Newton Raphson’s method is another method for root finding. The Newton-Raphson expression of root-finding utilizes the linear approximation which we have discussed.
Newton-Raphson method video.
The video includes the steps we use to come closer to the root point, for a function, as we can see. where the function value=0 or close to zero, by starting with an initial point say x0, coming up to get the f(x0) value, then creating a slope at the curve, then hitting the x-axis at another point say x1.
Check the f(x1), if it is zero then this is the root point, else, take a vertical line and hit the curve, then create a new slope line, get another point say x2, and so on till you find a point that is the root point.
Two solved examples are introduced. The video has a closed caption in English.
You can watch the video from start till time 7.30 for the content of this post.
From the next slide image. L(xb)=f(xa)+f'(xa)*(xb-xa) as a is the starting point and the xb is the ending point Now if we consider that L(xb) approximately equals (xb) and create a little modification can be done on the previous equation by letting (xb-xa) on the left side, and then rewrite the equation.
We could say x final=x intial+(1/ slope at the ix initial)+(1/ slope at x intial)*(f(x fina)l- f(x initial).
If we are looking for xb where the root =0, or saying f(xb)=0.
The formula will be xb=xa+(1/f'(xa)(0-f(xa)=xa-(1/f'(xa)(f(xa).
The formula can be used to get the distance x for the root point b for which we are looking.
This will create another form of the equation as (xb-xa)=(1/f'(xa))*(f(xb)-f(xa). The next step is to find the value of xb which is=xa+(1/f'(xa))*(f(xb)-f(xa).
Suppose we have a curve and that curve we are looking for the root of that curve at a certain point. We want to find what is the x value of that root point. So we are saying that if we have x1 point, we go up and then we make a tangent at the curve at that point.
So we get another point which we call x 2 and we get a relation between x2 the new point and the old point. This relation will be x2=x1– f(x1) /f'(x1) as shown in the next slide image to continue this process till a point where we have f(x) close to or =0. This is the Newton-Raphson method.
Solved problem using the Newton-Raphson method.
First, the equation of Newton-Raphson is written. followed by a solved example #4 Example number 4. Use the Newton method to find the roots of the √29.
The solution will be made through the next steps.
We put x= √ 29 or it could be expressed at X^2= to 29 then let X^2-29 =0.
1-We readjust the formula a for the function and we equate it to 0.
2-We put x0=5 as starting point after that get f(5) = 5^2-29=-4.
The negative sign will change the relationship as we will see later.
3-Estimate the f'(x0=5) =(2*x0)-0=2*5.0=10.00.
4- Estimate X1 value by using the Newton-Raphson method, X1=5-(-4/10.00)=5.40.
5-This is for the first iteration. We started from X0 we get X1 then again we are substituting by this new value which we are getting which is = 5.40. For the second iteration.
6- We put x1=5.40 as obtained from the previous iteration, f(5.40) = (5.40)^2-29=+0.16.
7-Estimate the f'(x1=5.40) =(2*x1)-0=2*5.40=10.80.
8- Estimate X2 value by using the Newton-Raphson method, X2=5.40-(+0.16/10.80)=5.3852.
9-x2=5.3852, check (5.385)^2-29=-0.001775 not zero, so proceed to get the next point x3.
For the third iteration.
10- We put x2=5.3852 as obtained from the previous iteration, f(5.3852) = (5.3852)^2-29=–0.00022
11-Estimate the f'(x1=5.3852) =(2*x2)-0=2*5.3852=10.7703.
12- Estimate X3 value by using the Newton-Raphson method, X2=5.3852-(-0.00022/10.7703)=5.38516.
13-x3=5.38516, check (5.38516)^2-29=4.2E-10 close to zero.
This is the excel sheet showing the iterations and the different f(x) values and the first derivative values.
This is a useful link for a numerical analysis calculator.