Question

Solve the recurrence relation a_{n+2}=sqrt{a_{n+1} * a_n} w/ a0=2 a1=8 & find the lim as n approaches infinity of a_n

Answer 1

\[a_{n+2}=\sqrt{a_{n+1}*a_n} \]

Answer 2

so I made a table of values for it and it looks like it converges between 5.03937 and 5.0397989 also I simplified the expression down to \[\log (a_n) = 2\log(a_{n+2}) -\log(a_{n+1}) \]

Answer 3

any help would be appriciated

Answer 4

lets define a new variable\[a_{n+2}=\sqrt{a_{n+1} \ a_n}\]divide by \(a_{n+1}\)\[\frac{a_{n+2}}{a_{n+1}}=\frac{\sqrt{a_{n+1} \ a_n}}{a_{n+1}}=\sqrt{\frac{a_n}{a_{n+1}}}=\frac{1}{\sqrt{\frac{a_{n+1}}{a_{n}}}} \ \ (\star)\]let \(b_n=\frac{a_{n+1}}{a_n}\) from \((*)\) u have\[b_{n+1}=\frac{1}{\sqrt{b_n}}\]\[b_0=\frac{a_1}{a_0}=\frac{8}{2}=4\]\[b_1=\frac{1}{\sqrt{b_0}}=\frac{1}{\sqrt{4}}=4^{-\frac{1}{2}}\]\[b_2=\frac{1}{\sqrt{b_1}}=\frac{1}{\sqrt{4^{-\frac{1}{2}}}}=4^{\frac{1}{4}}\]\[b_3=\frac{1}{\sqrt{b_2}}=\frac{1}{\sqrt{4^{\frac{1}{4}}}}=4^{-\frac{1}{8}}\]\[. \\ . \\ . \]\[b_n=4^{\frac{(-1)^n}{2^n}}\]ok now u can find general term og \(a_n\) using this relation\[a_{n+1}=4^{\frac{(-1)^n}{2^n}} a_n\]let me know if its not clear..hope itll help