 # 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.

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.

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

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.

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:

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.

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.

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.

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.

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.

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.

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

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.

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.

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.

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.

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.

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

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.

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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