# How many decimal digits of accuracy gained on each iteration of the bisection method

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

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.