Download Iteration diagrams and convergence

Iteration Diagrams and
Convergence
© Christine Crisp
There are some equations that we can’t solve.
e.g.
x  1 x
3
However, we can find an approximate solution to some
of these equations.
The approximation can be very accurate, say to 6 or
more decimal places.
There are several methods of finding approximate
solutions and in this presentation we will study one of
them.
There are 2 stages to getting a solution:
Stage 1. Find a 1st estimate
Stage 2. Find a formula to improve the estimate.
Sometimes we can just spot an approximate solution
to an equation.
Can you spot the approximate value of the solution to
x3  2x  1 ?
Ans: It’s quite close to 0 as the l.h.s. is then 0 and
the r.h.s. is 1.
If we can’t quickly spot an approximation, we can use
a method involving finding bounds for the solution.
Bounds are numbers which lie on either side of the
solution.
If these are integers we call them integer bounds.
e.g. To find integer bounds for x  1  x 3 we
can sketch y  x and y  1  x 3
y  1  x3
This is the point . . . where
y
x
y
x
and
so the x coordinate gives the solution to
y  1  x3
x  1  x3
Rearrange the equation to get zero on one side.
e.g. For x  1  x 3 , get x 3  1  x  0
 Call this
f ( x)  0

To show how the method works I’m going to sketch
y  f ( x ) ( but you won’t usually have to do this ).
The solution, a, is
f ( x)  x 3  1  x
now where
f ( x)  0
a
At a,
f ( x)  0
To the left of a, e.g. at x = 0, f ( x )  1  0
To the right of a, e.g. at x = 1, f ( x )  1  0
Iteration diagrams and Convergence
Once we have the bounds we may try and
get a formula to find the root
x  1 x

3
Using the formula x n1  1 
we got
xn

1
3
with
x0  0  5
x 0  0  5 , x1  0  66 , x 2  0  57 , . . .
The solution to the equation we called a.
We can illustrate the iteration process on a diagram.
Iteration diagrams and Convergence

x n 1  1  x n


1
3

1
3
we draw y  1  x
and
yx
So, a is the x-coordinate of the point of intersection.
For

y  1 x

1
3
yx
a
a  0  605423
( to 6 d.p. )
We’ll zoom in on the graph to enlarge the part near
the intersection.
Iteration diagrams and Convergence

We start the iteration with
x1  1  x 0
and we substituted x 0  0  5 to get x1 .
On the diagram
we draw from x 0
to the curve . . .

y  1 x

1
3

1
3
yx
Iteration diagrams and Convergence

We started the iteration with x1  1  x 0
and we substituted x 0  0  5 to get x1 .
On the diagram
we draw from x 0
to the curve . . .
which gives a
y-value, y  0  66

y  1 x

1
3
x
(0  5, 0  66)
x0

1
3
yx
Iteration diagrams and Convergence

We started the iteration with x1  1  x 0
and we substituted x 0  0  5 to get x1 .
On the diagram
we draw from x 0
to the curve . . .
which gives a
y-value, y  0  66

y  1 x


1
3
yx
1
3
x
x
x0
x1
(0  5, 0  66)
(0  66, 0  66)
By drawing across to y = x this y-value gives the point
where x  0  66 . This is x1 .
Iteration diagrams and Convergence

In the iteration, x1 is now substituted into x 2  1  x1
to give x 2 .
On the diagram
we draw from
(0  66, 0  66)
to the curve . . .
giving the next
y-value.

y  1 x

yx
1
3
x
(0  5, 0  66)
x
(0  66, 0  66)
x(0  66,
x0
0  57)
x1
The next line converts this y-value to x 2 and so on.

1
3
Iteration diagrams and Convergence

In the iteration, x1 is now substituted into x 2  1  x1
to give x 2 .
On the diagram
we draw from
(0  66, 0  66)
to the curve . . .
giving the next
y-value.

y  1 x

yx
1
3
x
x
(0  66, 0  66)
(0  5, 0  66)
x(0  66,
x0

1
3
x2
0  57)
x1
The next line converts this y-value to x 2 and so on.
x 0  0  5 , x1  0  66 , x 2  0  57 , . . .
The lines are drawn to the curve and y = x alternately,
starting by joining x 0 to the curve.
Iteration diagrams and Convergence
The diagram we’ve drawn illustrates a convergent,
oscillating sequence.

y  1 x

yx
1
3
a
x0
x2 x4 x3
x1
This is called a cobweb diagram.
Iteration diagrams and Convergence
Looking at the original graph, we have

y  1 x

1
3
yx
a
x0
Iteration diagrams and Convergence
SUMMARY
To draw a diagram illustrating
 Draw
x  g( x ) iteration:
y  g( x ) and y  x on the same axes.
 Mark x 0 on the x-axis and draw a line parallel to
the y-axis from x 0 to y  g ( x ) ( the curve ).
 Continue the cobweb line, going parallel
to the x-axis to meet y  x .
 Continue the cobweb line, going
parallel to the y-axis to meet
y  g( x ) .
Repeat
Iteration diagrams and Convergence
Iteration does not always give an oscillating sequence.
We can also draw a diagram for sequences which
iterate directly towards the solution.
I am going to use an example presentation: Solve
x  2x
3
Iteration diagrams and Convergence
Before we zoom in, the graphs look like this.
yx
y  3 2x
The solution has 2 roots.
We will find the one nearest to the origin.
Iteration diagrams and Convergence
xn1  2
So, to
3
x0  1  5
xn
yx
This is called a
staircase diagram.
y  3 2x
a x2
x1
x0
Iteration diagrams and Convergence
Exercise
Copy the following graphs and sketch the diagrams
which illustrate the convergence of the iterative
process, showing x 0 , x1 , x 2 and a
(a)
(b)
y  g( x )
yx
yx
y  g( x )
x0  2  5
x0  1  5
Write on the names of the diagrams.
Iteration diagrams and Convergence
Solution: (a)
yx
y  g( x )
a
x0
x2 x3
A cobweb diagram.
x1
Iteration diagrams and Convergence
Solution: (b)
yx
y  g( x )
a x2
x1
A staircase diagram.
x0
Iteration diagrams and Convergence
In the next example we’ll look at an equation which
has 2 roots and the iteration produces a surprising
result.
The equation is
e  x20.
x
We’ll try the simplest iterative formula first :

ex  2  x
x n1  e xn  2
Iteration diagrams and Convergence
Let’s try x n1
positive root.
e
x
n
 2 with x 0 close to the
y  ex  2
yx

Let
x0  1  5
a
Iteration diagrams and Convergence
Let’s try x n1
positive root.
e
x
n
 2 with x 0 close to the
The staircase moves
away from the root.

y  ex  2
yx
a
x0
x1
The sequence diverges rapidly.
Using the iterative formula, x1  2  48 , x 2
 9  96
Iteration diagrams and Convergence
Suppose we try a value for
x0
on the left of the root.
y  ex  2
yx

a
x0
Iteration diagrams and Convergence
Suppose we try a value for
x0
on the left of the root.
y  ex  2
yx

a
x1
Iteration diagrams and Convergence
Suppose we try a value for
x0
on the left of the root.
y  ex  2
yx

x2
a
Iteration diagrams and Convergence
Suppose we try a value for
x0
on the left of the root.
y  ex  2
yx

a
x3
Iteration diagrams and Convergence
Suppose we try a value for
x0
on the left of the root.
y  ex  2
yx

a
The sequence now converges but to the other root !
Iteration diagrams and Convergence
It is possible to use our iterative method to find a in
the previous example but not with the arrangement we
had.
The equation was e x  x  2  0 .
The rearrangement we used was x
x
formula was x n1  e n  2.
With
x0  1
 ex  2
so the
this gives the negative root:
  1 8414 (
4 d.p. )
We can also arrange the equation as follows:
ex  x  2  0  ex  x  2
Change to log form:  x  ln( x  2)
giving x n1  ln( x n  2)
With x 0  1 ,a  1 1462 ( 4 d.p. )
Iteration diagrams and Convergence
We will now look at why some iterative formulae give
sequences that converge whilst others don’t and others
converge or diverge depending on the starting value.
Collecting the diagrams together gives us a clue.
I’ve included the 4th type of diagram that we haven’t
yet met: a diverging cobweb.
See if you can spot the important difference once
you can see the 4 diagrams
Iteration diagrams and Convergence
Staircase: converging
Cobweb: converging
yx
yx
y  g( x )
y  g(x )
a
x0
The gradients
of
for the
converging
x2 x3
x1 g ( x )
a x 2 x1
sequences
are shallow
yx
y  g( x )
y  g( x )
x3
x1 a x 0
x2
Cobweb: diverging
yx
a
x 0 x1
x0
x2
Staircase: diverging
Iteration diagrams and Convergence
It can be shown that x n1  g( x n ) gives a
convergent sequence if the gradient of g ( x ) . . .
is between 1 and +1 at the root.
g / ( x)
We write

 1  g / (a )  1
or,
g (a )  1
/
Unfortunately since we are trying to find a we don’t
know its value and can’t substitute it !
In practice, to test for convergence we use a value
close to the root.
The closer g / (a ) is to zero, the faster will be the
convergence.

Iteration diagrams and Convergence
e.g. By using calculus, determine which of the
following arrangements of the equation
ln x  4  x will give convergence to a root
near x = 3 and which will not.
(a) x  4  ln x
Solution:
(b)
x  e 4 x
(a) g(x)  4  ln x
1
 g (x)  
x
1
/
 g (3) 
  1  g /( 3 )  1
3
/
The sequence will converge.
Iteration diagrams and Convergence
e.g. By using calculus, determine which of the
following arrangements of the equation
ln x  4  x will give convergence to a root
near x = 3 and which will not.
(a) x  4  ln x
Solution:
(b)
x  e 4 x
(b) g(x)  e 4 x
 g / (x)   e 4 x
 g / ( 3 )  e   2  7
The sequence will not converge.
Iteration diagrams and Convergence
We can now see why we had the strange result when
we tried to solve e x  x  2  0 with x n1  e xn  2
At  , the gradient is
less than 1 ( so the
curve here is shallower
than y = x ): the
iteration converges.

y  ex  2
yx
a
At a , the gradient is
greater than 1 ( so the
curve here is steeper
than y = x ): the
iteration diverges.
Iteration diagrams and Convergence
We can now see why we had the strange result when
we tried to solve e x  x  2  0 with x n1  e xn  2
y  ex  2
yx

a
Both iterations started close to a but the one that
converged to  started on the left of a .
Iteration diagrams and Convergence
SUMMARY
To show that a formula of the type
will give a convergent sequence,

find

show that
x n1  g( x n )
g / ( x)
g / ( x )  1 , where x is close
to the solution.