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Revision:Iteration

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Iteration

This is a way of solving equations. An iteration formula might look like the following:

\displaystyle x_{n+1} = 2 + \frac{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 xn.

This will give you x1.

(This is because if \displaystyle n = 0,\ x_1 = 2 + \frac{1}{x_0} and \displaystyle x_0 = 3).


\displaystyle x_1 = 2 + \frac{1}{3} = 2.333 333 (by substituting in 3).


To find x2, substitute the value you found for x1.

\displaystyle x_2 = 2 + \frac{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

\displaystyle x = 1 + \frac{11}{x - 3}

is a rearrangement of the equation:

\displaystyle x^2 - 4x - 8 = 0.


(b) - Use the iterative formula:

\displaystyle x_{n+1} = 1 + \frac{11}{x_n - 3}

together with a starting value of x1 = -2 to obtain a root of the equation:

\displaystyle x^2 - 4x - 8 = 0

accurate to one decimal place.


(a) multiply everything by (x - 3):


\displaystyle x(x - 3) = 1(x - 3) + 11

so \displaystyle x^2 - 3x = x + 8

so \displaystyle x^2 - 4x - 8 = 0


(b) x1 = -2


\displaystyle x_2 = 1 + \frac{11}{-2 - 3} (substitute -2 into the iteration formula)

\displaystyle = -1.2


\displaystyle x_3 = 1 + \frac{11}{-1.2 - 3} (substitute -1.2 into the above formula)

\displaystyle = -1.619


\displaystyle x_4 = -1.381

\displaystyle x_5 = -1.511

\displaystyle x_6 = -1.439

\displaystyle x_7 = -1.478


therefore, to one decimal place, x = 1.5.


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