Last updated on August 10th, 2025
Newton's Method, also known as the Newton-Raphson method, is an iterative numerical technique used to find approximate solutions to real-valued functions. In this topic, we will learn the formula for Newton's Method and how it is used to find the roots of equations.
Newton's Method is a powerful tool to approximate the roots of a real-valued function. Let's learn the formula used in Newton's Method to calculate these roots.
Newton's Method is used to approximate the roots of a real-valued function f(x).
The formula is: x_(n+1) = x_n - f(x_n)/f'(x_n) where x_n is the current approximation, f(x_n) is the function value at x_n, and f'(x_n) is the derivative value at x_n.
In mathematics and applied sciences, Newton's Method is crucial for solving equations and finding roots of functions.
Here are some important aspects of Newton's Method:
Students often find the Newton's Method formula challenging.
Here are some tips to master it:
In real life, Newton's Method is applied in various fields to solve practical problems.
Here are some applications:
Students often make errors when applying Newton's Method. Here are some mistakes and how to avoid them to master Newton's Method.
Apply Newton's Method to approximate the root of f(x) = x^2 - 2 starting with x_0 = 1.
The approximate root is 1.4142
First, compute f'(x) = 2x.
Using x_0 = 1, f(x_0) = 12 - 2 = -1, f'(x_0) = 2(1) = 2.
Apply the formula: x_1 = 1 - (-1)/2 = 1.5
Repeat: f(x_1) = 1.52 - 2 = 0.25, f'(x_1) = 2(1.5) = 3 x_2 = 1.5 - 0.25/3 ≈ 1.4167
Continue until convergence or desired precision.
Use Newton's Method to find an approximation to the cube root of 27, starting with x_0 = 3.
The approximate cube root is 3
Let f(x) = x3 - 27, then f'(x) = 3x2.
Starting with x_0 = 3, f(x_0) = 33 - 27 = 0, f'(x_0) = 3(3)2 = 27. x_1 = 3 - 0/27 = 3
The method quickly confirms the cube root is 3.
Approximate the root of f(x) = cos(x) - x using Newton's Method with x_0 = 1.
The approximate root is 0.7391
First, compute f'(x) = -sin(x) - 1.
Using x_0 = 1, f(x_0) = cos(1) - 1 ≈ -0.4597, f'(x_0) = -sin(1) - 1 ≈ -1.8415. x_1 = 1 - (-0.4597)/(-1.8415) ≈ 0.7504
Repeat until convergence.
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