3.5 Spline Interpolation 87
b.Repeat (a)with theHermite interpolating polynomial ofdegree atmost 5,using x0=1,
=1.05 ,andx2=1.07.
5. Usetheerror formula andMATLAB tofindabound fortheerrors intheapproximations of/(x)in
(a)and(c)ofExercise 2.
6. Thefollowing table listsdata forthefunction described by/(x)=e01*2.Approximate /(1.25)by
using //5(1.25 )and//3(1.25 ),where uses thenodes Xo=1,xj=2,andx2=3and uses the
nodes x0=1and X!=1.5.Find error bounds forthese approximations .
X f i x)=e01*2f i x )=0.2xe01*2
x0=x0=11.105170918 0.2210341836
X!=1.5 1.252322716 0.3756968148
*1=2 1.491824698 0.5967298792
x2=3 2.459603111 1.475761867
7. Acartraveling along astraight road isclocked atanumber ofpoints .Thedatafrom theobservations
aregiven inthefollowing table ,where thetime isinseconds ,thedistance isinfeet,andthespeed is
infeetpersecond .
Time 0 3 5 8 13
Distance 0225 383 623 993
Speed 75 77 80 74 72
a.UseaHermite polynomial topredict theposition ofthecaranditsspeed when t=10s.
b.Usethederivative oftheHermite polynomial todetermine whether thecareverexceeds a55-mi/h
speed limit ontheroad.Ifso,what isthefirsttime thecarexceeds thisspeed ?
c.What isthepredicted maximum speed forthecar?
8. Letzo=x0.Z\=x0,Z2=X|,andz3=xi.Form thefollowing divided -difference table.
Zo=*o/[zo]=/(*o)
/[zo.Zt]=/'(*o)
Z i=X0/lZ|]=/(*o) /[Zo,Zi,Z2]
/[ZI.Z2] /[Zo,Zl,Z2,Z3]
Z2=*l/[Z2]=/(*l) /[ZI.Z2.Z3]
/[Z2.Z3]=/'(*l)
Z3=*l/[Z3]=/(*l)
Show if
P(x)=f[z0)+/[z0.Zi](x-x0)+/[z0,Z|,z2](*-xo)2+/[z0,zi,z2,z3](x-x0)2(x-x,).
then
P(xo)=fixo ),PM=/(X,).P'ixo)=f\xo),andP'(x.)=/'(*.).
which implies thatP(x)=H3(x).
3.5 Spline Interpolation
The previous sections usepolynomials toapproximate arbitrary functions .However ,rela-
tively high-degree polynomials areneeded foraccurate approximation andthese have some
serious disadvantages .They canhave anoscillatory nature ,andafluctuation over asmall
portion oftheinterval caninduce large fluctuations over theentire range .Wewillseean
example ofthislater inthissection .
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Analternative approach istodivide theinterval into acollection ofsubintervals and
construct adifferent approximating polynomial oneach subinterval .This iscalled piecewise
polynomial approximation .
Piecewise -Polynomial Approximation
The simplest piecewise polynomial approximation joins thedata points (xo,/(xo)),
(*i,f(x\)),...,(x„ ,/(*„ ))byaseries ofstraight lines ,such asthose shown inFigure 3.7.
Adisadvantage oflinear approximation isthattheapproximation isgenerally notdif-
ferentiable attheendpoints ofthesubintervals ,sotheinterpolating function isnot“ smooth ” 
atthese points.Itisoften clear from physical conditions thatsmoothness isrequired ,and
theapproximating function must becontinuously differentiable .
Figure 3.7
y
m V
H 1 1 1 1 1 1 1
*0 *2    X j Xj+i X j+2 ... A
Isaac Jacob Schoenberg
(1903-1990 )developed hiswork
onsplines during World WarIIat
theArmy ’sBallistic Research
Laboratory inAberdeen ,
Maryland ,while onleave from
theUniversity ofPennsylvania .
Hisoriginal work involved
numerical procedures forsolving
differential equations .The much
broader application ofsplines to
theareas ofdata fitting and
computer -aided geometric design
became evident with the
widespread availability of
computers inthe1960 s.One remedy forthisproblem istouseapiecewise polynomial ofHermite type.For
example ,ifthevalues of/and/'areknown ateach ofthepoints xo<x\<   <x„ ,
acubic Hermite polynomial canbeused oneach ofthesubintervals [xo,xj],[x j,X2],...,
,xn]toobtain anapproximating function thathasacontinuous derivative ontheinterval
[xo,xn].Todetermine theappropriate cubic Hermite polynomial onagiven interval ,we
simply compute thefunction H$(x)forthatinterval .
The Hermite polynomials arecommonly used inapplication problems tostudy the
motion ofparticles inspace.Thedifficulty with using Hermite piecewise polynomials for
general interpolation problems concerns theneed toknow thederivative ofthefunction being
approximated .The remainder ofthissection considers approximations using piecewise
polynomials thatrequire noderivative information ,except perhaps attheendpoints ofthe
interval onwhich thefunction isbeing approximated .
Cubic Splines
The most common piecewise -polynomial approximation uses cubic polynomials between
pairs ofnodes andiscalled cubic spline interpolation .Ageneral cubic polynomial involves
four constants ,sothere issufficient flexibility inthecubic spline procedure toensure that
theinterpolant hastwocontinuous derivatives ontheinterval .Thederivatives ofthecubic
spline donot,ingeneral ,however ,agree with thederivatives ofthefunction ,even atthe
nodes.(SeeFigure 3.8.)
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Figure 3.8
S/,xj+1)~/(*>+1)— Sj+faj+i)
SXxJ +])= iC*,+i)
H 1 1 1 1 1 1—I—I
*0 X2    Xj Xj+\ Xj+2    xn-2Xn-1XnX
Cubic Spline Interpolation
Given afunction /defined on [a,b)andasetofnodes ,a=xo<x\<    <xn=b,
acubic spline interpolant ,S,for/isafunction thatsatisfies thefollowing conditions :
(a)Foreach j=0,1,...,n— 1,S(;t)isacubic polynomial ,denoted byS j(x),
onthesubinterval [x j,X j+j).
(b)S j(X j)=f(x j)andS j(x j+1)=f(x j+1)foreach j=0,1,...,n-1.
(c)S,+i(*j+i)=S;(*,+i)foreach j=0,1,. . .,n-2(Implied by(b).)
(d)S'j+l(Xj+1)=S j(X j+j)foreach j=0,1,. . .,n-2.
(e)SJ+1(x,+i)=SJ(*y+i)foreach;=0,1 n-2.
(f)One ofthefollowing setsofboundary conditions issatisfied :
(i)S"(*o)=$"(*„ )=0(natural orfree boundary );
(ii)S'(xo)=f(x0)andS'(x„ )=f\xn)(clamped boundary ).Six)
SI
So2
s
S
SjXxj +\)— Sj+x(Xj+,)
T h e root oftheword “ spline ” is
thesame asthatofsplint .Itwas
originally asmall strip ofwood
thatcould beused tojoin two
boards .Later theword wasused
torefer toalong flexible strip,
generally ofmetal,thatcould be
used todraw continuous smooth
curves byforcing thestrip topass
through specified points and
tracing along thecurve .
Example 1Although cubic splines aredefined with other boundary conditions ,theconditions
given in(0aresufficient forourpurposes .When thenatural boundary conditions areused ,
thespline assumes theshape thatalong flexible rodwould take ifforced togothrough
thepoints {(*o,/(*o)).(*i./C*i)),.. .,(x„ ,/(*„ ))}.This spline continues linearly when
x<XQandwhen x>xn.
Construct anatural cubic spline thatpasses through thepoints (1,2),(2,3),a n d(3,5).
Solution This spline consists oftwocubics .The firstfortheinterval (1,2),denoted
So(*)=ao+bo(x-1)+cQ(x-l)2-fdQ(x-1)\
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Anatural spline hasnoconditions
imposed forthedirection atits
endpoints ,sothecurve lakes the
shape ofastraight lineafter it
passes through theinterpolation
points nearest itsendpoints .The
name derives from thefactthat
thisisthenatural shape aflexible
strip assumes ifforced topass
through specified interpolation
points with noadditional
constraints .(SeeFigure 3.9.)and theother for[2,3],denoted
s,(*)=a,+M*-2)+ci(x-2)2+</,(*-2)\
There are8constants tobedetermined ,which requires 8conditions .Four conditions come
from thefactthatthesplines must agree with thedata atthenodes .Hence
2=/(1)=a0,3=/(2)=oo+*>0+0)+do,3=/(2)=au and
5=/(3)=a\+b\+C]4-d\.
Two more come from thefactthat5Q(2)=5J(2)and SQ(2)=SJ'(2).These are
So(2)=51(2):bo+2c0+3do=bx and S£(2)=$;'(2):2c0+6d0=2c,.
Thefinal twocome from thenatural boundary conditions :
$o(l)=0:2c0=0 and $;'(3)=0:2c\+6d\=0.
Figure 3.9 Solving thissystem ofequations gives thespline
S(x)2+?(*-l)+j(*-l)\for*e[1,2]4 4
3+\{x2)+|(JC-2)2-l-(x-2)\for*[2,3],2 4 4
Construction ofaCubic Spline
Clamping aspline indicates that
theends oftheflexible stripare
fixed sothatitisforced totakea
specific direction ateach ofits
endpoints .This isimportant ,for
example ,when twospline
functions should match attheir
endpoints .This isdone
mathematically byspecifying the
values ofthederivative ofthe
curve attheendpoints ofthe
spline .Ingeneral ,clamped boundary conditions leadtomore accurate approximations because they
include more information about thefunction .However ,forthistype ofboundary condition ,
weneed values ofthederivative attheendpoints oranaccurate approximation tothose
values .
Toconstruct thecubic spline interpolant foragiven function /,theconditions inthe
definition areapplied tothecubic polynomials
S j(x)=a}+b j(x-X j)+C j(x-x j)2+d j(x-x j)3
foreach j=0,1,...,n— I.
Since
S j l x j )=d j=f(x j),
condition (c)canbeapplied toobtain
tfy+i=Sj+ilxj+i)=S j(xj+1)=aj+bj(xj+1-x j)+C j(x j+1-X j)2+djlxj +i-X j)3
foreach j=0,1,...,n— 2.
Since theterm xj+j-xjisused repeatedly inthisdevelopment ,itisconvenient to
introduce thesimpler notation
hi=xi+1“ xb
foreach j=0,1,.. ..n— 1.Ifw ealsodefine a n=/(*„ ),then theequation
a j+i=d j+b j h j+C j h2+d j h j (3.1)
holds foreach j=0,1,...,n— 1.
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Inasimilar manner ,define bn=S'(xn)andobserve that
S'W=b j+2C j(x-x j)+3d j(x-X j)2(3.2)
implies thatS'(*y)=bjforeach j=0,1,...,n— I.Applying condition (d)gives
b j+1=b j4-2C j h j 4-3d j h2
j, (3.3)
foreach;=0,1,...,«-1.
Another relation between thecoefficients ofSjisobtained bydefining cn=S"(xn)/2
andapplying condition (e).Inthiscase,
C j+1=C j4-3d j h j t (3.4)
foreach j=0,1,...,n—1.
Solving fordjinEq.(3.4)andsubstituting thisvalue intoEqs.(3.1)and (3.3)gives
thenewequations
andh2:
aj+i=Q j+b j h j+-y-(2C j+cj+i) (3.5)
b j+i=b j+h j(c j4-C j+j) (3.6)
foreach j=0,1,...,n— 1.
Thefinal relationship involving thecoefficients isobtained bysolving theappropriate
equation intheform ofEq.(3.5)forbj,
1 hbj=— (fly+1-fly)-~/(2c;H-cy+i), (3.7)
h j 3
andthen,with areduction oftheindex ,forbj-uwhich gives
bj-1=7 (fly~Oj-\)4^(2Cy_
i+Cy).
";-l J
Substituting these values intotheequation derived from Eq.(3.6),when theindex isreduced
by1,gives thelinear system ofequations
3 3hj-\Cj-\4-2(/»y_
i4-hj)cj4-/»yCy+i=— (ay+i-ay)--— (ay-ay_
i) ( 3.8)
hj hj-\
foreach j=1,2,...,/i—1.This system involves only {cy}” =0asunknowns since the
values of{hj}*!,$and {ay}'*=0aregiven bythespacing ofthenodes {*y}y«oandthevalues
o-Once thevalues of{cy}"=0aredetermined ,itisasimple matter tofindtheremainder
oftheconstants {bj)*Z^from Eq.(3.7)and {dy)"~
Qfrom Eq.(3.4)andtoconstruct the
cubic polynomials {Sy(*)}yli.Inthecase oftheclamped spline ,wealso need equations
involving the{cy}thatensure thatS'(JCO)=/'(*o)andS'(xn)=/'(*„ ).InEq.(3.2)we
have S'j(x)interms ofbj,cy,anddj.Since wenow know bjanddjinterms ofcy,wecan
usethisequation toshow thattheappropriate equations are
32h0c04-h0ci=— (a,-oo)-3f'(x0) (3.9)
ho
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Program NCUBSP 34
creates aNatural Cubic
Spline
Example 2
Program CCUBSP 35
creates aClamped Cubic
Splineand
3hn-\cn-\42hn^c„=3/'(*„ )-7— ifln~an-1). (3.10 )
hn-\
Thesolution tothecubic spline problem with thenatural boundary conditions S"(*o)=
S"(xn)=0canbeobtained byapplying theprogram NCUBSP 34.Theprogram CCUBSP 35
determines thecubic spline with theclamped boundary conditions S'(xo)=/'(JCO)and
S'(xn)=/'(*„ ).
Determine theclamped cubic spline forf(x)=xsin4xusing thenodes XQ=0,JCJ=0.25,
x2=0.4,and*3=0.6.
Solution Wefirstdefine thefunction /(x)anditsderivative f p(x)s/'(x)inMATLAB
with
f=inline(’x*sin(4*x)*,*x*)
fp=inline(,sin(4*x)+4*x*cos(4*x)J,’x’)
Wedefine thenodes using theMATLAB capability ofdefining subscripted variables within
square brackets ,where ablank isused toseparate entries .Thesubscripts inMATLAB begin
with 1so JC(1)=0,x(2)=0.25,x(3)=0.4,andx(4)=0.6.This isentered inMATLAB
as
x=[00.250.40.6]
Thestepsizes arcdefined by
h=Cx(2)-x(l)x(3)-x(2)x(4)-x(3)]
andthevalues ofthefunction atthenodes by
a=[f(x(l))f(x(2))f(x(3))f(x(4))]
MATLAB responds tothislastcommand with
a=00.210367746201974 0.399829441216602 0.405277908330691
A4x4array Aisdefined whose rows aredefined bythesystem ofequations used to
determine thequadratic coefficients ,thatis,thec’s.These aregiven inEqs.(3.8),(3.9),
and(3.10 ),butEq.(3.9)involves anindex of0,which MATLAB does notpermit .Sothe
indices must allbeincreased by1tocompensate .SoAisdefined asfollows :
Row 1:theleft-hand side ofEq.(3.9),with alltheindices increased by1;thatis:
2h\C\4h\c240 C340 C4.
Row 2:theleft-hand sideofEq.(3.8)when j=2:
h\C\42{h\4h2)c2-fh2 C340 c4.
Row 3:theleft-hand side ofEq.(3.8)when j=3:
0 c\4h2c242{h2«f/*3)^34/23^4-
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Row 4:theleft-hand sideofEq.(3.10)when n=4:
0 Ci+0  C2 +h3c3-f-2/13C4.
This gives
A=[2*h(l)h(l)00;h(l)2*(h(l)+h(2))h(2)0;0h(2)
2*(h(2)+h(3) ) h(3);0 0 h(3)2*h(3)]
The right -hand sides ofthesame equations arestored inthe4x1array B.
Row 1:theright -hand sideofEq.(3.9),with alltheindices increased by1;thatis:
^-(a2-ai)-3/'(*i).
Row 2:theright -hand sideofEq.(3.8)when j=2:
3 3— (a)-a2)~-(a2-at).h2 hi
Row 3:theright -hand sideofEq.(3.8)when j=3:
3 3r-(a4-as)-r-(a3-a2).«3 n2
Row 4:theright -hand sideofEq.(3.10)when n=4:
SoBisdefined by
B=[3*(a(2)-a(l))/h(l)-3*fp(x(l));
3*(a(3)-a(2))/h(2)-3*(a(2)-a(l))/h(l);
3*(a(4)-a(3))/h(3)-3*(a(3)-a(2) )/h(2);
3*fp(x(4))-3*(a(4)-a(3))/h(3)]
MATLAB gives these as
A=1:1:111101010 :111:1:1 0.5(
0.250000000000000
0
00.250000000000000
0.800000000000000
0.150000000000000
00
0.150000000000000
0.700000000000000
0.2000000000000000
0
1:1in 0.2(
0.400000000000000
and
2.524412954423689’
1.264820945868869“ -3.707506893581232
_-3.364572216954841 _
Wenow canhave MATLAB solve thesystem using thelinsolve command .
c=linsolve (A,B)
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Table 3.13
IllustrationMATLAB responds with solution consisting ofc(l),c(2),c(3),andc(4)as
4.649673230468573'
0.798305356757612
C“ -3.574944314362422_-6.623958385205892
Now useEq.(3.7)toobtain thevalues of6(1),6(2),and6(3)with thecommand
b=[(a(2)-a(l) )/h(l)-h(l)*(2*c(l)+c(2) )/3;
(a(3)-a(2) )/h(2)-h(2)*(2*c(2)+c(3) )/3;
(a(4)-a(3) )/h(3)-h(3)*(2*c(3)+c(4))/3]
which MATLAB gives as
6=0.000000000000000
1.361994646806546
0.945498803165825
Finally ,thevalues ofd(l),d(2),andd(3)areobtained using Eq.(3.4)andthecommand
d-[(c(2)-c(l))/(3*h(l));(c(3)-c(2))/(3*h(2));
(c(4)-c(3) )/(3*h(3))]
producing
d=-5.135157164947948-9.718332602488962-5.081690118072451
This implies thatthecubic spline ,tothree decimal places ,isasshown inTable 3.13.
j o j b j C j d j
0 0.000 0.000 0.000 4.650 -5.135
1 0.250 0.210 1.362 0.798 -9.718
2 0.400 0.400 0.945 -3.575 -5.082
3 0.600 0.405 -6.624
Inthefollowing Illustration theshape ofthecurve ismuch more complex .Placing a
minimal number ofdata points along thecurve togetagood representation would require
some experimentation .
Figure 3.10 shows aruddy duck inflight.Toapproximate thetopprofile oftheduck ,we
have chosen points along thecurve through which wewant theapproximating curve topass.
Table 3.14 lists thecoordinates of21data points relative tothesuperimposed coordinate
system shown inFigure 3.11.Notice thatmore points areused when thecurve ischanging
rapidly than when itischanging more slowly .
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Figure 3.10
c)-VJ
Figure 3.11
fix)n
J*
<to
*: 5'-^101II213
r
j
.FW
Table 3.14
X 0.9 1.3 1.9 2.1 2.6 3.0 3.9 4.4 4.7 5.0 6.0 7.0 8.0 9.2 10.5 11.3 11.6 12.0 12.6 13.0 13.3
f i x)1.3 1.5 1.85 2.1 2.6 2.7 2.4 2.15 2.05 2.1 2.25 2.3 2.25 1.95 1.4 0.9 0.7 0.6 0.5 0.4 0.25
Using theprogram NCUBSP 34togenerate thenatural cubic spline forthisdata pro-
duces thecoefficients shown inTable 3.15.This spline curve isnearly identical totheprofile ,
asshown inFigure 3.12.
Forcomparison purposes ,Figure 3.13 gives anillustration ofthecurve thatisgen-
erated using aLagrange interpolating polynomial tofitthedata given inTable 3.14.The
interpolating polynomial inthiscase isofdegree 20andoscillates wildly .Itproduces a
very strange illustration oftheback ofaduck ,inflight orotherwise .
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Table 3.15 Figure 3.12
jX J a j b j C J d j
00.91.3 5.40 0.00-0.25
11.3 1.5 0.42-0.30 0.95
21.9 1.85 1.09 1.41-2.96
32.12.1 1.29-0.37-0.45
42.62.6 0.59-1.04 0.45
53.02.7-0.02-0.50 0.17
63.9 2.4-0.50-0.03 0.08
74.4 2.15-0.48 0.08 1.31
84.7 2.05-0.07 1.27-1.58
95.0 2.1 0.26-0.16 0.04
10 6.0 2.25 0.08-0.03 0.00
11 7.02.3 0.01-0.04-0.02
12 8.0 2.25-0.14-0.11 0.02
13 9.2 1.95-0.34-0.05-0.01
1410.5 1.4-0.53-0.10-0.02
1511.3 0.9-0.73-0.15 1.21
1611.6 0.7-0.49 0.94-0.84
1712.0 0.6-0.14-0.06 0.04
1812.6 0.5-0.18 0.00-0.45
1913.0 0.4-0.39-0.54 0.60
2013.3 0.25fix)
JV.
/$- -4^10111213
/
//i
Figure 3.13
f i x)
rr\——f\\j\ft
*2t56'£ >-l01112\
f
j
y
Touseaclamped spline toapproximate thiscurve wewould need derivative approx -
imations fortheendpoints .Even ifthese approximations were available ,wecould expect
little improvement because oftheclose agreement ofthenatural cubic spline tothecurve
ofthetopprofile.
Cubic splines generally agree quite wellwith thefunction being approximated ,provided
thatthepoints arenottoofarapart andthefourth derivative ofthefunction iswell behaved .
Forexample ,suppose that/hasfour continuous derivatives on [a,b]andthatthefourth
derivative onthisinterval hasamagnitude bounded byAf.Then theclamped cubic spline
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S i x)agreeing with f i x)atthepoints a— xo<X|<   <x n=bhastheproperty that
forallxin[a,b\.
5M
Asimilar— butmore complicated — result holds forthenatural cubic splines .
E X E R C I S E S E T 3 . 5
1.
2.
3.Determine thenatural cubic spline Sthatinterpolates thedata/(0)=0,/(1)=1,and/(2)=2.
Determine theclamped cubic spline sthatinterpolates thedata/(0)=0,/(1)=1,/(2)=2and
satisfies s'(0)=s'(2)=1.
Construct thenatural cubic spline forthefollowing data.
a.
c.X f i x)
8.3 17.56492
8.6 18.50515
X f i x)
-0.5-0.0247500-0.25 0.3349375
0 1.1010000X f i x)
0.8 0.22363362
1.0 0.65809197
X f i x)
0.1-0.62049958
0.2-0.28398668
0.3 0.00660095
0.4 0.24842440
4. Thedata inExercise 3were generated using thefollowing functions .Usethecubic splines constructed
inExercise 3forthegiven value ofxtoapproximate /(x)and/'(x),andcalculate theactual error.
a./(x)=xlnx ;approximate /(8.4)and/'(8.4).
b./(x)=sin(e*-2);approximate /(0.9)and/'(0.9).
c./(x)=x3+4.001 x2+4.002*+1.101 ;approximate f i— \)and/'(— 5).
d./(x)=xcosx-2x2+3x— 1;approximate /(0.25)and/'(0.25).
5. Construct theclamped cubic spline using thedata ofExercise 3andthefactthat
a./'(8.3)=3.116256 and/'(8.6)=3.151762
b./'(0.8)=2.1691753 and/'(1.0)=2.0466965
c./'(-0.5)=0.7510000 and/'(0)=4.0020000
d./'(0.1)=3.58502082 and/'(0.4)=2.16529366
6. Repeat Exercise 4using theclamped cubic splines constructed inExercise 5.
7. a.Construct anatural cubic spline toapproximate /(x)=COSTTX byusing thevalues given by
/(x)atx=0.0.25,0.5,0.75,and1.0.
b.Integrate thespline over [0,1],andcompare theresult tof(!cosn xd x=0.
c.Use thederivatives ofthespline toapproximate /'(0.5)and/"(0.5),andcompare these ap-
proximations totheactual values.
8. a.Construct anatural cubic spline toapproximate /(x)=e~xbyusing thevalues given by/(x)
atx=0,0.25,0.75,and1.0.
b.Integrate thespline over [0,1],andcompare theresult tof Q]e~xd x=1-1/e.
c.Usethederivatives ofthespline toapproximate /'(0.5)and/"(0.5),andcompare theapproxi -
mations totheactual values .
9. Repeat Exercise 7,constructing instead theclamped cubic spline with/'(0)=/'(1)=0.
10. Repeat Exercise 8,constructing instead theclamped cubic spline with/'(0)=-1,/'(1)=-e~*.
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11. Anatural cubic spline Son[0,2]isdefined by
S(x)=
12.
13.
14.SoCO=14*2x— x3, if0<x<1,
S,(x)=a+b{x-1)+c(x-l)2+d(x-l)3,if1<x<2.
Finda,b,c,andd.
Aclamped cubic spline sforafunction /isdefined on[1,3]by
s(x)_(s0(x)=3\si(x)=a=3(x-1)+2(x-l)2-(x-l)3, if1<x<2,
+b(x-2)+c(x-2)2+d(x-2)\ if2<x<3.
Given/'(1)=/'(3),finda,b,c,andd.
Anatural cubic spline Sisdefined by
S(x)=So(x)=1+B(x-1)-D(x-l)3, if1<x<2,
St(x)=l+b(x-2)-\(x-2)2+d(x-2)\ if2<x<3.
IfSinterpolates thedata (1,1),(2,1),and(3.0),find B.D,b,andd.
Aclamped cubic spline sforafunction /isdefined by
s(x)5o(x)=1+Bx+2x2-2x\ if0<x<1,
5,(x)=1+b(x-1)-4(x-l)2+7(x-l)3,if1<x<2.
Find/'(0)and/'(2).
15. Suppose that/(x)isapolynomial ofdegree 3.Show that/(x)isitsown clamped cubic spline but
thatitcannot beitsown natural cubic spline .
16. Suppose thedata {x,,/(x,)),lieonastraight line.What canbesaid about thenatural andclamped
cubic splines forthefunction /?[Hint:Take acuefrom theresults ofExercises 1and2.]
17. The data inthefollowing table give thepopulation oftheUnited States fortheyears 1960 to2010
andwere considered inExercise 16ofSection 3.2andExercise 6ofSection 3.3.
Year 1960 1970 1980 1990 2000 2010
Population (thousands )179,323 203,302 226,542 249,633 281,442 307,746
a.Find anatural cubic spline agreeing with these data,andusethespline topredict thepopulation
intheyears 1950 ,1975 ,and2020.
b.Compare your approximations with those previously obtained .Ifyou hadtomake achoice ,
which interpolation procedure would youchoose ?
18. Acartraveling along astraight road isclocked atanumber ofpoints .Thedata from theobservations
aregiven inthefollowing table,where thetime isinseconds ,thedistance isinfeet,andthespeed is
infeetpersecond .
Time 0 3 5 8 13
Distance 0 225 383 623 993
Speed 75 77 80 74 72
a.Useaclamped cubic spline topredict theposition ofthecaranditsspeed when t=10s.
b.Usethederivative ofthespline todetermine whether thecareverexceeds a55-mi/hspeed limit
ontheroad;ifso,what isthefirsttime thecarexceeds thisspeed ?
c.What isthepredicted maximum speed forthecar?
19. The 2011 Kentucky Derby waswon byahorse named Animal Kingdom (at20:1odds )inatime of
2:02.04 (2minutes and2.04 seconds )forthe1
^-mile race.Times atthequarter -mile ,half-mile,and
mile poles were 0:24.26 ,0:59.68 ,and1:47.95.
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a.Usethese values together with thestarting time toconstruct anatural cubic spline forAnimal
Kingdom ’srace.
b.Usethespline topredict thetime atthethree-quarter -mile pole,andcompare thistotheactual
time of1:24.40 .
c.Usethespline toapproximate Animal Kingdom ’sspeed atthefinish line.
20. Itissuspected thatthehigh amounts oftannin inmature oakleaves inhibit thegrowth ofthewinter
moth (Operophtera bromata L,Geometridae )larvae thatextensively damage these trees incertain
years.Thefollowing table liststheaverage weight oftwosamples oflarvae attimes inthefirst28days
after birth.Thefirstsample wasreared onyoung oakleaves ,whereas thesecond sample wasreared
onmature leaves from thesame tree.
a.Useanatural cubic spline toapproximate theaverage weight curve foreach sample .
b.Find anapproximate maximum average weight foreach sample bydetermining themaximum
ofthespline .
Day 0 6 10 13 17 20 28
Sample 1average
weight (mg)6.67 17.33 42.67 37.33 30.10 29.31 28.74
Sample 2average
weight (mg)6.67 16.11 18.89 15.00 10.56 9.44 8.89
3.6 Parametric Curves
None ofthetechniques wehave developed canbeused togenerate curves oftheform shown
inFigure 3.14,because thiscurve cannot beexpressed asafunction ofonecoordinate
variable interms oftheother .Inthissection wewillsechow torepresent general curves by
using aparameter toexpress both the*-andy-coordinatc variables .This technique canbe
extended torepresent general curves andsurfaces inspace .
Figure 3.14
V
1
*-
1 1
1
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