Iteration
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This is a way of solving equations.
An iteration formula might look like the following:
x_n+1 = 2  + 1
                  x_n .

You are usually given a starting value, which is called x₀. If x₀ = 3, substitute 3 into the original equation where it says x_n. 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 X_n+1 = 1 +   11  
                                                                 x_n - 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 Pinkey

Other Notes in this Category

1. Errors
2. Iteration

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