1 chapter 3 numerical methods roots of nonlinear equations interpolation (in -dimension) numerical...

CHAPTER 3 NUMERICAL METHODS

Roots of Nonlinear equations

Interpolation (In -Dimension)

Numerical Integration

Numerical Solution of Differential Equations

3.1 Roots of Nonlinear Equations

Let say we want to find the solution of f (x) 0. For example:

These equations can not be solved directly.

0sin2)

We need numerical methods to compute the approximate solutions.

3.1.1 Iteration Methods Let x0 be an initial value that is close to the

solution of f (x) 0. From f (x) 0, another equation is produced such that we can use x0 to compute x.

This step is repeated to obtain values of x2, x3,...This method is called Iteration Method.

We hope that every new xi converges to the solution of f (x) 0.

3.1.2 Fixed Point Iteration First we write f (x) 0 in the form x F (x). Note that F (x) is not unique. For instance,

see the following.

xxxfExample 3.1

more.many and

We shall use these forms (x F (x)) in our next example, denoted by

We can say that the solution of x F (x) is the intersection of two graphs y x and y F (x).

For example, see the following figure:

Figure 3.1 Fixed Point Iteration

xx0 x1 x2 x3

y0 = x1

y1 = x2

y2 = x3

)(xFy xy

1. Start the computation with initial value x0.

2. From y F (x), we have y0 F (x0).

3. Then, from graph y x, we may assume x1 y0.

From here, we have y1 F(x1) and x2 y1.

4. Similarly, we will obtain x3, x4, … and so on.

Solution steps

Conclusion

Fixed point iteration is of the form

ii xFx

We hope that the neighborhood denoted by the dashed line converges to the intersection point of the two graphs y x and y F(x).

Example 3.2

Approximation by Fixed Point Iteration

(Based on Example 3.1)

In Example 3.1, we have obtained various forms of x F(x). Now referring to xi1 F(xi), we put subscripts as follows.

The distance between xi1 and xi increases, i.e. xi1 – xi xi – xi1 This iteration fails (since it diverges).

1.185138

1.367163

1.458333

:computecan we(1) From

value.initial thebe 1.5Let

The distance between xi+1 and xi decreases, i.e. xi+1 – xi xi – xi1 This iteration converges (succeeds).

0.347296

0.347294

0.347271

0.346992

0.343643

: computecan we(2) From 0.3.Put

We have |xi+1 – xi | | xi – xi-1 |. This shows that this iteration diverges (Iteration fails).

1.915679

1.848431

1.906730

1.855956

: computecan we(3) From

1.9. Take

(3) 1313

1 1i i i ix x x x

.347296.0x

000001.0

1)347296.0(3)347296.0( 3

is 0133 xx

REMARK

An iteration converges if it satisfies

From the convergent iteration, we conclude that one of the solutions of

CHECK :

How to make sure that initial value x0 can give a convergent iteration?

Answer:

We differentiate F(x) to get F ′(x).The iteration converges if

.0)( ofsolution the toclose is )

and ,1)(1or 1)()

Example 3.3

divergent. is valueinitial with thisiteration theand

satisfiednot is 1)('1 Condition

.125.2)(' have we,5.1For

1)( i)

: that3.1 Examplein determined have we

,013For

converge.possibly couldit since examined, becan

valueinitial with thisiteration theTherefore,

.1)('1 condition satisfies

that 07.0)(' have we,3.0For

1)( ii)

diverges. valueinitial with thisiteration theThus,

satisfied.not is 1)('1 Condition

.11.1)(' have we,9.1At

13)( iii)

converge.possibly coulditeration thesuch that

1)('1condition thesatisfiesthat

valueinitial thechoosemust We

Conclusion

3.1.3 Newton-Raphson Method Tangent line of f (x) is f ′(x). This is the basic of

Newton-Raphson Method described below.

xx0 x1 x2O

f (x0)

f (x1)

f (x2)

Figure 3.2 Newton-Raphson Method

Gradient of the tangent line of f (x) at (x0, f (x0)) is given by f ′(x0). If we continue the tangent line at (x0, f (x0)) to (x1,0), then the gradient of this tangent line can be determined by

Equalizing these two values, we have

(4) .)(

This means that from the initial value x0, using formula (4), we can compute x1 which is close to the solution of f (x) 0.

Using the same formula, we can compute x2

which is a better approximation to the solution compare to x1.

We can also compute x3 , x4 , … and so on. We hope that our iteration converges to the

solution of f (x) 0.

In general, Newton-Raphson method can be formulated as

iii xf

Example 3.4

i xi f (xi) f ′ (xi)

0 0 -1 -3

1 -0.333333 -0.037037 -2.666667

2 -0.347222 -0.000196 -2.638310

3 -0.347296 -0.000000 -

:formula sRaphson'-Newtonon based tablea Make )

.point starting a as 0 Choose )

.33)(obtain to)( ateDifferenti )

xxfxfi

.013)( 3 xxxfSolve

Solution

From the table we conclude that:

* xi converges to the solution of f (x) 0, that is

x - 0.347296.

* Values in column f (xi) go to zero.

(Remark: this shows that we have done a

good approximation)

i xi f (xi) f ′(xi)

0 0.99 -2.999701 -0.0597

1 -49.256248 -119 358.65 -

table.following at theLook

?99.0 choose weifWhat 0 x

fails. methodRaphson -Newton that means This

. )( than zero fromfarer is )(that

table thefrom seecan We

01 xfxf

IMPORTANT!!! Newton-Raphson method fails if the

approximate root of equation, for instance x = x0, close to roots of f ′(x) = 0.

Therefore, we have to make sure that the starting point x0 does not give f ′(x0) close to zero.

The reason is that dividing a number by a value close to zero will give a number with large absolute value.

3.1.4 Secant Method This method is a revision of Newton-Raphson

method as described in the following figure. f (x)

x x0 x1 x2 x3O

f (x0)

f (x1)

f (x2)

f (x3)

Figure 3.3 Secant Method

In this method, we begin with two initial values x0

and x1. The straight line from (x0, f (x0)) to (x1,f (x1)) is continued to the x-axis.

Let x2 be the value where this straight line intersects the x-axis.

By equalizing gradient from (x0, f (x0)) to (x1, f (x1)) and gradient from (x1, f (x1)) to (x2,0), we obtain

Using the same formula, we compute x3, x4, ... and so on.

We hope that xi converges to the solution of f (x) 0.

)()()(0

xxxfxx

In general, Secant method can be formulated as

iiiii xfxf

xxxfxx

i xi f (xi)

0 0 -1

1 1 -3

2 -0.5 0.375

3 -0.333333 -0.037037

4 -0.348315 0.002685

5 -0.347302 0.000016

6 -0.347296 -0.000000

7 -0.347296 -0.000000

.013)( 3 xxxfSolve

Example 3.5

methodSecant of formula on the based tablea Make )

.1 and 0 valuesinitial Choose ) 10

xxi Solution

From the table, notice that:

* Iteration xi converges to x - 0.347296.

* Values in column f (xi) go to zero.

(Remark as in Newton-Raphson method) If f (xi) is farer from zero than f (xi -1), then the

iteration fails. This means that we made a mistake in estimating xi using this iteration.

3.1.5 How to choose initial values?

Besides conditions of the method we use, we may also see sign of f (x) for the tested x.

For instance, see values of f (x) x3 – 3x – 1 used in our previous example.

We can make the following sign table:

x f (x) Sign f (x) Remark

-5 -111 -ve

The sign changes

-4 -53 -ve

-2 -3 -ve *

-1 1 +ve *

0 -1 -ve *

1 -3 -ve *

2 1 +ve *

From the sign table, we know that the solution of f (x) 0 is clearly in between

x -2 and x -1,

x -1 and x 0,

and x 1 and x 2.

If we use Secant Method, we may choose the following

x0 -2 and x1 -1,

or x0 -1 and x1 0,

or x0 1 and x1 2.

If we use Newton-Raphson Method, we focus on the first derivative of f (x) x3 – 3x – 1,

i.e. f ′(x0) 3x2 – 3.

We may not choose x0 -1 or x0 1, since this gives f ′(x0) 0.

But, from the sign table, we may choose x0 -2 or x0 0 or x0 2 as long as f ′(x0) is not too close to zero.

If we use Fixed Point Method, we can choose x0 that satisfies

.1)(1or 1)( 0'

0' xFxF

3.2 Interpolation (In -dimension)

The meaning of interpolation in 2-dimension is to define a curve that passes through given data points. As an example, see the following:

Figure 3.4 Points interpolation

3.2.1 Interpolation Polynomial

unique. and exists ,,for

system theofsolution that theimplies This

equations.t independen and

consistent ofnumber )1( of system a build to

of valuesdifferent with ,, 1, ,0 ),,(

points )1( need we,,, find To

s.coeficientunknown are ,, where

:polynomial following heConsider t

xCxCxCCy

Given the data points and what is the unique polynomial that passes through the data points?

Solution

Since we are given three different points, then we can only find unique value for three coefficients: C0, C1 and C2. Then, our interpolation polynomial is quadratic:

y = C0 + C1x + C2x2.

Example 3.6

If we substitute the given data points and to the polynomial, we get the following three equations:

3 C0 + C1 + C2

3 C0 + 3C1 + 9C2.

This is a system of linear equations. The solution is.

4,2 210 CCC

Thus, the unique interpolation polynomial that passes through the given data points is

42 2xxy

Theorem (Interpolation polynomial)

.,,1,0,)(

),,(,),,(),,( )1( 1100

nkyxPthat such

nto equal or than less order ofxPpolynomial

unique a produce can wexof value different with

yxyxyxpoints datanGiven

There are many methods to find this unique interpolation polynomial. In this part, we only consider two methods: Lagrange Method and Newton’s Divided Difference Interpolation.

3.2.2 Lagrange Method

.)())(())((

)())(())(()(

)()()()(

is Method Lagrangeby built polynomialion Interpolat

).,(, ),,( ),,(

:points data following given the are say weLet

221100

niiiiiii

xxxxxxxxxx

xxxxxxxxxxxL

yxLyxLyxLyxLy

yxyxyx

Given the data points and Apply Lagrange Method to find interpolation polynomial passing through these points.

221100

)()()(

bygiven is polynomialion interpolat Lagrange Then,

(3,3).),(dan (1,3)),( (0,2),),(Let

yxLyxLyxLy

yxyxyx

Example 3.7

Solution

)3)(10(1

)3)(0(

))(()(

1)31)((

)30)(10(

)31)((

))(()(

xxxxxL

polynomialion interpolat the

obtain weequation, sLagrange' the to)( ngSubstituti

)13)(03(

)1)(0(

))(()(

xxxxxxy

xxxxxL

3.2.3 Newton’s Divided Difference Interpolation

Lagrange method has the following weaknesses:

1) Lagrange Interpolation Polynomial is hard to find if we have many data points.

2) If we want to add data, we have to restart our computation from the beginning.

In this part, we discuss a method that can handle the weaknesses of Lagrange Method, called Newton’s Divided Difference Interpolation. Before that, we first need to produce a table of divided difference.

3.2.3.1 Table of Divided Difference

3.5. Table

in listed formulas by the defined areon so and

three two,one, orders of difference divided The

)).(,(, )),(,( )),(,(

: points following given the are weSuppose

1100 nn xfxxfxxfx

xxfxxf

110 ],,[],,[

211 ],[],[

xxfxxf

Order Symbol Definition

0 f [xi] f (xi)

1 f [xi , xi +1]

2 f [xi , xi +1 , xi +2]

3 f [xi , xi +1 , xi +2 , xi +3]

n f [x0 , x1 ,…, xn - 1 , xn]

32121 ],,[],,[

iiiiii

xxxfxxxf

Table 3.5 Orders of Divided Difference

x f (x)

x0 f (x0)

x1 f (x1)

x2 f (x2)

x3 f (x3)

Table 3.6 Divided Difference Formula2 31

],[],[

xxfxxf

321210

],,[],,[

xxxfxxxf

],[],[

xxfxxf

Based on Table 3.5, we can produce table of divided difference as in Table 3.6.

Though Table 3.6 shows an estimation for four data points, a similar way can be used for different number of data points.

Example 3.8

Produce a table of divided difference for the following data points:

SolutionTable of divided difference for this data is

2 0 7/3

4 29/6

4 8 12 65/72

16 17/8

3 -8 21/6

3.2.3.2 Newton’s Divided Difference Interpolation

].,,,[)())((

],,[))((

as written is polynomialion interpolat difference

divided sNewton' ,difference divided of definition theFrom

nn xxxfxxxxxx

xxxfxxxx

Example 3.9

Find interpolation polynomial that passes through (0,2), (1,3) dan (3,3) using Newton’s divided difference interpolation.

Solution

Table of divide difference for these three points is

1)1)(0()1)(0(2

],,[))((],[)()()(

210101000

bygiven is example for this polynomialion interpolat difference divided sNewton' theThus,

xxxfxxxxxxfxxxfxPy

REMARKS

42)3(,3 when )

42)1(,1 when )

.2)0(,0 when )

3.3 Integral Solution Using Numerical Methods

In this part, we discuss the following numerical integration methods:

1) Rectangular Rule

2) Trapezoidal Rule

3) Simpson’s Rule

Before doing the approximation, we first have to give the number of subintervals that we want to consider and find the length of each subinterval.

3.3.1 Number of subintervals and the length of each subinterval

We suppose that the interval from x a to xb is divided into n number of subintervals, where the length of each subinterval is

If x0 a and xn b, then in general, xi a ih.

The positions of x0, x1, … ,xn are shown in the following figure.

Figure 3.5 Number of subintervals

3.3.2 Rectangular Rule

In this method, we suppose that every subinterval forms a rectangle, where the height of each subinterval is

Areas of all rectangles are computed. Total area of these rectangles is the approximate area

below the curve of the function, which is also the estimation of the integral that we want to determine.

1),( 1

** iiii xxxxf

Figure 3.6 Rectangular Rule

x1* x2

We have given the length of each subinterval denoted by h.

If the height of the ith rectangle is

then the area of the ith rectangle is Total area of all rectangles

),( *ixf

).( *ixfh

)].()()([

)()()(**

xfxfxfh

xfhxfhxfh

RECTANGULAR RULE

. lssubinterva ofnumber theincrease we

if accurate more be will)( of Estimation

1 where 1

])()()([)(

axfxfxfhdxxf

Example 3.10

als subintervten usingdxeEstimate x

.747131.0

)471308.7(1.0 table, thefrom Thus,

is lsubinterva

each oflength then the1, and 0Given

Solution

i xi xi* f (xi

0 0 - -

1 0.1 0.05 0.997503

2 0.2 0.15 0.977751

3 0.3 0.25 0.939413

4 0.4 0.35 0.884706

5 0.5 0.45 0.816686

6 0.6 0.55 0.738968

7 0.7 0.65 0.655406

8 0.8 0.75 0.569783

9 0.9 0.85 0.485537

10 1 0.95 0.405555

TOTAL 7.471308

3.3.3 Trapezoidal Rule

A revision of the Rectangular Rule. While in the Rectangular Rule we make a rectangle

for each subinterval, in Trapezoidal Rule we make a trapezoid for each subinterval.

Areas of all trapezoids are computed. Total area of all trapezoids is the estimation of the

area below the curve of the function, which is also the approximation of the integral of the function.

Figure 3.7 Trapezoidal Rule

x0 x1 x2 xn

Thus, If the length of each subinterval is h, Height of the left hand side of the ith trapezoid = Height of the right hand side of the ith trapezoid =

Then the area of the ith trapezoid is

),( 1ixf

),( ixf

)).()((2

11 ii xfxfh

Total area of all trapezoids

1)()()()(

))()((2

1))()((

xfxfxfxfxfh

xhfxfh

xhfxfhxhfxfh

xfxfhxfxfh

TRAPEZOIDAL RULE

1)()()()(

121 bfxfxfxfafhdxxf n

where x0 a and xn b

. intervalin )(graph for the

minimum a )( and maximum a )(

give which and Here,

:Rule lTrapezoidafor boundError

btabta

Example 3.11

0.1 10

is lsubinterva

each oflength then the1, and 0Given

dxeestimate to

als subintervten withRule lTrapezoida Use

Solution

i xi f (a) & f (b ) f (xi ), i=,…, n

.)23(4)(

)12(2)(

2)(Then ,)(Given

ation.differenti following the

do toneed webound,error thecompute To

.746211.0

)778167.6()367879.1(2

table thefrom Thus,

xexfexf

Figure 3.7 Trapezoidal Rule

below. described as 10

intervalin increasing is )(graph that means This

.10 intervalin every for ,0)( that Note"

.lyrespective ,2 )0( and 735759.0 )1(

are 10 intervalin )(

of valuesminimum and maksimum the, So

735759.0)1)1(2(2 )1(

and 2)1)0(2(2 )0(

have we,)12(2)( From

1.0 intervalin maximum is )1( and

minimum is )0( , versus)( ofgraph theFrom

.001667.0000614.0 or

)2()10(12

)01()735759.0(

)10(12

)01(or

:Rule lTrapezoida of formulaerror the

to valuesminimum and maximum thengsubstitutiby

ion approximat thisof bounderror theestimate weNow

ion.approximat integral the tobounderror theaddingby

estimationour of accuration theincreasecan We

0.747878. 745597.0or

0.0016670.746211 000614.0 0.746211

interval following in the isresult theThus,

3.3.4 Simpson’s Rule

Basic of Rectangular Rule Using constant to estimate the area of each subinterval.

Basic of Trapezoidal Rule Using linear equation to estimate the area of each subinterval.

Basic of Simpson’s Rule Using quadratic equation to estimate the area of each subinterval.

.2or 2

gives This

. and between in is of valueThe

integrate. want tohat wefunction t in the

lssubinterva successive twodefine that points thebe

)),(,( and ))(,()),(,(Let

221100

xfxxfxxfx

:figure following in the described as ),( parabola

below area theis )( that know We

.))(,( and ))(,()),(,(

esinterpolat that parabola a be )(Let

221100

xfxxfxxfx

cbxaxxP

Figure 3.7 Simpson’s Rule

x0 x1 x2 h h

P(x) = ax2 + bx + c

)).()()()((2

))()()()((4

)()((3

have we, and Taking

:follows as Rule sSimpson' formulatecan wecurve, a

below area estimate toparabola using ofresult theFrom

xfxfxfxf

lyrespective , intervalin )( of values

minimum and maximum theare )( and )(

and , where

bygiven is Rule sSimpson'for bounderror The

btabta

following thehave Then we

. bounderror with

is Rule sSimpson' using

integralan of valueeapproximat that theSuppose

ESdxxfES

Example 3.12

integral. thecompute tous help that will tablea make We

.1.010

01 intervaleach oflength The

.1 to0 from respect to

with )( integrate want to We.)(Given 2

xfexf x

dxecompute to

als subintervten withRule sson'Apply Simp

Solution

i xi f (a) & f (b) f (xodd) f (xeven)

.10 intervalin )( of valuesminimum and

maximum thefind toneed webound,error theestimate To

.746825.0

)037901.3(2)740266.3(4367879.13

have we table, theFrom

.0)15204(8)(

writeWe

.0)( ofsolution thecalculate toneed we

),( of points extrimumget To

.)15204(8)(

)3124(4)(

have we,)( From

exxxxf

.357589.7)1(

and ,419481.7)958572.0(

,12)0(

computecan we

,)3124(4)( from Now,

958572.0or 015204 ii)

or ,0or 08 i)

haveonly we

10 intervalin then ,0 sincebut

tested,becan that factors threeexists There

1.0 )( 3.8Figure )4( xinterval in x versusxfGraph

7.419481

7.357589

minimum

maximum

f (4)(x)

x1 0.958572

.000005.0000007.0

00000412.000000667.0

)419481.7()10(180

)01()12(

)10(180

bygiven is bounderror theThus,

419481.7)958572.0( is valueminimum the

and 12)0( is valuemaximum the

,10 intervalin )(for

: thatconcludecan wen,computatioour on Based

lue.biggest va the takealways we

error, maximumfor and

alue,smallest v the takealways we

error, minimum for the example,For

lue.closest va not the

interval,biggest a takealways we

them,ofion approximat the

take toneed weif that boundserror for Note

error bounds

approximate error bounds

.746830.0)(746818.0

000005.0746825.0)(000007.0746825.0

Conclusion

3.4 Numerical Solution of Differential equations

In this part, we discuss two numerical methods to compute approximate solutions of differential equations:

1) One-step Euler method 2) The fourth order Runge-Kutta method

NOTATION

i.e. , and in function

a asnotation thisuse also wepart, In this

by denoted isoperation aldifferenti theUsually,

yxfydx

3.4.1 One-step Euler Method

This method is the most basic numerical method used to solve differential equation.

We shall discuss this Euler Method using the following example:

Example 3.13

(1,1). )2(

xatyof value the eApproximat

through passingyxxdx

dyGiven

Figure 3.9 One-step Euler Method with h = 1.

Solution

See the following figure

(x0, y0)(x1, y1)

. of value theofion approximat theis

that assume wefigure, previous In the

.'by given is curve a of

line tangent theofgradient that theknow We

l.subinterva considered theoflength theis and

),,(at line tangent theofgradient theis ' where

have we together,line tangent theof

gradient theandequation gradient thePutting

3)2)(1(1

is 2at of valueeapproximat theThus,

21)12(1

that Note

tablefollowing in the explained as

2at 3 eapproximat that weisresult The xy

Notice that in general, the formula we use to estimate y can be written as

' 1 iii hyyy

The formula of One-step Euler Method

i xi yi yi’

0 1 1 2

1 2 3 -

2.0. 1.5, ,0.1 and

is lsubintervaeach oflength Then the

ls.subinterva twointo interval thedivide weSuppose

ls?subinterva moreor twointo divided is 2 to

1 from interval theifion approximat

our ofaccuracy theincrease Can we

See the following figure:

Figure 3.10 One-step Euler Method with h 0.5.

(x, y)

1 1.5 2

)2)(5.0(1

obtain we,5.0 with nowbut before, did weAs

.5.1at of valueeapproximat theis Here

. compute tohave we, computing Before

. is 2at of value

eapproximat that thesuppose Now,

2)512)(51(

)(' Thus,

2. usinggradient the

eapproximat tois docan What we.5.1at

line tangent theofgradient exact thecomputenot can we

,5.1at of eexact valu theknownot do weSince

i xi yi yi′

0 1 1 2

1 1.5 2 2.75

2 2 3.375 -

: tablefollowing in the listed are results The

before. one n thebetter tha ision approximat This

.375.3

)75.2)(5.0(2

3.2.8,,.4, 11.2, , 1

are of Values

3)2.0(1

Example 3.14

NOTATION

.at last value thereaching until lsubinterva of

n calculatioevery in of value the to up add

and begin with that wemeansnotation This

by denoted becan considered be to of valuesThe

3.2.8,,.4, 11.2, , 1

are of Values

3)2.0(1

Example 3.15

.2 , (1,1)

xatyeapproximat through passing

dygiven i.e. problem previous our

solveto als subintervfive take weSuppose

.2)2.0(1 and

.2.0 5

12 Here,

Solution

i xi yi yi’

0 1.0 1 2

1 1.2 1.4 2.36

2 1.4 1.872 2.712

3 1.6 2.4144 3.0544

4 1.8 3.02528 3.38528

5 2.0 3.702336 -

The results of our approximation are listed in the following table:

Approximation by One-step Euler method with h 0.2.

If we take ten subintervals, we have

i xi yi yi’

0 1.0 1 2

1 1.1 1.2 2.19

2 1.2 1.419 2.379

3 1.3 1.6569 2.5669

4 1.4 1.91359 2.75359

5 1.5 2.188949 2.938949

6 1.6 2.482844 3.122844

7 1.7 2.795128 3.305128

8 1.8 3.125641 3.485641

9 1.9 3.474205 3.664205

10 2.0 3.840626 -

Approximation by One-step Euler method with h 0.1.

3.4.2 : The fourth order Runge-Kutta Method

Runge-Kutta Method is more accurate compare to Euler Method.

As usual, given y′ f (x, y) and the initial condition (x0, y0) and we want to find the value of y at other values of x.

).22(6

: follows as valuesteintermediafour

calculate toneed weMethod, Kutta-RungeIn

kkkkyy

kyhxhfk

Example 3.16

2.)5.0(1

(1,1). )2(

xforyof value the eApproximat

through passingyxxdx

dyGiven

have we(1,1),),(

Using values.teintermediafour following the

calculate we,5.1at estimate To

.2,5.1,1 have weexample In this

Solution

.273438.1

]609375.1)25.12(25.1)[5.0(

)609375.1 ,25.1((0.5)

21875.1

]5.1)25.12(25.1)[5.0(

)5.1 ,25.1((0.5)

1 ]1)12(1)[5.0(

)1 ,1((0.5)

1.5. at 249349.2 :Result

.249349.2

)511719.1)273438.1(2)21875.1(2(16

511719.1

]273438.2)5.12(5.1)[5.0(

)273438.2,5.1((0.5)

432101

kkkkyy

kyhxhfk

have We

2.249349). (1.5,),( using nowbut n,calculatio

previousour repeat we,2at compute To

.718343.1

]2.999186)75.12(75.1)[5.0(

)2.999186 ,75.1((0.5)

499674.12.249349,

5.05.1(0.5)

499674.1

]2.249349)5.12(5.1)[5.0(

)2.249349 ,5.1((0.5)

.011180.2

]022359.4)22(2[)5.0(

)022359.4,2()5.0(

)773010.1249349.2,5.05.1((0.5)

),( 3114

kyhxhfk

773010.1

]108521.3)75.12(75.1)[5.0(

)108521.3,75.1()5.0(

718343.1249349.2,

5.05.1(0.5)

i xi yi k1 k2 k3 k4

0 1 1 1 1.21875 1.273438 1.511719

1 1.5 2.249349 1.499674 1.718383 1.773010 2.011180

2 2 3.998276 - - - -

:ncalculatioour ofresult for the tablefollowing theSee

2. when 998276.3 :Result

.998276.3

432112

kkkkyy

Conclusion

If we increase the number of subintervals(i.e. with smaller value of h),then we obtain a more accurate approximation

.2at 999997.3obtain we1(0.1)2, i.e.

ls,subinterva by taking continue, weIf

.2at 999948.3 have we

2, )2.0(1 i.e. ls,subinterva take weIf

1 chapter 3 numerical methods roots of nonlinear equations interpolation (in -dimension) numerical...

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solution of x f x

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