Theo_14
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a) Show that x/(1-x)^3=x+3x^2+6x^3
b) Using this result, together with a suitable value of x, obtain a decimal estimate of the value of 100/729.
Part a was fine, but I'm stuck on b. In the markscheme, they use x=0.1, but I don't understand how they chose 0.1.
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BuddyYo
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I think it's a question about numerical approximation in which case, you can use either the sign-change rule or the iterative process to show that x=0.1371 correct to the specified d.p

Also, suitable x_0 is really up to you. It's arbitrary. It should ideally be close to the root though which is easy to do since it's been given already.
Last edited by BuddyYo; 1 week ago
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mqb2766
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(Original post by Theo_14)
a) Show that x/(1-x)^3=x+3x^2+6x^3
b) Using this result, together with a suitable value of x, obtain a decimal estimate of the value of 100/729.
Part a was fine, but I'm stuck on b. In the markscheme, they use x=0.1, but I don't understand how they chose 0.1.
A bit late, but part a) isn't quite an equality. The right hand side is the expansion of the left up to the cubic term. Really there are other higher order terms as well (quartic, quintic, ...).
So you want to estimate
100/729 = 0.1/0.729
which is x/(1-x)^3 when x = 0.1.

So sub x=0.1 in the right hand side to get a decent approximation (~3 d.ps, as it a cubic function) for 100/729.
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