Newton’s Raphson method
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The Newton Raphson Method is referred to as one of the most commonly used techniques for finding the roots of given equations. It can be efficiently generalised to find solutions to a system of equations. Moreover, we can show that when we approach the root, the method is quadratically convergent. In this article, you will learn how to use the Newton Raphson method to find the roots or solutions of a given equation, and the geometric interpretation of this method.
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
This is called Newton Raphson formula.
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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.
Solved Examples
Example 1:
Find the cube root of 12 using the Newton Raphson method assuming x0 = 2.5.
Solution:
We know that, the iterative formula to find nth root of a is given by:
From the given, a = 12, b = 3
Let x0 be the approximate cube root of 12, i.e., x0 = 2.5.
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 method 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 – [cos 0.5/ (-4 – sin 0.5)]
= 0.5 – [0.8775/ (-4 – 0.4794)]
= 0.5 + 0.1958
= 0. 6958
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