Newton Raphson Method
Table of Contents
One of the most popular methods for locating the roots of given equations is the Newton Raphson Method. To efficiently find answers to a set of equations, it can be generalized. Additionally, we can demonstrate that the method is quadratically convergent as we get closer to the root. This article explains the geometric interpretation of the Newton-Raphson method for locating the roots or solutions of a given equation.
Newton Raphson Method Formula
Let x0 be the approximate root of f(x) = 0 and let x1 = x0 + h be the correct root. Then f(x1) = 0
⇒ f(x0 + h) = 0….(1)
By expanding the above equation using Taylor’s theorem, we get:
f(x0) + hf1(x0) + … = 0
⇒ h = -f(x0) /f’(x0)
Therefore, x1 = x0 – f(x0)/ f’(x0)
Now, x1 is the better approximation than x0.
Similarly, the successive approximations x2, x3, …., xn+1 are given by
xn+1 = xn – f(xn)/f'(xn)
This is called Newton Raphson formula.
Geometrical Interpretation of Newton Raphson Formula
The geometric meaning of Newton’s Raphson method is that a tangent is drawn at the point [x0, f(x0)] to the curve y = f(x).
It cuts the x-axis at x1, which will be a better approximation of the root. Now, drawing another tangent at [x1, f(x1)], which cuts the x-axis at x2, which is a still better approximation and the process can be continued till the desired accuracy is achieved.
Convergence of Newton Raphson Method
The order of convergence of Newton Raphson method is 2 or the convergence is quadratic. It converges if |f(x).f’’(x)| < |f’(x)|2. Also, this method fails if f’(x) = 0.
Newton Raphson Method Examples
Example 1:
Find the cube root of 12 using the Newton Raphson method assuming x0 = 2.5.
Solution:
From the given, a = 12, b = 3
Let x0 be the approximate cube root of 12, i.e., x0 = 2.5.
We know that, the iterative formula to find bth root of a is give by:
So, x1 = (⅓) [2x0 + 12/x02]
= (⅓) [2(2.5) + 12/(2.5)2]
(⅓) [5 + 12/6.25]
= (⅓)(5 + 1.92)
6.92/3
= 2.306
Now,
x2 = (⅓)[2x1 + 12/x12]
= (1/3) [2(2.306) + 12/(2.306)2]
(⅓) [4.612 + 12/5.3176]
= (⅓) [4.612 + 2.256]
6.868/3
= 2.289
Therefore, the approximate cube root of 12 is 2.289.
Example 2:
Find a real root of the equation -4x + cos x + 2 = 0, by Newton Raphson up to four decimal places, assuming x0 = 0.5.
Solution:
Given equation: -4x + cos x + 2 = 0
x0 = 0/5
Let f(x) = -4x + cos x + 2
f’(x) = -4 – sin x
Now,
f(0) = -4(0) + cos 0 + 2 = 1 + 2 = 3 > 0
f(1) = -4(1) + cos 1 + 2 = -4 + 0.5403 + 2 = -1.4597 < 0
Thus, a root lies between 0 and 1.
Let us find the first approximation.
x1 = x0 – f(x0)/f’(x0)
= 0.5 – [-4(0.5) + cos 0.5 + 2]/ [-4 – sin 0.5]
0.5 – [(-2 + 2 + cos 0.5)/ (-4 – sin 0.4)]
= 0.5 – [cos 0.5/ (-4 – sin 0.5)]
0.5 – [0.8775/ (-4 – 0.4794)]
0.5 – (0.8775/-4.4794)
= 0.5 + 0.1958
= 0. 6958
Frequently Asked Questions
Where is the Newton Raphson method used?
This method for numerically solving the equations is effective. In terms of solutions, it provides us with better approximations.
Which is the correct formula of the Newton Raphson method?
xn+1 = xn – f(xn)/f'(xn)
Is the Newton Raphson method always convergent?
Not always, the RPM is not convergent. That implies that it cannot always ensure that the condition is met. However, if f'(x) equals 0, this method fails.