566 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES b = x1; } x1 = (a + b)/2; printf(“%f”,x1); } The root of the given equation is 1.5121. 13.4 ALGORITHM FOR FALSE POSITION METHOD Step 1. Start of the program to compute the real root of the equation Step 2. Input the value of x 0 , x 1 and e Step 3. Check f(x 0 ) × f(x 1 ) < 0 Step 4. If no, print “Error” and exit Step 5. If yes, compute x 2 = xfx xfx fx fx 01 10 10 >C >C >C >C − − Step 6. Compute f(x 2 ) Step 7. Again, if f(x 2 ) × f(x 0 ) < 0 Step 8. Set x 1 = x 2 Step 9. Else, set x 0 = x 2 Step 10. Continue the process step 5 to step 9 till to get required accuracy Step 11. Print output Step 12. End of the program. 13.5 PROGRAMMING FOR FALSE POSITION METHOD (1) Find the Real Root of the Given Equation x 3 2x 5 = 0 #include<conio.h> #include<stdio.h> #include<math.h> void false(float, float); void main() { float x0 = 3, x1 = 4; clrscr(); false(x0, x1); getch(); } void false(float x0, float x1) { COMPUTER PROGRAMMING IN ‘C’ LANGUAGE 567 int i; float x2 = 0, a = 0, b = 0, c = 0; for(i = 0; i < 12; i++) { a = pow(x0, 3)–2*x0–5; b = pow(x1, 3)–2*x1–5; x2 = x0–(x1–x0)/(b–a)*a); c = pow(x2, 3)–2*x2–5; if(c < 0) x0 = x2; else x1 = x2; } printf(“%f”, x2); } The root of the given equation is 2.094. (2) Find the Real Root of the Given Equation 3x + sin x e x = 0 #include<conio.h> #include<stdio.h> #include<math.h> float flase(float, float); void main() { float a, x0 = 0, x1 = 1, b, x2; clrscr(); a = false(x0, x1); b = false(x2); printf(“%f”, b); getch(); } float false(float x0, float x1) { float x2; int i; for(i = 1; i <= 13; i++) { y0 = 3*x0 + sin(x0)–pow(2.7187, x0); y1 = 3*x1 + sin(x1)–pow(2.7187, x1); x2 = x0–(x1–x0)/(y1–y0)*y0; y2 = 3*x2 + sin(x2) – pow(2.7187, x2); if(y2 < 0) 568 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES x0 = x2; else x1 = x2; } return (x2); } The root of the given equation is 36042. 13.6 ALGORITHM FOR ITERATION METHOD Step 1. Start of the program to compute the real root of the equation Step 2. Input the value of x 0 (initial guess) Step 3. Input the value of required alllowed error e Step 4. Input the total iteration to be allowed n Step 5. Compute φ(x 0 ), x 1 ← φ(x 0 ) (step 7 to 8 are repeated until the procedure converges to a root) Step 6. For i = 1 to n, in step 2 to step 4 do Step 7. x 0 ← x 1 , x 1 ← φ(x 0 ) Step 8. If xx x 10 1 − ≤ e then GOTO step 11 end for Step 9. Print “does not converge to a root”, x 0 , x 1 Step 10. Stop Step 11. Print “converge to a roof ”, i, x 1 Step 12. End of the program. 13.7 PROGRAMMING FOR ITERATION METHOD (1) Find the Real Root of the Given Equation xe x = 1 #include<conio.h> #include<stdio.h> #include<math.h> void main() { void iterat(float); int i, j; float x0, a; clrscr(); COMPUTER PROGRAMMING IN ‘C’ LANGUAGE 569 a = 0.5; x0 = a; iterat(x0); getch(); } void iterat(float x0) { int i; float x1, x2; for(i = 1; i <= 12; i++) { x1 = 1/pow(e, x0); x2 = 1/pow(e, x1); x0 = x2; } print(“The result is: %f”, x2); } The root of the given equation is 0.5671. (2) Find the Real Root of the Given Equation 2x log 10 x = 7 #include<conio.h> #include<stdio.h> #include<math.h> void main() { void iterat(float); int i, j; float x0, a; clrscr(); a = 3.7; x0 = a; iterat(x0); getch(); } void iterat(float x0) { int i; float x1, x2; for(i = 1; i <= 12; i++) { 570 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES x1 = (7 + log(x0)/2; x2 = (7 + log(x1)/2; x0 = x2; } print(“The result is: %f”, x2); } The root of the given equation is 4.2199. (3) Find the Real Root of the Given Equation x sin x = 1 #include<conio.h> #include<stdio.h> #include<math.h> void main() { void iterat(float); int i, j; float x0, a; clrscr(); a = 1.5; x0 = a; iterat(x0); getch(); } void iterat (float x0) { int i; float x1, x2; for (i = 1; i <= 12; i++) { x1 = 1/sin (x0); x2 = 1/sin (x1); x0 = x2; } printf(“The result is: %f”, x2); } The root of the given equation is 1.114. (4) Find the Real Root of the Given Equation 2x log 10 x = 7 #include<stdio.h> #include<conio.h> #include<math.h> COMPUTER PROGRAMMING IN ‘C’ LANGUAGE 571 float iteration(float); void main() { float a; float x = 3.7; clrscr(); a = iteration(x); printf(“%f”, a); getch(); } float iteration(float x) { int i; float s = 0; for(i = 0; i < 15; i++) { s = 0.5*(7 + log(x)); x = s; } return(s); } The root of the given equation is 4.2199. 13.8 ALGORITHM FOR NEWTON’S RAPHSON METHOD Step 1. Start of the program to compute the real root of the equation Step 2. Input the value of x 0 , n and e Step 3. For i = 1 and repeat if i < = n Step 4. f 0 = f(x 0 ) Step 5. df 0 = df(x 0 ) Step 6. Compute x1 = x 0 – (f 0/df 0) Step 6a. If xx x 10 1 − < e Step 6b. Print “convergent” Step 6c. Print x1, f(x1), i Step 7. End of the program Step 8. Else, Set x 0 = x1 Step 9. Repeat the process until to get required accuracy Step 10. End of the program. 572 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 13.9 PROGRAMMING FOR NEWTON RAPHSON METHOD (1) Find the Real Root of the Given Equation x 2 = 12 //program-netwon raphson #include<conio.h> #include<stdio.h> #include<math.h> float f(float x) { return((x*x)–(12)); } float d(float x) { return((2*x)); } void main() { float x, y, s; int i, c = 0; clrscr(); for(i = 0; ; i++) { if(f(i) > 0) break; } x = i; aa: { ++c; y = x–(f(x)/d(x)); x = y; s = (y*10000); printf(“\nthe position of iteration %d”, c); printf(“\nthe root is %f”, y); y = (y*10000); if(y != s) goto aa; } printf(“\nreal root is %f”,y); getch(); } The root of the given equation is 3.4641. COMPUTER PROGRAMMING IN ‘C’ LANGUAGE 573 (2) Find the Real Root of the Given Equation x 2 5x + 2 = 0 //program-netwon raphson #include<conio.h> #include<stdio.h> #include<math.h> float f(float x) { return((x*x) – (5*x) + 2; } float d(float x) { return((2*x) – 5); } void main() { float x, y = 0; int i; clrscr(); for(i = 0;; i++) { if(f(i) > 0) break; } x = i; for(i = 0; i < 10; i++) { y = x–f(x)/d(x); x = y; printf(“\nreal root is %f”, y); } getch(); } The root of the given equation is 0.438447. 13.10 PROGRAMMING FOR MULLER’S METHOD (1) Find the Real Root of the Given Equation x 3 x 2 x 1 = 0 #include<conio.h> #include<stdio.h> 574 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES #include<math.h> void main() { int i, j; flat x0, x1, x2, x3, y0, y1, y2, a, b; float val(float); clrscr(); x0 = 1.9; x1 = 2.0; x2 = 2.1; for(i = 0; i < 3; i++) { y0 = val(x0); y1 = val(x1); y2 = val(x2); a = ((x0–x1)*(y1–y2)–(x1–x2)) (y0–y2))/((x1–x0)*(x1–x2)*(x0–x2)); b = (pow((x0–x1), 2)*(y1–y2)–pow((x1–x2), 2)*(y0–y2))/((x0–x1)*(x1–x2)*(x0–x2)); x3 = x2–((2*y2)/(b + pow((pow(b, 2)–4*a*y2), .5))); x0 = x1; x1 = x2; x2 = x3; } printf(“%f”, x3); getch(); } float val(float x) { float y; y = pow(x, 3) – pow(x, 2) – x – 1; return(y); } The root of the given equation is 1.8382067. (2) Find the Real Root of the Given Equation x 3 3x 5 = 0 #include<conio.h> #include<stdio.h> #include<math.h> void main() { int i, j; float x0, x1, x2, x3, y0, y1, y2, a, b; COMPUTER PROGRAMMING IN ‘C’ LANGUAGE 575 float val(float); clrscr(); x0 = 1.9; x1 = 2.0; x2 = 2.7; for(i = 0; i < 3; i++) { y0 = val(x0); y1 = val(x1); y2 = val(x2); a = ((x0–x1)*(y1–y2)–(x1–x2)*(y0–y2))/((x1–x0)*(x1–x2)*(x0–x2)); b = (pow((x0–x1), 2)*(y1–y2)–(pow(x1–x2), 2)*(y0–y2))/((x0–x1)*(x1–x2)*(x0–x2)); x3 = x2–((2*y2)/(b + pow((pow(b, 2)–4*a*y2),. 5))); x0 = x1; x1 = x2; x2 = x3; } printf(“%f”, x3); getch(); } float val(float x) { float y; y = pow(x, 3)–(3 * x) –5; return(y); } The root of the given equation is 2.417728. 13.11 ALGORITHM FOR NEWTON’S FORWARD INTERPOLATION METHOD Step 1. Start of the program to interpolate the given data Step 2. Input the value of n (number of terms) Step 3. Input the array ax for data of x Step 4. Input the array ay for data of y Step 5. Compute h = ax[1] – ax[0] Step 6. For i = 0; i < n–1; i++ Step 7. diff[i] [1] = ay[i+1]–ay[i] Step 8. End of the loop i Step 9. For j = 2; j <= 4; j++ . i Step 7. End of the program Step 8. Else, Set x 0 = x1 Step 9. Repeat the process until to get required accuracy Step 10. End of the program. 572 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 13.9. 0 #include<conio.h> #include<stdio.h> 574 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES #include<math.h> void main() { int i, j; flat x0, x1, x2, x3, y0, y1, y2, a, b; float val(float); clrscr(); x0 =. PROGRAMMING IN ‘C’ LANGUAGE 571 float iteration(float); void main() { float a; float x = 3.7; clrscr(); a = iteration(x); printf(“%f”, a) ; getch(); } float iteration(float x) { int i; float s = 0; for(i