Square Roots and Sequences
in [Math]
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From: Maury Barbato on 14 Apr 2010 02:34 Hello, let {s_n} be the sequence defined by s_0 = 2^(1/2), s_{n+1} = (2 + s_n)^(1/2), and {t_n} the sequence defined by t_0 = 2^(1/2), t_{n+1} = (2 + [(s_n)^(1/2)])^(1/2). Then it's easy to see that {s_n} and {t_n} are two strictly increasing sequences and that t_n <= s_n < 2, for every n. So they are both convergent. My question is: what are the limits of {s_n} and {t_n}? Thank you very much for your attention. My Best Regards, Maury Barbato
From: José Carlos Santos on 14 Apr 2010 06:56 On 14-04-2010 11:34, Maury Barbato wrote: > let {s_n} be the sequence defined by > > s_0 = 2^(1/2), > > s_{n+1} = (2 + s_n)^(1/2), > > and {t_n} the sequence defined by > > t_0 = 2^(1/2), > > t_{n+1} = (2 + [(s_n)^(1/2)])^(1/2). > > Then it's easy to see that {s_n} and {t_n} are two > strictly increasing sequences and that t_n<= s_n< 2, > for every n. So they are both convergent. > My question is: what are the limits of {s_n} and {t_n}? Since (s_n)_n converges to some limit _s_, then s = (2 + s)^(1/2). Therefore, s = 2. OTOH lim_n t_n = (2 + (lim_n s_n)^(1/2))^(1/2) = (2 + 2^(1/2))^(1/2). Best regards, Jose Carlos Santos
From: William Elliot on 14 Apr 2010 07:13 On Wed, 14 Apr 2010, Maury Barbato wrote: > Hello, > let {s_n} be the sequence defined by > > s_0 = 2^(1/2), > > s_{n+1} = (2 + s_n)^(1/2), > > and {t_n} the sequence defined by > > t_0 = 2^(1/2), > > t_{n+1} = (2 + [(s_n)^(1/2)])^(1/2). > > Then it's easy to see that {s_n} and {t_n} are two > strictly increasing sequences and that t_n <= s_n < 2, I'll take your word for it about s and ignore your observation about t. > for every n. So they are both convergent. > My question is: what are the limits of {s_n} and {t_n}? > Let s_n -> a. Notice 0 < a. a = sqr(2 + a) a^2 = 2 + a a^2 - a - 2 = (a - 2)(a + 1) = 0 a = 2 t_n -> sqr(2 + sqr a) = sqr(2 + sqr 2)
From: Maury Barbato on 14 Apr 2010 05:45 Jose Carlos Santos wrote: > On 14-04-2010 11:34, Maury Barbato wrote: > > > let {s_n} be the sequence defined by > > > > s_0 = 2^(1/2), > > > > s_{n+1} = (2 + s_n)^(1/2), > > > > and {t_n} the sequence defined by > > > > t_0 = 2^(1/2), > > > > t_{n+1} = (2 + [(s_n)^(1/2)])^(1/2). > > > > Then it's easy to see that {s_n} and {t_n} are two > > strictly increasing sequences and that t_n<= s_n< > 2, > > for every n. So they are both convergent. > > My question is: what are the limits of {s_n} and > {t_n}? > > Since (s_n)_n converges to some limit _s_, then > > s = (2 + s)^(1/2). > > Therefore, s = 2. > > OTOH lim_n t_n = (2 + (lim_n s_n)^(1/2))^(1/2) = (2 + > 2^(1/2))^(1/2). > > Best regards, > > Jose Carlos Santos Ops, I had forgotten this trick, whcih allows us to calculate very easily the limit of recursive sequences .. Anyhow, there was a typo in the definition of {t_n}. The sequence I had in mind is the following: t_0 = 2^(1/2), t_{n+1} = (2 + [(t_n)^(1/2)])^(1/2). But, of course the same reasoning applies: {t_n} is strictly increasing and t_n < 2 for every n. So {t_n} is convergent, and the limit t is the unique solution of the equation t^2 = 2 + t^(1/2). Thank you very very much, Jose, once again. Friendly Regards, Maury Barbato The future influences the present just as much as the past. (F. Nietzsche)
From: Ludovicus on 15 Apr 2010 04:18
On Apr 14, 9:45 am, Maury Barbato <mauriziobarb...(a)aruba.it> wrote: > So {t_n} is convergent, and the limit t is the unique > solution of the equation t^2 = 2 + t^(1/2). The solution of the equation: t = [2 + t^(1/2)]^(1/2) is 1.83117720722624... I don't know if it is possible to express it algebraically. |