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When using linear interpolation, with similar triangles, to find the root of a function you narrow down the interval the root is in.

If $f(1) < 0$ and $f(2) > 0$ then the root is in $[1, 2]$

Then you do linear interpolation to find $1.460, f(1.460) < 0$, then the root is in $[1.460, 2]$

Then again linear interpolation is done to find $1.467$, say $f(1.467) < 0$, so the root is in $[1.467, 2]$

Why is it that you stop and find the answer because $1.460 = 1.467$ (to 1 dp)? (ie the last two attempts are equal when rounded to 1 dp)

Why don't you stop when in the interval $[a, b], a = b$ when $a$ and $b$ are rounded to $1$ d.p.?

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    Don't the answers to your [previous question](http://math.stackexchange.com/questions/142399/interval-bisection-to-find-a-root-of-fx) apply here too?2012-05-08
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    No it is a different method, and you seem to stop in the exactl the opposite circumstances. (there you stop when [a,b] a = b, and not when you get 2 consecutive (a+b)/2)2012-05-08

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