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I am trying to solve the recurrence: $$ a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}} $$ but here is a problem for me. After few steps I have this: $$ a_n^2 = a_{n-1}\cdot a_{n-2} $$ and I don't now what to do further. I can solve a recurrence like that $$ a_{n+2} + a_{n+1} - a_n = 5 \cdot 2^n, $$ but I can't find any information about this case (when I have some degree or
square root in a recurrence).

1 Answers 1

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Hint: Let $b_n=\log a_n$ and solve a recurrence for $b_n$.

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    $$ b_n=\log a_n , a_n^{b_n} = k ? $$ What do you mean? How it works?2012-11-28
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    Where did you get $k$? If $b_n=\log a_n$ then $e^{b_n}=a_n$.2012-11-28
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    :) Now I've the same question. Where did you get that $ e $ ?2012-11-28
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    Sorry, mathematicians usually use $\log$ to mean the natural logarithm, which sometimes is written as $\ln$ instead. $e$ is just Euler's constant, the base for the natural logarithm.2012-11-28
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    You don't need to use natural logarithms, you can use any logarithm base you want with this problem. If you define $b_n=\log_{10} a_n$ then you know that $10^{b_n}=a_n$.2012-11-28
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    Take the log of both sides of $a_{n+2}^2=a_{n+1}a_n$. What do you get? It is not $b_{n+2}=b_{n+1}+b_n$...2012-11-28
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    oh, thank you very much! Now I got it, I think... $ a_n = 10^{b_n} $, $ 10^{b_n} = 10^{b_{n-1}} * 10^{b_{n-2}} $, $ 10^{b_n} = 10^{b_{n-1} + b_{n-2}} $, $ b_n = b_{n-1} + b_{n-2} $ is this a rigth way?2012-11-28
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    The last statement, $b_n=b_{n-1}+b_{n-2}$ is wrong. You forgot that the original expression was $a_n^2$ on the left hand side...2012-11-28
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    yeap... it must be $ 2b_n = b_{n-1} + b_{n-2} $2012-11-28
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    is it the linear recurrence? Will be right if I'll do something like this $ 2\lambda^2 - \lambda - 1 = 0 $ or I must to do this $ \lambda^2 - 1/2 \lambda - 1/2 = 0 $2012-12-01
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    @Buga1234 Those two equations are the same - it yields the same values for $\lambda$.2012-12-01