The document describes the Newton-Raphson method for finding the roots of nonlinear equations. It provides the derivation of the method, outlines the algorithm as a 3-step process, and gives an example of applying it to find the depth a floating ball submerges in water. The advantages are that it converges fast if it converges and requires only one initial guess. Drawbacks include potential issues with division by zero, root jumping, and oscillations near local extrema.

1. Roots of a Nonlinear
Equation
Topic: Newton-Raphson Method
Major: General Engineering
07/20/09 http://numericalmethods.eng.usf.edu 1

2. Newton-Raphson Method
f(x)
f(xi )
f(xi)
[x f( x )] xi +1 = xi -
i, i ′
f (xi )
f(xi-1)
θ
xi+2 xi+1 xi X
2

3. Derivation
f(x)
AB
tan(α ) =
AC
f(xi) B
f ( xi )
f ' ( xi ) =
xi − xi +1
f ( xi )
xi +1 = xi −
C α A X f ' ( xi )
xi+1 xi
3

4. Algorithm for Newton-
Raphson Method
4

5. Step 1
Evaluate f′(x) symbolically
5

6. Step 2
Calculate the next estimate of the root
f(xi )
xi +1 = xi -
f'(xi )
Find the absolute relative approximate error
xi +1- xi
∈a = x 100
xi +1
6

7. Step 3
 Find if the absolute relative approximate error is
greater than the pre-specified relative error
tolerance.
 If so, go back to step 2, else stop the algorithm.
 Also check if the number of iterations has exceeded
the maximum number of iterations.
7

8. Example
 You are working for ‘DOWN THE TOILET COMPANY’ that
makes floats for ABC commodes. The ball has a specific
gravity of 0.6 and has a radius of 5.5 cm. You are asked
to find the distance to which the ball will get submerged
when floating in water.
8

9. Solution
The equation that gives the depth ‘x’ to which the ball is
submerged under water is given by
f ( x ) = x 3-0.165 x 2+3.993x10- 4
f ( x ) = 3x 2-0.33x
Use the Newton’s method of finding
roots of equations to find the depth
‘x’ to which the ball is submerged
under water. Conduct three
iterations to estimate the root of the
above equation.
9

10. Graph of function f(x)
f ( x ) = x -0.165 x +3.993x10
3 2 -4
10

11. Iteration #1
x0 = 0.02
f ( x0 )
x1 = x0 −
f ' ( x0 )
3.413x10−4
x1 = 0.02 −
− 5.4 x10−3
= 0.08320
∈ = 75.96%
a
11

12. Iteration #2
x1 = 0.08320
f ( x1 )
x2 = x1 −
f ' ( x1 )
−1.670x10 −4
x2 = 0.08320 −
− 6.689 x10 −3
= 0.05824
∈ = 42.86%
a
12

13. Iteration #3
x2 = 0.05824
f ( x2 )
x3 = x2 −
f ' ( x2 )
3.717 x10−5
= 0.05284 −
− 9.043x10−3
= 0.06235
∈ = 6.592%
a
13

14. Advantages
 Converges fast, if it converges
 Requires only one guess
14

15. Drawbacks
10
f(x)
5
0 x
-2 -1 2 0 3 1 2
1
-5
f ( x ) = ( x − 1) = 0
-10
3
-15
-20
Inflection Point
15

16. Drawbacks (continued)
1.00E-05
f(x)
7.50E-06
5.00E-06
2.50E-06
0.00E+00 x
-0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04
-2.50E-06
0.02
-5.00E-06
f ( x ) = x 3 − 0.03 x 2 + 2.4 x10 −6 = 0
-7.50E-06
-1.00E-05
Division by zero
16

17. Drawbacks (continued)
1.5
f(x)
1
0.5
x
0
-2
-0.063069 0.54990
0 2 4
4.462 6
7.53982
8 10
-0.5
-1 f ( x ) = Sin x = 0
-1.5
Root Jumping
17

18. Drawbacks (continued)
6
f(x)
5
4
3
3
2
2 f ( x) = x2 + 2 = 0
11
4
x
0
-2 -1 0 1 2 3
-1.75 -0.3040 0.5 3.142
-1
Oscillations near Local Maxima or Minima
18