586 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES for(i = 0; i < n–j; i++) { diff[i][j] = diff[i+1][j–1]–diff[i][j–1]; } } i = 0; do { i++; }while(ax[i]<x); i– –; p = (x–ax[i])/h; y1 = p*diff[i–1][1]; y2 = p*(p+1)*diff[i–1][2]/2; y3 = (p+1)*p*(p–1)*diff[i–2][3]/6; y4 = (p+1)*p*(p-1)*(p+2)*diff[i–3][4]/24 y = ay[i] + y1 + y2 + y3 + y4; printf(“when x = %6.4f, y = %6.8f”, x, y); printf(“press enter to exit”); getch(); } OUTPUT Enter the no. of term –7 Enter the value in form of x– Enter the value of x1 – 1.00 Enter the value of x2 – 1.05 Enter the value of x3 – 1.10 Enter the value of x4 – 1.15 Enter the value of x5 – 1.20 Enter the value of x6 – 1.25 Enter the value of x7 – 1.30 Enter the value in the form of y– Enter the value of y1 – 2.7183 Enter the value of y2 – 2.8577 Enter the value of y3 – 3.0042 Enter the value of y4 – 3.1582 Enter the value of y5 – 3.3201 Enter the value of y6 – 3.4903 Enter the value of y7 – 3.6693 COMPUTER PROGRAMMING IN ‘C’ LANGUAGE 587 Enter the value of x for Which you want the value of y – 1.35 When x = 1.35, y = 3.8483 Press enter to exit 13.20 ALGORITHM FOR STIRLING’S 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++ Step 10. for i = 0; i < n–j; i++ Step 11. diff [i][j] = diff [i+1] [j–1] – diff [i] [j–1] Step 12. End of the loop i Step 13. End of the loop j Step 14. i = 0 Step 15. Repeat step 16 until ax[i] < x Step 16. i = i+1 Step 17. i = i–1 Step 18. p = (x – ax[i])/h Step 19. y1 = p * (diff [i][1] + diff [i – 1][1])/2 Step 20. y2 = p * p * diff[i – 2][2]/2 Step 21. y3 = p * (p *p–1)* (diff[i – 1][3] + diff[i–2][3])/6 Step 22. y4 = p * p * (p * p–1) * diff [i – 2] [4]/24 Step 23. y = ay[i] + y1 + y2 + y3 + y4 Step 24. Print the output x, y Step 25. End of the program. 13.21 PROGRAMMING FOR STIRLING’S METHOD #include<stdio.h> #include<conio.h> #include<math.h> #include<string.h> #include<process.h> 588 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES void main() { int n; int i, j; float ax[10]; float ay[10]; float h; float p; float x, y; float diff[20][20]; float y1, y2, y3, y4; clrscr(); printf(“enter the no. of term–“); scanf(“%d”, & n); printf(“enter the no. in the form of x–”); for(i = 0; i < n; i++) { printf(enter the value of x%d”, i+1); scanf(“%f”, & ax[i]); } printf(“enter the value in the form of y”); for(i = 0; i < n; i++) } printf(“enter the value of y%d “, i+1); scanf(“%f”, & ay[i]; } printf(“enter the value of x for”); printf(“which you want the value of y”); scanf(“%f”, %x); h = ax[1] –ax[0]; for(i = 0; i < n–1; i++) { diff[i][1] = ay[i+1]–ay[i]; } for(j = 2; j <= 4; j++) { for(i = 0; i < n–j; i++) { diff[i][j] = diff[i+1][j–1]–diff[i][j–1]; } } COMPUTER PROGRAMMING IN ‘C’ LANGUAGE 589 i = 0; do { i++; }while(ax[i] < x); i– –; p = (x–ax[i])/h; y1 = p*(diff[i][1]+diff[i–1][1]/2; y2 = p*(p)*diff[i–1][2]/2; y3 = p*(p*p–1)*(diff[i–1][3]+diff[i–2][3])/6; y4 = p*p*(p*p–1)*diff[i–2][4]/24; y = ay[i] + y1 + y2 + y3 + y4; printf(“when x = %6.4f, y = %6.8f”, x, y); printf(“press enter to exit”); getch(); } OUTPUT Enter the no. of term –7 Enter the value in form of x Enter the value of x1 – .61 Enter the value of x2 – .62 Enter the value of x3 – .63 Enter the value of x4 – .64 Enter the value of x5 – .65 Enter the value of x6 – .66 Enter the value of x7 – .67 Enter the value in the form of y– Enter the value of y1 – 1.840431 Enter the value of y2 – 1.858928 Enter the value of y3 – 1.877610 Enter the value of y4 – 1.896481 Enter the value of y5 – 1.915541 Enter the value of y6 – 1.934792 Enter the value of y7 – 1.954237 Enter the value of x for Which you want the value of y – 0.6440 When x = 0.6440, y = 1.90408230 Press enter to continue. 590 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 13.22 ALGORITHM FOR BESSEL’S 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++ Step 10. for i = 0; i < n–j; i++ Step 11. diff [i][j] = diff [i+1] [j–1] – diff [i] [j–1] Step 12. End of the loop i Step 13. End of the loop j Step 14. i = 0 Step 15. Repeat step 16 until ax[i] < x Step 16. i = i+1 Step 17. i = i–1 Step 18. p = (x – ax[i])/h Step 19. y1 = p * (diff [i][1]) Step 20. y2 = p * (p–1) * (diff [i] [2] + diff [i – 1] [2]/4 Step 21. y3 = p * (p-1) * (p–0.5) * (diff [i – 1] [3])/6 Step 22. y4 = p * (p+1) * (p–2) * (p–1) * (diff [i – 2] [4] + diff [i – 1][4])/48 Step 23. y = ay[i] + y1 + y2 + y3 + y4 Step 24. Print the output x, y Step 25. End of the program. 13.23 PROGRAMMING FOR BESSEL’S METHOD #include<stdio.h> #include<conio.h> #include<math.h> #include<string.h> #include<process.h> void main() { int n; int i, j; COMPUTER PROGRAMMING IN ‘C’ LANGUAGE 591 float ax[10]; float ay[10]; float h; float p; float x, y; float diff[20][20]; float y1, y2, y3, y4; clrscr(); printf(“enter the no. of term–“); scanf(“%d”, & n); printf(“enter the no. in the form of x–”); for(i = 0; i < n; i++) { printf(enter the value of x%d”, i+1); scanf(“%f’, & ax[i]); } printf(“enter the value in the form of y”); for(i = 0; i < n; i++) { printf(“enter the value of y%d “, i+1); scanf(“%f”, & ay[i]); } printf(“enter the value of x for”); printf(“which you want the value of y”); scanf(“%f”,%x); h = ax[1]–ax[0]; for(i = 0; i < n–1; i++) { diff[i][1] = ay[i+1]–ay[i]; } for(j = 2; j <= 4; j++) { for(i = 0; i < n–j; i++) { diff[i][j] = diff[i+1][j–1]–diff[i][j–1]; } } i=0; do { 592 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES i++; }while(ax[i] < x); i– –; p = (x–ax[i])/h; y1 = p*(diff[i][1]); y2 = p*(p–1)*(diff[i][2]+diff[i–1][2])/4/2; y3 = p*(p–1)*(p–.5)*(diff[i–1][3])/6; y4 = (p+1)*p*(p–1)*(p–2)*(diff[i–2][4]+diff[i–1][4])/48; y = ay[i] + y1 + y2 + y3 + y4; printf(“when x = %6.4f, y = %6.8f”, x, y); printf(“press enter to exit”); getch(); } OUTPUT Enter the no. of term –7 Enter the value in form of x– Enter the value of x1 – .61 Enter the value of x2 – .62 Enter the value of x3 – .63 Enter the value of x4 – .64 Enter the value of x5 – .65 Enter the value of x6 – .66 Enter the value of x7 – .67 Enter the value in the form of y– Enter the value of y1 – 1.840431 Enter the value of y2 – 1.858928 Enter the value of y3 – 1.877610 Enter the value of y4 – 1.896481 Enter the value of y5 – 1.915541 Enter the value of y6 – 1.934792 Enter the value of y7 – 1.954237 Enter the value of x for Which you want the value of y – 0.6440 When x = 0.6440, y = 1.90408230 Press enter to continue. 13.24 ALGORITHM FOR LAPLACE EVERETT METHOD Step 1. Start of the program to interpolate the given data Step 2. Input the value of n (number of terms) COMPUTER PROGRAMMING IN ‘C’ LANGUAGE 593 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++ Step 10. for i = 0; i < n–j; i++ Step 11. diff [i][j] = diff [i+1] [j-1] – diff [i] [j-1] Step 12. End of the loop i Step 13. End of the loop j Step 14. i = 0 Step 15. Repeat step 16 until ax[i] < x Step 16. i = i+1 Step 17. i = i–1 Step 18. p = (x –ax[i])/h Step 19. q = 1–p Step 20. y1 = q * (ay [i]) Step 21. y2 = q * (q * q–1) * (diff [i–1] [2])/6 Step 22. y3 = q * (q * q–1) * (q *q–4) * (diff [i–2] [4])/120 Step 23. py 1 = p * ay [i + 1] Step 24. py2 = ay[i] + y1 + y2 + y3 + y4 Step 25. Print the output x, y Step 26. End of the program. 13.25 PROGRAMMING FOR LAPLACE EVERETT METHOD #include<stdio.h> #include<conio.h> #include<math.h> #include<string.h> #include<process.h> void main() { int n; int i, j; float ax[10]; float ay[10]; float h; 594 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES float p, q; float x, y = 0; float nr, dr; float diff[20][20]; float y1, y2, y3, y4; float py1, py2, py3, py4; clrscr(); printf(“enter the no. of term–“); scanf(“%d”, & n); printf(“enter the value in the form of x–”); for(i = 0; i < n; i++) { printf(“enter the value of x%d”, i+1); scanf(“%f”, & ax[i]); } printf(“enter the value in the form of y”); for(i = 0; i < n; i++) { printf(“enter the value of y%d “, i+1); scanf(“%f”,& ay[i]); { printf(“enter the value of x for”); printf(“which you want the value of y”); scanf(“%f”, %x); h = ax[1]–ax[0]; for(i = 0; i < n–1; i++) { diff[i][1] = ay[i+1]–ay[i]; } for(j = 2; j < = 4; j++) { for(i = 0; i < n–j; i++) { diff[i][j] = diff[i+1][j–1]–diff[i][j–1]; } } i = 0; do { i++; COMPUTER PROGRAMMING IN ‘C’ LANGUAGE 595 }while(ax[i] < x); i– –; p = (x–ax[i])/h; q = 1–p; y1 = q*(ay[i]); y2 = q*(q*q–1)*(diff[i–1][2])/6; y3 = q*(q*q–1)*(q*q–4)*(diff[i–2][4])/120; py1 = p*ay[i+1]; py2 = p*(p*p–1)*diff[i][2]/6; py3 = p*(p*p–1)*(p*p–4)*(diff[i–1][4])/120 y = y1 + y2 + y3 + y4 + py1 + py2 + py3; printf(“when x = %6.4f, y = %6.8f”, x, y); printf(“press enter to exit”); getch(); } OUTPUT Enter the no. of term –7 Enter the value in form of x– Enter the value of x1 – 1.72 Enter the value of x2 – 1.73 Enter the value of x3 – 1.74 Enter the value of x4 – 1.75 Enter the value of x5 – 1.76 Enter the value of x6 – 1.78 Enter the value of x7 – 1.79 Enter the value in the form of y– Enter the value of y1 – .1790661479 Enter the value of y2 – .1772844100 Enter the value of y3 – .1755204006 Enter the value of y4 – .1737739435 Enter the value of y5 – .1720448638 Enter the value of y6 – .1703329888 Enter the value of y7 – .1686381473 Enter the value of x for Which you want the value of y – 1.7475 When x = 1.7475, y = .17420892 Press enter to exit. . 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. i, j; float ax[10]; float ay[10]; float h; 594 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES float p, q; float x, y = 0; float nr, dr; float diff[20][20]; float y1, y2, y3, y4; float py1,. METHOD #include<stdio.h> #include<conio.h> #include<math.h> #include<string.h> #include<process.h> 588 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES void main() { int n; int i, j; float ax[10]; float ay[10]; float h; float p; float x, y; float diff[20][20]; float y1, y2, y3,