3.Newton’s Forward Difference Interpolation: Example 1 | Numerical Analysis

In this video, we solve a numerical analysis problem using Newton’s Forward Difference Interpolation method. The problem involves finding the degree and form of a function given a table of values for X and f(x). We also use the method to find the values of f(3.5) and f(5.5). This tutorial is perfect for students and professionals looking to improve their understanding of numerical analysis and interpolation methods. Watch now to learn how to solve this problem step-by-step! Given the table of values for X and f(x), we can use Newton’s Forward Difference method to find the degree and form of the function f(x). First, we construct a table of forward differences by subtracting consecutive values of f(x) to obtain the first forward difference. We repeat this process with the first forward difference to obtain the second forward difference, and so on, until we obtain a row of constant values. In this case, the second forward difference is constant, indicating that the degree of the function is 2. Next, we can use the formula for Newton’s Forward Difference Interpolation to find the form of f(x). The formula is given by: f(x) = f(0) + (x * Δf(0)) / 1! + ((x * (x-1)) * Δ²f(0)) / 2! Substituting the values from the table into this formula, we obtain: f(x) = 3 + (x * 3) / 1! + ((x * (x-1)) * 2) / 2! Simplifying this expression, we find that f(x) = x² + 2x + 3. Finally, we can use this expression for f(x) to find the values of f(3.5) and f(5.5). Substituting x = 3.5 into the expression for f(x), we obtain f(3.5) = (3.5)² + 2(3.5) + 3 = 22.25. Similarly, substituting x = 5.5 into the expression for f(x), we obtain f(5.5) = (5.5)² + 2(5.5) + 3 = 44.25. Therefore, the degree of the function is 2, its form is f(x) = x² + 2x + 3, and its values at x = 3.5 and x = 5.5 are f(3.5) = 22.25 and f(5.5) = 44.25.