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).
How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$?
6
$\begingroup$
discrete-mathematics
recurrence-relations
1 Answers
13
Hint: Let $b_n=\log a_n$ and solve a recurrence for $b_n$.
-
0$$ b_n=\log a_n , a_n^{b_n} = k ? $$ What do you mean? How it works? – 2012-11-28
-
0Where did you get $k$? If $b_n=\log a_n$ then $e^{b_n}=a_n$. – 2012-11-28
-
0:) Now I've the same question. Where did you get that $ e $ ? – 2012-11-28
-
0Sorry, 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
-
1You 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
-
0Take 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
-
0oh, 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
-
0The 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
-
1yeap... it must be $ 2b_n = b_{n-1} + b_{n-2} $ – 2012-11-28
-
0is 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
-
0@Buga1234 Those two equations are the same - it yields the same values for $\lambda$. – 2012-12-01