feminizmiki

2022-01-19

Prove an algorithm for logarithmic mean $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=\underset{n\to \mathrm{\infty }}{lim}{b}_{n}=\frac{{a}_{0}-{b}_{0}}{\mathrm{ln}{a}_{0}-\mathrm{ln}{b}_{0}}$

nghodlokl

Expert

You have several exact identities, such as ${a}_{n+1}-{b}_{n+1}=\frac{{a}_{n}+{b}_{n}}{2}⇒{a}_{n}-{b}_{n}={2}^{-n}\left({a}_{0}-{b}_{0}\right),$ $\frac{{a}_{n+1}}{{b}_{n+1}}=\sqrt{\frac{{a}_{n}}{{b}_{n}}}⟹\frac{{a}_{n}}{{b}_{n}}=\left(\frac{{a}_{0}}{{b}_{0}}{\right)}^{2-n}$ The first tells us that if one of the sequences has a limit the other has the some limit. In combination bn can be eliminated to get, via mean value theorem, ${a}_{n}=\frac{{2}^{-n}\left({a}_{0}-{b}_{0}\right)}{1-\left(\frac{{b}_{0}}{{a}_{0}}{\right)}^{2-n}}=\frac{{a}_{0}-{b}_{0}}{\mathrm{ln}{a}_{0}-\mathrm{ln}{b}_{0}+\left({2}^{-n}\right)}$