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

Last Updated on September 23, 2026 by Maged kamel

Solved Problem-8 by the Modified Newton-Raphson method.

Solved Problem 8 using the Modified Newton-Raphson method. The function f(x)=X^3-5*X^2+7x-3 with the first choice of x₀=0. The following slide includes the formula for the Modified Newton-Raphson method.

From the graph (next slide), we can see three roots: x1 = 3, x2 = 1, and x3 = 1, as shown in the Excel sheet for Solved problem-8 using the Modified Newton-Raphson method. 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. Substituting into the modified Newton-Raphson equation, we get x1 = 1.10526. Please refer to the next slide image for a detailed estimation of the various parameters.

We plug in x = 1.105263 and get the corresponding values of f(1.105263), f'(1.105263), and f”(1.105263).

4- Substitute into the modified Newton-Raphson method. We get a new Point with x2 = 1.00. Then 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 in the following slide.

This Table shows the number of iterations and the corresponding f(x), f'(x), and f ” (x) for each case. 

Solved Problem 8 by the 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 the Value of f(4)=9, and the slope Value at x=4, 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 image on the next slide.

3-Plug the previous calculation into the Modified Newton-Raphson equation and substitute the Value of X2. Substitute x2 = 2.8202 and get the values for f(2.6363), f'(2.8202) & f”^2(2.8202), f”(2.8202), and x3 = 2.9617.

This is the Excel sheet for the calculation based on the Modified Newton-Raphson Method, starting from x0 = 4 to x5 = 3.00.

Solve the Problem using the Newton-Raphson method with x0 = 0.

This compares the Newton-Raphson method without modification for Problem 8, based on the Newton formula. Starting as before with x0=0 and using the equation of xi=x0-(F(x0)/f'(x0)

We get x1 = 0.4285, then substitute to get x2 = 0.6857.

This is an Excel sheet for the points that are obtained by using the Newton-Raphson method to solve Problem 8

The initial Point is (0). The Table lists all function values up to x4. That Point will be equal to 0.95578.

If we start with x0 = 4.00 to find the other root using the Newton-Raphson method, the Excel sheet with more details shows the different x values.

This Excel sheet shows the points obtained, starting with x0 = 4.00 and ending with x6 = 3.

This Excel sheet shows a comparison of the modified Newton-Raphson method for Problem 8.

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The next Post is Structural analysis numerically by the Newton-Raphson method.

This Link is useful for a numerical analysis calculator.