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I'm trying to solve the following problem.

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I've proven the two equalities, but can't figure out how to prove that last bit about the rate of convergence being equal to $\log_{10}(2)$.

John Smith
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  • This is an alternative definition of "rate". It tells you how fast the amount of correct decimal digit grows per iteration. – Lutz Lehmann Oct 19 '22 at 05:42
  • @LutzLehmann Thank you, but how do I see that that is what the rate is? – John Smith Oct 19 '22 at 20:44
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    Because $\frac1{2^k}=10^{-k·\log{10}(2)}$. If the error is smaller than $10^{-d}$, then in a majority of cases the first $d$ decimal digits after the dot are fixed. – Lutz Lehmann Oct 19 '22 at 21:04

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