Using Excel for Iteration Formula
Suppose we want to solve x 2  3x  1 0 , we know there is a root between 0 and 1
because if we write
then
f  x   x 2  3x  1
f 0  1 0
and
f 1 12  3 1  1 1  0
we can rearrange the given equation into the iteration formula
xn2  1
with a start value of x0  0.5
3
This can be entered into Excel as shown below:
xn1 
a) enter the start value in a cell
b) enter the iteration formula in another cell. This MUST be entered exactly as
shown
=(cell number of start value^2+1)/3
c) copy this formula into cell below
d) repeat this until you are satisfied with your answer
Your spreadsheet formula will look this
1 Solve x^2-3x+1 = 0
n
0
1
2
3
4
5
6
7
8
9
10
11
x
0.5
=(B2^2+1)/3
=(B3^2+1)/3
=(B4^2+1)/3
=(B5^2+1)/3
=(B6^2+1)/3
=(B7^2+1)/3
=(B8^2+1)/3
=(B9^2+1)/3
=(B10^2+1)/3
=(B11^2+1)/3
=(B12^2+1)/3
Your answers will look like this
1 Solve x^2-3x+1 = 0
n
0
1
2
3
4
5
6
7
8
9
10
11
X
0.5
0.416666667
0.391203704
0.384346779
0.382574149
0.382120993
0.382005484
0.381976063
0.381968571
0.381966663
0.381966177
0.381966054
2a) Find the other solution to x 2  3x  1 0 , by making another iteration formula
xn1  3xn  1
This has to be entered into Excel as
=SQRT(3* cell number-1)
Find a root to 3dp.
using a start value of x0  2
b)
Change your start value to see what happens.
c)
Format your results to 3 decimal places by using the
3.
Show that x 3  2 x  4  0 has a root near x = 1.2.
.00
icon
 .0
1
Solve the equation using the iteration formula x n1  (4  2 x n ) 3
4.
Given that f x   4 x  e x show that the equation f x   0 has a root in the interval
0.3  x  0.4 . Find the root using the iteration formula x n1 
entered into Excel as
5a)
b)
c)
6.
ex
. This has to be
4
=EXP(cell number)/4
1
on the same diagram.
x
1
How many roots are there to the equation e x 
x
Sketch the graphs of y  e x and y 
Show that the equation x  e x  0 has a root in the interval 0.5 < x < 0.6 Find the
root to 2dp. Use the iteration formula xn1  e  xn
Show that the equation x 3  6 x 2  9 x  2  0 has a root lying between 0 and 0.5. Use
the iteration formula x n1 
6 x n2  2
9  x n2
with x0  0 to find this root to 3dp.
1
7.
Using the iteration formula x n1  (22 x n  50) 4 and x0  3.5 find a root to the
equation x 4  22 x  50  0 to 4sf. Check that your answer is correct to 4sf.