Last Updated on January 11, 2024 by Maged kamel
Solved problem-8 by Modified Newton-Raphson method.
For the given Solved problem-8 by the Modified Newton-Raphson method. The function f(x)=X^3-5*X^2+7x-3 with the first choice of x0=0.
Again I make a graph by plugging different values and getting the corresponding values of f(x), I have started from x=0 to x=3.50.
From the graph, as we can see from the next slide image, the roots are three roots x1=3 & x2=1 and x3=1 as shown in the Excel sheet for the Solved problem-8 by Modified Newton-Raphson method. the x=0 will give a negative value of -3.
![](https://magedkamel.com/wp-content/uploads/2022/01/Pict-5-num-7-numerical-posts.jpg)
1-We start to use the modified Newton-Raphson method, consider the initial point x0 equals 0, and we start to find the expressions for f(x),f'(x) , f’^2(x), and f”(x) .
2- Substitute at x=0 and get the values for f (0), f'(0) & f’^2(0), and f”(0). The value of f(0)=-3. The value of f'(0)=7.
The value of f”(0)=-10. Substitute in the equation of the modified Newton-Raphson method, we get the value of x1, which will be =1.105. Please refer to the next slide image for a detailed estimation of the various parameters.
![The detailed calculation of how to get the value of x1 for solved problem #8.](https://magedkamel.com/wp-content/uploads/2022/01/Pict-6-num-7-numerical-posts.jpg)
We plug in with the value of x=1.105263 and get the corresponding values of f(1.105263),f'(1.105263), and”(1.105263).
![The detailed calculation of how to get the value of x2 for the solved problem #8 from an esmated x1 value.](https://magedkamel.com/wp-content/uploads/2022/01/Pict-7-num-7-numerical-posts.jpg)
4- Substitute in the modified Newton-Raphson method. We get a new point with x2=1.00. and again continue to estimate f(1.00) and f'(1.00) and then apply them in the equation to get a new point that will be point x3.
![The detailed calculation of how to get the value of x2 for solved problem #8.](https://magedkamel.com/wp-content/uploads/2022/01/Pict-8-num-7-numerical-posts.jpg)
5-Continue the process until x converges to 1.00. The Excel sheet for the various values of x, based on the modified Newton-Raphson method from x0=0 till x4=1.00 is shown.
![An excel sheet to show the values of xi till we will get the f(1.00)=0.](https://magedkamel.com/wp-content/uploads/2022/01/Pict-9-num-7-numerical-posts.jpg)
This table shows how many iterations and the corresponding f(x),f'(x), and f”(x) for each case.
Solved problem-8 by Modified Newton-Raphson method using x0=4
Now if we consider the starting point as x0=4.00, and proceed to get the x value for f(x)& f'(x) , f’^2(x) and f”(x) for x0=4.00. We have f(4)=9,f'(4)=15 and f”(4)=14. The value of x1 is found to be=2.6363.
![Starting as the second choice by letting x0=4.00 and get the x1 value.](https://magedkamel.com/wp-content/uploads/2022/01/Pict-10-num-7-numerical-posts.jpg)
2- Substitute at x1=2.6363 and get the values for f (2.6363), f'(2.6363) & f”(2.6363). the values are shown in the next slide image.
![Substitute by the value of x1=2.6363 in the solved problem #8.](https://magedkamel.com/wp-content/uploads/2022/01/Pict-11-num-7-numerical-posts.jpg)
3-Plug in with previous calculation to the Modified Newton Raphson equation and d out the value of X2 Substitute at x2=2.8202 and get the values for f (2.6363), f'(2.8202) & f”^2(2.8202) and f”(2.8202) and get the value of x3, it will be=2.9617.
![Get the value of x2 by the modified newton-raphson method.](https://magedkamel.com/wp-content/uploads/2022/01/Pict-12-num-7-numerical-posts.jpg)
This is the Excel sheet for the calculation based on the Modified Newton Raphson Method starting from x0=4 till x5=3.00.
![Excel table for the different values of x for solved problem #8.](https://magedkamel.com/wp-content/uploads/2022/01/Pict-13-num-7-numerical-posts.jpg)
Solved problem-8 by Newton Raphson method with x0=0.
This is a comparison by the Newton Raphson method without modification for the same problem-8 but based on the Newton formula. Starting as before with x0=0 and using the equation of xi=x0-(F(x0)/f'(x0), we can get the x1 value 0.4285, then substitute and get the x2 value=0.6857.
![Resolve problem #8 by the Newton method.](https://magedkamel.com/wp-content/uploads/2022/01/Pict-14-num-7-numerical-posts.jpg)
This is an Excel sheet for the points that are obtained by using the Newton-Raphson method for the Solved problem-8
The initial point is (0) the table lists all the values of the function till point x4. That point will be equal to 0.95578.
![Excel table for the different values of x for solved problem #8.](https://magedkamel.com/wp-content/uploads/2022/01/Pict-15-num-7-numerical-posts.jpg)
If we start with x0=4.00 to get the other root but based on the Newton-Raphson method. The Excel sheet with more details shows the different values of x.
![The steps to estimate x1 and x2 from starting x0=4.](https://magedkamel.com/wp-content/uploads/2022/01/Pict-16-num-7-numerical-posts.jpg)
This is an Excel sheet for the points obtained while considering the initial point x0=4.00 ending with point x6 with a value of 3.
![Excel table for the different values of x for the solved problem #8 for x0=4.00.](https://magedkamel.com/wp-content/uploads/2022/01/Pict-17-num-7-numerical-posts.jpg)
This is a comparison between the Newton Raphson and Modified Newton Raphson methods for the same solved problem-8 using an Excel sheet.
![Excel table for the different values of x for solved problem #8.](https://magedkamel.com/wp-content/uploads/2022/01/Pict-13-num-7-numerical-posts.jpg)
This is the PDF file used for the illustration of this post.
The next post is Structural analysis numerically by the Newton-Raphson method.
This is a useful link for a numerical analysis calculator.