Unit 2.4 Non linear curve fitting Geometry curve y = a * x^b

// Alogorithm  y = a * x^b
start
Read number of data(n)
for i=1 to n:
    1. read xi and yi
    2. Next I
Initializ
    1. sumx=0
    2. sumx2 = 0
    3. sumy = 0
    4. sum xy =0

calculate required sum
for i =1 to n:
    1. sumx= sumx + log(xi)
    2. sumx2 = sumx2 + log(xi)*log(xi)
    3. sumy = sumy + log(yi)
    4. sum xy =sumxy + log(xi) * log(yi)

Next i
calculate required constant A and b of Y = A + bx
b = (n*sumxy - sumx * sumy)/(n*sumx2 - sumx * sumx)
A = (sumy - b*sumx)/n

display the value of and b
stop

//program mam wala
#include <stdio.h>
#include<math.h>
#include<conio.h>
void main()
{
    int i,n;
    float A,B,a,b,x[10],y[10],sumx=0,sumy=0,sumx2=0,sumxy=0;
    printf("Enter the value of n\n");
    scanf("%d",&n);
    printf("Enter the value of x and y\n");
    for(i=1 ; i<=n ; i++){
    scanf("%f",&x[i]);
    }
    for(i=1 ; i<=n ; i++){
    scanf("%f",&y[i]);
    }
    for(i=1 ; i<=n ; i++){
    sumx = sumx + x[i];
    sumx2 = sumx2 + (x[i]*x[i]);
    sumy = sumy + log(y[i]);
    sumxy = sumxy + (x[i] * log(y[i]));
    }

     printf("\nsumx = %f , sumx2 = %f , sumy = %f , sumxy = %f",sumx,sumx2,sumy,sumxy);
     B =  ((n*sumxy) - (sumx * sumy))/((n*sumx2) - (sumx * sumx));
     A = (sumy - (B*sumx))/n;
 // a=exp(A);
   a = exp(A);
   b = exp(B);
    printf("\n y= %.2f*x^%.2f",a,b);
    getch();
}
// program
#include <stdio.h>
#include<math.h>
int main()
{
    int n;
    float a,x[10],y[10],sumx=0,sumy=0,sumx2=0,sumxy=0;
    printf("Enter the value of n\n");
    scanf("%d",&n);
    printf("Enter the value of x and y\n");
    for(int i=1 ; i<=n ; i++){
        scanf("%f",&x[i]);
    }
    for(int i=1 ; i<=n ; i++){
        scanf("%f",&y[i]);
    }
    for(int i=1 ; i<=n ; i++){
        sumx = sumx + log(x[i]);
        sumx2 = sumx2 + (log(x[i])*log(x[i]));
        sumy = sumy + log(y[i]);
        sumxy = sumxy + (log(x[i]) * log(y[i]) );
    }

    // printf("\nsumx = %f , sumx2 = %f , sumy = %f , sumxy = %f",sumx,sumx2,sumy,sumxy);
    float b =  ((n*sumxy) - (sumx * sumy))/((n*sumx2) - (sumx * sumx));
    float A = (sumy - (b*sumx))/n;

    a = exp(A);
    printf("\n y= %.2fx + %.2f",a,b);
    return 0;
}