Solved problems for Newton-Raphson method.
The first problem of the two Solved problems for Newton-Raphson method
Another solved problem for the Newton-Raphson method for root extraction: find the roots of x^3-3x-5=0.
1- starting by the initial point x0=3.00.
First, we try to find that root value by plotting the curve by plugging different values of x for example starting from 0 to 3.0 and negative values from x= -0.50 to -4.00.
We are getting the corresponding values of y accordingly, to find roots, we find that the function value between one positive and one negative review lies between x=2 and x =2.50.
1-So our first iteration is selecting x0 value =3.
2-Get f(x0=3) and f'(x0=3) f (x) is =x^3-3x-5=0 and f'(x) = 3*x^2-3.
At x0=3 then f(3) = 3^3-3*3-5 =13 and f'(x0=3.00)=3*(3)^2-3=24.00.
3- X1 =3-(13/24)=2.45833.
For the second iteration
4- Get the value of f(2.45833) and f'(2.45833), by using the excel sheet, we get a new value x2=2.2943 and the corresponding f(x) for each iteration, shown in the next slide.
We keep making iterations till we get x4=2.279, the error value is small.
Two videos are introduced.
The first video includes solved problem #5 for the Newton Raphson method. The illustrations start from time 7.35 till the end. The video has a closed caption in English.
The second solved Problem of the two Solved problems for Newton-Raphson method
The second video includes an illustration of the second solved problem Number #6. This is the second solved problem of the two Solved problems for the Newton-Raphson method which is problem number 6: Use the Newton method for root extraction to find the roots of this function.
If you wish to review the pdf data used in the illustration, please continue reading.
This is the first derivative of the function.
Solution: He did not give us a starting point so it is good to make a casing for the root range by substituting
1-We make the table we put different values of X and we find the corresponding values of our function.
2-We’ll find that when we plug in X=0 then (0*e^0)-2, we get -2
3-For the next point where x=0.25, we get (-1.678994).
4-For x=0.50 we get -1.175639.
5- For x=0.750, we get. (-0.41225).
6- For x=1, we get +0.7182818.
So between x=0 and x=1.0 , x=0 is giving minus and x=1 is giving a positive value.
So our root should exist between 0 and 1. You have the choice either to start point x=0 or X =1. We put X=0 f(0)=(0)*e^0-2=-2.00 , for the slope value , check the next relation, it will be=1.
x1=x0-f(x0)/f'(x0)=0-(-2/1)=+2.00.
For f(2), we have the value of 12.778, and f'(2)=22.16716. So X2 will be =+2-f(x1=2)/f'(x1=2).
X2 will be =+2-(12.778/22.16716)=1.4325. After substitution, we get the value of the function as well as the derivative value. We use the expression of X3=x2- f(x2)/f'(x2), we get x3=1.034936.
It is better to make the calculation by using an excel sheet this includes a starting point with a selected value followed by a column that represents the function value f(x) and another new column for the value of f'(x), a column for the numerator which I f(x), and the denominator which is ‘(x), then a column for numerator/ denominator.
We have two starting points the first starting point is x0=0, and the second starting point is x0=1.
This is a function value f'(x) =4.34575 and we go on until we find that f(x) is coming closer to (0) at x value=0.8526.
When X =0.853003, the f(x) is zero, if our starting point x0=1.
This is the pdf file used for the illustration of this post and the previous post.
The next post is about the Modified-Newton-Raphson method, for which there is a modification of the Newton-Raphson method.
This is a useful link for a numerical analysis calculator.
