j=1,
…,
n
i=1
i≠j
 st
r
i
ct
l
ydi
agonal
l
ydomi
nanti
f
i=n
∑
ajj|> |
aij|
,
|
j=1,
…,
n
i=1
i≠j
 i
r
r
educi
bl
ydi
agonal
l
ydomi
nanti
fAi
si
r
r
educi
bl
e,
and
i=n
∑
ajj|≥ |
aij|
,
|
j=1,
…,
n
i=1
i≠j
wi
t
hst
r
i
cti
nequal
i
t
yf
oratl
eastonej
.
Conv
er
genceoft
heJacobi
andGaussSei
delMet
hods
I
fAi
sst
r
i
ct
l
ydi
agonal
l
ydomi
nant
,t
hent
hesy
st
em ofl
i
nearequat
i
ongi
v
enbyAx=b hasa
uni
quesol
ut
i
ont
owhi
cht
heJacobimet
hodandt
heGaussSei
delmet
hodwi
l
lconv
er
gef
orany
i
ni
t
i
al
appr
oxi
mat
i
on.
 Jacobimet
hod(
‘
si
mul
t
aneousdi
spl
acement
’
)
TheJacobimet
hodi
st
hesi
mpl
esti
t
er
at
i
v
emet
hodf
orsol
v
i
nga(
squar
e)l
i
nearsy
st
em Ax=b.
I
ti
snamedaf
t
erCar
l
Gust
avJacobJacobi
(
1804–1851)
.
Twoassumpt
i
onsmadeonJacobi
Met
hod:
1.Thesy
st
em gi
v
enby
47
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
a11x1 +a12x2 +… +a1nxn =b1
a21x1 +a22x2 +… +a2nxn =b2
⋮
an1x1 +an2x2 +… +annxn =bn
hasauni
quesol
ut
i
on
2.Thecoef
f
i
ci
entmat
r
i
xAhasnozer
osoni
t
smai
ndi
agonal
,
namel
y
,
….
,
e
a11 ,a22 ,
ann ar
nonzer
os.
Themat
r
i
xAi
swr
i
t
t
enasL +U +Dwher
eL,UandDar
el
owert
r
i
angul
ar
,uppert
r
i
angul
arand
di
agonal
par
t
sofAr
espect
i
v
el
y
.
Consi
dert
he3×3sy
st
em
a11x1 +a12x2 +a13x3 =b1
}
a21x1 +a22x2 +a23x3 =b2
-(
1)
a31x1 +a32x2 +a33x3 =b3
I
nt
hi
scase
a11 0 0
0 0 0
0 a12 a13
L=a21 0 0,U =0 0 a23 and D = 0 a22 0
a31 a32 0
0 0 a33
0 0 0
t
h
Thi
sf
i
r
stst
agei
st
or
ear
r
anget
hei equat
i
oni
nt
hesy
st
em (
1)sot
hatonl
yt
er
msi
nv
ol
v
i
ngt
he
di
agonal
coef
f
i
ci
entofxioccuront
hel
ef
thandsi
de:
a11x1 =b1a12x2a13x3
a22x2 =b2a21x1a23x3
a33x3 =b3a31x1 +a32x2
ori
nmat
r
i
xnot
at
i
on
(
L +U)
Dx=bx
-(
2)
(
) (
(
)
)
t
h
Thenextst
agei
st
odi
v
i
debot
hsi
desoft
hei
equat
i
onbyt
hedi
agonal
el
ementaiit
oobt
ai
n
1
x1 = (
ba xa x)
a11 1 12 2 13 3
1
x2 = (
ba xa x)
a22 2 21 1 23 3
1
x3 = (
ba x +a x)
a33 3 31 1 32 2
Thi
si
sequi
v
al
entt
opr
emul
t
i
pl
y
i
ngbot
hsi
desofequat
i
on(
2)byt
hei
nv
er
seofDandso
1
(
)
L
+
U
bx]
x=D [
m
n ax
1
b∑
or xm +1 = i i=0 ij j
aii
j=1
i=1,
2,
3,
…,
n
wher
em i
st
henumberofi
t
er
at
i
on
Thi
si
scal
l
edt
heJacobi
met
hodormet
hodofsi
mul
t
aneousdi
spl
acement
.
(
)
Exampl
e1:
Sol
v
et
hesy
st
em ofequat
i
onbyGaussJacobi
met
hod
27x+6yz=85
6x+15y+2z=72
x+y+54z=110
48
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
Sol
ut
i
on
Toappl
yt
hi
smet
hod,
f
i
r
stwehav
et
ocheckt
hedi
agonal
el
ement
sar
edomi
nant
i
.
e.27>6 +1; 15>6 +2; 54>1 +1
Soi
t
er
at
i
onmet
hodcanbeappl
i
ed
1
x= (
856y+z)
27
1
y= (
726x2z)
15
1
z= (
110xy
)
54
Fi
r
sti
t
er
at
i
on:
Fr
om t
heabov
eequat
i
ons,
west
ar
twi
t
hx=y=z=0
8
5
x(1) = 27 =3.
14815
(
1) 7
= 215 =4.
8
y
110
z(1) = 54 =2.
03704
(
1)
Secondi
t
er
at
i
on:
Consi
dert
henewv
al
uesofy
03704
=4.
8 andz(1) =2.
1
4.
8)+2.
856(
03704]=2.
x(2) = [
15693
27
1
(
2)
726(
2(
2.
03704)
3.
14815)y
= [
26913
]=3.
15
1
1103.
148154.
8]=1.
z(2) = [
88985
54
(
2)
(
2)
Thi
r
di
t
er
at
i
on:
Consi
dert
henewv
al
uex =2.
15693,y
88985
=3.
26913 and z(2) =1.
1
3.
26913)+1.
856(
88985]=2.
x(3) = [
49167
27
1
(
3)
726(
2(
1.
88985)
2.
15693)y
= [
68525
]=3.
15
1
1102.
156933.
26913]=1.
z(3) = [
93655
54
Thus,
wecont
i
nuet
hei
t
er
at
i
onandr
esul
ti
snot
edbel
ow
I
t
er
at
i
onNo.
x
y
z
4
2.
40093
3.
54513
1.
92265
5
2.
43155
3.
58327
1.
92692
6
2.
42323
3.
57046
1.
92565
7
2.
42603
3.
57395
1.
92604
8
2.
42527
3.
57278
1.
92593
9
2.
42552
3.
57310
1.
92596
10
2.
42546
3.
57300
1.
92595
11
2.
42548
3.
57302
1.
92595
12
2.
42548
3.
57301
1.
92595
49
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
13
2.
42548
3.
57301
1.
92595
Fr
om t
heabov
et
abl
e12th and13th i
t
er
at
i
onsar
eequal
;byconsi
der
i
ngt
hef
ourdeci
malpl
aces.
Hencet
hesol
ut
i
onoft
heequat
i
oni
s
x=2.
4255 y=3.
5730 z=1.
9260
Exampl
e2:
Sol
v
ebyJacobi
met
hod,
t
heequat
i
on
20x
+
2
17
x
x
1
2
3 =
3x1 +20x218
x3 =2x13x2 +20x3 =25
Sol
ut
i
on:
Wewr
i
t
et
hegi
v
enequat
i
oni
nt
hef
or
m
1
x1m +1 = (
17x2m +2x3m)
20
1
m
x2m +1 = (183x
x3m)
1 +
20
1
x3m +1 = (
2x1m +3x3m)
2520
Usi
ngx10 =x20 =x30 =0,
weobt
ai
n
17
x11 = =0.
85
20
18
=0.
9
x21 =
20
25
25
x31 = =1.
20
Put
t
i
ngt
hesev
al
uesont
her
i
ghtofequat
i
on(
*
)
,
weobt
ai
n,
1
17x21 +2x31)=1.
x12 = (
02
20
1
x22 = (0.
965
3x11 +x31)=1820
1
x32 = (
1515
2x11 +3x31)=1.
2520
Theseandf
ur
t
heri
t
er
at
esar
el
i
st
edi
nt
het
abl
ebel
ow:
x1m
x2m
m
-(
*
)
x3m
0
0
0
0
1
0.
85
–0.
90
1.
25
2
1.
02
–0.
965
1.
1515
3
1.
0134
9954
-0.
1.
0032
4
1.
0009
0018
-1.
0.
9993
5
1.
0000
-1.
0002
0.
9996
6
1.
0000
-1.
0000
1.
0000
Thev
al
uesi
n5thand6thi
t
er
at
i
onsbei
ngpr
act
i
cal
l
yt
hesame,
west
op,
hencet
hesol
ut
i
onsar
e
x1 =1, x2 =1 and x3 =1
50
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
Exampl
e3:Uset
heJacobimet
hodt
oappr
oxi
mat
et
hesol
ut
i
onoft
hef
ol
l
owi
ngsy
st
em ofl
i
near
equat
i
ons
5x12x2 +3x3 =1
3x1 +9x2 +x3 =2
2x17x3 =3
x2Cont
i
nuet
hei
t
er
at
i
onsunt
i
lt
wosuccessi
v
eappr
oxi
mat
i
onsar
ei
dent
i
calwhenr
oundedt
ot
hr
ee
si
gni
f
i
cantdi
gi
t
s.
Sol
ut
i
on:
Tobegi
n,
wr
i
t
et
hesy
st
em i
nt
hef
or
m
1 2 3
x1 =- + x2-x3
5 5 5
2 3 1
x2 = + x1-x3
9 9 9
3 2 1
x3 =- + x1-x2
7 7 7
Becausey
oudonotknowt
heact
ualsol
ut
i
on,
choose x1 =0,x2 =0,x3 =0 asaconv
eni
ent
i
ni
t
i
al
appr
oxi
mat
i
on.So,
t
hef
i
r
stappr
oxi
mat
i
oni
s
1 2 3
x1 =- + (
0.
200
0)-(
0)=5 5 5
2 3 1
222
x2 = + (
0)-(
0)≈0.
9 9 9
3 2 1
x3 =- + (
0.
429
0)-(
0)≈7 7 7
Thet
abl
eshowst
hesequenceofappr
oxi
mat
i
on
n
0
1
2
3
4
5
6
7
x1
0.
0000
0.
200
0.
146
0.
192
0.
181
0.
185
0.
186
0.
186
x2
0.
0000
0.
222
0.
203
0.
328
0.
332
0.
329
0.
331
0.
331
x3
0.
0000
0.
429
0.
517
0.
416
0.
421
0.
424
0.
423
0.
423
Becauset
hel
astt
wo col
umnsi
nt
heTabl
ear
ei
dent
i
cal
,y
oucanconcl
udet
hatt
ot
hr
ee
si
gni
f
i
cantdi
gi
t
s:
186, x2 =0.
331, x3 =0.
423
x1 =0.
Exer
ci
se
Sol
v
et
hesy
st
em ofequat
i
onbyGaussJacobi
met
hod
5x2y+z=4
1.
0
I
.
Ans:x=x+6y2z=1
y=1
3x+y+5z=13
z=3
I
I
.
3.
15x1.
96y+3.
85z=12.
95
2.
13x5.
12y2.
892z=8.
61
5.
92x+3.
051y+2.
15z=6.
88
x=1.
7089y=1.
8005z=
0488
Ans:
,
, 1.
I
I
I
.
8x3y+2z=20
x=3.
0168
Ans:
51
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
6x+3y+12z=35
4x+11yz=33
y=1.
9859
z=0.
9118
 GaussSei
delmet
hod(
‘
Successi
v
edi
spl
acement
s’
)
Youwi
l
lnow l
ookatamodi
f
i
cat
i
onoft
heJacobimet
hodcal
l
edt
heGaussSei
delmet
hod,
namedaf
t
erCar
lFr
i
edr
i
chGauss(
1777– 1855)andPhi
l
i
ppL.Sei
del(
1821– 1896)
.Thi
s
modi
f
i
cat
i
oni
snomor
edi
f
f
i
cul
tt
ouset
hant
heJacobimet
hod,andi
tof
t
enr
equi
r
esf
ewer
i
t
er
at
i
onst
opr
oducet
hesamedegr
eeofaccur
acy
.
Wi
t
ht
heGaussSei
delmet
hod,ont
heot
herhand,y
ouuset
henewv
al
uesofeachxias
soonast
heyar
eknown.Thati
s,oncey
ouhav
edet
er
mi
nedx1 f
r
om t
hef
i
r
stequat
i
on,i
t
sv
al
ue
i
st
henusedi
nt
hesecondequat
i
ont
oobt
ai
nt
henewx2.Si
mi
l
ar
l
y
,
t
henewx1andx2ar
eusedi
n
t
het
hi
r
dequat
i
ont
oobt
ai
nt
henewx3,
andsoon.
Recal
l
Jacobi
met
hodf
ort
hesy
st
em ofequat
i
on:
1
ba x(k)a13x3(k))
x1(k+1) = (
a11 1 12 2
1
x2(k+1) = (
ba x(k)a23x3(k))
a22 2 21 1
1
x3(k+1) = (
ba x(k)a32x2(k))
a33 3 31 1
TheGaussSei
delmet
hodi
mpl
ement
st
hest
r
at
egyofal
way
susi
ngt
hel
at
estav
ai
l
abl
ev
al
ueof
apar
t
i
cul
arv
ar
i
abl
e.
Nowi
mmedi
at
el
yuseev
er
ynewi
t
er
at
e
1
(
k+1)
ba x(k)a13x3(k))
x1
= (
a11 1 12 2
1
x2(k+1) = (
ba x(k+1)a23x3(k))
a22 2 21 1
1
x3(k+1) = (
ba x(k+1)a32x2(k+1))
a33 3 31 1
f
ork=0,
1,
2,
…
Thi
si
scal
l
edt
heGaussSei
del
i
t
er
at
i
onmet
hodort
hemet
hodofsuccessi
v
edi
spl
acement
s
(
r
epl
acement
s)
.
Exampl
e4:Uset
heGaussSei
deli
t
er
at
i
onmet
hodt
oappr
oxi
mat
et
hesol
ut
i
ont
ot
hesy
st
em of
equat
i
onsgi
v
eni
nExampl
e3.
Sol
ut
i
on
x1,x2,
x3)=(
Thef
i
r
stcomput
at
i
oni
si
dent
i
cal
t
ot
hatgi
v
eni
nExampl
e3.Thati
s,
usi
ng(
0,
0,
0)
ast
hei
ni
t
i
al
appr
oxi
mat
i
on,
y
ouobt
ai
nt
hef
ol
l
owi
ngnewv
al
ueofx1.
1 2 3
x1 =- + (
0.
200
0)-(
0)=5 5 5
Nowt
haty
ouhav
eanewv
al
uef
orx1,
howev
er
,
weusei
tt
ocompl
et
eanewv
al
uef
orx2.Thati
s
2 3
1
0)≈0.
156
x2 = + (0.
200)-(
9 9
9
52
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
0.
200andx2 =0.
156t
Si
mi
l
ar
l
y
,
usex1 =ocomput
eanewv
al
uef
orx3.Thati
s,
3 2
1
0.
508
x3 =- + (0.
200)-(
0.
156)≈7 7
7
0.
200,x2 =0.
156,and x3 =0.
508.Cont
Sot
hef
i
r
stappr
oxi
mat
i
oni
sx1 =i
nuedi
t
er
at
i
ons
pr
oducet
hesequenceofappr
oxi
mat
i
onsshowni
nt
heTabl
ebel
ow
n
0
1
2
3
4
5
x1
0.
000
0.
200
0.
167
0.
191
0.
186
0.
186
x2
0.
000
0.
156
0.
334
0.
333
0.
331
0.
331
x3
0.
000
0.
508
0.
429
0.
422
0.
423
0.
423
Not
et
hataf
t
eronl
yf
i
v
ei
t
er
at
i
onsoft
heGaussSei
delmet
hod,
y
ouachi
ev
edt
hesameaccur
acy
aswasobt
ai
nedwi
t
hsev
eni
t
er
at
i
onsoft
heJacobi
met
hodi
nExampl
e3.
Exampl
e5:
Sol
v
et
hesy
st
em ofequat
i
onbyGaussSei
del
met
hod
28x+4yz=32
4x+3y+10z=24
2x+17y+4z=35
Sol
ut
i
on:
Toappl
yt
hi
smet
hod,f
i
r
stwehav
et
or
ewr
i
t
et
heequat
i
oni
nsuchwayt
hatt
of
ul
f
i
l
ldi
agonal
el
ement
sar
edomi
nant
.
28x+4yz=32
2x+17y+4z=35
4x+3y+10z=24
i
.
e.28>4 +1;17>2 +4; 10>4 +3
Soi
t
er
at
i
onmet
hodcanbeappl
i
ed
1
x= (
324y+z)
28
1
y= (
352x4z)
17
1
z= (
24x3y
)
10
Fi
r
sti
t
er
at
i
on:
Fr
om t
heabov
eequat
i
ons,
west
ar
twi
t
hx=y=z=0,
weget
3
2
x(1) = 28 =1.
1429
Newv
al
ueofxi
susedf
orf
ur
t
hercal
cul
at
i
oni
.
e.
,
x=1.
1429
1
(
1)
352(
1.
1429)4(
0)
)=1.
y
= (
9244
17
x=1.
1429andy=1.
9244
Newv
al
uesofxandyi
susedf
orf
ur
t
hercal
cul
at
i
oni
.
e.
,
1
(
1)
241.
14293(
1.
9244)
]=1.
z = [
7084
10
Secondi
t
er
at
i
on:
(
1)
Consi
dert
henewv
al
uesofy
7084
=1.
9244and z(1) =1.
1
1.
9244)+1.
324(
7084]=0.
x(2) = [
9290
28
53
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
1
(
2)
352(
4(
1.
7084)
0.
9290)y
= [
5476
]=1.
17
1
240.
92903(
1.
5476)
]=1.
z(2) = [
8428
10
Thi
r
di
t
er
at
i
on:
(
2)
y
Consi
dert
henewv
al
uesofx(2) =0.
9290,
8428
=1.
5476andz(2) =1.
1
1.
5476)+1.
324(
8428]=0.
x(3) = [
9876
28
1
(
3)
352(
4(
1.
8428)
0.
9876)y
= [
5090
]=1.
17
1
240.
98763(
1.
5090)
]=1.
z(3) = [
8485
10
Thus,
wecont
i
nuet
hei
t
er
at
i
onandr
esul
ti
snot
edbel
ow
I
t
er
at
i
onNo.
x
4
0.
9933
1.
5070 1.
8486
5
0.
9936
1.
5070 1.
8485
6
0.
9936
1.
5070 1.
8485
y
z
9936,y=1.
5070,z=1.
8485
Ther
ef
or
ex=0.
Exer
ci
se
(
i
)
UseGaussSei
del
t
osol
v
et
hesy
st
em ofequat
i
ons
2x+y+z=4
I
.
x+2y+z=4
Ans:
x=1y=
, 1z=
, 1
x+y+2z=4
2xy5z+w =19
4x10y+z+4w =5
2x+y+z+5w =25
x=0.
5y=
3w
Ans:
, 2z=,
, =5
12x2y+2z+w =1
(
i
i
) Sol
v
et
hef
ol
l
owi
ngsy
st
ems
10x+2y+z=9
2x+20y2z=44
2x+3y+10z=22
by(
a)Jacobi
’
smet
hod,
and
(
b)GaussSei
del
met
hod.
I
neachcase,
car
r
yy
ourcomput
at
i
ont
o10i
t
er
at
i
ons.
2z=
Ans:
;
; 3
x=1y= Met
hodofRel
axat
i
on
Thi
smet
hodi
snott
r
eat
ed.Youcani
nv
est
i
gat
eabouti
tony
ourown.
I
I
.
⌂3⌂
FI
NI
TEDI
FFERENCES
54
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
Lety=f
(
x)
Thev
al
ues,
whi
cht
hei
ndependentv
ar
i
abl
ext
akes,
ar
ecal
l
edar
gument
sandt
hecor
r
espondi
ng
v
al
uesoff
ecal
l
edent
r
i
es.Thedi
f
f
er
encebet
weenconsecut
i
v
ev
al
uesofxi
scal
l
edt
he
(
x)ar
i
nt
er
v
al
ofdi
f
f
er
enci
ng.
I
ft
hei
nt
er
v
al
ofdi
f
f
er
enci
ngbehandt
hef
i
r
star
gumentbea,
t
hen
Ar
gument
sx
: a,
a +h,
a +2h,
a +3h, …
Ent
r
i
esf
:f
, f
, f
, f
, …
(
x)
(
a)
(
a +h)
(
a +2h)
(
a +3h)
Forbr
ev
i
t
y
,
t
heseent
r
i
esar
edonat
edby
y
, y
, y
, y
3,…
1
0
2
Fi
ni
t
eDi
f
f
er
ence
Supposet
haty=f
st
abul
at
edf
ort
heequal
l
yspacedv
al
ues,
t
hati
sx=x0 +i
i=0,
1,
2,
…,
n
h,
(
x)i
x0 +2h,
x0 +3h,
x
y=
y
y
y
y
orx=x0 ,x0 +h,
…,
g
i
v
i
n
g
,
,
,
…,
t
of
i
n
dt
h
ev
a
l
u
e
s
o
f
+
n
h
f
(
x
)
n
0
1 2
0
'
orf
(
x)f
orsomei
nt
er
medi
at
ev
al
uesofx,t
hef
ol
l
owi
ngt
hr
eet
y
pesofdi
f
f
er
encesar
ef
ound
usef
ul
:
(
1)For
war
dDi
f
f
er
ence
Thedi
f
f
er
ence
y
-,
y y
-,
y …y
y
n1 0
n1
2 1
ar
ecal
l
edt
hef
i
r
stf
or
war
ddi
f
f
er
enceoft
hef
unct
i
onyandar
edenot
edby
∆y
, ∆y
…,
∆y
1
n1
0
Ther
ef
or
e,
a)=f
y
or ∆f
∆y
=y
(
(
f
(
a)
a +h)1 0
0
y
y
∆y
=
∆
f
(
)
=
f
(
)f
(
a +h)
a
+
h
a
+
2
h
1
2 1
y
∆y
=y
∆f
(
(
f
(
a +2h)
a +2h)=f
a +3h)2
3 2
x
or ∆f
∆y
y
y
()=f
(
f
(
x)
x+h)n =
nn1
wher
e∆i
scal
l
edt
hef
or
war
ddi
f
f
er
enceoper
at
or
.
Thedi
f
f
er
encesoft
hef
i
r
stf
or
war
ddi
f
f
er
encesar
ecal
l
edsecondf
or
war
ddi
f
f
er
ences.
y
y
∆y
Thus ∆2y
=∆(
∆(
or ∆2y
=∆y
∆y
1 0)
0)=
1
0
0
0
2
2
y
y
y
∆
∆y
∆
y
y
y
=
∆
=
∆
=
∆
∆
(
)
(
)
1
2
1
1
1
1
2
2
y
y
y
∆
∆2y
=
∆
=
∆
∆
y
=
∆
y
∆
y
(
)
(
)
3
2
2
2
2
3
2
y
y
∆y
∆2y
∆2y
∆(
∆(
∆y
∆y
nn1)
n)=
n =
n =
nn1
Si
mi
l
ar
l
yonecandef
i
net
hi
r
df
or
war
ddi
f
f
er
ences,
f
our
t
hf
or
war
ddi
f
f
er
encesandsoon.
Thus
y
y ∆y
∆y
∆2y
=∆(
∆(
∆y
1 0)=
0)=
1
0
0
=y
y
yy
2 1 (1 0)
yy +y
=y
2 1 1
0
=y
2
+
y
y
1
2
0
2y
+y
y
∆3y
=∆(
∆2y
=∆(
1
2
0)
0
0)
55
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
=∆y
2∆y
+∆y
1
2
0
y
y
y
y
=y
2
+y
(
)
1
2
1 0
3 2
=y
2
+
2
+
y
y
y
y
y
1
1 0
3 2
2
=y
3y
+3y
y
1 0
3
2
and,
3y
+3y
y
y
∆4y
=∆(
=∆(
∆3y
1 0)
3
2
0
0)
=∆y
3∆y
+3∆y
∆y
1
3
2
0
y
y
y
y -y
y
y
=y
3
+
3
(
1 0)
2 1)(
4 3 (3 2)
=y
3y
+3y
+3y
3y
yy +y
4 3
1 1
3
2
2
0
y
y
y
y
=y
4
+
6
4
+
4
1
3
2
0
I
ti
st
her
ef
or
ecl
eart
hatanyhi
gheror
derdi
f
f
er
encecaneasi
l
ybeexpr
essedi
nt
er
msoft
he
or
di
nat
essi
ncet
hecoef
f
i
ci
ent
soccur
r
i
ngont
her
i
ghtsi
dear
et
hebi
nomi
al
coef
f
i
ci
ent
s.
Thet
abl
eshowi
ngt
hev
ar
i
ousf
or
war
ddi
f
f
er
encesi
scal
l
edf
or
war
ddi
f
f
er
encest
abl
eandi
s
gi
v
enbel
ow.
Ar
gument
Ent
r
y
Fi
r
stDi
f
f
.
SecondDi
f
f
.
Thi
r
dDi
f
f
.
x
y
a
y
0
∆2y
∆y
∆3y
y
y =∆y
1 0
0
a +h
∆y
∆y
=∆2y
1
0
0
y
1
∆2y
∆2y
=∆3y
1
0
0
y
y =∆y
1
2 1
a +2h
∆y
∆y
=∆2y
1
1
2
y
2
y
y =∆y
3 2
2
a +3h
y
3
y
i
scal
l
edt
hel
eadi
ngt
er
m and∆y
,∆2y
,∆3y
,
…ar
ecal
l
edt
hel
eadi
ngdi
f
f
er
ences.
0
0
0
0
Theoper
at
or∆hast
hef
ol
l
owi
ngpr
oper
t
i
es:
∆c=0c
(
i
)
, bei
ngaconst
ant
x)=c∆f
(
i
i
) ∆cf
(
(
x)
x)+bg(
x)=b∆g(
(
i
i
i
) ∆[
af
(
x)
(
x)
]=a∆f
t
h
t
h
coef
f
i
.
ofxn)
n!.hnand
(
i
v
) Then di
f
f
er
enceofann degr
eepol
y
nomi
ali
saconst
ant =(
hencehi
gheror
derdi
f
f
er
encesar
ezer
o.
I
l
l
ust
r
at
i
v
eExampl
es
1.Pr
ov
et
hat
:
x)
x
x)
(
i
)
f
(
g(
x)
(
∆g(
)+g(
∆f
(
x)
x+h)
[
]=f
x)
x)x)
f
(
x) g(
∆f
(
f
(
∆g(
x)
(
i
i
) ∆
=
g(
x)
g(
g(
x)
x+h)
[ ]
56
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
Sol
ut
i
on
x)
x)
(
i
)∆[
f
(
g(
x)
(
gx+h)f
(
g(
x)
x+h)(
]=f
x)
=f
(
gx+h)f
(
gx)+f
(
gx)f
(
g(
x)
x+h)(
x+h)(
x+h)(
=f
(
g(
g(
x)
x)
f
(
f
(
x)
x+h)
x+h)x+h)[
]+g(
[
]
x)+g(
x)
=f
(
∆g(
∆f
(
x)
x+h)
x)
f
(
x) f
(
x+h) f
(
x) f
(
gx
)f
(
g(
x+h)
x+h)(
(
i
i
)∆
=
=
g(
x) g(
x+h)g(
x)
g(
g(
x)
x+h)
[ ]
x)(
x)(
x)
x)
f
(
gx)f
(
gx)+f
(
gx)f
(
g(
x+h) g(
f
(
f
(
x)
f
(
x)
g(
g(
x)
x+h)(
x+h)x+h)[
][
]
=
=
g(
g(
x)
g(
g(
x)
x+h)
x+h)
x)
x
x)
g(
∆f
(
)f
(
∆g(
x)
=
g(
g(
x)
x+h)
2.Ev
al
uat
et
hef
ol
l
owi
ng,
i
nt
er
v
al
ofdi
f
f
er
enci
ngbei
nguni
t
y
.
x
x
e
2
og3x)
e2xl
an-1ax (
(
i
)∆t
i
i
)∆
(
i
i
i
)∆ x
(
i
v
)∆(
e +e-x
(
!
x+1)
(
)
(
)
Sol
ut
i
on
x+1)(
i
)∆t
an-1ax=t
an-1a(
t
an-1ax
an
=t
(
i
i
)
a(
ax
x+1)-
a
an-1
=t
1 +a2x+a2x2
1 +a(
.
ax
x+1)
1
x
x
x
x+2) 2x+12x+1 -2(
2x+1
x.2
x.2 ∆
=
=
=
=(
! (
x+1)
(
(
(
!(
!
!
!
!
x+1)
x+2)
x+2)
x+2)
x+2)
(
2x
)
2x+1
2x
ex +e-x]ex+1 +e-(x+1)]
ex[
ex+1[
(
i
i
i
)∆ x
=
=
e +e-x ex+1 +e-(x+1) ex +e-x
ex +e-x]
ex+1 +e-(x+1)]
[
[
e-e-1
e2x+1 +e-e2x+1-e-1
=
=
ex +e-x] [
ex +e-x]
ex+1 +e-(x+1)]
ex+1 +e-x-1)]
[
[
[
(
ex
)
ex+1
ex
og3x)=e2(x+1)l
e2xl
(
i
v
)∆(
og3(
-2xl
og3x
x+1)e
1
og3x
=e2xe2l
og3x1 + e2xl
x
1 2x
og3x
el
=e2xe2l
og3x+l
og1 + x
1
l
og3x
og3x+e2l
og1 + =e2xe2l
x
( )
( )]
[
( )
[
57
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
]
1
og3xl
og3x+e2l
og 1 +
=e2x exl
x
1
2x 2
2
l
og3x
og 1 + +(
1)
e=e el
x
{
{ ( )
( )}
}
3.Ev
al
uat
et
hef
ol
l
owi
ng,
i
nt
er
v
al
ofdi
f
f
er
enci
ngbei
ngh:
x2
si
n2x
( )
x2 +si
nx) (
si
n2xcos4x) (
(
i
)∆(
i
i
)∆(
i
i
i
)∆cotax (
i
v
)∆
Sol
ut
i
on
2
x2 +si
x2 +si
nx)=[
nx]
(
i
)∆(
+si
n (
x+h)
(
x+h)
[
]Or
2 2
x2 +si
nx)=∆x2 +∆si
si
n(
-i
nx]
∆(
nx=[
(
x+h)s
x+h)
x]+[
-
Recal
l
t
het
r
i
gonomet
r
i
ci
dent
i
t
y
:
A +B
AB
si
nAsi
nB =2cos
si
n
2
2
h h
2
=2hx+h +2cos x+ si
n
2 2
h
h
=h(
n cosx+
h +2x)+2si
2
2
( )
( )
( )( )
(
i
i
)∆(
si
n2xcos4x)=si
n2(
cos4(
-i
n2xcos4x
x+h)
x+h)s
Usi
ngt
het
r
i
gonomet
r
i
ci
dent
i
t
y
:
1
si
n(
n(
AB)
si
nAcosB = [
A +B)+si
]
2
1
1
=[
si
n6(
n (2x2h)
si
n6x+si
n(2x)
x+h)+si
]-[
]
2
2
Not
e:si
n(si
nθ
θ)=si
n6(
-i
n2(
-i
n6x+si
n2x
x+h)s
x+h)s
=
2
1
-si
=[
si
n6(
-i
n6x
n2(
-i
n2x}
x+h)s
x+h)s
{
}{
]
2
Recal
l
t
het
r
i
gonomet
r
i
ci
dent
i
t
y
:
A +B
AB
si
nAsi
nB =2cos
si
n
2
2
1
=[
2cos(
si
n3h2cos(
si
nh]
6x+3h)
2x+h)
2
=si
n3hcos(
-i
nhcos(
2x+h)
6x+3h)s
( )( )
(
i
i
i
)
naxcosax+h -cosaxsi
nax+h
cosax+h cosax si
∆cota =cota - x=
cota =
na
nax
si
nax+h si
si
nax+hsi
si
n(
nax(
1-ah)
axax+h) si
=
=
nax si
nax
si
nax+hsi
nax+hsi
x
x+h
x
58
Numer
i
cal
Anal
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si
sMat
hemat
i
cs&Engi
neer
i
ng
2
2
2
si
n2xx
-2si
(
(
n2(
x+h)
x+h)
x+h)
x2
x
(
i
v
)∆
=
=
si
n2x si
n2x
si
n2x
si
n2(
n2(
x+h)si
x+h)
2
2
2
2
n2x -xsi
n2x +xsi
n2x-xsi
n2(
x+h)
(
x+h)si
=
si
n2x
si
n2(
x+h)
( )
2
n2x] h(
si
n2(
-i
si
n2x-x2[
-x2]
(
si
n2x2x2si
nhcos(
2x+h)
x+h)s
x+h)
h +2x)
[
=
si
n2x
si
n2(
x+h)
=
si
n2x
si
n2(
x+h)
4.Ev
al
uat
et
hef
ol
l
owi
ng,
i
nt
er
v
al
ofdi
f
f
er
enci
ngbei
ngh:
x
2
2
n cx+d
n
c
o
s
2
x
(
)
(
)
(
i
)∆
(
i
i
)∆ ab (
i
i
i
)∆ a
(
i
v
)∆ cos(
cx+d)
Sol
ut
i
on
(
i
)∆cos2x=cos2(
-os2x
x+h)c
=2si
n(
si
nh2si
nhsi
n(
2x+h)
2x+h)
2
cos2x)=∆(
∆cos2x)
∆(
=∆(2si
nhsi
n(
2x+h)
2si
nh∆si
n(
2x+h)
)==2si
nh[
si
n{
si
n(
2(
x+h)+h}2x+h)
]
n2hcos(
=2si
nh.
2cos(
si
nh=hsi
2x+2h)
2x+2h)
x h
(
i
i
)∆(
abx)=a∆(
bx)=a(
bx
bx)=ab(
1)=a(
1)
bx+hbbhx
x
x
h
2
ab)=∆(
∆ab)=∆[
a(
b]
∆(
1)
b2 x
=a(
∆ x =a(
.
bx(
b
1)b
1)
1)=a(
1).
bhbhbhbh-
1)=(
1)
achach(
i
i
i
)∆acx+d =ac(x+h)+dacx+d =acx+d(
acx+d
2 cx+d
cx+d
ch
c
x+
d
∆a )=∆(
1)
a =∆(
∆a
a
2 cx+d
cx+d
ch
ch
(
=(
∆
=(
1)a
1)
1)
1)
acx+d =(
a
a a achach3 cx+d
3 cx+d
ch
1).
a ∆a
Si
mi
l
ar
l
y
,
=(
a andsoon
n cx+d
Gener
al
i
zi
ng,
∆nacx+d =(
1).
acha
n
ch
ch +π
∆ncos(
cos cx+d +n
Ans:
n
cx+d)=2si
2
2
(
(
i
v
)Sol
v
ei
tasat
r
y–
) [
(
)]
Di
f
f
er
enceofPol
y
nomi
alofDegr
een
Weshal
lpr
ov
eher
et
hatt
henth di
f
f
er
encesofapol
y
nomi
aloft
henth degr
eear
e
x)=2x3 +3x2 +4x+3 i
(
const
ant
.For i
nst
ance,f
s a pol
y
nomi
alof degr
ee 3 and on
di
f
f
er
ent
i
at
i
ng3t
i
meswegetaconst
ant
.
Gener
al
l
y
,
l
et
y=axn +bxn-1 +cxn-2 +… +kx+1
-(
1)
t
h
beapol
y
nomi
al
ofn degr
ee(
a≠0)
n
n2
n1
y+∆y=a(
+c(
+… +k(
x+h)
x+h) +b(
x+h)
x+h)+1
-(
2)
wher
eh=∆x.Nowsubt
r
act
i
ngEquat
i
on(
1)f
r
om Equat
i
on(
2)
,
wehav
e,
n
n1 nn2 n∆y=a{
xn}+b{
(
(
(
x+h)
x 1}+c{
x+h)
x+h)x 2}+… +kh
-
59
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
n
Nowexpandi
ng(
ngbi
nomi
al
expansi
on,
weobt
ai
n
x+h)usi
{
}
n(
n1) 2 n(
(
n2) 3
n1)
n
∆y=ax
xn-2h +
xn-3h +… +hnxn +b
+nxn-1h +
2!
3!
(
(
n2) 2
n1)
n2
n1)
xn-1 +(
x
xn-3h +… +hn-1xn-1 +c
h+
2!
(
(
n3) 2
n2)
n3
n2)
xn-2 +(
x
xn-4h +… +hn-2xn-2
h+
2!
{
{
}
}
Ther
ef
or
e
h}+xn-3{
h}+… +kh
anc2h2 +b(
anc3h3 +bn-1c2h2 +c(
∆y=anxn-1h +xn-2{
n1)
n2)
Now i
f∆x=h i
saconst
ant
,t
hent
hebr
acket
edcoef
f
i
ci
ent
sar
eal
lconst
ant
,sot
hatwemay
1 1 1
d et
r
epl
acet
hem bysi
mpl
econst
antcoef
f
i
ci
ent
sb,
c.Thus,
c,
∆y=anxn-1h +b1xn-2 +c1xn-3 +… +kh
-(
3)
Hencet
hef
i
r
stdi
f
f
er
enceofapol
y
nomi
aloft
henth degr
eei
sanot
herpol
y
nomi
alof(
n1)
degr
ee.
Now t
of
i
ndt
heseconddi
f
f
er
ence,wegi
v
exani
ncr
ement∆x=h i
nEquat
i
on(
3)
,we
hav
e
n2
n3
n1
y+∆y
)=anh(
x+h)
x+h)
x+h)
∆(
+b1(
+c1(
+… +kh
Thati
s,
∆y+∆2y=…
-(
4)
Subt
r
act
i
ngEquat
i
on(
3)f
r
om (
4)
,
weobt
ai
n,
n1 nn2 nn3 n(
(
(
∆2y=anh{
x+h)
x 1}+b1{
x+h)
x+h)
x 2}+c1{
x 3}+… +kh
Agai
nexpandi
ngbybi
nomi
al
t
heor
em andr
epl
aci
ngbycoef
f
i
ci
entasbef
or
e,
weobt
ai
n,
n1)
∆2y=an(
h2xn-2 +b11xn-3 +c11xn-4 +… +kh
Ther
ef
or
e,t
heseconddi
f
f
er
encei
sapol
y
nomi
alofdegr
ee(
.Bycont
i
nui
ngi
nt
hi
smanner
n1)
wear
r
i
v
eatapol
y
nomi
al
ofdegr
eezer
of
ort
henthdi
f
f
er
ence.
Thedi
f
f
er
ence
(
2)Backwar
dDi
f
f
er
ence
y
y y
y …y
y
-,
-,
n1 0
n1
2 1
ar
ecal
l
edt
hef
i
r
stf
or
war
ddi
f
f
er
enceoft
hef
unct
i
onyandar
edenot
edby
∇y
, ∇y
…,
∇y
1
n1
0
r
espect
i
v
el
y
,
sot
hat
y-1
∇y
=y
0
0
y
∇y
=y
1
1 0
y
∇y
=y
2
2 1
a)=f
a)y
y
or ∇f
(
(
f
(
ah)
∇y
=
n
n n1
wher
e∇i
scal
l
edt
hebackwar
ddi
f
f
er
enceoper
at
or
.
I
nasi
mi
l
arway
,
onecandef
i
nebackwar
ddi
f
f
er
encesofhi
gheror
der
s.Thus,
60
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
y
y ∇y
=∇(
∇(
∇y
∇y
∇2y
2 1)
2)=
1
2
2
y
y
=y
=
+y
y
y
y
y
(
)
1 0
2 1
2 1 1
0
=y
2
y
+
y
1
2
0
y
=∇(
=y
3y
+3y
andsoon
∇3y
∇2y
1 0
3
3)
3
2
TheDi
spl
acement(
orShi
f
t
)oper
at
orE
Theoper
at
orEi
ncr
easest
hev
al
ueoft
hear
gumentbyonei
nt
er
v
al
.
xa,a +h,a +2h,
y
y
I
f:
…andy
,
,
,
…
=f
(
a)
=f
(
a +h)
=f
(
a +2h)
1
0
2
a)=f
y
;
=
Ef
(
(
a +h) or Ey
1
0
=y
i
.
e.Ey
Ef
(
(
a +2h) or Ey
a +h)=f
y
1
2
n =
n +1
Whent
heoper
at
orEi
sappl
i
edt
wi
ce,
t
hev
al
ueoft
hear
gumenti
ncr
easesbyt
woi
nt
er
v
al
s.
2
2
y
y
y
y
y
E2y
=
,
E
=
,
E
=
n
1
0
2
3
n +2
r
I
ngener
alEry
,
=
y
y
=
y
E
n
n
nr
n +r
Theoper
at
orEhast
hef
ol
l
owi
ngpr
oper
t
i
es:
x
x
x)+bEg(
(
i
)Ecf
(
i
i
)E[
()=cEf
(
x)
af
()+bg(
x)
(
x)
]=aEf
m +n
m n
x)=∆Ef
(
i
i
i
)E [
;(
i
v
)Eand∆ar
ecommut
at
i
v
e,
i
.
e.
,
E∆f
Ef
(
x)
(
x)
(
(
x)
]=E f
Rel
at
i
onsbet
ween∆,∇andE
(
i
)
E ≡1 +∆ and ∆ ≡E1
y
y
y
y
∆y
=
=Ey
(
E1)
n
nn =
n
n +1 n
∆ ≡E1andE ≡1 +∆
n
1 +∆)
En ≡(
Not
e.I
ngener
al
1
(
i
i
) ∇ ≡1E
y
y
y
E-1y
E-1
=y
∇ ≡1∇y
(
E-1)
1n =
nnn =
n
n1
1
(
i
i
i
) ∇ ≡∆E
y
y
∆y
=∆y
=∆E-1y
∇ ≡∆E-1
n =
nn
n1
n1
I
l
l
ust
r
at
i
v
eExampl
es
1.Pr
ov
et
hat∇E =E∇ =∆ =E1
Sol
ut
i
on ∇Ey
y =∆y
∇y
=y
n =
n
n +1
n +1 n
y
y
y
y
E∇y
=
E
=
=
(n n-1) n +1 n ∆y
n
n
y
y
y
y
=
∆
(
E1)
n =
n
n
n +1
Hence∇E =E∇ =∆ =E1
2 2
2.Ev
al
uat
e(
,h =1
(
)
∇ +∆)
x +x
2 2
2 2
Sol
ut
i
on (
∇ +∆)(
x +x)=(
x +x)
E-1 +E11)(
2 2
1
=(
=(
)
(
)
(
x
x2 +x)
+
x
EE
E22 +E-2)
2 2
2
x +x)x2 +x)+E (
x2 +x)
=E(
2(
22
2
22
2
x +x)+E x +E-2x
=Ex +Ex2(
2
2
x+2) +x+2x2)
=(
2(
+x2
x2 +x)+(
Putx=h=1
2 2
2
2
(
∇ +∆)
1 +2)
12)
+1 +22(
+12 =8
∴ (
x +x)=(
12 +1)+(
61
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
2
∆2
x)
u(
x
)and∆ u(
E
Eu(
x)
()
3.Expl
ai
nt
hedi
f
f
er
encebet
ween
2
(
E1)
∆2
E22E +1
x)=
x)=
u(
u(
u(
x)
E
E
E
=(
u(
x)
E2 +E-1)
x
x
=Eu()2u()+E-1u(
x)
x)+u(
=u(
2u(
x1)
x+1)2
u(
x) (
x) (
∆2u(
u(
x) u(
2u(
x)
E1)
x+1)+u(
x+2)2E +1)
E2=
=
=
Eu(
x) u(
x+1)
u(
x+1)
u(
x+1)
Thedi
f
f
er
encei
sev
i
dent
.
Sol
ut
i
on
[ ] (
()
[
)
]
∆f
(
x)
x)=l
4.Pr
ov
et
hat
:
(
i
)∆l
ogf
og 1 +
(
f
(
x)
∆2 3
(
i
i
) x =6xi
nt
er
v
al
ofdi
f
f
er
enci
ngbei
nguni
t
y
E
Sol
ut
i
on
f
(
x+1)
Ef
(
x)
x)=l
x)=l
(
i
)
∆l
ogf
ogf
-ogf
og
=l
og
(
(
(
x+1)l
f
(
x)
f
(
x)
x)+∆f
(
f
(
x)
f
(
(
x)
1 +∆)
=l
og
=l
og
f
(
x
)
f
(
x)
∆f
(
x)
=l
og 1 +
f
(
x)
2
2
2
(
)
E1
E
2E +1 3
∆ 3
x=
x3 =
x =(
x3
(
i
i
)
E2 +E-1)
E
E
E
3
=Ex32x3 +E-1x
3
3
=(
2x3 +(
=6x
x+1)
x1)
()
()
[ ] [ ]
] [
]
]
[
[
1
1 1
5.I
fDst
andsf
ort
hedi
f
f
er
ent
i
al
oper
at
ord ,
pr
ov
et
hatD = ∆-∆2 + ∆3…
3
h 2
dx
Sol
ut
i
on
h2 ''
h3 '''
'
x
x
x
x
x)+…
(
)
(
)
(
)
(
)
(
)
Weknowt
hatEf =fx+h =f +hf + f + f(
2!
3!
[
Tay
l
or
’
sExpansi
on]
2
3
(
hD) (
hD)
x)=ehDf
=1 +hD +
(
x)
(
+
+…f
2!
3!
1 +∆ =ehD l
E =ehD
og(
1 +∆)=hD
1 12 13
1
D =l
og(
1 +∆)= ∆-∆ + ∆ …
3
h 2
h
[
[
]
(
6.Gi
v
ent
hat
:
x:1
2
3
y
:2
5 10
Fi
ndt
hev
al
ueof∇2y
.
5
)
4
17
5
26
62
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
]
2
Sol
ut
i
on ∇2y
=1=1y
E-1)y
2E-1 +E-2)
5 (
5 (
5
=y
2E-1y
+E-2y
=y
2y
+y
5
5
5
5
4
3
17)+10 =2
=262(
Exer
ci
se
(
i
)
∆∇)
x2,
Fi
nd(
wher
ehi
st
hei
nt
er
v
al
ofdi
f
f
er
enci
ng.
(
i
i
)
Ev
al
uat
e
Ans:
2h2
∆2
f
(
x)
,
wher
ehi
st
hei
nt
er
v
al
ofdi
f
f
er
enci
ng
E
()
x)Ans:
f
(
2f
(
(
x2h)
xh)+f
(
i
i
i
)
5
1
Pr
ov
et
hat∇y
=h1 + ∇ + ∇2 +…y
n
n +1
2 12
(
i
v
)
Pr
ov
et
hat∆
(
v
)
(
1 +∆)
1∇)=1
Pr
ov
et
hat(
a)(
[
Ux
]
Vx∆UxUx∆Vx
VxVx+1
(V)=
x
∆∇
(
b)∆ +∇ = ∇∆
(
c)∇2 =12E-1 +E-2
(
v
i
)
Ev
al
uat
et
hef
ol
l
owi
ng,
t
hei
nt
er
v
al
ofdi
f
f
er
enci
ngbei
nguni
t
y
:
(
a)∆l
ogx
23
(
e)∆ x
Ex3
∆2
(
c) x2
E
()
(
b)(
x
E-1∆)
2
(
f
)∆2E3x
2
2
(
g)(
(
2∆ +1)
x+2)
(
(
1x)
12x)
13x)
(
i
)∆3(
(
h)(
ex +x)
(
(
E1)
E +2)
(
3)Cent
r
al
Di
f
f
er
ence
Thecent
r
al
di
f
f
er
enceoper
at
orδi
sdef
i
nedbyt
her
el
at
i
ons:
y
y =δ1
1 0
2
y
y =δ3
2 1
2
63
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
22
(
d) ∆ x
x+l
ogx]
E[
………………
y
y
=δn-1
nn1
2
Si
mi
l
ar
l
y
,
hi
gheror
derdi
f
f
er
encesar
edef
i
nedas
2
y
δ
δ
δ
δ2y
δ2y
δ3δ1 =δy
,
=
,
=δ3y
5 3
1
1
2
2
3
2
2
2
2
2
andsoon.
Thecent
r
al
di
f
f
er
enceoper
at
ori
sal
sodef
i
nedas
h
h
x)=fx+ (
δf
fx2
2
( )( )
Rel
at
i
onsbet
weent
heoper
at
or
s
1
2
1
2
(
i
)
δ =E E
(
i
i
)
δ =∆E 2
(
i
i
i
)
δ =∇E
1
-
1
2
Exer
ci
se
∇2x3
1
(
a) 2 =31x
δx
( )
(
b)δ2y
=y
2y
+y-1
1
0
0
3
1
1
2
2
1
1
2
2
(
c)δ
y-1
=y 3y +3y
0
(
e)δ(
E +E
δ2
1
+
(
d)E =
4
1
2
1
2
( )
δ
+2
)=∆E-1 +∆
⌂4⌂
I
NTERPOLATI
ONANDEXTRAPOLATI
ON
x2,…,xn.Thent
Lety
,y,y
,….,y
easetofv
al
uesoff
unct
i
ony=f
he
(
x)atx0,x
nb
1,
0 1
2
pr
ocessoff
i
ndi
ngt
hev
al
uesofycor
r
espondi
ngt
oanyv
al
ueofx=x1 bet
weenx0 andxn i
s
cal
l
edi
nt
er
pol
at
i
on.Thus,
i
nt
er
pol
at
i
oni
st
het
echni
queofest
i
mat
i
ngt
hev
al
ueofaf
unct
i
onf
or
anyi
nt
er
medi
at
ev
al
ueoft
hei
ndependentv
ar
i
abl
ewhi
l
et
hepr
ocessofcomput
i
ngt
hev
al
uesof
t
hef
unct
i
onout
si
det
hegi
v
enr
angei
scal
l
edext
r
apol
at
i
on.
Newt
onGr
egor
yFor
mul
aef
orEqualI
nt
er
v
al
s
(
a)Newt
on’
sFor
war
dI
nt
er
pol
at
i
onFor
mul
a
(
b)Newt
on’
sBackwar
dI
nt
er
pol
at
i
onFor
mul
a
 Newt
on’
sFor
war
dI
nt
er
pol
at
i
onFor
mul
a
Supposet
hef
ol
l
owi
ngt
abl
er
epr
esent
sasetofv
al
uesofxandy
x: x0
x1
x2
x3
xn
⋯
y
: y
y
y
y
y
⋯
n
1
0
2
3
OR
x: a
a +h
a +2h
a +3h
a +nh
⋯
y
: f
(
a)
f
(
a +h)
f
(
a +2h)
f
(
a +3h) ⋯
f
(
a +nh)
n
n
a)=y
f
(
a)
(
(
(
1 +∆)f
a +nh)=Ef
n =
64
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
[
]
n(
n1) 2 n(
(
n2) 3
n1)
=1 +n∆ +
(
a)
∆ +
∆ +…f
2!
3!
n(
n(
(
n1) 2
n1)
n2) 3
∴ (
a)+n∆f
a)+
a)+
a)+…
(
(
(
(
∆f
∆f
fa +nh)=f
2!
3!
OR
n(
n(
(
n1) 2
n1)
n2) 3
∆y
∆y
f
+n∆y
+
+
+…
(
a +nh)=y
0
0
0
0
2!
3!
wher
ey
sapol
y
nomi
al
ofdegr
een
x)i
n(
xa
n=
h
or
xx0
n=
h
x0 =a=f
i
r
stt
er
m
h=i
nt
er
v
al
ofdi
f
f
er
enci
ng;
x1x0 =h,x2x0 =2h andsoon
x=gi
v
env
al
ueoft
hei
ndependentv
ar
i
abl
e
2
3
,
∆y
,
∆y
ar
eobt
ai
nedf
r
om di
f
f
er
encet
abl
e
∆y
0
0
0
Vi
t
alpoi
nt
:
I
fa +nh i
sneart
hebegi
nni
ng,
usef
or
war
di
nt
er
pol
at
i
onf
or
mul
a.
I
l
l
ust
r
at
i
on1:
I
nacer
t
ai
nexper
i
ment
,
t
hev
al
uesofxandywer
ef
oundasf
ol
l
ows:
x: 0 1
2
3
4
5
6
y
: 0 1
16 81
256
625
1296
5,
Fi
ndt
hev
al
ueofywhenx=2.
usi
ngNewt
on’
sf
or
war
di
nt
er
pol
at
i
onf
or
mul
a.
Sol
ut
i
on:
Her
ea=0h=
, 1,a +nh =2.
snear
ert
hebegi
nni
ngt
hanend
5i
2.
50
Son=
=2.
5
1
Thef
ol
l
owi
ngi
st
hef
or
war
ddi
f
f
er
encet
abl
e:
x
y
∆y
∆4y
∆2y
∆3y
0
∆5y
0 =y
0
1 =∆y
0
1
14 =∆2y
0
1
36 =∆3y
0
15
2
16
65
3
81
24
110
256
0 =∆5y
0
60
175
4
24 =∆4y
0
50
0
84
24
194
369
65
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
108
5
625
302
671
6
1296
Usi
ngNewt
on’
sf
or
war
di
nt
er
pol
at
i
onf
or
mul
a
n(
n(
(
n(
(
(
n3) 4
n1) 2
n1)
n1)
n2) 3
n2)
∆y
∆y
∆y
f
+n∆y
+
+
+
(
a +nh)=y
0
0
0
0
0
4!
2!
3!
weget
(
(
1.
5)
(
(
(
0.
5)
(
(
(
(0.
5)
1.
5)
1.
5)
2.
5)
2.
5)
2.
5)
0.
5)
f
(
24)
(
(
(
(
1)+
14)+
2.
5)=0 +(
2.
5)
36)+
6
2
24
=0 +2.
5 +26.
25 +11.
250.
9375
=39
I
l
l
ust
r
at
i
on2:Appl
y
i
ngNewt
on’
sf
or
war
di
nt
er
pol
at
i
onf
or
mul
a,
comput
et
hev
al
ueof 5.
5,
gi
v
en
t
hat 5 =2.
828,
cor
r
ectupt
ot
hr
eepl
acesofdeci
mal
.
449, 7 =2.
236, 6 =2.
646and 8 =2.
Sol
ut
i
on
Let
’
spr
esenti
tt
hi
sway
:
x: 5
6
7
8
y
: 2.
236
2.
449
2.
646
2.
828
Her
ea=5h=
, 1a
, +nh =5.
snear
ert
hebegi
nni
ngt
hanend
5i
5.
55
Son=
=0.
5
1
Thef
ol
l
owi
ngi
st
hef
or
war
ddi
f
f
er
encet
abl
e:
x
y
∆y
∆2y
5
∆3y
2.
236 =y
0
0.
213 =∆y
0
6
–0.
016 =∆2y
0
2.
449
0.
001 =∆3y
≈0
0
0.
197
7
2.
646
015
-0.
0.
182
8
2.
828
Usi
ngNewt
on’
sf
or
war
di
nt
er
pol
at
i
onf
or
mul
a
n(
n1) 2
∆y
f
+n∆y
+
(
a +nh)=y
0
0
0
2!
(
(0.
5)
0.
5)
f
236 +(
(0.
016)
(
(
5.
5)=2.
0.
5)
0.
213)+
2
=2.
236 +0.
1065 +0.
002 =2.
3445
∴ 5.
5 =2.
345t
o3deci
mal
pl
aces.
66
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
I
l
l
ust
r
at
i
on3:
Thef
ol
l
owi
ngar
edat
af
r
om t
hest
eam t
abl
e:
Temper
at
ur
150
160
170
180
e 0C: 140
2
/
Pr
essur
ekgf
685 4.
854 6.
302 8.
076
10.
225
cm : 3.
Usi
ngsui
t
abl
ef
or
mul
a,
f
i
ndt
hepr
essur
eofst
eam f
orat
emper
at
ur
e1420C.
Sol
ut
i
on:
Her
ea=140,h=10,a +nh =142 i
snear
ert
hebegi
nni
ngt
hanend.
142140
So,n=
=0.
2
10
Thef
ol
l
owi
ngi
st
hef
or
war
ddi
f
f
er
encet
abl
e:
x
y
∆y
∆2y
∆3y
140
∆4y
3.
685 =y
0
1.
169 =∆y
0
150
0.
279 =∆2y
0
4.
854
0.
047 =∆3y
0
1.
448
160
6.
302
1.
774
170
0.
002 =∆4y
≈0
0
0.
326
8.
076
0.
049
0.
375
2.
149
180
10.
225
Usi
ngNewt
on’
sf
or
war
di
nt
er
pol
at
i
onf
or
mul
a
n(
n(
(
n1) 2
n1)
n2) 3
∆y
∆y
f
+n∆y
+
+
(
a +nh)=y
0
0
0
0
2!
3!
(
(0.
8)
(
((1.
8)
0.
2)
0.
2)
0.
8)
f
685 +(
(
0.
047)
(
(
(
142)=3.
0.
2)
1.
169)+
0.
279)+
6
2
=3.
685 +0.
23380.
02232 +0.
002256
=3.
899Kgf
/
cm2
TestYourKnowl
edge
1.Usi
ngNewt
on’
sf
or
war
di
nt
er
pol
at
i
onf
or
mul
a,
f
i
ndyatx=8f
r
om t
hef
ol
l
owi
ngt
abl
e:
5
10
15
20
25
x: 0
y
: 7
11 14
18
24
32
Ans:
12.
77
x)i
(
2.Af
unct
i
onf
sgi
v
enbyt
hef
ol
l
owi
ngt
abl
e.Fi
ndf
t
abl
ef
or
mul
a:
(
0.
2)byasui
x: 0
1
2
3
4
5
6
: 176 185
194
203
212
220
229 Ans:
177.
75
y
3.Gi
v
en:
n500 =0.
7660,si
n600 =0.
8660,
f
i
ndsi
n520,
si
n450 =0.
7071,si
n550 =0.
8192,si
usi
ngNewt
on’
sf
or
war
di
nt
er
pol
at
i
onf
or
mul
a.
Ans:
0.
7880
4.Usi
ngNewt
on’
sf
or
war
di
nt
er
pol
at
i
on,
f
i
ndt
hev
al
ueofl
gi
v
en
og10π,
l
og3.
141 =0.
4970679364; l
og3.
142 =0.
4972061807;
l
og3.
143 =0.
4973443810;
l
og3.
144 =0.
4974825374; l
og3.
145 =0.
4976206498
67
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
1416]
[
Hi
nt
.π =3.
Ans:
0.
4971508883
 Newt
on’
sBackwar
dI
nt
er
pol
at
i
onFor
mul
a
Newt
onGr
egor
ybackwar
di
nt
er
pol
at
i
onf
or
mul
ai
sgi
v
enas
n
n
a)=(
f
(
a)=(
(
a)
(
(
1∇)f
anh)=E-nf
E-1)f
[
]
n(
n1)2 n(
(
n2)3
n1)
=1(
a)
n∇ +
∇∇ +…f
2!
3!
n(
n(
(
n1)2
n1)
n2)3
∴ (
a)a)+
a)a)+…
(
(
(
(
nh)=f
fan∇f
∇f
∇f
2!
3!
OR
n(
(
n1)2 n(
n1)
n2)3
y
f
n
∇
+
+…
∇y
∇y
(
anh)=y
0
0
0
0
2!
3!
ax
anh =x i
Al
so,
.
e. n=
h
Vi
t
alpoi
nt
:
I
fasneart
heend,
usebackwar
di
nt
er
pol
at
i
onf
or
mul
a.
nhi
I
l
l
ust
r
at
i
on4:
Fr
om t
hegi
v
ent
abl
e,
comput
et
hev
al
ueofsi
n380
x
0
10
20
30
40
0
si
nx
0
0.
17365
0.
34202
0.
50000
0.
64279
Sol
ut
i
on:
Her
ea=40,h=10,asnear
ert
heendt
hanbegi
nni
ng
nh=38 i
38
ax 40=
=0.
2
∴ n=
10
h
Thef
ol
l
owi
ngi
st
hebackwar
ddi
f
f
er
encet
abl
e:
si
nx0
x
y
0
0
∇2y
∇y
∇3y
∇4y
0.
17365
10
0.
17365
0.
00528
0.
16837
20
0.
34202
0.
00511
-
0.
0048 =∇3y
0
0.
15798
30
0.
00031 =∇4y
0
0.
01039
-
0.
01519 =∇2y
0
0.
50000
0.
14279 =∇y
0
68
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
40
0.
64279 =y
0
Usi
ngNewt
on’
sbackwar
di
nt
er
pol
at
i
onf
or
mul
a,
n(
(
n(
(
(
n3)4
n1)2 n(
n1)
n1)
n2)3
n2)
f
n∇y
+
+
∇y
∇y
∇y
(
anh)=y
0
0
0
0
0
4!
2!
3!
(
(0.
8)
(
((1.
8)
0.
2)
0.
2)
0.
8)
(
(
(
((38)=0.
0.
2)
0.
14279)+
0.
01519)0.
0048)+
f
642796
2
(
(((2.
8)
0.
2)
0.
8)
1.
8)
(
0.
00031)
24
=0.
642790.
028556 +0.
0012152 +0.
00023040.
000010416
0
n38 =0.
6157
∴ si
I
l
l
ust
r
at
i
on5:Thef
ol
l
owi
ngt
abl
egi
v
est
hepopul
at
i
onofat
owndur
i
ngt
hel
astsi
xcenses.
Est
i
mat
e,
usi
ngNewt
on’
si
nt
er
pol
at
i
onf
or
mul
a,
t
hei
ncr
easei
nt
hepopul
at
i
ondur
i
ngt
heper
i
od
1946t
o1948.
Year
: 1911 1921 1931 1941 1951
1961
Popul
at
i
on(
i
nt
housands)
: 12
15
20
27
39
52
Sol
ut
i
on:
Bot
ht
hey
ear
s1946and1948ar
enear
ert
heendt
hanbegi
nni
ng.
Her
ea=1961,h=10
Thef
ol
l
owi
ngi
st
hebackwar
ddi
f
f
er
encet
abl
e:
Year Popul
at
i
o
n
x
∇y
∇4y
∇2y
∇3y
y
1911
∇5y
12
3
1921
15
2
5
1931
20
0
2
7
1941
27
4 =∇3y
0
1 =∇2y
0
13 =∇y
0
1961
(
i
)
7 =∇4y
0
5
39
10 =∇5y
0
3
12
1951
3
52 =y
0
Fort
hey
ear1946,anh=1946
69
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
ax 19611946
∴ n=
=
=1.
5
10
h
Usi
ngNewt
on’
sbackwar
di
nt
er
pol
at
i
onf
or
mul
a
n(
(
n(
(
(
n3)4 n(
(
(
(
n4)
n1)2 n(
n1)
n1)
n1)
n2)3
n2)
n2)
n3)
f
n∇y
+
+
∇y
∇y
∇y
(
anh)=y
0
0
0
0
0
4!
5!
2!
3!
∇5y
0
weget
(
(
0.
5) (
(
(0.
5)
(
(
((1.
5)
1.
5)
1.
5)
1.
5)
0.
5)
0.
5)
0.
5)
(
(
(
(
((4)+
7)1.
5)
1)1946)=5213)+
f
6
2
24
(
(
(((2.
5)
1.
5)
1.
5)
0.
5)
0.
5)
(10)
120
=5219.
5 +0.
3750.
250.
16406250.
1171875
(
1946)=32.
f
344t
o3deci
mal
pl
aces
(
i
i
)
Fort
hey
ear1948,anh=1948
1948
ax 1961∴ n=
=
=1.
3
10
h
Usi
ngNewt
on’
sbackwar
di
nt
er
pol
at
i
onf
or
mul
a,
weget
(
(
0.
3) (
(
(0.
7)
(
(
((1.
7)
1.
3)
1.
3)
0.
3)
1.
3)
0.
3)
0.
7)
(
(
(
(
((4)+
7)1)1948)=521.
3)
13)+
f
6
2
24
(
(
(((2.
7)
1.
7)
1.
3)
0.
3)
0.
7)
(10)
120
=5216.
9 +0.
1950.
1820.
13536250.
1044225
(
1948)=34.
f
873cor
r
ectt
o3deci
mal
pl
aces.
equi
r
edi
ncr
easei
npopul
at
i
ondur
i
ng1946t
o1948
∴Ther
=f
(
f
(
1946)
1948)=34.
87332.
344 =2.
529t
housand.
3
2
x
()=x I
l
l
ust
r
at
i
on6:Fr
om t
hedi
f
f
er
encet
abl
eoff
3x +5x+7f
ort
hev
al
uesof0,2,4,6,8
andext
endt
het
abl
ef
ort
hecal
cul
at
i
onoff
(
10)
x)=x3Sol
ut
i
on:Her
ey=f
3x2 +5x+7
(
(
0)=7
f
(
2)=8f
12 +10 +7 =13
(
4)=64f
48 +20 +7 =43
(
6)=216f
108 +30 +7 =145
(
8)=512f
192 +40 +7 =367
Thef
ol
l
owi
ngi
st
hebackwar
ddi
f
f
er
encet
abl
e:
x
y
∇y
∇4y
∇2y
∇3y
0
7
6
2
13
24
30
4
43
48
72
70
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
0
48 =∇3y
0
102
6
120 =∇2y
0
145
222 =∇y
0
8
367 =y
0
810
∴ n=
=1
2
Usi
ngNewt
on’
sbackwar
di
nt
er
pol
at
i
onf
or
mul
a,
n(
(
n1)2 n(
n1)
n2)3
f
n∇y
+
∇y
∇y
(
anh)=y
0
0
0
0
2!
3!
((2)
(((3)
1)
1)
2)
f
(
48)
(
((
(
1)
10)=367222)+
120)6
2
=367 +222 +120 +48 =757
nh=10
Her
e a=8,h=2,a-
1.Gi
v
en:
TestYourKnowl
edge
x: 1
2
3
4
5
6
7
8
y
: 1
8 27
64
125
216
343
512
Est
i
mat
ef
Ans:
421.
875
(
7.
5)
2.Thef
ol
l
owi
ngdat
agi
v
est
hemel
t
i
ngpoi
ntofanal
l
oyofl
eadandzi
nc,wher
eti
st
he
0
t
emper
at
ur
ei
n Candpi
st
heper
cent
ageofl
eadi
nt
heal
l
oy
.
: 60
70
80
90
p(
%)
:
226 250
276
304
t
Fi
ndt
hemel
t
i
ngpoi
ntoft
heal
l
oycont
ai
ni
ng84%ofl
ead,usi
ngNewt
on’
si
nt
er
pol
at
i
on
f
or
mul
a.
Ans:
286.
96
Lagr
ange’
sFor
mul
af
orUnequalI
nt
er
v
al
s
Lety=f
unct
i
onwhi
cht
akest
hev
al
uesy
,y,y
,….,y
enxassumest
he
(
x)beaf
n wh
0 1
2
x2,…,xn r
v
al
ues x0,x
espect
i
v
el
y
.I
ft
hev
al
uesofxar
eatequali
nt
er
v
al
s,weuseNewt
on’
s
1,
f
or
war
dorbackwar
df
or
mul
a.I
ft
hev
al
uesofxar
enotatequali
nt
er
v
al
s,weuset
hef
ol
l
owi
ng
f
or
mul
a:
…x…xxx2)
xx3)(
xx0)
xx2)
xx3)(
xx1)
xn)
xn)
(
(
(
(
(
(
x)=y
y
f
+
+y
(
1
0
…x0…x1x0x1)
x0x2)
x0x3)(
xn)
x1x0)
x1x2)
x1x3)(
xn) 2
(
(
(
(
(
(
…x…xxx0)
xx3)(
xx0)
xx2)(
xx
xx1)
x
xn)
(
(
(
(
(
(
1)
n1)
+… +y
n
…
…
x2x0)
x2x
x2x3)(
x2x
xnx0)
xnx2)(
xnx1)
xnxn-1)
n)
(
(
(
(
(
(
1)
Thi
si
scal
l
edLagr
ange’
si
nt
er
pol
at
i
onf
or
mul
af
orunequal
i
nt
er
v
al
s
I
l
l
ust
r
at
i
on7:
Fi
ndy
v
eny
(
10)gi
(
(
(
(
5)=12,y
6)=13,y
11)=16.
9)=14andy
Sol
ut
i
on
Forunder
st
andi
ng,
pr
esenti
tasf
ol
l
ows:
x: 5
6
9
11
y
: 12
13
14
16
Si
ncet
hei
nt
er
v
al
ar
enotequal
,
weuseLagr
ange’
si
nt
er
pol
at
i
onf
or
mul
a;
71
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
=12,y
=13,y
Hence x0 =5,x1 =6,x2 =9, x3 =11; y
=14, y
=16
1
0
2
3
∴
xx2)
xx3)
xx0)
xx2)
xx3)
xx0)
xx3)
xx1)
xx1)
(
(
(
(
(
(
(
(
(
x)=y
y
y
f
+
+
+y
(
0
x0x1)
x0x2)
x0x3) 1(
x1x0)
x1x2)
x1x3) 2(
x2x0)
x2x1)
x2x3) 3
(
(
(
(
(
(
(
xx0)
xx2)
xx1)
(
(
(
x3x0)
x3x1)
x3x2)
(
(
(
(
(
(
(
(
(
(
(
(
106)
109)
1011)
105)
109)
1011)
105)
106)
1011)
=(
+(
+(
+(
16)
14)
12)
13)
(
(
(
(
(
(
(
(
(
56)
65)
511)
611)
59)
69)
95)
96)
911)
(
(
(
105)
106)
109)
(
(
(
115)
116)
119)
13 35 16 44
=2- + + = =14.
7
3 3 3 3
I
l
l
ust
r
at
i
on8:
UseLagr
ange’
sf
or
mul
at
of
i
tapol
y
nomi
al
t
ot
hedat
aandhencef
i
ndy
(
1)
x: 1
0
2
3
: 8
3
1
12
y
(
Hi
nt
:
keepxi
nt
hef
or
mul
aasi
ti
s)
Sol
ut
i
on
=8,y
=3,y
=12
Her
e x0 =1,x1 =0,x2 =2, x3 =3; y
=1, y
1
0
2
3
Usi
ngLagr
ange’
sf
or
mul
a,
xx2)
xx3)
xx0)
xx2)
xx3)
xx0)
xx3)
xx1)
xx1)
(
(
(
(
(
(
(
(
(
x
y
y
y
f
=
+
+
+y
() 0
x0x1)
x0x2)
x0x3) 1(
x1x0)
x1x2)
x1x3) 2(
x2x0)
x2x1)
x2x3) 3
(
(
(
(
(
(
(
xx0)
xx2)
xx1)
(
(
(
x3x0)
x3x1)
x3x2)
(
(
(
x(
(
(
(
x(1)
)
(
(
(
()
(
(
1)
x0)
x2)
x3)
x2)
x3)
x0)
x3)
x)=(f
+(
+(
+(
12)
(
1)
8)
3)
((((
()
(
(
(
()
(
(
1)
1)
10)
12)
13)
02)
03)
20)
23)
02x(
()
(
(
1)
x0)
x2)
(
()
(
(
1)
30)
32)
32 3 2
1 3 2
13 2
=(
xxxx3x25x +6x)+ (
4x +x+6)-(
2x 3x)+(
2x)
6
3
2
=2x36x2 +3x+3
|
onsi
mpl
i
f
i
cat
i
on
x)=2x3(
y
nomi
al
i
ff
6x2 +3x+3
∴ Thepol
Forx=1
3
2
(
1)=2(
1)+3
y=f
6(
+3(
1)
1)
∴ y=2
TestYourKnowl
edge
1.Usi
ngLagr
ange’
sf
or
mul
a,
f
i
ndt
hef
or
m oft
hef
unct
i
onf
v
ent
hat
:
(
x)gi
x: 0
2
3
6
y
: 659
705
729 804
72
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
Ans:1(x3 +29x2 +1602x+47448)
72
2.Gi
v
ent
hev
al
ues:
x: 0
2
3
6
: 4 2
14 158
y
Uset
heappr
opr
i
at
ef
or
mul
af
ori
nt
er
pol
at
i
ont
of
i
ndt
hev
al
ueoff
(
4)
Ans:
40
3.Appl
yLagr
ange’
sf
or
mul
at
of
i
ndf
r
om t
hef
ol
l
owi
ngdat
a:
(
x)f
1
4
5
x: 0
y
: 4
3
24 39
Hencef
i
ndywhenx=6
x)=2x2(
Ans:
f
3x+4;y=58
NEWTONDI
VI
DEDDI
FFERENCEFORMULA
x
x2)
x1)
xn)bet
(
(
(
(
Letf
,f
,f
,…,f
hev
al
uesoft
hef
unct
i
onf
r
espondi
ngt
ot
he
(
x)cor
0)
ar
gument
sx0,
…,
x1,
x2,
xn.
The f
i
r
stdi
v
i
ded di
f
f
er
ence f
ort
he ar
gument
si
s def
i
ned as x0 ,x1 i
s def
i
ned as
∆
x
x
(1)(0)andi
f
f
x0)or
(
(
x0,x1)orxf
sdenot
edbyf
x0,x1]
[
1
x1x0
Thus
x0)
x
(
(
∆
f
f
1)
x0)=
x0,x1)=xf
(
f
(
x1x0
x2)x1)
(
(
∆
f
f
x1)=
x1,x2)=xf
(
Si
mi
l
ar
l
y
,f
(
2
x2x1
x
x2)
(
(
∆
f
f
3)
x
x
x
,
(
)
and
f
et
c
(2 3)=x3f 2 =
x3x2
1
x1,
x2 i
Theseconddi
v
i
deddi
f
f
er
encef
ort
hear
gument
s x0,
sdef
i
nedas
2
x1,
x2)x0,
x1)
∆
(
(
f
f
x0)=
x0,
x1,
x2)=x,
(
f
f
(
x
x2x0
1 2
Thet
hi
r
ddi
v
i
deddi
f
f
er
encef
ort
hear
gument
sx0,
sdef
i
nedas
x1,
x2,
x3i
x1,,
x2x3)x0,,
x1x2)
∆3
(
(
f
f
x
x0,
x1,
x2,
x3)=x,,
(
)
f
f
=
andsoon
(
0
xx
1x
2x
3
3
0
Newt
on’
sdi
v
i
deddi
f
f
er
encef
or
mul
ai
s
∆
∆2
x
x
x
x
x
x
x0)+… +
x
x
x
()=f
(0)+( 0)
(0)+( 0)
( 1)
(
f
xf
x,
xf
1
1 2
n
∆
xx0)
xx1)(
xn-1)
x0)
(
(
(
…xx,
…,
xf
1
n
(
(
5)=120,
3)=24,f
I
l
l
ust
r
at
i
on9:UseNewt
on’
sdi
v
i
deddi
f
f
er
encef
or
mul
at
of
i
nd f
ff
(
7) i
(
(
(
8)=524,f
9)=720 and f
12)=1716
f
Bet
t
erwr
i
t
ei
tt
hi
sway
Sol
ut
i
on
73
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
x:
3
5
8
9
12
y
:
24
120
524
720
1716
Thedi
v
i
deddi
f
f
er
encet
abl
ei
sasf
ol
l
ows:
x f
(
x)
∆f
(
x)
3
(
x)
∆2f
24
∆
12024
(
3)=
=48
5f
53
5
∆2
134.
6748
(
3)=
=17.
33
5,
8f
83
120
∆
524120
(
5)=
=134.
67
8f
85
8
∆2
196134.
67
(
5)=
=15.
33
8,
9f
95
524
∆
720524
(
8)=
=196
9f
98
9
∆2
332196
(
8)=
=34
9,
12f
128
720
∆
1716720
(
9)=
=332
12f
129
12
1716
∆3f
(
x)
74
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
∆4f
(
x)
∆3
15.
3317.
33
(
3)=
=0.
33
5,
8,
9f
93
∆4
(2.
670.
33)
(
3)=
=0.
33
5,
8,
9,
12f
123
∆3
3415.
33
(
5)=
=2.
67
8,
9,
12f
125
Newt
on’
sdi
v
i
deddi
f
f
er
encef
or
mul
ai
s
n
∆
∆
∆2
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x0)
x)=f
(
(0)+( 0)
(
)
(
)
(
)
(
)
(
)
(
)(
)
1
1
n1
0
0
0
0
(
f
f
+
f
+
…
+
…
x1
x1,
x2
x,
…,
xnf
1
∆
∆2
∆3
x
()=f
(
(
(
(
(
(
(
(
(
(
x5) f
x5)
x5)
3)+(
x3)
3)+(
x3)
3)+(
x3)
x8)
3)+(
x3)
x8)
x9)
f
f
5f
5,
8
5,
8,
9
4
∆
(
3)
f
5,
8,
9,
12
(
(
(
(
((
(
(
7)=24 +(
75)
75)
75)
73)
73)
73)
78)
0.
33)+(
73)
78)
79)
f
48 +(
17.
33 +(
0.
33
=24 +192 +138.
64 +2.
64 +5.
28
(
7)=362.
f
56 ≈363
I
l
l
ust
r
at
i
on10:
Usi
ngNewt
on’
sdi
v
i
dedf
or
mul
a,
ev
al
uat
ef
,
gi
v
en
(
x)andf
(
15)
x:
4
5
7
10
11
13
y
:
48
100
294
920
1210
2028
Thedi
v
i
deddi
f
f
er
encet
abl
ei
sasf
ol
l
ows:
x f
(
x)
∆f
(
x)
4
48
75
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
∆2f
(
x)
∆
10048
(
4)=
=52
5f
54
5
∆2
9752
(
4)=
=15
5,
7f
74
100
∆
294100
(
5)=
=97
7f
75
7
∆2
208.
6797
(
5)=
=22.
33
7,
10f
105
294
∆
920294
(
7)=
=208.
67
10f
107
10
∆2
290208.
67
(
7)=
=20.
33
10,
11f
117
920
∆
1210920
(
10)=
=290
11f
1110
11
1210
13
2028
∆2
409290
(
10)=
=39.
67
11,
13f
1310
∆
20281210
(
11)=
=409
13f
1311
(
x)
∆3f
(
x)
∆4f
∆3
22.
3315
(
4)=
=1.
22
5,
7,
10f
104
76
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
∆5f
(
x)
∆4
0.
331.
22
(
4)=
=0.
22
5,
7,
10,
11f
114
∆5
0.
44(0.
22)
(
4)=
=0.
07
5,
7,
10,
11,
13f
134
∆3
20.
3322.
33
(
5)=
=0.
33
7,
10,
11f
115
∆4
3.
22(0.
33)
(
5)=
=0.
44
7,
10,
11,
13f
135
∆3
39.
6720.
33
(
7)=
=3.
22
10,
11,
13f
137
Newt
on’
sdi
v
i
deddi
f
f
er
encef
or
mul
ai
s
n
∆
∆
∆2
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x0)
x
()=f
(0)+( 0)
(0)+( 0)
( 1)
(0)+… +( 0)
( 1)(
n1)
(
f
…
x1f
x1,
x2f
x,
…,
xf
1
n
∆
∆2
∆3
x
()=f
(
(
(
(
(
(
(
(
(
4)+(
x4)
4)+(
x4)
x5) f
4)+(
x4)
x5)
x7)
4)+(
x4)
x5)
x7)
f
f
5f
5,
7
5,
7,
10
4
5
∆
∆
(
(
(
(
(
(
(
4)+(
x4)
x5)
x7)
4)
x11)
x10)
x10)
f
f
5,
7,
10,
11
5,
7,
10,
11,
13
(
(
(
(
(
(
(x4)
x4)
x5)
x4)
x5)
x7)
x4)
x5)
x7)
x4)
x10)
0.
22)+(
=48 +(
52 +(
15 +(
1.
22 +(
(
(
(
(
x5)
x7)
x11)
x10)
0.
07
122 3
22 4
x4)+15(
=48 +52(
x2xx9x+20)+ (
16x2 +83x140)- (
26x3 +243x2970x+1400)
100
100
7 5
+ (
x37x4 +529x33643x2 +12070x15400)
100
7
281
4397 3 31299 2 26914 7084
xx +
= x5- x4 +
x5
100 100
100
100
25
1
4
x)= (
(
7x5281x
+4397x331299x2 +107656x141680)
f
100
77
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
∴
1
3
2
5
4
(
15)= [
15)
15)
15)
15)
15)7(
281(
+4397(
31299(
+107656(
141680]
f
100
1
360760)=3607.
= (
6
100
TestYourKnowl
edge
1.UseNewt
on’
sdi
v
i
deddi
f
f
er
encef
or
mul
at
of
i
ndf
r
om t
hef
ol
l
owi
ngdat
a:
(
x)f
x: 0
1
2
4
5
6
y
: 1
14
15
5
6
19
Al
sof
i
ndf
(
3)
9x2 +21x+1;30
x3Ans:
2.UnderI
l
l
ust
r
at
i
on9,
f
i
ndf
(
x)
Ans: 1 (
33x4858x3 +9620x239094x+53595)
100
⌂5⌂
NUMERI
CALDI
FFERENTI
ATI
ONANDI
NTEGRATI
ON
A.Numer
i
cal
Di
f
f
er
ent
i
at
i
on
y,
y,
y
x1,
Lety=f
unct
i
ont
aki
ngt
hev
al
uesy
,
…,
or
r
espondi
ngt
ot
hev
al
ues x0,
(
x)beaf
nc
0 1 2
x2,
…,
xn oft
hei
ndependentv
ar
i
abl
ex.Nowwear
et
r
y
i
ngt
of
i
ndt
heder
i
v
at
i
v
ev
al
ueofy=y
f
or
k
t
hegi
v
enx=xk.,i
.
e.I
ft
hev
al
ueoccurbet
weenx0 t
o x1 orbegi
nni
ngoft
het
abl
e,weuse
Newt
on’
sf
or
war
ddi
f
f
er
encef
or
mul
awhi
l
e,i
fi
toccur
satt
heend,weuseNewt
on’
sbackwar
d
di
f
f
er
encef
or
mul
a.
 Newt
on’
sFor
war
dDi
f
f
er
enceFor
mul
a
Supposet
hef
ol
l
owi
ngt
abl
er
epr
esent
sasetofv
al
uesofxandy
x: x0
x1
x2
x
… xn
3
y
: y
0
y
1
y
2
y
… y
n
3
Fr
om t
heabov
ev
al
ues,
wewantt
of
i
ndt
heder
i
v
at
i
v
eofy=f
ngt
hr
ough(
nt
s,
(
x)passi
n +1)poi
atapoi
ntcl
osert
ot
hest
ar
t
i
ngv
al
uex=x0
Remembert
heNewt
on’
sf
or
war
ddi
f
f
er
encei
nt
er
pol
at
i
onf
or
mul
a:
n(
n(
(
n1) 2
n1)
n2) 3
∆y
∆y
f
+n∆y
+
+
+…
(
a +nh)=y
0
0
0
0
2!
3!
Di
f
f
er
ent
i
at
i
ngwi
t
hr
espectt
on,
weget
78
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
2n12
3n26n +2 3
4n318n2 +22n64
'
y
y
hf
+
+
+
+…
∆
∆
∆y
(
a +nh)=∆y
0
0
0
0
6
2
24
Di
f
f
er
ent
i
at
i
ngagai
nw.
r
.
t
.n,weget
6n63
12n236n +22 4
'
2'
2
+
+
+…
∆y
∆y
hf(
a +nh)=∆ y
0
0
0
6
24
Di
f
f
er
ent
i
at
i
ngagai
nw.
r
.
t
.n,
24n36 4
'
'
'
3
+
+…
∆y
h3f
(
a +nh)=∆ y
0
0
24
Put
t
i
ngn=0andsi
mpl
i
f
y
i
ng
1
1
1
'
a)=∆y
hf
-∆2y
+ ∆3y
-∆4y
+…
(
0
0
0
4 0
2
3
11
5
'
'
a)=∆2y
∆3y
+ ∆4y
-∆5y
+…
h2f
(
0
0
0
6 0
12
3
'
'
'
a)=∆3y
-∆4y
+…
h3f
(
0
2 0
Si
mi
l
ar
l
y
,
wecanf
i
ndf
or
mul
aef
orhi
gheror
derder
i
v
at
i
v
es
d
y
1
1
1
1
∴
= ∆y
-∆2y
+ ∆3y
-∆4y
+…
0
0
0
4 0
2
3
dx h
{
}
d2y 1 2
11
5
= 2 ∆y
∆3y
+ ∆4y
-∆5y
+…
0
0
0
2
6 0
12
dx h
{
}
d3y 1 3 3 4
= ∆y
-∆ y
+…
0
2 0
dx3 h3
{
}
ar
et
hef
i
r
st
,
secondandt
hi
r
dder
i
v
at
i
v
e.
Exampl
e1:Fi
ndt
hef
i
r
st
,secondandt
hi
r
dder
i
v
at
i
v
esoft
hef
unct
i
ont
abul
at
edatt
hepoi
nt
x=15:
x: 15
17
19
21
23
25
y
: 3.
873 4.
123
4.
359
4.
583
4.
796
5.
000
Sol
ut
i
on:
Const
r
uctt
hedi
f
f
er
encet
abl
e
2
x
y=x
∆y
y
∆
0
0
15
∆3y
0
∆4y
0
∆5y
0
3.
873
0.
25
17
4.
123
0.
014
0.
236
19
4.
359
0.
002
0.
012
-
0.
224
21
4.
583
0.
001
0.
001
0.
011
0.
213
79
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
0.
002
0.
001
0.
002
23
4.
796
0.
009
0.
204
25
5.
000
xx0 1515
Her
e a=15,h=2,n=
=
=0
2
h
Byappl
y
i
ngNewt
on’
sf
or
war
df
or
mul
a:
dy
1
1
1
1
1
= ∆y
-∆2y
+ ∆3y
-∆4y
+ ∆2y
0
0
0
0
4
5 0
2
3
dxn=0 h
1
1
1
1
1
= 0.
0.
014)+ (
0.
002)-(0.
001)+ (
0.
002)
25-(4
5
2
3
2
=0.
129
2
11
5
1
dy
= 2∆2y
∆3y
+ ∆4y
-∆5y
0
0
0
2
6 0
12
dx n=0 h
11
5
1
0.
001)-(
0.
02) =0.
0140.
002 + (=2 0.
005
6
12
2
3
7
1
d3y
= 3∆3y
-∆4y
+ ∆5y
0
0
3
4 0
2
dx n=0 h
3
1
7
0.
001)+ (
0.
002) =0.
= 30.
002-(000875
4
2
2
[] [
]
[
[]
[]
[
[
[
[
]
]
]
]
]
Exampl
e2:Thepopul
at
i
onofacer
t
ai
nt
owni
sgi
v
enbel
ow.Fi
ndt
her
at
eofgr
owt
hoft
he
popul
at
i
oni
n1931and1941.
x
Year:
1931
1941
1951
1961
1971
Popul
at
i
ony
: 40.
62 60.
80 79.
95
103.
56
132.
65
Sol
ut
i
on:
Const
r
uctt
hedi
f
f
er
encet
abl
e
x
y
∆y
0
1931
∆2y
0
∆3y
0
∆4y
0
40.
62
20.
18
1941
60.
80
1.
03
19.
15
1951
79.
95
5.
49
4.
46
23.
61
1961
103.
56
1971
1.
02
5.
48
29.
09
132.
65
80
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
4.
47
-
Byappl
y
i
ngNewt
on’
sf
or
war
df
or
mul
a
'
Tof
i
nd(
i
)(
f1931)
xx0 19311931
wher
en=
=
=0
10
h
dy
dy
1
1
1
1
=
= ∆y
-∆2y
+ ∆3y
-∆4y
0
0
0
4 0
2
3
dxx=1931 dxn=0 h
1
1
1
1
4.
47) =2.
1.
03)+ (
5.
49)-(18-(= 20.
36425
4
2
3
10
'
'
Tof
i
nd(
i
i
)f
(
1941)
xx0 19411931
wher
en=
=
=1
10
h
dy
dy
1
1
1
1
=
= ∆y
-∆2y
+ ∆3y
-∆4y
0
0
0
4 0
2
3
dxx=1941 dxn=1 h
1
2n12
3n26n +2 3
4n318n2 +22n64
= ∆y
+
+
+
∆y
∆y
∆y
0
0
0
0
6
2
24
h
21
418 +226
1
36 +2
= 20.
((4.
47)
1.
03)+
18 +
5.
49 +
6
10
2
24
1
20.
180.
5150.
9150.
3725)=1.
= (
83775
10
[]
[] [
]
[]
[] [
[
]
[
[
( )
(
]
)
(
)
]
]
Exampl
e3:Asl
i
deri
namachi
nemov
esal
ongast
r
ai
ghtr
od.I
t
sdi
st
ancexcm al
ongt
her
odi
s
gi
v
enbel
owf
orv
ar
i
ousv
al
uesoft
het
i
metsec.Fi
ndt
hev
el
oci
t
yandaccel
er
at
i
onoft
hesl
i
der
whent=0.
3sec.
: 0
0.
1
0.
2
0.
3
0.
4
0.
5
0.
6
t
x: 30.
13
31.
62
32.
87
33.
64
33.
95 33.
81
33.
24
'
'
'
(
(
0.
3)andaccel
0.
3)
3i
3i
Sol
ut
i
on:
Vel
oci
t
yatt=0.
sf
er
at
i
onatt=0.
sf
,
wher
ex=f
(
t
)
2t
6
Weuset
hev
al
uesofxf
r
om t=0.
ot=0.
Thef
or
war
ddi
f
f
er
encet
abl
ei
s
x
∆x
t
0.
2
∆2x
∆3x
∆4x
32.
87
0.
77
0.
3
33.
64
0.
46
0.
31
0.
4
33.
95
0.
01
0.
45
-
0.
41
0.
5
33.
81
0.
01
0.
02
0.
43
0.
57
-
81
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
0.
6
33.
24
2,h=0.
1,a +t
h =0.
3
Her
e a=0.
t=1
Newt
on’
sf
or
war
di
nt
er
pol
at
i
onf
or
mul
ai
s
t
t
(
(
(
t
(
(
(
t3) 4
t1) 2
t1)
t1)
t2) 3
t2)
a)+t
a)+
a)+
a)+
(
(
(
(
(
∆f
∆f
∆f
f
∆f
(
a)
a +t
h)=f
4!
2!
3!
2
3
2
3
2
4
t
3t
+2t 3
6t
+11t
6t 4
t
t
t
a)+t
a)+ ∆2f
a)+
a)+
(
(
(
(
=f
∆f
(
a)
∆f
∆f
4!
2!
3!
Di
f
f
er
ent
i
at
i
ngw.
r
.
t
.‘’
t,
weget
2
3
2
2t12
3t
6t+2 3
4t
18t
+22t64
'
a)+
a)+
a)+
(
(
(
(
a +t
h)=∆f
hf
(
a)
∆f
∆f
∆f
6
2
24
Di
f
f
er
ent
i
at
i
ngagai
nw.
r
.
t'
,
weget
t
'
2
12t
36t+22 4
6t63
'
'
a)
a)+
a)+
(
(
(
(
a +t
h)=∆2f
∆f
∆f
h2f
6
24
Subst
i
t
ut
i
ngt
hev
al
ues
1
1
1
'
(
(
0.
1)
0.
3)=0.
0.
46)-(
0.
01)+ (
0.
01)
f
77 + (6
2
12
'
(
0.
3)=5.
34
f
3i
oci
t
ywhent=0.
s5.
34cm/
sec
∴ Vel
1
2'
'
(
(
0.
1)
0.
3)=0.
01)
f
0.
46 +0- (
12
'
'
(
0.
3)=46.
08
f
3i
46.
08cm/
er
at
i
onwhent=0.
s∴ Accel
sec2
 Newt
on’
sBackwar
dDi
f
f
er
enceFor
mul
a
Her
e,
wei
nt
endt
of
i
ndt
heder
i
v
at
i
v
eofy=f
orx=xncl
osert
ot
heendv
al
ue.
(
x)f
Di
f
f
er
ent
i
at
et
heNewt
on’
sbackwar
ddi
f
f
er
encei
nt
er
pol
at
i
onf
or
mul
ai
nasi
mi
l
arwayt
of
i
ndt
he
f
i
r
st
,
second,
t
hi
r
det
cder
i
v
at
i
v
es.
n(
(
n(
(
(
n3)4
n1)2 n(
n1)
n1)
n2)3
n2)
f
n∇y
+
+
…
∇y
∇y
∇y
(
anh)=y
0
0
0
0
0
4!
2!
3!
n 2 n33n2 +2n 3
6n3 +11n26n 4
n2n4y
y
y
=y
n
∇
+
+
…
∇
∇
∇y
0
0
0
0
0
4!
2!
3!
Di
f
f
er
ent
i
at
i
ngw.
r
.
tn,
weget
2n1 2 3n26n +2 3
4n318n2 +22n64
'
∇y
+
+
hf
∇y
∇y
∇y
(
anh)=0
0
0
0
6
2
24
Di
f
f
er
ent
i
at
i
ngagai
nw.
r
.
t
.n,
weget
12n236n +22 4
6n63
'
2
h2hf
+
…
∇y
∇y
(
anh)=∇y
0
0
0
6
24
Put
t
i
ngn=0 andsi
mpl
i
f
y
i
ng
1
1
1
'
a)=∇y
hf
+ ∇2y
+ ∇3y
+ ∇4y
+…
(
0
0
0
4 0
2
3
82
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
11
5
'
'
a)=∇2y
h2f
+∇3y
+ ∇4y
+ ∇5y
+…
(
0
0
0
6 0
12
Si
mi
l
ar
l
y
,
wecanf
i
ndf
or
mul
aef
orhi
gheror
derder
i
v
at
i
v
es.
1
1
1
1
2
3
4
'
∴ f
a)= ∇y
+ ∇y + ∇y + ∇y +…
(
h 0 2 0 3 0 4 0
{
}
11
5
1
'
'
a)= 2 ∇2y
+∇3y
+ ∇4y
+ ∇5y
+…
(
f
0
0
0
6 0
12
h
{
}
0
Exampl
e4:
Fi
ndt
hef
i
r
standsecondder
i
v
at
i
v
eoft
hef
unct
i
ont
abul
at
edbel
owatx=4.
x:
3.
0
3.
2
3.
4
3.
6
3.
8
4.
0
: 14
10.
32
5.
296
0.
256
6.
672
14
f
(
x)
Sol
ut
i
on:
Thebackwar
ddi
f
f
er
encet
abl
ei
s
2
x
y
∇y
∇4y
∇5y
y
∇
∇3y
0
0
0
0
0
3.
0
14
3.
68
3.
2
10.
32
-
1.
344
5.
024
3.
4
5.
296
-
1.
328
0.
016
5.
04
3.
6
0.
256
-
1.
872
6.
56
-
1.
888
6.
928
3.
8
3.
2
6.
672
3.
36
1.
488
-
0.
4
7.
328
4.
0
14
x0x 4.
04.
0
0,h=0.
2 and n=
Her
e a=4.
=
=0
0.
2
h
Byappl
y
i
ngNewt
on’
sbackwar
df
or
mul
a:
dy
dy
12
1
1
1
1
=
= ∇y
+ ∇y
+ ∇3y
+ ∇4y
+ ∇5y
0
0
0
0
4
5 0
2
3
dxx=4 dxn=0 h
1
1
1
1
1
6.
56) =37.
0.
4)+ (1.
488)+ (3.
36)+ (
328 + (
= 7.
52
4
5
2
3
0.
2
5
11
1
d2y
d2y
=
= 2∇2y
+∇3y
+ ∇4y
+ ∇5y
0
0
0
2
2
6 0
12
dx x=4 dx n=0 h
11
5
1
6.
56) =3.
36)+ (0.
41.
488 + (=
240.
87
2
6
12
(
0.
2)
[] [] [
[] []
[
[
]
[
]
]
]
Exampl
e5:Thepopul
at
i
onofacer
t
ai
nt
owni
sgi
v
enbel
ow.Fi
ndt
her
at
eofgr
owt
hoft
he
83
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
popul
at
i
oni
n1961and1971.
Year:
1931
1941
1951
x
Popul
at
i
ony
:40.
62 60.
80 79.
95
Sol
ut
i
on:
Const
r
uctt
hedi
f
f
er
encet
abl
e
x
y
∇y
0
1931
1961
103.
56
1971
132.
65
∇2y
0
∇4y
0
∇3y
0
40.
62
20.
18
1941
60.
80
1.
03
19.
15
1951
5.
49
79.
95
4.
46
4.
47
-
23.
61
1961
1.
02
103.
56
5.
48
29.
09
1971
132.
65
'
(
1961)
For(
i
)f
x0x 19711961
wher
e a=1971,h=10,n =
=
=1
10
h
dy
dy
1
1
1
1
=
= ∇y
+ ∇2y
+ ∇3y
+ ∇4y
0
0
0
4 0
2
3
dxx=1961 dxn=1 h
2
1
2n12
3n 6n +2 3 4n318n2 +22n64
= ∇y
y
y
∇
+
∇
∇y
0
0
0
0
6
2
24
h
1
21
418 +226
36 +2
= 29.
(4.
47)
095.
48 +
1.
026
10
2
24
1
29.
092.
740.
17 +0.
3725]=2.
= [
65525
10
'
(
1971)
For(
i
i
)f
x0x 19711971
wher
ea=1971,h=10,n =
=
=0
10
h
dy
dy
1
1
1
1
=
= ∇y
+ ∇2y
+ ∇3y
+ ∇4y
0
0
0
4 0
2
3
dxx=1971 dxn=0 h
1
1
1
1
4.
47) =3.
5.
48)+ (
1.
02)+ (09 + (
= 29.
10525
4
2
3
10
[]
[] [
[
[
[]
]
( )
) (
(
[] [
]
[
)
]
]
Exer
ci
se
1.Fi
ndt
hev
al
uesofcos(
r
om t
hef
ol
l
owi
ngt
abl
e.
1.
74)f
x:
1.
7
1.
74
1.
78
1.
82
1.
86
si
nx: 0.
9917 0.
9857 0.
9782 0.
9691
0.
9585
[
Hi
nt
:
Angl
esar
et
akeni
nr
adi
ans]
Ans:0
1684
-.
2.Fi
ndt
hef
i
r
standsecondder
i
v
at
i
v
eoft
hef
unct
i
ont
abul
at
edbel
owatx=3
84
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
]
x:
3.
0
3.
2
3.
4
x)
(
f
: 14 10.
32 5.
296
3.
6
0.
256
-
3.
8
6.
672
4.
0
14
'
'
'
(
(
3)=f
3)=18
Ans:
f
3.Ar
odi
sr
ot
at
i
ngi
napl
aneaboutoneofi
t
sends.I
ft
hef
ol
l
owi
ngt
abl
egi
v
est
heangl
eθ
r
adi
anst
hr
oughwhi
cht
her
odhast
r
av
elf
ordi
f
f
er
entv
al
uesoft
i
metseconds,f
i
ndi
t
s
7seconds.
angul
arv
el
oci
t
yandangul
araccel
er
at
i
onwhent=0.
0
0.
2
0.
4
0.
6
0.
8
1.
0
tseconds: 0.
θr
adi
ans: 0.
0 0.
12
0.
48
1.
10
2.
0
3.
20
Ans:ω =4.
50r
adi
ans/
sec
α =7.
25r
adi
an/
sec2
4.Thef
ol
l
owi
ngt
abl
egi
v
est
hev
al
uesofaf
unct
i
onatequal
i
nt
er
v
al
s
x: 0.
0
0.
5
1.
0
1.
5
2.
0
y
: 0.
3989 0.
3521 0.
2420 0.
1295 0.
0540
'
(
)
1
.
5
Ev
al
uat
ef
st
at
i
ngt
hef
or
mul
aused.
Ans:0
3911
-.
B.Numer
i
cal
I
nt
egr
at
i
on
b
∫
Thepr
ocessofcomput
i
ngt
hev
al
ueofadef
i
ni
t
ei
nt
egr
al
dxf
r
om asetofnumer
i
cal
v
al
uesof
ay
t
hei
nt
egr
andyi
scal
l
ednumer
i
cal
i
nt
egr
at
i
on.
 Tr
apezoi
dalRul
e
Supposewehav
e:
x: x0
x1
x2
x3 … xn
y
y
y
y
: y
… y
n
1
0
2
3
Fr
om t
heabov
ev
al
ues,
wewantt
of
i
ndt
hei
nt
egr
at
i
onofy=f
t
ht
her
angex0 andx0 +nh
(
x)wi
x +nh
h
y
y
+y
+y
+y
+… +y
f
(
x)
(
2(
n)+
dx= [
]
1
n1)
0
2
3
x
2
sum oft
hef
i
r
standl
astor
di
nat
es)+
h(
=
2 2(
sum oft
her
emai
ni
ngor
di
nat
es)
whi
chi
sknownasTr
apezoi
dal
r
ul
e.
∫
0
0
[
]
3 4
Exampl
e6:
Ev
al
uat
e∫
xbyusi
ngTr
apezoi
dal
r
ul
e.Ver
i
f
yt
her
esul
tbyact
ual
i
nt
egr
at
i
on.
3xd
Sol
ut
i
on
x)=x4
(
St
ep1:
Wear
egi
v
enf
85
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
(3)=6
I
nt
er
v
al
l
engt
hba=3-
6
Sowedi
v
i
de6equal
i
nt
er
v
al
s;h = =1
6
Tabul
at
et
hev
al
uesasbel
ow:
x: 3
2
1
0
1
2
3
: 81
16
1
0
1
16
81
y
St
ep2:
Wr
i
t
edownt
het
r
apezoi
dal
r
ul
eandputt
her
espect
i
v
ev
al
uesi
nt
hatr
ul
e
b
h
y
y
+y
+y
+y
+… +y
f
(
x)
(
2(
n)+
dx= [
]
1
n1)
0
2
3
a
2
1
=[
(
81 +81)+2(
16 +1 +0 +1 +6)
]
2
3 4
∫
x=115
3xd
5
243 243
3)
35 -(3 4
b
+
∫
f
(
x
)
Byact
ual
i
nt
egr
at
i
on∫
d
x=
d
x=
=
=97.
2
x
a
3
5
5
5
5
∫
[( )( )] [( ) ( )]
13
Exampl
e7:
UseTr
apezoi
dal
r
ul
et
oev
al
uat
e∫
xconsi
der
i
ngf
i
v
esubi
nt
er
v
al
s.
0xd
Sol
ut
i
on
10
Di
v
i
di
ngt
hei
nt
er
v
al(
nt
of
i
v
eequalpar
t
s,eachofwi
dt
hh =
=0.
2,t
hev
al
uesof
0,1) i
5
x)=x3ar
(
f
egi
v
enbel
ow:
x: 0
0.
2
0.
4
0.
6
0.
8
1.
0
: 0 0.
008 0.
064 0.
216 0.
512 1.
000
f
(
x)
y
y
y
y
y
y
4
5
1
0
2
3
Byt
r
apezoi
dal
r
ul
e,
wehav
e
1
h
y
y
+y
+y
+y
+y
(
2(
x3dx= [
]
5)+
4)
1
0
2
3
0
2
0.
2
(
0 +1.
000)+2(
0.
008 +0.
064 +0.
216 +0.
512)
= [
26
]=0.
2
∫
Exer
ci
se
1
1
2
1.Ev
al
uat
e∫
dxbyusi
ngTr
apezoi
dal
r
ul
ewi
t
hh =0.
0
1 +x2
Ans:
0.
783732
6 1
2.Ev
al
uat
e∫
dxbyusi
ngTr
apezoi
dal
r
ul
e
0
1 +x
Ans:
3.
69366
5.
2
ogexdxbyusi
3.Ev
al
uat
e∫
ngTr
apezoi
dal
r
ul
e
4l
Ans:
1.
83
π
s
i
n
x
4.Ev
al
uat
e∫
d
x
b
y
u
s
i
n
gT
r
a
p
e
z
o
i
d
a
l
r
u
l
e
,
b
y
d
i
v
i
d
i
n
gt
h
e
r
a
ngei
nt
ot
enequal
par
t
s.
0
Ans:
1.
9843
 Si
mpson’
sonet
hi
r
dr
ul
e
b
(
x)
Let∫
dx r
epr
esent
st
hear
eabet
weeny=f
,wi
t
hx=a and x=b.Thi
si
nt
egr
at
i
oni
s
(
x)
af
possi
bl
eonl
ywhenf
sexpl
i
ci
t
l
ygi
v
enorot
her
wi
sei
ti
snotpossi
bl
et
oev
al
uat
e.
(
x)i
b
h
y
y
y
+y
+y
+… +y
+y
+… +y
f
(
x)
(
2(
4(
n)+
dx= [
]
4
1
n1)
0
2
n2)+
3
a
3
sum oft
hef
i
r
standl
astor
di
nat
es)
(
h
= +2(
sum ofr
emai
ni
ngev
enor
di
nat
es)
3
sum ofr
emai
ni
ngoddor
di
nat
es)
+4(
∫
[
86
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
]
Theabov
eequat
i
oni
scal
l
edSi
mpson’
sonet
hi
r
dr
ul
eandi
ti
sappl
i
cabl
eonl
ywhennumberof
or
di
nat
esmustbeodd(
no.ofpai
r
s)
.
3 4
Exampl
e8:
Ev
al
uat
e∫
xbySi
mpson’
sonet
hi
r
dr
ul
e.Ver
i
f
yr
esul
tbyact
ual
i
nt
egr
at
i
on.
3xd
x)=x4;
(
(3)=6
St
ep1:
Gi
v
enf
I
nt
er
v
al
l
engt
h(
ba)=36
Sowedi
v
i
de6equal
i
nt
er
v
al
s;
h = =1
6
Tabl
et
hev
al
uesasbel
ow
x: 3
2
1
0
1
2
3
y
: 81
16
1
0
1
16
81
St
ep2:
Usi
ng
b
h
y
y
y
+y
+y
+… +y
+y
+… +y
f
(
x)
(
2(
4(
n)+
dx= [
]
4
1
n1)
0
2
n2)+
3
a
3
h
y
y
y
+y
+y
+y
+y
=[
(
2(
4(
]
4)+
5)
6)+
1
0
2
3
3
1
(
1 +1)+4(
16 +1 +16)
81 +81)+2(
=[
]=98
3
Byact
ual
i
nt
egr
at
i
on,
i
ti
s97.
2
∫
 Si
mpson’
st
hr
eeei
ght
hr
ul
e
I
ti
sgi
v
enas
3h
y +y)+2(
y +y +… +y )+3(
y +y +y +y +… +y
f
(
x)
(
dx= [
∫
8
b
a
0
n
3
6
1
n3
[
]
2
4
5
n2
+y
]
n1)
sum oft
hef
i
r
standl
astor
di
nat
es)+
(
3h
= 2(
sum ofmul
t
i
pl
esoft
hr
eeor
di
nat
es)
8
sum ofr
emai
ni
ngor
di
nat
es)
+3(
Theabov
eequat
i
oni
scal
l
edSi
mpson’
s3r
ul
ewhi
chi
sappl
i
cabl
eonl
ywhenni
smul
t
i
pl
eof3.
8
1
x usi
1 d
Exampl
e9:
Ev
al
uat
e∫
ngSi
mpson’
s3r
ul
et
aki
ngh =
0
6
8
1 +x2
Sol
ut
i
on
1
1
1
2
5
1
x: 0
6
6
3
2
3
y
: 1
0.
9730 0.
9000 0.
8000 0.
6923 0.
5902 0.
5000
y
y
y
y
y
y
y
4
5
6
1
0
2
3
BySi
mson’
s3r
ul
e,
8
1 d
x 3h
y
y
+y
+y
+y
+y
2y
+3(
= [
(
]
4
5)
6)+
1
0
2
3
01 +
x2 8
1
= [
(
1 +0.
5)+2(
0.
8)+3(
0.
9730 +0.
9 +0.
6923 +0.
5902)
]
16
=0.
7854
Si
mpson’
sRul
ei
sawei
ght
edav
er
aget
hatr
esul
t
si
nanev
enmor
eaccur
at
eappr
oxi
mat
i
on.
∫
Exer
ci
se
87
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
42
xusi
1.Est
i
mat
e∫
ngSi
mpson’
s1r
ul
eandh =4
0xd
3
1
x usi
1 d
2.Ev
al
uat
e∫
ngSi
mpson’
s1r
ul
et
aki
ngh =
0
4
3
1 +x2
x usi
6 d
3.Ev
al
uat
e∫
ngSi
mpson’
s3r
ul
et
aki
ng
0
2
8
1 +x
88
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
Ans:
21.
333
Ans:
0.
7854
Ans:
1.
3571
MATH309TUTORI
ALQUESTI
ONS
2013/
2014SESSI
ON
1.Leta=0.
471×10-2 andb= –0.
185×10-4.Uset
hr
eedi
gi
t
sf
l
oat
i
ngpoi
nt
ar
i
t
hmet
i
c
t
ocomput
ea+b,a–b,abanda/
b.Fi
ndt
her
oundi
nger
r
ori
neachcase.
4
Fur
t
her
,
i
fc=0.
869×10,
showt
hatt
hecomput
edv
al
ueofa+bi
sequal
t
oc.
2 Wr
i
t
et
heNewt
on’
spr
ocedur
ef
orf
i
ndi
ng
,
wher
eNi
sr
eal
number
.Usei
tt
o
f
i
nd
cor
r
ectt
o2deci
mal
pl
aces.
3 I
ff
(
)andf
(
)ar
eofopposi
t
esi
gns,
t
henf
(
x)=0hasatl
eastoner
ootbet
ween
andpr
ov
i
dedonl
yi
f
………………………………
4 Expl
ai
nhowt
hef
al
seposi
t
i
onorr
egul
ar
f
al
si
scheme
i
sder
i
v
edandhenceusei
tt
of
i
ndt
hef
our
t
hr
ootof32cor
r
ectt
ot
hr
eedeci
mal
pl
aces.
5 Uset
hei
t
er
at
i
v
eorf
i
xedpoi
ntmet
hodt
osol
v
et
heequat
i
onx=exp(
–x)
,
st
ar
t
i
ng
wi
t
h
x=1.
00.Per
f
or
my
ouri
t
er
at
i
ont
aki
ngt
her
eadi
ngupt
of
ourdeci
mal
pl
aces
3
2
6 Fi
ndt
her
eal
r
ootoft
heequat
i
onx +x –1=0byt
hei
t
er
at
i
v
emet
hodi
nt
he
i
nt
er
v
al
[
0.
7,
0.
8]i
nf
i
v
ei
t
er
at
i
ons.
7 Whati
st
hebi
sect
i
onmet
hodf
orf
i
ndi
ngt
her
oot
sofanequat
i
onf
(
x)=0?
8 Whati
st
her
egul
arf
al
si
emet
hodf
orf
i
ndi
ngt
her
ootoff
(
x)=0andi
nwhatway
i
si
tbet
t
ert
hant
hebi
sect
i
onmet
hod?
9 Whatdoy
ouunder
st
andbyi
nt
er
pol
at
i
on?
10Dot
heJacobi
andGaussSei
del
met
hodsbot
hconv
er
gef
orasy
st
em ofl
i
near
equat
i
ons?Expl
ai
ncl
ear
l
y
.
11Fr
om t
hef
ol
l
owi
ngt
abl
e,
est
i
mat
et
henumberofst
udent
swhoobt
ai
nedmar
ks
bet
ween40and50.
Mar
ks
30–40 40–50 50–60 60–70
70-80
No.of
St
udent
31
42
51
35
31
(
Hi
nt
:
For
m acumul
at
i
v
ef
r
equencyt
abl
ebef
or
econst
r
uct
i
ngt
hedi
f
f
er
encet
abl
e
wi
t
h
x0=40,
x=45andh=10)
12UseTay
l
or
’
sser
i
esorot
her
wi
seder
i
v
et
heNewt
on-Raphsonmet
hodi
t
er
at
i
v
e
89
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
schemeandusei
tt
of
i
ndt
her
ealr
ootoft
heequat
i
onx4 +x3 –7x2 –x+5=
0whi
chl
i
esbet
ween2and3cor
r
ectt
ot
hr
eedeci
mal
pl
aces.
13I
nt
het
abl
ebel
ow,
t
hev
al
uesofyar
econsecut
i
v
et
er
msofaser
i
esofwhi
ch23.
6
t
h
i
st
he6 t
er
m.Fi
ndt
hef
i
r
sandt
ent
ht
er
msoft
heser
i
es.
x
3
4
5
6
7
8
9
y
4.
8
8.
4 14.
5 23.
6 36.
2 52.
8 73.
9
(
Hi
nt
:For1stt
er
m,useNewt
onf
or
war
di
nt
er
pol
at
i
onandNewt
onbackwar
df
or
t
h
t
he10 t
er
m)
14Assumi
ngani
ni
t
i
albr
acket[
1,5]
,f
i
ndt
hesecond(
att
heendof2i
t
er
at
i
ons)
–t
i
t
er
at
i
v
ev
al
ueoft
her
ootoft
e –0.
3=0usi
ngt
hebi
sect
i
onmet
hod.
15Whati
st
henexti
t
er
at
i
v
ev
al
ueoft
her
ootofx2 – 4=0usi
ngt
heNewt
onRaphsonmet
hodi
ft
hei
ni
t
i
al
guessi
s3?
16Whati
s(
ar
e)t
henumberofr
oot
(
s)oft
heequat
i
on3x3 –12x2 +8x–1i
nt
he
r
ange0x5?
17Whati
st
heor
derofconv
er
gencei
nNewt
onRaphsonmet
hod?
18Di
scusst
hemet
hodofGaussSei
del
i
t
er
at
i
v
emet
hod.Henceorot
her
wi
sesol
v
e
t
hef
ol
l
owi
ngsy
st
em:
19Li
stal
l
t
hebr
acket
i
ngmet
hody
ouknowanddi
scussext
ensi
v
el
yonanyoneof
t
hem.
20Byusi
ngbi
sect
i
onmet
hod,
f
i
ndanappr
ox
i
mat
er
ootoft
heequat
i
onsi
n(
x)=1/
x,
t
hatl
i
esbet
weenx=1andx=1.
5(
measur
edi
nr
adi
ans)
.Car
r
youtcomput
at
i
ons
upt
ot
he7thst
age.
21Uset
hemet
hodoff
al
seposi
t
i
ont
of
i
ndt
hef
our
t
hr
ootof32cor
r
ectt
ot
hr
ee
deci
mal
pl
aces.
22Appl
yGaussSei
deli
t
er
at
i
onmet
hodt
ot
hef
ol
l
owi
ngsy
st
em ofl
i
nearequat
i
ons
i
nsi
xi
t
er
at
i
ons.
23Sol
v
et
hesy
st
em
4x+y+3z=17
x+5y+z=14
2x–y+8z=12
byJacobi
’
si
t
er
at
i
v
emet
hod.
act
ual
v
al
ues
Car
r
yy
ourcomput
at
i
onst
o6i
t
er
at
i
ons.(
The
90
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
ar
ex=3,
y=2andz=1)
24Consi
dert
hesy
st
em
x+3y–z=6
4x–y+z=5
x+y–7z=–9
(
a)Canweuseei
t
herofJacobi
orGaussSei
del
met
hodst
of
i
ndt
heappr
oxi
mat
e
sol
ut
i
onst
ot
hesy
st
em?
(
b)Sol
v
et
hesy
st
em usi
ngbot
hi
t
er
at
i
v
eschemes.Car
r
youty
ourcomput
at
i
ons
t
h
upt
ot
he8 i
t
er
at
i
ons.
91
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
UNVERSI
TYOFMAI
DUGURI
(
Facul
t
yofSci
ence)
Depar
t
mentofMat
hemat
i
csandSt
at
i
st
i
cs
MATH309
NUMERI
CALANALYSI
S
2UNI
TS
2013/
2014SESSI
ON
I
NSTRUCTI
ON:Answeranyf
i
v
e(
5)Quest
i
onsi
nTwoandahal
f(
2½)hour
s
1.(
i
) Der
i
v
et
he Newt
onRaphson’
si
t
er
at
i
v
e scheme usi
ng Tay
l
or
’
s ser
i
es or
ot
her
wi
se
3
(
i
i
) Wr
i
t
et
heNewt
on’
spr
ocedur
ef
orf
i
ndi
ng N,wher
eNi
sr
eal
number
3
(
i
i
i
) Uset
her
esul
ti
n(
i
i
)t
of
i
nd 18 cor
r
ectt
o2deci
mal
pl
aces
2.Expl
ai
nhowt
hef
al
seposi
t
i
onorr
egul
af
al
si
scheme
x0)
x1)(
(
f
f
x0)=
xx0)
yf
(
(
x1x0
i
sder
i
v
edandhenceusei
tt
of
i
ndt
hef
our
t
hr
ootof32cor
r
ectt
ot
hr
eedeci
mal
pl
aces.
3.(
a)Li
stal
lt
hebr
acket
i
ngmet
hody
ouknowanddi
scussext
ensi
v
el
yonanyone
oft
hem.
1 =0 byt
(
b)Fi
ndt
her
ealr
ootoft
heequat
i
on x3 +x2hei
t
er
at
i
v
emet
hodi
nt
he
i
nt
er
v
al[
nf
i
v
ei
t
er
at
i
ons.
0.
7,0.
8]i
4.Fr
om t
hef
ol
l
owi
ngt
abl
e,est
i
mat
et
henumberofst
udent
swhoobt
ai
nedmar
ks
bet
ween40and50.
92
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
Mar
ks
3040
4050
5060
6070
7080
No.ofSt
udent
31
42
51
35
31
(
Hi
nt
:Fr
om acumul
at
i
v
ef
r
equencyt
abl
ebef
or
econst
r
uct
i
ngt
hedi
f
f
er
encet
abl
e
wi
t
h x0 =40,x=45 and h =10)
5.I
nt
het
abl
ebel
ow,t
hev
al
uesof y ar
econsecut
i
v
et
er
msofaser
i
esofwhi
ch
t
h
23.
6i
st
he6 t
er
m.Fi
ndt
hef
i
r
standt
ent
ht
er
msoft
heser
i
es.
x
3
4
y
4.
8
8.
4
5
6
7
8
9
14.
5 23.
6 36.
2 52.
8 73.
9
(
Hi
nt
:For1stt
er
m,useNewt
onf
or
war
di
nt
er
pol
at
i
onandNewt
onbackwar
df
or
t
h
t
he10 t
er
m)
6.Appl
yGaussSei
deli
t
er
at
i
onmet
hodt
ot
hef
ol
l
owi
ngsy
st
em ofl
i
nearequat
i
ons
i
nsi
xi
t
er
at
i
ons.
10x12x2x3x4 =3
x3x4 =15
2x1 +10x2x1x2 +10x32x4 =27
x1x22x3 +10x4 =9
7.Sol
v
et
hesy
st
em
4x+y+3z=17
x+5y+z=14
2xy+8z=12
93
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
byJacobi
’
si
t
er
at
i
v
emet
hod.Car
r
yy
ourcomput
at
i
onst
o6i
t
er
at
i
ons.
GOODLUCK
94
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
UNVERSI
TYOFMAI
DUGURI
FACULTYOFSCI
ENCE
DEPARTMENTOFMATHEMATI
CSANDSTATI
STI
CS
FI
RSTSEMESTERENDOFCOURSEEXAMI
NATI
ON2014/
2015SESSI
ON
MTH309:NUMERI
CALANALYSI
S
Ti
me:2.
30hour
s
I
nst
r
uct
i
ons:
Answeranyf
our(
4)quest
i
ons
1.(
a)Ev
al
uat
et
hef
ol
l
owi
ng,
t
hei
nt
er
v
al
ofdi
f
f
er
enci
ngbei
nguni
t
y
()
∆2 2
x
(
i
i
)
E
(
i
)∆l
ogx
ex
((x+1)!)
(
i
i
i
)∆
Ux Vx∆UxUx∆Vx
=
Vx
VxVx+1
()
(
b)Pr
ov
et
hat∆
2.(
a)Di
st
i
ngui
shbet
weeni
nt
er
pol
at
i
onandext
r
apol
at
i
on
(
b)Fr
om t
hegi
v
ent
abl
e,
comput
et
hev
al
ueofsi
n380
x0
0
10
20
30
40
si
nx0
0
0.
17365
0.
34202
0.
50000
0.
64279
3.(
a)Usi
ngNewt
on’
sdi
v
i
dedf
or
mul
a,
ev
al
uat
ef
,
gi
v
en
(
x)and f
(
15)
X
4
5
7
10
11
13
Y
48
100
294
920
1210
2028
(
(
5)=120,
3)=24,f
(
7)i
(
b)UseNewt
on’
sdi
v
i
deddi
f
f
er
encef
or
mul
at
of
i
nd f
ff
(
(
(
8)=524,f
9)=720 and f
12)=1716
f
x11 =0,cor
4.(
a)Fi
ndt
her
ootoft
heequat
i
on x3r
ectt
o4deci
mal
susi
ng
bi
sect
i
onmet
hod
95
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
1
1 1
(
b)Ev
al
uat
e∫
usi
ngSi
mpson’
s3r
ul
et
aki
ng h=
0
6
8
1 +x2
5.Fi
ndar
ootoft
hef
ol
l
owi
ngbyf
al
seposi
t
i
onmet
hod;
(
i
) x=cosx
(
i
i
)xl
og10x =1.
2
6.Sol
v
et
hef
ol
l
owi
ngequat
i
onsbyGaussJor
danmet
hod
3x+4yz=8
2x+y+z=3
x+2yz=2
96
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
UNI
VERSI
TYOFMAI
DUGURI
(
Facul
t
yofSci
ence)
Depar
t
mentofMat
hemat
i
csandSt
at
i
st
i
cs
MTH309 NUMERI
CALMETHODS 2UNI
TS2015/
2016SESSI
ON TI
ME:2HRS
I
NSTRUCTI
ON:
AnswerQUESTI
ON1(
One)andAnyThr
ee(
3)Quest
i
ons
1.(
a)Def
i
neapol
y
nomi
al
andl
i
stsomemet
hodsofpol
y
nomi
al
appr
oxi
mat
i
ons.
(
b)Whatar
et
headv
ant
agesanddr
awbacksofusi
ngt
hei
nt
er
v
al
hal
v
i
ngmet
hod?
(
c)Tof
i
ndt
hei
nv
er
seofa,
onecanuset
heequat
i
on
1
c)=a- =0
(
f
c
wher
ec i
st
hei
nv
er
se a.Uset
hebi
sect
i
onmet
hodoff
i
ndi
ngt
her
oot
sof
5.Conductt
equat
i
onst
of
i
ndt
hei
nv
er
seofa=2.
hr
eei
t
er
at
i
onst
oest
i
mat
e
t
her
ootoft
heabov
eequat
i
on.Fi
ndt
heabsol
ut
er
el
at
i
v
eappr
oxi
mat
eer
r
or
att
heendofeachi
t
er
at
i
on.
2.Di
scussext
ensi
v
el
yoneachoft
hef
ol
l
owi
ng,
gi
v
i
ngexampl
ei
neachcase.
(
i
)Roundof
fer
r
or
(
i
i
)Tr
uncat
i
oner
r
or
x)=x3(
3.Ar
ealr
ootoft
heequat
i
on f
5x+1 =0 l
i
esi
nt
hei
nt
er
v
al(
0,
1)
.Per
f
or
m
f
ouri
t
er
at
i
onsoft
heSecantmet
hodandRegul
aFal
si
met
hodt
oobt
ai
nt
hi
sr
oot
4.Der
i
v
et
heNewt
on’
smet
hodofi
t
er
at
i
onbyTay
l
or
’
smet
hodorot
her
wi
seand
6 asi
hencef
i
ndt
her
ealr
oott
heequat
i
on 3x=cosx +1.Take x0 =0.
ni
t
i
al
guess
5.Show,usi
ngNewt
on’
smet
hodt
hatt
hei
t
er
at
i
v
ef
or
mul
af
orf
i
ndi
ngt
hesquar
e
r
ootofNi
s
1
nd
xn +1 = xn +N .Hencefi
x
2
n
(
)
28
6.(
a)Whyi
sGaussSei
del
met
hodi
spr
ef
er
r
edov
ert
heJacobi
met
hod?
97
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
2x+y+6z=9
(
b)Sol
v
ebyGaussSei
del
met
hod
8x+3y+2z=13
x+5y+z=7
7.I
nt
het
abl
ebel
ow,t
hev
al
uesofy ar
econsecut
i
v
et
er
msofaser
i
esofwhi
ch
t
h
23.
6i
st
he6 t
er
m.Fi
ndt
hef
i
r
standt
het
ent
hoft
heser
i
es.
x
3
y
4.
8
4
5
6
7
8.
4 14.
5 23.
6 36.
2
98
Numer
i
cal
Anal
y
si
sMat
hemat
i
cs&Engi
neer
i
ng
8
9
52.
8
73.
9
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