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# Iteration Q... Help! watch

1. I need help in solving part c... i've managed to do a and b but am stugglin to find out how to do part c...

f(x) = x^3 - 2x - 5

a) Show that there is a root (alpha) of f(x) = 0 for x in the interval [2,3]

The root (alpha) is to be estimated using the iterative formula

x{n+1} = sqr rt ( 2 + (5 / x{n})), x{0} = 2

b) calculate the values of x{1}, x{2}, x{3}, x{4} to four significant figures.

c) Prove that, to 5 sig fig, (alpha) is 2.0946

Help greatly appreciated...
2. if you carry on with the iteration, the fifth decimal place stabilises to 5.
hence, you have 2.09455, which to 5sf = 2.0946
3. thanks

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Updated: February 27, 2006
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