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03.05 
Newton-Raphson 
Method of Solving 
Nonlinear Equations 
 
 
 
After reading this chapter, you should be able to: 
 
1. Understand how to use the Newton-Raphson method for solving nonlinear 
equations. 
 
 
 
What is the Newton-Raphson method and how does it differ from other 
methods to solve nonlinear equations? 
 
Methods such as bisection method and the false position method of 
finding roots of a nonlinear equation f(x=0 require bracketing of the root by two 
guesses. Such methods are called bracketing methods. These methods are always 
convergent since they are based on reducing the interval between the two guesses 
to zero in on the root. 
03.05.1 
03.05.2 Chapter 03.05 
 
In the Newton-Raphson method, the root is not bracketed. Only one initial 
guess of the root is needed to get the iterative process started to find the root of an 
equation. Hence, the method falls in the category of open methods. 
Newton-Raphson method is based on the principle that if the initial guess 
of the root of f(x)=0 is at xi, then if one draws the tangent to the curve at f(xi), the 
point xi+1 where the tangent crosses the x-axis is an improved estimate of the root 
(Figure 1). 
 Using the definition of the slope of a function, at ixx = 
 θ) = (xf i tan′
 
1
0
+−
−′
ii
i
i xx
)f(x) = = (xf 
which gives 
 
)f'(x
)f(x - = xx
i
i
ii 1+ 
 (1) 
Equation (1) is called the Newton-Raphson formula for solving nonlinear 
equations of the form . So starting with an initial guess, xi, one can find 
the next guess, xi+1, by using equation (1). One can repeat this process until one 
finds the root within a desirable tolerance. 
( ) 0=xf
 
Algorithm 
The steps to apply Newton-Raphson method to find the root of an equation f(x) = 
0 are 
1. Evaluate f′(x) symbolically 
2. Use an initial guess of the root, xi, to estimate the new value of the root 
xi+1 as 
Newton-Raphson Method 03.05.3 
 
 
)f'(x
)f(x - = xx
i
i
ii 1+ 
3. Find the absolute relative approximate error, a∈ as 
010
1
1 x 
x
- xx = 
i
ii
a
+
+∈ 
4. Compare the absolute relative approximate error, a∈ with the pre-
specified relative error tolerance, s∈ . If a∈ > s∈ , then go to step 2, else 
stop the algorithm. Also, check if the number of iterations has exceeded 
the maximum number of iterations. 
 
 f(x) 
 f(xi) 
 f(xi+1) 
xi+2 xi+1 xi X 
 θ 
 ( )[ ]ii xfx , 
 
Figure 1. Geometrical illustration of the Newton-Raphson method 
 
 
 
03.05.4 Chapter 03.05 
 
Example 1 
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. 
 
 
Figure 2. Floating ball problem 
 
The equation that gives the depth ‘x’ to which the ball is submerged under water 
is given by 
010993.3165.0 423 =×+− −xx 
Use the Newton-Raphson method of finding roots of equations to find 
a) the depth ‘x’ to which the ball is submerged under water. Conduct three 
iterations to estimate the root of the above equation. 
b) find the absolute relative approximate error at the end of each iteration, 
and 
c) the number of significant digits at least correct at the end of each iteration. 
Solution ( ) 423 10993.31650 −×+−= x.xxf 
Newton-Raphson Method 03.05.5 
 
 ( ) x.xxf 3303 2 −=′
Let us assume the initial guess of the root of ( ) 0=xf is ..x 0500 =
Iteration #1 
 
The estimate of the root is 
 
( )
( )0
0
01 xf
xfxx ′−= 
 ( ) ( )( ) ( )0503300503
10993.30501650050050 2
423
...
.... −
×+−−=
−
 
 3
4
109
10118.1050 −
−
×−
×−= . 
 ( )01242.0050 −−= . 
 062420.=
The absolute relative approximate error, a∈ at the end of Iteration #1 is 
 100
1
01 ×−=∈
x
xx
a 
 100
062420
050062420 ×−=
.
.. 
 = 19.89% 
The number of significant digits at least correct is 0, as you need a absolute 
relative approximate error of less than 5% for one significant digit to be correct in 
your result. 
Iteration #2 
The estimate of the root is 
( )
( )1
1
12 xf
xfxx ′−= 
 
03.05.6 Chapter 03.05 
 
 ( ) ( )( ) ( )0624203300624203
10993.30624201650062420062420 2
423
...
.... −
×+−−=
−
 
 3
7
1090973.8
10977813062420 −
−
×−
×−−= .. 
 ( )5104645.4062420 −×−= . 
 062380.=
The absolute relative approximate error, a∈ at the end of Iteration #2 is 
 100
2
12 ×−=∈
x
xx
a 
 100
062380
062420062380 ×−=
.
.. 
 %06410.=
The number of significant digits at least correct is 2. 
Iteration #3 
The estimate of the root is 
( )
( )2
2
23 xf
xfxx ′−= 
 ( ) ( )( ) ( )0623803300623803
10993.30623801650062380062380 2
423
...
.... −
×+−−=
−
 
 3
11
1091171.8
1042937.4062380 −
−
×−
×−= . 
 ( )9109703.4062380 −×−−= . 
 062380.=
The absolute relative approximate error, a∈ at the end of Iteration #3 is 
Newton-Raphson Method 03.05.7 
 
 100
062380
062380062380 ×−=∈
.
..
a 
 0=
The number of significant digits at least correct is 4, as only 4 significant digits 
are carried through in all the calculations. 
 
 
03.05.8 Chapter 03.05 
 
Drawbacks of Newton-Raphson Method 
1. Divergence at inflection points: If the selection of a guess or an iterated value 
turns out to be close to the inflection point of f(x), that is, near where f”(x)=0, 
the roots start to diverge away from the root. For example, to find the root of 
( ) ( ) .01 3 =−= xxf 
Table 1. Divergence at inflection point 
 
Iteration 
Number 
xi a∈ % f(xi) 
0 
1 
2 
3 
4 
5 
6 
7 
8 
-1 
-0.33333 
 0.11111 
 0.40741 
 0.60494 
 0.73663 
 0.82442 
 0.88294 
 0.92196 
200 
400 
72.73 
32.65 
17.88 
10.65 
 6.629 
 4.232
-8 
-2.3704 
-0.70233 
-0.2081 
-0.06166 
-0.01827 
-5.4132x10-3 
-1.60389x10-3 
 -4.75226x10-3 
-20
-15
-10
-5
0
5
10
-2 -1 0 1 2
 1
 2 3
 f(x)
 x
 
Figure 3: Divergence at inflection point for ( ) ( ) 01 3 =−= xxf . 
Newton-Raphson Method 03.05.9 
 
What is an inflection point? 
For a function f(x) where the concavity changes from up-to-down or down-to-up 
are called inflection points of the graph. For example in the function, 
, the concavity changes at x=1. In fact, it also changes sign at x=1 
and thus brings up the Inflection Point Theorem for a function f(x) that states : “If 
f’(c) exists and f’’(c) changes sign at x = c , then the point (c, f(c)) is an inflection 
point of the graph of f. 
( ) ( )31−= xxf
2. Division of zero or near zero: The formula of Newton-Raphson method is 
)(xf
)f(xxx
i
i
ii ′−=+1 
Consequently if an iteration value, is such that ix ( ) 0≅′ ixf , then one can face 
division by zero or a near-zero number. This will give a large magnitude for the 
next value, xi+1. An example is finding the root of the equation 
( ) 623 1042030 −×+−= .x.xxf 
in which case 
( ) x.xxf 0603 2 −=′ 
For or , division by zero occurs (Figure 4). For an initial guess 
close to0.02, of , even after 9 iterations, the Newton-Raphson 
method is not converging. 
00 =x 02.00 =x
01999.00 =x
Table 2. Division by near zero in Newton Raphson method 
Iteration 
Number 
xi a∈ f(xi) 
0 
1 
2 
3 
4 
5 
 0.019990 
-2.6480 
-1.7620 
-1.1714 
-0.77765 
-0.51518 
100.75 
 50.282 
 50.422 
 50.632 
 50.946 
 -1.6000 x 10-6 
-18.778 
 -5.5638 
 -1.6485 
 -0.48842 
 -0.14470 
 
03.05.10 Chapter 03.05 
 
6 
7 
8 
9 
-0.34025 
-0.22369 
-0.14608 
-0.094490 
 51.413 
 52.107 
 53.127 
 54.602 
 -0.042862 
 -0.012692 
 -0.0037553 
 -0.0011091 
-1.00E-05
-7.50E-06
-5.00E-06
-2.50E-06
0.00E+00
2.50E-06
5.00E-06
7.50E-06
1.00E-05
-0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04
x
f(x)
0.02
 
Figure 4. Pitfall of division by zero or a near zero number. 
 
3. Root jumping: In some case where the function f(x) is oscillating and has a 
number of roots, one may choose an initial guess close to a root. However, the 
guesses may jump and converge to some other root. For example for solving the 
equation 
0sin =x 
if you choose ( 539822.74.20 == )πx as an initial guess, it converges to the root 
of x=0 as shown in Table 3. However, one may have chosen this an initial guess 
to converge to π2=x . 28318536.=
Table 3. Root jumping in Newton Raphson method 
Iteration 
Number 
ix f(xi) a∈ 
0 
1 
 7.539822 
 4.461 
 0.951 
-0.969 
 
 68.973 
Newton-Raphson Method 03.05.11 
 
2 
3 
4 
5 
 0.5499 
-0.06303 
6.376x10-4 
-1.95861x10-13 
 0.5226 
-0.06303 
8.375x10-4 
-1.95861x10-13 
711.44 
971.91 
7.54x104 
4.27x1010 
 
-1.5
-1
-0.5
0
0.5
1
1.5
-2 0 2 4 6 8 10
x
f(x)
 -0.06307 0.5499 4.461 7.539822
 
Figure 5. Root jumping from intended location of root for ( ) 0sin == xxf
 
03.05.12 Chapter 03.05 
 
Oscillations near local maximum and minimum: Results obtained from Newton-
Raphson method may oscillate about the local maximum or minimum without 
converging on a root but converging on the local maximum or minimum. 
Eventually, it may lead to division to a number close to zero and may diverge. 
For example for 
 ( ) 022 =+= xxf
the equation has no real roots 
 
Table 4. Oscillations near local maxima and minima in Newton Raphson 
method. 
Iteration 
Number 
ix f(xi) a∈ 
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
-1.0000 
 0.5 
-1.75 
-0.30357
 3.1423 
 1.2529 
-0.17166
 5.7395 
 2.6955 
 0.9770 
3.00 
2.25 
5.062 
2.092 
11.874 
3.57 
2.029 
34.942 
9.268 
2.954 
_____ 
300.00 
128.571 
 476.47 
109.66 
150.80 
829.88 
102.99 
112.927 
175.96 
Newton-Raphson Method 03.05.13 
 
-1
0
1
2
3
4
5
6
-2 -1 0 1 2 3
f(x)
x
 3
 4
 2
 1
 -1.75 -0.3040 0.5 3.142
 
Figure 6: Oscillations around local minima for ( ) 22 += xxf 
 
Derivation of Newton Raphson method from Taylor series 
Newton-Raphson method can also be derived from Taylor series. For a 
general function f(x), the Taylor series is 
( ) ( ) ( )( )iiiii xxxfxfxf −′+= ++ 11 + ( ) L+++ 212 iii xx!
)f"(x
 
As an approximation, taking only the first two terms of the right hand side, 
( ) ( ) ( )( )iiiii xxxfxfxf −′+≈ ++ 11 
and we are seeking a point where f(x) = 0, that is, if we assume 
( ) ,xf i 01 =+ 
( ) ( )( )iiii xxxfxf −′+≈ +10 
which gives 
 
03.05.14 Chapter 03.05 
 
.
)f'(x
)f(xxx
i
i
ii −=+1 
This is the same Newton-Raphson method formula series as derived previously 
using the geometric method. 
 
NONLINEAR EQUATIONS 
Topic Newton-Raphson Method of Solving Nonlinear Equations 
Summary Text book notes of Newton Raphson method of finding roots of nonlinear equation, 
including convergence and pitfalls. 
Major General Engineering 
Authors Autar Kaw 
Date June 25, 2007 
Web Site http://numericalmethods.eng.usf.edu 
 
	Newton-Raphson Method of Solving Nonlinear Equations
	What is the Newton-Raphson method and how does it differ from other methods to solve nonlinear equations?
	Algorithm
	Example 1
	Solution
	Drawbacks of Newton-Raphson Method
	What is an inflection point?
	Table 2. Division by near zero in Newton Raphson method
	Table 3. Root jumping in Newton Raphson method
	Table 4. Oscillations near local maxima and minima in Newton Raphson method.
	Derivation of Newton Raphson method from Taylor series
	This is the same Newton-Raphson method formula series as derived previously using the geometric method.
	Topic