I have the recursive sequence such as $x_{n+2}=\sqrt{x_{n+1}x_{n}}$ I've tried to solve it considering some solutions in the form $x_n=A* \lambda^n $ but with no results. Can someone explain me how to solve it, just some hints?
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1What are $x_1$ and $x_2$? Two arbitrary positive numbers? – Umberto P. Feb 13 '20 at 17:17
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1Let, $a_{n+1}=\frac{x_{n+2}}{x_{n+1}}$ then $a_{n+1}=(a_{n})^{-\frac{1}{2}}$ – Feb 13 '20 at 17:22
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3Try $y_n = \log x_n$. – copper.hat Feb 13 '20 at 17:22
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@copper.hat To have the equation $y_{n+1}-y_n-4=0$? – Anne Feb 13 '20 at 17:29
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@Anne: No, try again. – copper.hat Feb 13 '20 at 17:30
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Better still, look at @Saf_Swathi_Rosogolla's comment. – copper.hat Feb 13 '20 at 17:31
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I have a lot of confusion....can someone answer the question and get more information about the way to solve it, please? – Anne Feb 13 '20 at 17:39
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@Anne the answer is heavily dependent on what your initial choices of $x_1$ and $x_2$ are. – Cameron Williams Feb 13 '20 at 17:49
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x_1 and x_2 are two values important to find the constant of the general solution but not of absolute importance to find the general solution itself, otherwise my book would not report this exercize – Anne Feb 13 '20 at 17:53
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@Saf_Swathi_Rosogolla: You should add an answer based on your comment :-). – copper.hat Feb 13 '20 at 18:11
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I'm waiting for some answer from you instead, to understand why a solution for this exercize exists but no one seems to give me some valid hints to find it. – Anne Feb 13 '20 at 18:15
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Look at @Saf_Swathi_Rosogolla's comment!!! It is an excellent suggestion that lends itself to a simple solution. Plus it highlights issues of initial conditions appropriately. – copper.hat Feb 13 '20 at 18:18
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I understood Saf_swathi_Rosogolla's comment but how to continue from that point I have no idea! This is the reason why I wrote in a comment that evidently no one read to answer the question and write explicitly the solution or something to understand better – Anne Feb 13 '20 at 18:46
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No need to be snippy. You need to read the comments and do some work. This is not a homework solution service. If you follow @Saf_Swathi_Rosogolla's comment and take logs you can solve the problem. – copper.hat Feb 13 '20 at 18:55
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1Does this answer your question? Is it possible to utilize the convergence of the sequence $z_{n+1}=a/(1+z_n)$ to prove that the sequence $x_{n+2} = \sqrt{x_{n+1} x_n}$ is convergent? – rtybase Feb 13 '20 at 21:40
1 Answers
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This is a suggestion. I have not tried to complete all the details but it seems that it will work. If $a_n:=\log(x_n)$ you obtain
$$ a_{n+1}=\frac12 a_{n+1} + \frac12 a_n$$ this can be express as
$$ \begin{pmatrix} a_{n+2}\\ a_{n+1} \end{pmatrix} =\begin{pmatrix} \frac12 & \frac12\\ 1 & 0 \end{pmatrix} \begin{pmatrix} a_{n+1} \\ a_n \end{pmatrix} $$ Decomposing the matrix above as $P^{-1}\operatorname{diag}(-\frac12, 1) P$ for some appropriate $P$ will give a solution for $[a_{n+2}\, a_{n+1}]^\top$ in terms of initial conditions $[a_1\, a_0]^\top$. You then may recover an expression for $x_{n+2} = \exp(a_{n+2})$
Mittens
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excuse me, but this is an exercize from the Analisys I book. is it possible to have a solution without no more than analisys I notions? – Anne Feb 13 '20 at 18:11
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you don't have to use a little matrix algebra if you so prefer. Still, it is a standard thing to use when dealing with linear recursive sequences. It lightens the calculations a bit. – Mittens Feb 13 '20 at 18:16
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if I put $a_n=log (x_n)$ my result is $a_{n+1}-a_n-4=0$ and I don't see how it can be your system – Anne Feb 13 '20 at 18:23
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