Trial and Improvement
Any equation can be solved by trial and improvement (/error). However, this is a
tedious procedure.
Example:
Solve t³ + t = 17 by trial and improvement.
Firstly, select a value of t to try in the equation. I have selected t = 2. Put
this value into the equation. We are trying to get the answer of 17.
If t = 2, t³ + t = 2³ + 2 = 10 . This is lower than 17, so we try a higher
value for t.
If t = 2.5, t³ + t = 18.125 (too high)
If t = 2.4, t³ + t = 16.224 (too low)
If t = 2.45, t³ + t = 17.156 (too high)
If t = 2.44, t³ + t = 16.966 (too low)
If t = 2.445, t³ + t = 17.061 (too high)
So we know that t is between 2.44 and 2.445. So to 2 decimal places, t = 2.44.
Iteration
This is a way of solving equations.
An iteration formula might look like the following:
xn+1 = 2 +
1
xn
You are usually given a starting value, which is called x0. If x0 = 3, substitute 3 into the original equation where it says
xn. This will give you
x1. (This is because if n = 0, x1 = 2
+ 1/x0
and x0 =
3).
x1 = 2 + 1/3 = 2.333 333 (by substituting in 3).
To find x2,
substitute the value you found for x1.
x2 = 2 + 1/(2.333 333)
= 2.428 571
Repeat this until you get an answer to a suitable degree of accuracy. This may
be about the 5th value for an answer correct to 3s.f. In this example, x5
= 2.414...
Example:
a) Show that x = 1 + 11
x - 3
is a rearrangement of the equation x² - 4x - 8 = 0.
b) Use the iterative formula Xn+1
= 1 + 11
xn
- 3
together with a starting value of x1
= -2 to obtain a root of the equation x² - 4x - 8 = 0 accurate to one decimal
place.
a) multiply everything by (x - 3):
x(x - 3) = 1(x - 3) + 11
so x² - 3x = x + 8
so x² - 4x - 8 = 0
b) x1 = -2
x2 = 1 + 11
(substitute -2 into the
iteration formula)
-2 - 3
= -1.2
x3 = 1 + 11
(substitute -1.2 into the above formula)
-1.2 - 3
= -1.619
x4 = -1.381
x5 = -1.511
x6 = -1.439
x7 = -1.478
therefore, to one decimal place, x = 1.5 .
© Matthew Pinkney