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Newton Raphson Method or Newton Method is a powerful technique for solving equations numerically. It is most commonly used for approximation of the roots of the real-valued functions.
Newton Raphson Method or Newton's Method is an algorithm to approximate the roots of zeros of the real-valued functions, using guess for the first iteration (x0) and then approximating the next iteration(x1) which is close to roots, using the following formula.
where,
Note: f'(x0) should not be zero else the fraction part of the formula will change to infinity which means f(x) should not be a constant function.
In the general form, the Newton-Raphson method formula is written as follows:
Where,
Assume the equation or functions whose roots are to be calculated as f(x) = 0.
In order to prove the validity of Newton Raphson method following steps are followed:
Step 1: Draw a graph of f(x) for different values of x as shown below:
π Newton Raphson Method Calculation
Step 2: A tangent is drawn to f(x) at x0. This is the initial value.
Step 3:This tangent will intersect the X- axis at some fixed point (x1, 0) if the first derivative of f(x) is not zero i.e. f'(x0) β 0.
Step 4: As this method assumes iteration of roots, this x1 is considered to be the next approximation of the root.
Step 5: Now steps 2 to 4 are repeated until we reach the actual root x*.
Now we know that the slope-intercept equation of any line is represented as y = mx + c,
Where m is the slope of the line and c is the x-intercept of the line.
Using the same formula we, get
y = f(x0) + f'(x0) (x - x0)
Here f (x0) represents the c and f' (x0) represents the slope of the tangent m. As this equation holds true for every value of x, it must hold true for x1. Thus, substituting x with x1, and equating the equation to zero as we need to calculate the roots, we get:
0 = f(x0) + f'(x0) (x1 - x0)
Which is the Newton Raphson method formula.
Thus, Newton Raphson's method was mathematically proved and accepted to be valid.
The Newton-Raphson method tends to converge if the following condition holds true:
| f(x). f''(x) | < | f'(x) |2
It means that the method converges when the modulus of the product of the value of the function at x and the second derivative of a function at x is lesser than the square of the modulo of the first derivative of the function at x. The Newton-Raphson Method has a convergence of order 2 which means it has a quadratic convergence.
Let's consider the following example to learn more about the process of finding the root of a real-valued function.
Example 1: For the initial value x0 = 3, approximate the root of f(x)=x3+3x+1.
Solution:
Given, x0 = 3 and f(x) = x3+3x+1
f'(x) = 3x2+3
f'(x0) = 3(9) + 3 = 30
f(x0) = f(3) = 27 + 3(3) + 1 = 37Using Newton Raphson method:
= 3 - 37/30
= 1.767
Example 2: For the initial value x0 = 1, approximate the root of f(x)=x2-5x+1.
Solution:
Given, x0 = 1 and f(x) = x2-5x+1
f'(x) = 2x-5
f'(x0) = 2 - 5 = -3
f(x0) = f(1) = 1 - 5 + 1 = -3Using Newton Raphson method:
β x1 = 1 - (-3)/-3
β x1 = 1 -1
β x1 = 0
Problem 3: For the initial value x0 = 2, approximate the root of f(x)=x3-6x+1.
Given, x0 = 2 and f(x) = x3-6x+1
f'(x) = 3x2 - 6
f'(x0) = 3(4) - 6 = 6
f(x0) = f(2) = 8 - 12 + 1 = -3Using Newton Raphson method:
β x1 = 2 - (-3)/6
β x1 = 2 + 1/2
β x1 = 5/2 = 2.5
Problem 4: For the initial value x0 = 3, approximate the root of f(x)=x2-3.
Given, x0 = 3 and f(x) = x2-3
f'(x) = 2x
f'(x0) = 6
f(x0) = f(3) = 9 - 3 = 6Using Newton Raphson method:
β x1 = 3 - 6/6
β x1 = 2
Problem 5: Find the root of the equation f(x) = x3 - 5x + 3 = 0, if the initial value is 3.
Solution:
Given x0 = 3 and f(x) = x3 - 5x + 3 = 0
f'(x) = 3x2 - 5
f'(x0 = 3) = 3 Γ 9 - 5 = 22
f(x0 = 3) = 27 - 15 + 3 = 15Using Newton Raphson method
β x1 = 3 - 15/22
β x1 = 2.3181Using Newton Raphson method again:
x2 = 1.9705
x3 = 1.8504
x4 = 1.8345
x5 = 1.8342
Therefore, the root of the equation is approximately x = 1.834.
Problem 1: Find the root of f(x) = x2-2 using the Newton-Raphson method starting with x0=1.
Problem 2: Find the root of f(x) = x3-2x+1 using the Newton-Raphson method starting with x0=0.
Problem 3: Find the root of f(x) = cos(x)-x using the Newton-Raphson method starting with x0=0.5.
Problem 4: Find the root of f(x) = ex-3x using the Newton-Raphson method starting with x0=1.
Problem 5: Find the root of f(x) = x3-4x2+6 using the Newton-Raphson method starting with x0=2.
Problem 6: Find the root of f(x) = lnβ‘(x)-1 using the Newton-Raphson method starting with x0=2.
Problem 7: Find the root of f(x) = x4-8x2+16 using the Newton-Raphson method starting with x0=2.5.
Problem 8: Find the root of f(x) = xsinβ‘(x)-1 using the Newton-Raphson method starting with x0=1.
Problem 9: Find the root of f(x)=x5-3x3+2 using the Newton-Raphson method starting with x0=1.
Problem 10: Find a root of f(x) = x3-6x2+11x-6 using the Newton-Raphson method starting with x0=3.