Newton Forward Difference Interpolation C++

Newton’s forward difference interpolation is used when the function is tabulated at equal intervals.

If the data point to be interpolated lies in the upper half or in the beginning of the table then Newton’s forward difference interpolation is used because it gives the better approximation.

In order to interpolate at any point (say x) between X0 and Xn, Newton’s forward interpolation formula takes the form,

Formula

Newton Forward Difference Interpolation C++ formula y(x)=
Advertisements

Where,

U = ( x- x0 ) / h

‘h’ is the interval between data points

’∆‘ is the forward difference symbol

C++ Program

Newton Forward Difference Interpolation C++ Program

//Newton's Forward Difference interpolation //wikkihut.com #include<iostream> #include<bits/stdc++.h> //for specifying format using namespace std; int factorial(int n){ int fact=1; while(n){ fact=fact*n; n--; } return fact; } int main(){ float x[10],y[10],p[10],diff[10]; float X,f,f2=0,u,h; int i, j=1,n,k=1; cout<<"Enter the number of observations: "; cin>>n; cout<<"\nEnter the values of xi's and f(xi's)\n"; for(i=1; i<=n; i++){ cout<<"x["<<i<<"]: "; cin>>x[i]; cout<<"f["<<i<<"]: "; cin>>y[i]; } cout<<"\nx[i's]\tf[i's]\n"; for(i=1; i<=n; i++){ cout<<x[i]<<"\t"<<y[i]<<"\n"; } cout<<"\nEnter the value of x at which // you want to interpolate: "; cin>>X; int m=n; h=x[2]-x[1]; u=(X-x[1])/h; //saving (y[1]) because the array y[] changes // and we have add it at the last result f=y[1]; cout<<fixed<<setprecision(2); cout<<"Forward Difference table for the given set of points is\n"; do{ cout<<"diff.("<<j++<<") f(x)\n"; for(i=1; i<n; i++){ p[i]=( ( y[i+1]-y[i] ) ); cout<<p[i]<<"\n"; //this assignment is done because we are computing the new //difference array using previous one (delta f(x)'s) y[i]=p[i]; } diff[k++]=p[1]; n--; cout<<"\n"; }while(n!=1); float df=u; float l=1; for(int i=1; i<m; i++ ){ if(i>1) df=df/factorial(i); f2=f2+(df*diff[i]); for(int j=0; j<=i; j++){ l=l*(u-j); } df=l; } f=f+f2; cout<<"\nf("<<X<<") = "<<f; return 0; }
Code language: C++ (cpp)

Formation of Forward Difference table

Before studying the Newton’s Forward Difference interpolation, we should know how to form a forward difference table. Therefore, first we will see how to form a forward difference table.

General Example

Consider the following general set of data points.

Xx0x1x2x3x4x5
Yy0y1y2y3y4y5
Advertisements

The forward difference table for the above set of data points is given below:

Newton Forward Difference Interpolation C++ formula for tabulation of data points of x and y

Example

Newton Forward Difference Interpolation C++ Example:

Q. Consider the following set of data points and form a forward difference table.

X2.02.252.502.753.0
Y9.010.0611.2512.5614.0
Newton Forward Difference Interpolation C++ Program Example & Formula

Find y (2.35)?

Ans. The forward difference table for the given set of data points is given below:

Ans. We have,

 x = 2.35, ( given )

h = x1 – x0 = 2.25 – 2.0 = 0.25

 u = (x-x0) / h = (2.35-2.0)/0.25 = 0.35/0.25 = 1.4

Now, the forward difference table is as:

Newton Forward Difference Interpolation C++ formula and table

Now you can calculate y( x )

Newton Forward Difference Interpolation C++ formula

y (2.35) = 10.484 + 0.0364 + 0.00056 + 0.000448

y(2.35) = 10.52141

Suggested Read:

Advertisements