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=Welcome to Mr Gill's A2 Maths wiki=

12th June 2012

(i) we looked over the rearrangement method for the coursework. IF YOU WERE ABSENT YOU MUST SEE ME TO GO THROUGH THIS!

Basically this is a method for 'homing in' on the solution of a 'nasty' equation

eg Solve x^3 + 2x^2 - 3x + 1 = 0

The basic idea is to rearrange this equation into a form x = g(x).

There are lots of ways of doing this - here are just a couple...

A) x = (3x - 1 - 2x^2)^(1/3) ie the cube root of (3x - 1 - 2x^2)

B) x = (3x - 1)/(x^2 + 2x)

What we then do is use an iterative process (easy on Autograph) to (hopefully) find the solution given a starting point that is quite close to the root.

Sometimes this works and sometimes it doesn't - it all depends on the magnitude (size) of g'(x) at the solution.

So long as -1 < g'(x) < 1 at the solution this method will work - otherwise it will either find a different solution or find no solution at all.

(ii) In order to look at the magnitude of g(x) we may need to use a new method for differentiation - the CHAIN RULE

In short - to differentiate something like y = (3x - 1 - 2x^2)^(1/3), differentiate from the 'outside in', leaving the 'inner function' the same and then, finally, multiply by the derivative of what's inside the bracket.

So if g(x) = (3x - 1 - 2x^2)^(1/3) then g'(x) = (1/3)(3x - 1 - 2x^2)^(-2/3) x (3 - 4x)

PLEASE LOOK OVER THE ABOVE AND TALK TO ME IF YOU'RE UNSURE ABOUT ANY OF IT!