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On each iteration of the bisection method the error is halved. So we gain one binary digit of precision on each iteration. I want to find how many decimal digits of precision are gained. So does this look alright -

$$E_{k+1} = \frac{1}{2}E_k = (\frac{1}{10})^xE_k$$

$$\frac{1}{2} = \frac{1}{10^x}$$

$$2= 10^x$$

$$x = \log_{10} 2$$

$$x = 0.30103$$

So $0.30103$ decimal digits of precision are gained on each step?

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    $\begingroup$ That looks about right. Ten steps yields $1024 = 2^10$ times th precision, which is a little better than 3 decimal points. $\endgroup$ – Thomas Andrews Oct 11 '12 at 14:51
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Yes.

More generally $n$ bisections multiply the uncertainty by $$2^{-n} = 10^{-\log_2 (10) \, n } \approx 10^{-0.3\, n }$$ so give you about $0.3\,n$ more decimal digits of precision.

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