Brief content of Numerical post 7.

7- An Easy guide to Modified Newton-Raphson method.

Modified Newton-Raphson method.

Introduction to bracketing method. The bracketing method is a  Numerical method, that represents two values of a function having opposite signs, the root will be in -between.


The modified Newton -Raphson method is another method for root finding.  A simple modification to the previous method of Newton -Raphson was introduced.

The video I used for the illustration.

You can click on any picture to enlarge, then press the small arrow at the right to review all the other images as a slide show.

I have introduced the Modified Newton Raphson method by applying the new equation for two solved problems. the first solved problem is solved problem7. I have used the Modified newton raphson method to solve. I have introduced an excel table to show the iterations starting from x0=0.50.

Example 7- data for solution.

This is adetailed solution by using excel sheet for the Modified Newton raphson method shown in the next slide image.

For the second solved problem #8, I have included an excel sheet for the comparison between the results obtained by using the Newton Raphson method and the Modified newton raphson method.

Example 8- data and comparision tables.

This is a comparison sheet between the Newton raphson method and the Modified Newton raphson method shown in the next slide image.

Modified Newton-Raphson method

This formula of the Modified Newton-Raphson method is shown in the next slide image.

Solved problem#7 by using the modified Newton-Raphson Method.

We start by solving problem #6. For the modified Newton Raphson method, I used an excel sheet to determine the root value of the given f(x)=e^x-3*x^2.

Solved problem #7 by the modified Newton- Raphson method.

I make a table between x and the corresponding value of f(x), by selecting several values of x, starting from 0 to a value of 1.10.

After drawing the function by excel, the point of f(x)=0 was detected to be at x=0.91 or the root value is at x= 0.91, please refer to the excel sheet. at an initial point 0.50, it is required to estimate the root point, the steps are as follows:

Pict 3 num 7 numerical posts

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)  and f”(0.50) and get the value of x1, it will be=0.7117.

 3- Substitute  at x1=0.7117 and get the values for f (0.7117), f'(0.7117)  & f’^2(0.7117)  and f”(0.71112), now we consider i=1, we want to get the x value at I when=2. The value of x2, it will be=0.876.

An excel sheet to show the values of xi till we will get the f(0.91)=0.

4-continue the process till x converges to 0.91 as shown in the next table. the next 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. Check the f(x)=0.

Solved problem #8 for the Modified Newton-Raphson method.

For the given example of

\begin{equation} f(x)=X^{3}-5 X^{2}+7 x-3 \end{equation}


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.

Solved problem #8 by the modified Newton- Raphson method.

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 Solved problem No.8.

The detailed calculation of how to get the value of x1 for solved problem #8.

1- we start to use the modified Newton-raphson method, we estimate f(x),f'(x) , f’^2(x) and f”(x) as x0=0.                                 
2- Substitute  at x=0 and get the values for f (0), f'(0)  & f’^2(0) and f”(0) We can get the value of x1,which will be =1.105. please refer to the next slide image.

The detailed calculation of how to get the value of x2 for the solved problem #8.

3- Substitute  at x1=1.105263 and get the values for f (1.105263), f'((1.105263)  & f’^2((1.105263)  and f”((1.105263) and get the value of x2, it will be=1.000. The details of the calculations are shown in the next two slide images.

The detailed calculation of how to get the value of x2 for solved problem #8.

4- 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.

An excel sheet to show the values of xi till we will get the f(1.00)=0.

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.    

This table shows how many iterations and the corresponding f(x),f'(x), and f”(x) for each case.     

Solved problem#8 for the Modified Newton-Raphson method-with a starting point x0=4

Starting as the second choice by letting x0=4.00 and get the x1 value.

Now if we consider the starting point as x0=4.00, and proceed to get the x value for f(x)=0.1- Estimate f(x), f'(x) , f’^2(x) and f”(x) for x0=4.00. The value of x1 is found to be=2.6363.      

Pict 11 num 7 numerical posts

2- Substitute  at x1=2.6363 and get the values for f (2.6363), f'(2.6363)  & f”^2(2.6363)  and f”(2.6363) and get the value of x2, it will be=2.8202.

Get the value of x2.

3- 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.

Excel table for the different values of x for solved problem #8.

This is the excel sheet for the calculation based on the Modified Newton Raphson Method starting from x0=4 till x5=3.00.

Solved problem#8 for the Newton-Raphson method with a starting point x0=0.

Resolve problem #8 by the Newton method.

This is a comparison by the Netwon-Raphson method without modification for the same problem #8 but based on the Newton formula. Starting as before with x0=0 and get the x1 value 0.4285, then substitute and get the x2 value=0.6857.

Excel table for the different values of x for solved problem #8.

This is an excel sheet for the points that are obtained by using the Newton-Raphson method.

The steps to estimate x1 and x2 from starting x0=4.

If we start with x0=4.00 to get the other root but based on the Newton method. The excel sheet with more details shows the different values of x.

Excel table for the different values of x for the solved problem #8 for x0=4.00.

This is an excel sheet for the points obtained while considering x0=4.00

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.

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