METODE NUMERIK oget

TUGAS

METODE NUMERIK

DIKERJAKAN OLEH :

Nama : Sigit Rossandi Utomo

Nim : 360763013

Jurusan : Teknik Informatika

SEKOLAH TINGGI MANAJEMEN INFORMATIKA DAN KOMPUTER

INDONESIA MANDIRI

BANDUNG 2010

ALAMAT KAMPUS STMIK-IM Jl. Jakarta No. 79 Bandung 40272.

Telp. 022-7272672, 022-7208180. Fax. 022-7271693.

E-mail: [email protected]

mailto:[email protected]

Metode Setengah Selang

#include <stdio.h>#include <math.h>

double mutlak(double xk){if(xk<0) return xk * -1;else return xk;

}

double hasilFungsi(double x){return 5*x*x*x - 3*x*x*x*x*x;

}

void setengahSelang(double epsilon, double x1, double x2){

double xk;int k = 1;xk = (x1+x2)/2;

printf("Metode Setengah Selang\n\n");

while(mutlak(xk)>epsilon && k<=50 && hasilFungsi(xk) != 0){ printf("%d\t%lf\n", k, xk);

if( hasilFungsi(x1) * hasilFungsi(xk) < 0 ) x2 = xk;

else x1 = xk;k = k + 1;xk = (x1+x2)/2;

} }

main(){setengahSelang(0.0001,0.0,-2.5);

getch();return 0;

}

Hasil Run f(x) = 5x3 - 3x5

1 -1.2500002 -1.8750003 -1.5625004 -1.4062505 -1.3281256 -1.2890637 -1.3085948 -1.2988289 -1.29394510 -1.29150411 -1.29028312 -1.29089413 -1.29119914 -1.29104615 -1.29097016 -1.29100817 -1.29098918 -1.29099819 -1.29099420 -1.29099621 -1.29099522 -1.29099423 -1.29099524 -1.29099425 -1.29099526 -1.290994

Metode Setengah Selang



}

double hasilFungsi(double x){return exp(x) - 5*x*x;

}

void setengahSelang(double epsilon, double x1, double x2){

double xk;int k = 1;xk = (x1+x2)/2;

printf("Metode Setengah Selang\n\n");

while(mutlak(xk)>epsilon && k<=50 && hasilFungsi(xk) != 0){ printf("%d\t%lf\n", k, xk);

if( hasilFungsi(x1) * hasilFungsi(xk) < 0 ) x2 = xk;

else x1 = xk;k = k + 1;xk = (x1+x2)/2;

} }

main(){setengahSelang(0.0001,0.0,-2.5);

getch();return 0;

}

Hasil Run f(x) = ex - 5x2

1 -1.2500002 -0.6250003 -0.3125004 -0.4687505 -0.3906256 -0.3515637 -0.3710948 -0.3808599 -0.37597710 -0.37353511 -0.37231412 -0.37170413 -0.37139914 -0.37155215 -0.37147516 -0.37143717 -0.37141818 -0.37140819 -0.37141320 -0.37141621 -0.37141722 -0.37141723 -0.371418

Metode REGULA FALSI



}

double hasilFungsi(double x){return 5*x*x*x - 3*x*x*x*x*x;

}

void regulaFalsi(double epsilon, double a, double b){

double x1,x2,x3;int k = 2;x1 = a;x2 = b;

printf("Metode REGULA FALSI\n\n");

while(mutlak(x2 - x1)>epsilon && k<=50 && hasilFungsi(x2) != 0){ printf("%d\t%lf\n", k, x2); x3 = (a*hasilFungsi(b) - b*hasilFungsi(a)) / (hasilFungsi(b) - hasilFungsi(a));

if( hasilFungsi(x1) * hasilFungsi(x3) < 0 ) b = x3;

else a = x3;k = k + 1;x1 = x2;x2 = x3;

} }

main(){ regulaFalsi(0.0001,0.0,-2.5);

getch();return 0;

}


2 -2.500000

Metode REGULA FALSI



}

double hasilFungsi(double x){return exp(x) - 5*x*x;

}

void regulaFalsi(double epsilon, double a, double b){

double x1,x2,x3;int k = 2;x1 = a;x2 = b;

printf("Metode REGULA FALSI\n\n");

while(mutlak(x2 - x1)>epsilon && k<=50 && hasilFungsi(x2) != 0){ printf("%d\t%lf\n", k, x2); x3 = (a*hasilFungsi(b) - b*hasilFungsi(a)) / (hasilFungsi(b) - hasilFungsi(a));

if( hasilFungsi(x1) * hasilFungsi(x3) < 0 ) b = x3;

else a = x3;k = k + 1;x1 = x2;x2 = x3;

} }

main(){ regulaFalsi(0.0001,0.0,-2.5);

getch();return 0;

}


2 -2.5000003 -0.0777174 -0.1453345 -0.5230276 -0.3163667 -0.3633978 -0.3719179 -0.371387

Metode NEWTON RAPHSON



}

// fungsi aslidouble hasilFungsi(double x){

return 5*x*x*x - 3*x*x*x*x*x;}

// fungsi turunannyadouble hasilFungsi1(double x){

return 15*x*x - 15*x*x*x*x;}

void newtonRaphson(double epsilon, double x1){double x2,xk;int k = 2;x2 = x1 - hasilFungsi(x1) /

hasilFungsi1(x1);

printf("Metode NEWTON RAPHSON \n\n");

while(mutlak(x2 - x1)>epsilon && k<=50 && hasilFungsi(x2) != 0){ printf("%d\t%lf\n", k, x2); xk = x2 - hasilFungsi(x2) / hasilFungsi1(x2); k = k + 1; x1 = x2; x2 = xk; } } main(){ newtonRaphson(0.0001,-0.5);

getch();return 0;

}


2 -0.3111113 -0.2029624 -0.1341455 -0.0891036 -0.0593077 -0.0395108 -0.0263329 -0.01755210 -0.01170111 -0.00780012 -0.00520013 -0.00346714 -0.00231115 -0.00154116 -0.00102717 -0.00068518 -0.00045719 -0.00030420 -0.000203

Metode NEWTON RAPHSON



}

// fungsi aslidouble hasilFungsi(double x){

return exp(x) - 5*x*x;}

// fungsi turunannyadouble hasilFungsi1(double x){

return exp(x) - 10*x;}

void newtonRaphson(double epsilon, double x1){double x2,xk;int k = 2;x2 = x1 - hasilFungsi(x1) /

hasilFungsi1(x1);

printf("Metode NEWTON RAPHSON \n\n");

while(mutlak(x2 - x1)>epsilon && k<=50 && hasilFungsi(x2) != 0){ printf("%d\t%lf\n", k, x2); xk = x2 - hasilFungsi(x2) / hasilFungsi1(x2); k = k + 1; x1 = x2; x2 = xk; } } main(){ newtonRaphson(0.0001,-0.5);

getch();return 0;

}


2 -0.3852293 -0.3716144 -0.371418

Metode JACOBI


void jacobi(double x1,doub le x2,double x3){int a[3][3] = {{4,-1,1},{4,-8,1},{-2,1,5}};

int b[3] = {7,-21,15};int k = 1;double x,y,z;printf("Metode JACOBI\n\n");

while(k<=50){ x = x1;y = x2;z = x3; printf("%d\t%lf\t%lf\t%lf\n", k,x,y,z); x1 = (b[0] - a[0][1]*y - a[0][2]*z) / a[0][0]; x2 = (b[1] - a[1][0]*x - a[1][2]*z) / a[1][1]; x3 = (b[2] - a[2][0]*x - a[2][1]*y) / a[2][2]; k++; } }

main(){ jacobi(1.0,1.0,1.0);

getch();return 0;

}

Hasil Run JACOBI

1 1.000000 1.000000 1.0000002 1.750000 3.250000 3.2000003 1.762500 3.900000 3.0500004 1.962500 3.887500 2.9250005 1.990625 3.971875 3.0075006 1.991094 3.996250 3.0018757 1.998594 3.995781 2.9971878 1.999648 3.998945 3.0002819 1.999666 3.999859 3.00007010 1.999947 3.999842 2.99989511 1.999987 3.999960 3.00001112 1.999987 3.999995 3.00000313 1.999998 3.999994 2.99999614 2.000000 3.999999 3.00000015 2.000000 4.000000 3.000000

TRAPEZOID RULES Hasil Run

#include <iostream.h>#include <math.h>double f(double x){ return 1/(1+x);}main(){ double a, b, h, x, jumlah, integral; int N; a=0; b=1; N=10; h=(b-a)/N; jumlah=0; for ( int i=1; i<=N-1; i++) { x=a+i*h; jumlah = jumlah+2*f(x); } integral = h*(f(a)+jumlah+f(b))/2; cout << "Integralnya adalah " << integral << endl; getch();}

Integralnya adalah 0.693771

EULER n xn yn err

#include <iostream.h>#include <iomanip.h>#include <math.h>#include <stdio.h>#include <conio.h>double f(double x, double y){ return x+y;}main(){ double x, y, h; int n; x=0.0; y=1.0; h=0.2; n=0; cout<<" n xn yn err"<<endl; do { cout<<setw (4)<<n<< setiosflags ( ios ::showpoint |ios::fixed) <<setprecision(8)<<setw(15)<<x <<setw(15)<<y<<setw(15)<<(y-(2*exp(x)-x-1))<<endl; x=x+h; y=y+h*f(x,y); n++; } while (x<=1.0001); getch();}

0 0.00000000 1.00000000 0.00000000 1 0.20000000 1.24000000 -0.00280552 2 0.40000000 1.56800000 -0.01564940 3 0.60000000 2.00160000 -0.04263760 4 0.80000000 2.56192000 -0.08916186 5 1.00000000 3.27430400 -0.16225966

SIMPSON Hasil Run

#include <iostream.h>#include <math.h>#include <stdio.h>#include <conio.h>double f(double x){ return 1/(1+x);}main(){ double a, b, h, x, jumlah, integral; int N; a=0; b=1; N=10; h=(b-a)/N; jumlah = 0; for (int i=1; i<=N-1; i++) { x=a+i*h; if (i%2==1) { jumlah=jumlah+4*f(x); } else { jumlah=jumlah+2*f(x); } } integral=h*(f(a)+jumlah+f(b))/3; cout<<"integralnya adalah "<<integral<<endl; getch();}

integralnya adalah 0.69315

Polinom Newton Masukan banyaknya data yang akan di input : 4

#include<stdio.h>

double polinom(double xx,double x[100],int n){ if(n){ return (xx-x[n-1])*polinom(xx,x,n-1); } else return 1;}

double hasil(double xx,double y[100][100],double x[100],int n){ if(n){ return y[0][n-1]*polinom(xx,x,n-1)+hasil(xx,y,x,n-1); } else return 0; }

main(){ int i,n,j; printf("Masukan banyaknya data yang akan di input : "); scanf("%d", &n); double x[n],y[n][n],xx; printf("\nx\tf(x)\n"); for(i=0;i<n;i++){ scanf("%lf %lf", &x[i],&y[i][0]); } for(j=1;j<n;j++){ for(i=0;i<n;i++){ y[i][j] = (y[i+1][j-1]-y[i][j-1])/(x[i+j]-x[i]); } } printf("\nMasukan data yang ingin dicari : "); scanf("%lf", &xx); printf("\nf(%lf) = %lf\n", xx,hasil(xx,y,x,n)); getch(); return 0;}

x f(x)8 2.0794429 2.1972259.5 2.25129211 2.397895

Masukan data yang ingin dicari : 9.2

f(9.200000) = 2.219208

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METODE NUMERIK oget