3
$\begingroup$

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?

  • 2
    That looks about right. Ten steps yields $1024 = 2^10$ times th precision, which is a little better than 3 decimal points.2012-10-11

1 Answers 1

3

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.