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Consider the method of bisection applied to $f(x) = \arctan(x)$, with initial interval $x = [-4.9, 5.1].$

Are the hypotheses under which the method of bisection converges valid? If so, then how many iterations would it take to obtain the solution to within an absolute error of $10^{-2}$?

Please help!!

BlackAdder
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James
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    Well, what are the hypotheses for the method of bisection? Where are you having trouble checking them? –  Feb 12 '14 at 07:29
  • What is the width of the initial interval? How many times would you have to divide that by two to get $\leq 10^{-2}$? – Eric Towers Feb 12 '14 at 07:31

1 Answers1

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As a first working answer, I am getting 8 iterations are required.

$\left( \begin{array}{cccc} 0 & -4.9 & 0.1 & 5.1 \\ 1 & -4.9 & -2.4 & 0.1 \\ 2 & -2.4 & -1.15 & 0.1 \\ 3 & -1.15 & -0.525 & 0.1 \\ 4 & -0.525 & -0.2125 & 0.1 \\ 5 & -0.2125 & -0.05625 & 0.1 \\ 6 & -0.05625 & 0.021875 & 0.1 \\ 7 & -0.05625 & -0.0171875 & 0.021875 \\ 8 & -0.0171875 & 0.00234375 & 0.021875 \\ \end{array} \right)$

The middle number is the estimate of the root. Notice the sign change in the first and last columns. That is how interval bisection zeroes in on a root. Generally it takes about 3 iterations to get 1 correct digit. It is quite robust but quite slow.

bobbym
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