Transcript 19 Newton-Raphson Iteration

“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
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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
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.