7A-Solved problem-8 by Modified Newton-Raphson method

Solved problem-8 by Modified Newton-Raphson method.

For the given Solved problem-8 by 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.

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 detailed estimation of the various parameters.

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).

4- Substitute in the modified Newton raphson . 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.

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 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.      

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.

3-Plug in with previous calaculatyion to the Modified newton Raphson equation and d out the 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.

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 by Newton Raphson method with 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 using the equation of xi=x0-(F(x0)/f'(x0), then we can 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.

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.

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.

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.

This is a comparison between the Newton raphson and Modified Newton Raphson method for the same solved problem-8 using an excel sheet.

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.