“Teach A Level Maths”
Vol. 2: A2 Core Modules
19: Newton-Raphson
Iteration
© Christine Crisp
Newton-Raphson Iteration
Module C3
MEI/OCR
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Newton-Raphson Iteration
It isn’t always possible to find iterative formulae of
the type
x n1  g( x n )
that will find the solution of every equation.
Another iterative method that is useful is called the
Newton-Raphson method.
Newton-Raphson Iteration
Suppose we want to find an approximate solution to
the equation
f ( x)  0
To see how the method works, we’ll sketch
using f ( x )  x 3  x  1 .
The root 
lies between
1 and 2.
We’ll zoom in
near 
y  f ( x)
y  f ( x)

Newton-Raphson Iteration
y  f (x )

x2
x1
x0
Suppose our first estimate is given by x 0  2 .
We draw the tangent to the curve at x 0
The point where the tangent meets the x-axis we
call x1 .
Repeating . . .
Each point x1 , x 2 , . . . is closer to  .
Newton-Raphson Iteration
To carry out the iteration we need to find the points
where the tangents meet the x-axis.
( x 0 , y 0 )x
y0
y  f (x )
x 0  x1
x1
The grad. of the tangent 
x0
the change in y
the change in x
y0
dy

at x 0 
dx
x 0  x1
Newton-Raphson Iteration
y0
dy
and we need to find x .
at x 0 
1
dx
x 0  x1
dy
Using y 0 and
at x 0 in the formula isn’t very
dx
convenient, so, since y  f ( x ) we have
y
dy
0
and
y0  f ( x0 )
at x 0 
 f / ( x0 )
dx
x 0  x1
Then,
f ( x0 )
/
f ( x0 ) 
x 0  x1
f ( x0 )
f ( x0 )
Rearranging:
x 0  x1  /
 x1  x 0  /
f ( x0 )
f ( x0 )
We have
Newton-Raphson Iteration
So,
x1  x 0 
f ( x0 )
f / ( x0 )
We just need to alter the subscripts to find
x 2  x1 
Generalising gives
x n 1  x n 
x2 :
f ( x1 )
f / ( x1 )
f ( xn )
f / ( xn )
We don’t need a diagram to use this formula but
we must know how to differentiate f ( x ) .
Convergence is often very fast.
Newton-Raphson Iteration
e.g. Use the Newton-Raphson method with x 0  2
to find the root of the equation
x  x 1  0
3
correct to 4 d.p.
Solution:
Let
Differentiate:
x n 1  x n 
f ( x)  x 3  x  1
f / ( x)  3 x 2  1
f ( xn )
/
f ( xn )
3
 x n 1  x n 
( 3 x n  1)
2, ENTER
3
(ANS  ANS  1)
ANS 
( 3ANS 2  1)
x  1 6180 ( 4 d.p.)
Using a calculator we need:
Then,
( x n  x n  1)
2
Newton-Raphson Iteration
SUMMARY
To use the Newton-Raphson method to estimate a
root of an equation:
 rearrange the equation into the form f ( x )  0
 differentiate
 substitute
f ( x ) to find f / ( x )
f ( x ) and f / ( x ) into the
x n 1  x n 
formula
f ( xn )
f / ( xn )
 choose a suitable starting value for
 use a calculator to iterate
x0
Tip: It saves a lot of errors if, before you type the
formula into your calculator, you write the formula
with ANS replacing every x.
Exercise
Newton-Raphson Iteration
1. (a) Use the Newton-Raphson method to estimate
the root of the following equation to 6 d.p.
using the starting value given:
x3  2x2  2  0 ;
x0  1
(b) What happens if you use x 0  0 ?
(c) Use your calculator or a graph plotter to
sketch the graph of y  x 3  2 x 2  2 .
(d) What is special about the graph at x 0  0
and why does it explain the answer to (b) ?
2. Use the Newton-Raphson method to estimate one
root of 3 cos x  1  x to 4 d.p. using x 0  2
Newton-Raphson Iteration
(a)
x3  2x2  2  0 ;
x0  1
Solution: Let f ( x )  x 3  2 x 2  2

x n 1  x n 
x0  1 ,
f / ( x)  3 x 2  4 x
f ( xn )
f / ( xn )
3
2
( x n  2 x n  2)
 x n 1  x n 
2
(3 x n  4 x n )
(ANS3  2ANS 2  2)
 ANS 
2
( 3ANS  4ANS)
x1  0  857143 , x 2  0  839545 , . . .
x  0  839287 ( 6 d.p. )
Newton-Raphson Iteration
( x n  2 x n  2)
3
x n 1  x n 
2
(3 x n  4 x n )
(b) What happens if you use x 0  0 ?
(c)
2
The iteration fails immediately.
y  x  2x  2
3
2
At x = 0, there is a
stationary point.
At a stationary point
f / ( x )  0 so in the
formula we are
dividing by 0.
We also notice that the tangent never meets
the x-axis.
Newton-Raphson Iteration
2. Use the Newton-Raphson method to estimate one
root of 3 cos x  1  x to 4 d.p. using x 0  2
Solution:
Let
f ( x )  3 cos x  1  x
 f / ( x )  3 sin x  1
( 3 cos x n  1  x n )
x n 1  x n  /
 x n1  x n 
( 3 sin x n  1)
f ( xn )
( 3 cos ANS  1  ANS )
 x n1  ANS 
( 3 sin ANS  1)
x 0  2,
Radians!
x1  1  8562, x 2  1  8624
x  1 8624 ( 4 d.p. )
f ( xn )
Newton-Raphson Iteration
The Newton-Raphson method will fail if

f / ( x)  0
i.e. at a stationary point
It will also sometimes fail to give the expected root
if the initial value is close to a stationary point.
Can you draw a graph to show what could happen in
this case?
This is one example.
Newton-Raphson Iteration
y  x3  2x2  5x  1
x0
x 0  1  9 the
x  1 576 instead of
With
iteration gives the root
the closer root
x  0  187
.
Newton-Raphson Iteration
Newton-Raphson Iteration
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Newton-Raphson Iteration
It isn’t always possible to find iterative formulae of
the type
x n1  g( x n )
that will find the solution of every equation.
Another iterative method that is useful is called the
Newton-Raphson method.
Newton-Raphson Iteration
SUMMARY
To use the Newton-Raphson method to estimate a
root of an equation:
 rearrange the equation into the form
 differentiate
 substitute
f ( x)
to find
and
f / ( x)
f ( x)
f ( x)  0
f / ( x)
into the formula
 choose a suitable starting value for
x0
 use a calculator to iterate
Tip: It saves a lot of errors if, before you type the
formula into your calculator, you write the formula
with ANS replacing every x.
Newton-Raphson Iteration
e.g. Use the Newton-Raphson method with x 0  2
to find the root of the equation
x  x 1  0
3
correct to 4 d.p.
Solution:
Let
Differentiate:
x n 1  x n 
f ( x)  x 3  x  1
f / ( x)  3 x 2  1
f ( xn )
/
f ( xn )
3
 x n 1  x n 
3 xn  1
2, ENTER
3
ANS  ANS  1
ANS 
3ANS 2  1
x  1 6180 ( 4 d.p.)
Using a calculator we need:
Then,
xn  xn  1
2
Newton-Raphson Iteration
The Newton-Raphson method will fail if

f / ( x)  0
i.e. at a stationary point
It will also sometimes fail if the initial value is
close to a stationary point.