ITERATION METHOD * (SUCCESSIVE APPROXIMATION METHOD)* PREPARED BY”ARVIND KUMAR“MCA 2ND SEM STUDENT (JIM, GZB) SOME IMPORTANT TIPS FOR ITERATION METHOD

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There are various iterative method for solving the system of equation: ØBisection method ØIteration method ØRegularfalsi method ØNewton raphson method ØSecant method

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But we are more concern about the iteration method

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This method of iteration is useful for finding the real rootof an equation given in the form of infinite series

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By successive approximation x0,x1,x2,----we converge the root of f(x)=0 There are following steps are involving to find successive approximation:-

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We write the given equation f(x) =0 in this form x=Ф(x).

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We choose the initial approximation (guess) such a way that │ Ф’(x) │<1. Note =there are two methods for initial guess:- 1:-graphical method 2:-intermediate method Graphical method:- In this method we get the initial approximation of x with the help of Intersection point of y = x and y = Ф(x).and we get initially x0=a, where, ’a’ is nearest point to the intersection point. Intermediate method:- We get two values of f(x) at x=a, & x=b, at which they are in opposite sign, then we take the initially x0=a. Note:- 1) smaller the value of │Ф’(x) │ but positive then it is more rapidly convergence. 2) For rapidly convergence │Ф’(x) │=0. 3) For fixed point f(x) = x. Examples:- 1)x ex=1. 2)2x-lnx=7 3)x3+6x2+10x-20=0 4)Sin x=5x-2 5)x.tanh(x)=1. Solutions:- 1) x ex =1 So f(x) = x ex-1 x=e-x x

Ф(x), where Ф(x)

e-x Ф’(x) = -e-x Since │Ф’(x) │<1 It means roots are convergent. Since f(0) and f(1) are in opposite sign. So root of equation are lies in between them. Now We take initially x0=0. so x1= Ф(x0) x1 = e-0=>1 x2= Ф(x1) x2 = e-1=>0.3678794 similarly x3=0.6922006 . … ………….. ……….…….. ……………….. x20=0.567477 2)2x-lnx=7 So f(x) = 2x-lnx-7 x= (7+lnx)/2 x

Ф(x), where Ф(x)

(7+lnx)/2 Ф’(x) = 1/2x( ln e) Since │Ф’(x) │<1 It means roots are convergent. Since f(3) and f(4) are in opposite sign So root of equation are lies in between them. Now We take initially x0=3.4 So x1= Ф(x0) x1=3.77815 similarly x2= Ф(x1) x2=3.78863 ………….. …………… x3=3.78924 x4=3.78927 hence root of the equation is x=3.78927 3) x 3+6x 2+10x-20=0 f(x) =x3+6x2+10x-20 x=(20-x3-6x2)/10 x

Ф(x), where Ф(x)

(20-x3-6x2)/10 so Ф’(x) =(-3x2-12x)/10 Since f(1) and f(2) are in opposite sign So root of equation are lies in between them Since Ф’(x )<-1 so│Ф’(x) │>1 It means roots are not convergent . 4)sinx=(5x-2) f(x)=sinx-(5x-2) this equation can be write in two form as 1st – x=sin-1(5x-2) 2nd – x=1/5(sinx+2) Case 1st – x=sin-1(5x-2) x

Ф1(x), where Ф1(x)

sin-1(5x-2) Ф’1(x) = 5/ (1-(5x-2)2)1/2 since Ф1’(x)>1 so│Ф’1(x) │>1 It means roots are not convergent. Case 2nd – x=1/5(sinx+2) x

Ф2(x), where Ф2(x)

1/5(sinx+2) Ф’2(x) =1/5(cosx) Since │Ф’(x) │<1 It means roots are convergent. Now Since f(0) and f(1) are in opposite sign So root of equation are lies in between them. Now we take initially x0=0.5 x1= Ф2(x0) x1=0.4017 similarly x2=0.4014 x3=0.4014 Hence root of equation is x=0.4014 5) x.tanh(x)=1 f(x)=x.tanh(x)-1 this equation can be write as x=coth(x) x

Ф(x), where Ф(x)

coth(x) Ф’(x) = -4/sinh2x Since │Ф’(x) │<1 It means roots are convergent. Since f(1) and f(2) are in opposite sign. So root of equation are lies in between them. Now initially we take x0=1. x1= Ф(x0) x1=1.31303528549933 similarly x2=1.15601401811395 x3=1.21990402611162 x4=1.19100666638769 x5=1.20352756742564 x6=1.19799585895445 x7=1.20041926080065 x8=1.19935362719156 x9=1.19982145104824 x10=1.19961592438525 x11=1.19970618895035 x12=1.19966654047733 x13=1.19968395490739 x14=1.19967630592522 x15=1.19967966556677 hence The root is 1.1996786402 Q-for which value of k the equation x=x+k(x2-3) converges to x=31/2 ?. Sol :-let x= x= Ф(x), where Ф(x) =x+k(x2-3) Ф’(x) = (1+2kx) Since given equation will be converges to x = (3)1/2 if │Ф’(x) │<1 Means to say -1<Ф’(x) <1 So -1< (1+2kx) <1 The given equation will be rapidly convergence at x = 31/2 If f(31/2) =0 So 1+2kx=0 =>k= {-1/ (2*31/2} Thus we can take the value of k=(-1/4) which is nearly equal to k= {-1/ (2*31/2} ALGORITHM STEP 1ST:-read x0, e, n; Where x0=initial guess e=allowed error in root n=total iteration to be allowed for convergence STEP 2ND:- x1<-Ф(x0) STEP 3RD:- for i=1 to n in steps of 1 do STEP 4TH:-x0<-x1 STEP 5TH:-x1<-Ф(x0) STEP 6TH:-if│(x1-x0)/x1│<=e then GOTO 9th step End for. STEP 7TH:-WRITE ‘does not converge to a root’ STEP 8TH:-STOP NOTE:- step 4 to 6 is repeated until the procedure converges to a root or iteration reach n. FLOW CHART

STARTSTART

Define f(x)

Get the value of x0 and max -error

Set n=0

Xn+1=f ( Xn )

n=n+1

Is
│Xn+1-X n│>max error

Print the root is Xn

STOP

Y

N

STARTSTART

Define f(x)

Get the value of x0 and max -error

Set n=0

Xn+1=f ( Xn )

n=n+1

Is │Xn+1-X n│>max error

Print the root is Xn

STOP

Y

N

COMPUTER PROGRAME For example:-xex=1 #include<stdio.h> #include<conio.h> #include<math.h> float f(float x); float g(float x); float dg(float x); void main() { int i,n; float a, b,x0,x; char ch; clrscr(); do { printf("enter the interval\n"); scanf("%f%f",&a,&b); printf("\nthe interval is =(%f,%f)",a,b); if(f(a)*f(b)>0) printf("\nthe inteval is out of range"); else { x0=a; if(fabs(dg(x0))>1) printf("\niteration can not be convergece"); else { printf("\nenter the no.of iteration"); scanf("%d",&n); for(i=1;i<n;i++) { x=g(x0); printf("\nthe root after the %d th iteration is=%f",i,x); x0=x; } } } printf("\nif you want to continue then press y\n"); fflush(stdin); scanf("%c",&ch); }while(tolower(ch)=='y'); getch(); } float f(float x) { float f; f=x*exp(x)-1; return(f); } float g(float x) { float g; g=exp(-x); return(g); } float dg(float x) { float dg; dg=-exp(-x); return(dg); } OUTPUT:- enter the interval 0 1 the interval is =(0.000000,1.000000) enter the no.of iteration 20 the root after the 1 th iteration is=1.000000 the root after the 2 th iteration is=0.367879 the root after the 3 th iteration is= 0.697879 ………………………………………………… ………………………………………………… ………………………………………………… the root after the 19 th iteration is= 0.567157 the root after the 20 th iteration is=0.567136 if you want to continue then press y y the interval is out of range if you want to continue then press y n

ITERATION METHOD* (SUCCESSIVE APPROXIMATION METHOD)*PREPARED BY”

ARVIND KUMAR“MCA 2ND SEM STUDENT (JIM, GZB)SOME IMPORTANT TIPS FOR ITERATION METHODØ

Bisection methodØ

Iteration methodØ

Regularfalsi methodØ

Newtonraphson methodØ

Secant methoditeration methodreal rootof an equation given in the form of infinite seriesThere are following steps are involving to find successive approximation:-f(x) =0in this formx=Ф(x).│ Ф’(x) │<1.Note=there are two methods forinitial guess:-1:-graphical method2:-intermediate methodGraphical method:-In this method we get the initial approximation of x with the help of

Intersection point of y = x and y = Ф(x).and we get initially x0=a, where, ’a’ is nearest point to the intersection point.

Intermediate method:-We get two values of f(x) at x=a, & x=b, at which they are in opposite sign, then we take the initially x0=a.

Note:-1) smaller the value of │Ф’(x) │ but positive then it is more rapidly convergence.

2) For rapidly convergence │Ф’(x) │=0.

3) For fixed point f(x) = x.

Examples:-1) x ex=1.

2) 2x-lnx=7

3) x3+6x2+10x-20=0

4) Sin x=5x-2

5) x.tanh(x)=1.

Solutions:-1) x ex =1So f(x) = x ex-1

x=e-x

x

## Ф(x), where Ф(x)

e-xФ’(x) = -e-x

Since │Ф’(x) │<1

It means roots are convergent.

Since f(0) and f(1) are in opposite sign.

So root of equation are lies in between them.

Now

We take initially x0=0.

so x1= Ф(x0)

x1 = e-0=>1

x2= Ф(x1)

x2 = e-1=>0.3678794

similarly

x3=0.6922006

. … …………..

………. ……..

………………..

x20=0.567477

2)2x-lnx=7So f(x) = 2x-lnx-7

x= (7+lnx)/2

x

## Ф(x), where Ф(x)

(7+lnx)/2Ф’(x) = 1/2x( ln e)

Since │Ф’(x) │<1

It means roots are convergent.

Since f(3) and f(4) are in opposite sign

So root of equation are lies in between them.

Now

We take initially x0=3.4

So x1= Ф(x0)

x1=3.77815

similarly

x2= Ф(x1)

x2=3.78863

…………..

……………

x3=3.78924

x4=3.78927

hence root of the equation is

x=3.78927

3) x 3+6x 2+10x-20=0f(x) =x3+6x2+10x-20

x=(20-x3-6x2)/10

x

## Ф(x), where Ф(x)

(20-x3-6x2)/10so Ф’(x) =(-3x2-12x)/10

Since f(1) and f(2) are in opposite sign

So root of equation are lies in between them

Since Ф’(x )<-1

so│Ф’(x) │>1

It means roots are not convergent

.

4)sinx=(5x-2)f(x)=sinx-(5x-2)

this equation can be write in two form as

1st –

x=sin-1(5x-2)

2nd –

x=1/5(sinx+2)

Case 1st –x=sin-1(5x-2)

x

## Ф1(x), where Ф1(x)

sin-1(5x-2)Ф’1(x) = 5/ (1-(5x-2)2)1/2

since Ф1’(x)>1

so│Ф’1(x) │>1

It means roots are not convergent.

Case 2nd –x=1/5(sinx+2)

x

## Ф2(x), where Ф2(x)

1/5(sinx+2)Ф’2(x) =1/5(cosx)

Since │Ф’(x) │<1

It means roots are convergent.

Now

Since f(0) and f(1) are in opposite sign

So root of equation are lies in between them.

Now we take initially

x0=0.5

x1= Ф2(x0)

x1=0.4017

similarly

x2=0.4014

x3=0.4014

Hence root of equation is x=0.4014

5) x.tanh(x)=1f(x)=x.tanh(x)-1

this equation can be write as

x=coth(x)

x

## Ф(x), where Ф(x)

coth(x)Ф’(x) = -4/sinh2x

Since │Ф’(x) │<1

It means roots are convergent.

Since f(1) and f(2) are in opposite sign.

So root of equation are lies in between them.

Now initially we take x0=1.

x1= Ф(x0)

x1=1.31303528549933

similarly

x2=1.15601401811395

x3=1.21990402611162

x4=1.19100666638769

x5=1.20352756742564

x6=1.19799585895445

x7=1.20041926080065

x8=1.19935362719156

x9=1.19982145104824

x10=1.19961592438525

x11=1.19970618895035

x12=1.19966654047733

x13=1.19968395490739

x14=1.19967630592522

x15=1.19967966556677

hence

The root is 1.1996786402

Q-for which value of k the equation x=x+k(x2-3) converges to x=31/2 ?.Sol :-let x= x= Ф(x), where Ф(x) =x+k(x2-3)Ф’(x) = (1+2kx)

Since given equation will be converges to x = (3)1/2 if

│Ф’(x) │<1

Means to say

-1<Ф’(x) <1

So

-1< (1+2kx) <1

The given equation will be rapidly convergence at x = 31/2

If f(31/2) =0

So

1+2kx=0

=>k= {-1/ (2*31/2}

Thus we can take the value of k=(-1/4) which is nearly equal to

k= {-1/ (2*31/2}

ALGORITHMSTEP 1ST:-read x0, e, n;

Where x0=initial guess

e=allowed error in root

n=total iteration to be allowed for convergence

STEP 2ND:- x1<-Ф(x0)

STEP 3RD:- for i=1 to n in steps of 1 do

STEP 4TH:-x0<-x1

STEP 5TH:-x1<-Ф(x0)

STEP 6TH:-if│(x1-x0)/x1│<=e then GOTO 9th step

End for.

STEP 7TH:-WRITE ‘does not converge to a root’

STEP 8TH:-STOP

NOTE:-step 4 to 6 is repeated until the procedure converges to a root or iteration reach n.FLOW CHARTXn )│Xn+1-X n│>max error

Xn )│Xn+1-X n│>max error

COMPUTER PROGRAMEFor example:-

xex=1#include<stdio.h>#include<conio.h>#include<math.h>float f(float x);float g(float x);float dg(float x);void main(){int i,n;float a, b,x0,x;char ch;clrscr();do{printf("enter the interval\n");scanf("%f%f",&a,&b);printf("\nthe interval is =(%f,%f)",a,b);if(f(a)*f(b)>0)printf("\nthe inteval is out of range");else{x0=a;if(fabs(dg(x0))>1)printf("\niteration can not be convergece");else{printf("\nenter the no.of iteration");scanf("%d",&n);for(i=1;i<n;i++){x=g(x0);printf("\nthe root after the %d th iteration is=%f",i,x);x0=x;}}}printf("\nif you want to continue then press y\n");fflush(stdin);scanf("%c",&ch);}while(tolower(ch)=='y');getch();}float f(float x){float f;f=x*exp(x)-1;return(f);}float g(float x){float g;g=exp(-x);return(g);}float dg(float x){float dg;dg=-exp(-x);return(dg);}OUTPUT:-enter the interval01the interval is =(0.000000,1.000000)enter the no.of iteration 20the root after the 1 th iteration is=1.000000the root after the 2 th iteration is=0.367879the root after the 3 th iteration is= 0.697879………………………………………………………………………………………………………………………………………………………the root after the 19 th iteration is= 0.567157the root after the 20 th iteration is=0.567136if you want to continue then press yythe interval is out of rangeif you want to continue then press yn