The Chemistry Maths Book, Second Edition

20.3 Solution of ordinary equations 563

Then, iff(x

3

) 1 > 10 , so that the zero lies betweenx

1

andx

3

as in the figure, repeat for

the intervalx

1

tox

3

, and so on.

EXAMPLE 20.5Bisection method for

To find write f(x) 1 = 1 x

2

1 − 151 = 10. The root lies betweenx

<

1 = 12 and x

>

1 = 13 , with

f(x

<

) 1 = 1 − 1 andf(x

>

) 1 = 14. Thenx

3

1 = 1 (x

<

1 + 1 x

>

) 221 = 1 2.5. Table 20.2 shows the computation

to 4 significant figures on a standard pocket calculator.

Table 20.2 To find the square root of 5 correct to 4 significant figures

n 012345

x

<

2 2 2 2.125 2.1875 2.21875

x

>

3 2.5 2.25 2.25 2.25 2.25

2.5 2.25 2.125 2.1875 2.21875 2.23438

n 67891011

x

<

2.23438 2.23438 2.23438 2.23438 2.23535 2.23584

x

>

2.25 2.24219 2.23828 2.23633 2.23633 2.23633

2.24219 2.23828 2.23633 2.23535 2.23584

The values ofx

<

andx

>

after 11 iterations both round to 2.236, and this is the required

value (2.236

2

=4.999696).

0 Exercises 9–11

The bisection method is an example of a bracketing method, with the zero known to

lie in an interval of decreasing size. In the present case, the size of the interval is halved

at every iteration. The convergence is slow, but the method cannot fail. It is the method

of last resort when all else fails.

1

2

()xx

<>

1

2

()xx

<>

5 ,

5

.......... ........ .......

....

.......

........

..

...

.. ... ... .. ... ... .. ..

. ... ... .. ... ... .. ..

f(x)

f(x

2

)

f(x

3

)

f(x

1

)

x

1

x

2

x

3

••

0

..

....

.....

....

.....

....

.....

....

...

.....

....

.....

....

.....

....

.....

....

.....

....

.....

....

.....

......

.....

.......

.....

......

.......

......

.......

........

.......

........

.........

..........

...........

...............

..

Figure 20.1