Last Updated on September 23, 2026 by Maged kamel
Modified Newton-Raphson method.
Introduction to the bracketing method. The bracketing method is a Numerical method that uses two values of a function with opposite signs; the root lies between them.
The modified Newton-Raphson method is another root-finding method. I introduced a simple modification to the Newton-Raphson method.
I introduced the Modified Newton-Raphson method by applying it to two solved problems. The first solved problem is problem 7. I used the Modified Newton-Raphson method to solve the Problem. I introduced an Excel Table showing the iterations starting from x0 = 0.50. atItas a rt of the video.
Modified Newton-Raphson formula.
The next slide shows the formula for the Modified Newton-Raphson method.
Solved Problem #7 using the modified Newton-Raphson Method.
We start by solving Problem #7. Using the modified Newton-Raphson method, I used an Excel sheet to determine the root Value of the given f(x)=e^x-3* ^2. Make a Table of x and the corresponding values of f(x) by selecting several values of x, starting from 0 to 1.10.
After plotting the function in Excel, the Point where f(x)=0 was found at x=0.91, indicating that the root is at x=0.91. Please refer to the Excel sheet.
At an initial Point of 0.50, it estimates the root; the steps are as follows:
1- Estimate f(xi),f'(xi) , f’^2(xi) and f”(xi) at the starting Point of xi=0.50 for an initial i=0.
2- Substitute at xi=0.50 and get the values for f (0.50), f'(0.50), & f’^2(0.50,), an f ‘ ‘f ‘ ” (0.5050,), and get the Value of x1; it will be=0.7117.
3- Substitute x1 = 0.712and get the values for f (0.712), f'(0.712) & f’^2(0.712 and f”(0.712). We consider i = 1; wei = 1 to get the x-value at I when =2. The Value of x2 will be = 0. Please refer to the following slide image for more details.
4-Continue the process till x converges to 0.91, as shown in the next Table. The slide image shows the values of x starting from x0 = 0.50 to x1 = 0.711699, then x2 = 0.87601, then x3 = 0.909275, x4 = 0.91000, and then x5 = 0.91000. heck f(x) = 0.
The Table is shown more clearly in the slide image below.
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The next Post is 7A- Solved Problem 8 by the Modified Newton-Raphson method.
This is a useful Link for a numerical analysis calculator.