# Applications of Differentiation – Newton’s Method for Finding Roots of Functions

From previous math courses, finding the roots (or zeros) of simple functions, such as linear functions and quadratic functions, are quite simple by using basic formulas and procedures. However, for higher degree polynomials, such as cubic, quartic (4th degree), and quintic (5th degree), such formulas do not exist, or the calculations to find the roots are very laborious.

Newton’s Method is a recursive process for approximating the root of a differentiable function where other methods of finding roots may not exist or are too strenuous to do by using standard math procedures.

 Formula: Newton’s Method Suppose F(x) will be a differentiable function. Select a point x0 close to the root of the function. The Newton Method formula is given as: Where n = 0, 1, 2, 3, 4…… F(x) will be the given function and F’ (x) will be the First Derivative of the function.

The video below displays Newton’s Method in action:

The most important thing that should be remembered for this method is that this method will be invalid if the first derivative of any function is zero.

 Guidelines for Applying Newton’s Method Select an x value (x1) that is close to the root you want to approximate. Input x1 into the original function F(x) to obtain F(x1). Find the derivative of F(x). Input x1 into F’(x) to obtain F’(x1) Substitute the values of x1, F(x1), and F’(x1) into the formula to obtain the new x value xn+1 . Repeat steps 1 – 4 until the root is found to the desired accuracy. Important Notes: The more iterations you complete, the more accurate your approximation is. Although a function can have several roots, Newton’s Method is used to find only one root of a function at a time. To find the other roots of a function, you have to select another x value to start the process again. It will help you to create a table of xn, F(xn), F’(xn), and to keep track of your calculations.

Let’s solve some examples to apply this formula.

Example 1:

Find the root of the function F (x) = x3 by using Newton’s Method. Complete 4 iterations.

Solution:

Select an x value (x1) that is close to the root you want to approximate.

Input x1 into the original function F(x) to obtain F(x1).

Our x1 is 1, so we will find F(3):

F(1) = (1)3 = 1

Therefore, F(x1) = 1.

Find the derivative of F(x). Input x1 into F’(x) to obtain F’(x1).

Find the derivative of F (x) = x3

F’(x) = (3)x3-1=3x2(power rule)

Plugging in x1= 3 to F’(x),

F’(3) = 3(1)2= 3*1 = 3

Therefore, F’(x1) = 3

Substitute the values of x1, F(x1), and F’(x1) into the formula to obtain the new x value xn+1 .

Using Newton’s Method: The next x value to run through the process is x2= 0.666.

We are going to repeat the process. First, find F(0.666)

F(0.666) = (0.666)3 = 0.295

Find F’(0.666).

F’(0.666) = 2(0.666)2= 0.887

Applying the formula, The next x value to run through the process is x3= 0.333.

Once again, find F(0.333):
F(0.333) = (0.333)3= 0.110

Find F’(0.333).

F’(0.333) = 2(0.333)2= 0.221

Applying the formula, The next x value to run through the process is x4= -0.164

Once again, find F(-0.164):
F(-0.164) = (-0.164)3= -0.004

Find F’(-0.164).

F’(-0.164) = 2(-0.164)2= 0.053

Applying the formula, Let’s fill out the data table of our calculations.

 n xn F(xn) F’(xn) 1 1= x1 1 3 0.666 = x2 2 0.666 0.295 0.887 0.333 = x3 3 0.333 0.110 0.221 -0.164 = x4 4 -0.164 -0.004 0.053 -0.088 = x5

This completes the 4th and last iteration. x5= -0.088 and is our approximation of the root of the function. Even though our approximations oscillate from positive to negative, if we continue to do more iterations, the x value gets closer to zero, which is the real root of the function F(x) = x3. The graph of F(x) proves this: Example 2:

Suppose we have a function F(x) = x2-4x . Use Newton’s Method to find the root value of the given function.

Solution:

Select an x value (x1) that is close to the root you want to approximate.

Let’s first look at the graph of F(x): Let’s say we wanted to approximate what the root value that is highlighted in yellow. Let’s pick a number close to that value, such as 5.

Input x1 into the original function F(x) to obtain F(x1).

Our x1 is 5, so we will find F(5):

F(1) = (5)2-4(5) = 5

Therefore, F(x1) = 5.

Find the derivative of F(x). Input x1into F’(x) to obtain F’(x1).

Find the derivative of F (x) = x2-4x

F’(x) = 2x-4 (power rule)

Plugging in x1= 5 to F’(x),

F’(5) = 2(5)-4 = 6

Therefore, F’(x1) = 6

Substitute the values of x1, F(x1), and F’(x1) into the formula to obtain the new x value xn+1 .

Using Newton’s Method: The next x value to run through the process is x2= 4.16667.

We are going to repeat the process. First, find F(4.16667)

F(4.16667) = (4.1667)2-4(4.16667) = 0.6947

Find F’(4.16667).

F’(4.16667) = 2(4.16667)-4= 4.3333

Applying the formula, The next x value to run through the process is x3= 4.0064.

Once again, find F(4.0064):
F(4.0064) = (4.0064)2-4(4.0064)= 0.0256

Find F’(4.0064).

F’(4.0064) = 2(4.0064)-4= 4.0128

Applying the formula, Let’s fill out the data table of our calculations.

 n xn F(xn) F’(xn) 1 5 = x1 5 6 4.16667= x2 2 4.16667 0.6947 4.3333 4.0064 = x3 3 4.0064 0.0256 4.0128 4.0000 = x4

From three iterations, x4= 4.0000, which is our approximation of the root. If we continue to do iterations, it is clear that the root of the function near x = 5 is x = 4. Let’s graph the function F(x) and zoom in to prove the root is x = 4: Example 3:

Suppose we are given the function, F (x) = x3-3x2+1. Using the initial value x1 = 0.6, use the Newton Method to find a root of the given function.

Solution:

Select an x value (x1) that is close to the root you want to approximate.

Input x1 into the original function F(x) to obtain F(x1).

Our x1 is 1, so we will find F(0.6):

F(0.6) = (0.6)3-3(0.6)2+1= 0.136

Therefore, F(x1) = 0.136.

Find the derivative of F(x). Input x1 into F’(x) to obtain F’(x1).

Find the derivative of F (x) = x3-3x2+1

F’(x) = (3)x3-1-(2)32-1+0=3x2-6x(power rule)

Plugging in x1= 0.6 to F’(x),

F’(0.6) = 3(0.6)2-6(0.6)= -2.52

Therefore, F’(x1) = -2.52

Substitute the values of x1, F(x1), and F’(x1) into the formula to obtain the new x value xn+1 .

Using Newton’s Method: The next x value to run through the process is x2= 0.6539.

We are going to repeat the process. First, find F(0.6539)

F(0.6539) = (0.6539)3-3(0.6539)2+1= -0.0031

Find F’(0.6539).

F’(0.6539) = 3(0.6539)2-6(0.6539)= -2.6406

Applying the formula, The next x value to run through the process is x2= 0.6527.

We are going to repeat the process. First, find F(0.6527)

F(0.6527) = (0.6527)3-3(0.6527)2+1= 0

Find F’(0.6527).

F’(0.6539) = 3(0.6527)2-6(0.6527)= -2.6381

Applying the formula, Let’s fill out the data table with our calculations:

 n xn F(xn) F’(xn) 1 0.6 = x1 0.136 -2.52 0.6539= x2 2 0.6539 -0.0031 -2.6406 0.6527 = x3 3 0.6527 0 -2.6381 0.6527 = x4

From three iterations, x4= 0.6527, which is our approximation of the root. If we continue to do iterations, it is clear that the root of the function near x = 0.6 is x = 0.6527 or another value extremely close to this. Let’s graph the function F(x) and zoom in on the root in this region: As we see, the root is x = 0.6527.

 Summary of Section Newton’s Method is an algorithm used to create increasingly accurate approximations of roots of a function. Essentially, the steps are as follows: Select an x value that is close to the root of the function (x0). Find F(x0). Find the equation of the tangent line at (x0, F(x0)). Find the x intercept of the tangent line (x1). Plug in the x intercept (x1) back into F(x) to obtain F(x2). Find the equation of the tangent line at (x2, F(x2)). Find the x intercept of the tangent line (x3). And so on and so forth. This repeated interaction gives us the formula for Newton’s Method: Repeat the iteration until the desired root is found to the desired accuracy.