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Revision:Solving Equations

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Solving Equations

1 Trial and Improvement
2 Iteration
- 2.1 Example
3 Comments

Trial and Improvement

Any equation can be solved by trial and improvement (/error). However, this is a tedious procedure.

Example: Solve t^3 + t = 17 by trial and improvement.

Firstly, select a value of 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 , t^3 + t = 2^3 + 2 = 10 . This is lower than 17, so we try a higher value for t.

If t = 2.5 , t^3 + t = 18.125 (too high)

If t = 2.4 , t^3 + t = 16.224 (too low)

If t = 2.45 , t^3 + t = 17.156 (too high)

If t = 2.44 , t^3 + t = 16.966 (too low)

If t = 2.445 , t^3 + 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:

$\displaystyle x_{n+1} = 2 + \frac{1}{x_n}$

You are usually given a starting value, which is called x_0 .

If x_0 = 3 , substitute 3 into the original equation where it says x_n . This will give you x_1 . (This is because if n = 0 , $x_1 = 2 + \frac{1}{x_0}$ and x_0 = 3 ).

$\displaystyle x+1 = 2 + \frac{1}{3} = 2.333333$ (by substituting in 3).

To find x_2 , substitute the value you found for x_1 .

$\displaystyle x_2 = 2 + \frac{1}{2.333 333} = 2.428 571$

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, x_5 = 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 x_1 = -2 to obtain a root of the equation x^2 - 4x - 8 = 0 accurate to one decimal place.

Answers

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) $\displaystyle x_1 = -2$

$\displaystyle x_2 = 1 + \frac{11}{-2 - 3}$ (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)

$\displaystyle = -1.619$

Then:

x_4 = -1.381

x_5 = -1.511

x_6 = -1.439

x_7 = -1.478

therefore, to one decimal place, x = 1.5 .

Comments

This article is only concerned with looking at numerical methods - it could do with being renamed.

Another article is needed looking at general methods for solving (linear) equations. A link to the article on solving quadratic equations should be include as well as links to any other relevant pages.

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