Gauss’s Forward Interpolation

17 Jun, 2021
Interpolation refers to the process of creating new data points given within the given set of data. The below code computes the desired data point within the given range of discrete data sets using the formula given by Gauss and this method known as Gauss’s Forward Method.  
 
Gauss’s Forward Method:
The gaussian interpolation comes under the Central Difference Interpolation Formulae. Suppose we are given the following value of y=f(x) for a set values of x:X: x0 x1 x2 .......... xn Y: y0 y1 y2 ............ yn The differences y1 – y0, y2 – y1, y3 – y2, ......, yn – yn–1 when denoted by Δy0, Δy1, Δy2, ......, Δyn–1 are respectively, called the first forward differences. Thus the first forward differences are :Δy0 = y1 – y0and in the same way we can calculate higher order differences. 
And after the creating table we calculate the value on the basis of following formula: 
Now, Let’s take an example and solve it for better understanding. Problem: From the following table, find the value of e1.17 using Gauss’s Forward formula.
 
Solution: We have 
yp = y0 + pΔy0 + (p(p-1)/2!).Δy20 + ((p+1)p(p-1)/3!).Δy30 + ... where p = (x1.17 – x1.15) / h and h = x1 – x0 = 0.05so, p = 0.04Now, we need to calculate Δy0, Δy20, Δy30 ... etc.  
Put the required values in the formula- yx = 1.17 = 3.158 + (2/5)(0.162) + (2/5)(2/5 – 1)/2.(0.008) ... yx = 1.17 = 3.2246Code : Python code for implementing Gauss’s Forward Formula  
Python3
# Python3 code for Gauss's Forward Formula# importing libraryimport numpy as np # function for calculating coefficient of Ydef p_cal(p, n):     temp = p;    for i in range(1, n):         if(i%2==1):             temp * (p - i)         else:             temp * (p + i)    return temp;# function for factorialdef fact(n):    f = 1    for i in range(2, n + 1):        f *= i    return f # storing available datan = 7;x = [ 1, 1.05, 1.10, 1.15, 1.20, 1.25, 1.30 ]; y = [[0 for i in range(n)]        for j in range(n)];y[0][0] = 2.7183;y[1][0] = 2.8577;y[2][0] = 3.0042;y[3][0] = 3.1582; y[4][0] = 3.3201;y[5][0] = 3.4903;y[6][0] = 3.6693; # Generating Gauss's trianglefor i in range(1, n):    for j in range(n - i):        y[j][i] = np.round((y[j + 1][i - 1] - y[j][i - 1]),4); # Printing the Trianglefor i in range(n):    print(x[i], end = "\t");    for j in range(n - i):        print(y[i][j], end = "\t");    print(""); # Value of Y need to predict onvalue = 1.17; # implementing Formulasum = y[int(n/2)][0];p = (value - x[int(n/2)]) / (x[1] - x[0]) for i in range(1,n):    # print(y[int((n-i)/2)][i])    sum = sum + (p_cal(p, i) * y[int((n-i)/2)][i]) / fact(i) print("\nValue at", value,    "is", round(sum, 4));
Output :  
1       2.7183  0.1394  0.0071  0.0004  0.0     0.0     0.0001  
1.05    2.8577  0.1465  0.0075  0.0004  0.0     0.0001  
1.1     3.0042  0.154   0.0079  0.0004  0.0001  
1.15    3.1582  0.1619  0.0083  0.0005  
1.2     3.3201  0.1702  0.0088  
1.25    3.4903  0.179   
1.3     3.6693  

Value at 1.17 is 3.2246
 
adnanirshad158
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