6- Newton-Raphson method-Easy approach

Last Updated on September 23, 2026 by Maged kamel

Newton-Raphson method.

Newton-Raphson’s method is another root-finding method. The Newton-Raphson Expression for root-finding uses the linear approximation we discussed.

From the next slide image. L(x_b)=f(x_a)+f'(x_a)*(x_b-x_a), where xa is the starting Point, and xb is the ending Point.

Now, if we assume L(xb) is approximately equal to (xb), we can modify the previous equation by moving (x_b-x_a) to the left side and rewriting it.

We could say x final=x intial+(1/ slope at the ix initial)+(1/ slope at x initiall)*(f(x final)l- f(x initial).
If we are looking for x_b where the root =0, or saying f(x_b)=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 obtain the distance x to the root Point b we are looking for.

This will create another form of the equation as (x_b-x_a)=(1/f'(x_a))*(f(x_b)-f(x_a). The next step is to find the Value of x_b, which is=x_a+(1/f'(x_a))*(f(x_b)-f(x_a).

Suppose we have a curve and are looking for its root at a given Point. We want to find the x-value of that root Point. So we say that if we have a Point x1, we go up, then make a tangent to the curve at that Point.

So we get another Point, x2, and a relation between x2 and the old Point. T, his relation will be x₂=x₁-, to'(x1) as shown in the nuntilde point whereontinPointis process ununtil a pointwhere we have f(x) close to or =0. This is the Newton-Raphson method.

A solved problem using the Newton-Raphson method.

First, write the Newton-Raphson equation, then solve example #4. Use the Newton method to find the roots of √29. We will solve it in the next steps.
We set x = 29, or we can write it as x^2 = 29; then let x^2 – 29 = 0.
1-We readjust the function, and we set i equal to 0. The f(x) is to be written as f(x)=x^2-29.

We use an Excel graph for the given function to locate the Point where f(x)=0, which is between 5.3852 and 5.391.

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 with X0, got X1, then substituted this new Value, which we got as 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 nextPointt x₃.
For the third iteration.
10- We put   x₂=5.3852 as obtained from the previous iteration, f(5.3852) = (5.3852)^2-29=–0.000379
11-Estimate the f'(x₁=5.3852) =(2*x₂)-0=2*5.3852=10.7704.

12- Estimate X₃ value by using the Newton-Raphson method, X₂=5.3852-(-0.00022/10.7704)=5.38516.
13- The value of x3, x₃=5.3852, check (5.38516)^2-29=4.2E-10 close to zero.

This Excel sheet shows the iterations of the different f(x) values and their first derivatives.

You can review and download the PDF file for this Post and the following Post via the following document.

This is a useful Link for a numerical analysis calculator.

The next Post covers solved problems for the Newton-Raphson method.