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Define $$ S_{n+1} = \frac{S_n^2+x}{2S_n}$$ and $S_1 = k$, where x,k > 0. find an explicit formula for $S_n$ in terms of n.

I don't even know where to begin. I tried using algebraic manipulation to rid of $S_{n+1}$ but nothing is working. Since x needs to disappear I'm thinking we should use ratios.

Adam Staples
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  • They should be all small $s_n$. – science Mar 05 '15 at 05:06
  • Any starting value? i.e. $s_0$ – meta_warrior Mar 05 '15 at 05:07
  • Yes, $s_1 = k$ which is greater than 0, good catch, I'll edit the question. Also x is greater than 0. – Adam Staples Mar 05 '15 at 05:08
  • You should change the capital $S$ to $s$ in the recurrence relation. – science Mar 05 '15 at 05:12
  • Does this problem come from applications? – science Mar 05 '15 at 05:31
  • My friend asked me how to solve this. It was in his Advanced Calculus homework. We had proven earlier that $S_n \ge S_{n+1}$ – Adam Staples Mar 05 '15 at 05:34
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    The $x$ isn't very important because you can rewrite your recurrence relation as $\frac{S_{n+1}}{\sqrt{x}} = \frac{\left(\frac{S_n}{\sqrt{x}}\right)^2 + 1}{2\left(\frac{S_n}{\sqrt x} \right)}$ so a recurrence relation for $T_n = \frac{S_n}{\sqrt{x}}$ is just just $T_{n+1}=\frac{T_n^2 +1}{2T_n}$. In other words, you only need to deal with the case $x=1$. Note this could also be written as $T_{n+1} = \frac{T_n + \frac{1}{T_n}}{2}$ so that $T_{n+1}$ is the average of $T_n$ and its reciprocal. – Mike F Mar 05 '15 at 05:34

1 Answers1

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You can have the closed form

$$S_n = \sqrt {x}\coth \left( 2^{n-1}{ \coth^{-1}} \left( {\frac {k}{\sqrt {x}}} \right)\right) .$$

Added: If you want to find the limit without finding the closed form then you can advance as: assume $\lim_{n\to \infty } S_n = a $ then

$$ a=\frac{a^2+x}{2a} \implies a= \pm \sqrt{x}.$$

science
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  • That's a start but I think it wants it solely in terms of n. – Adam Staples Mar 05 '15 at 05:37
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    It is in terms of $n$! – science Mar 05 '15 at 05:38
  • Try $\frac{S_n}{S_{n+1}}$? or even $\frac{T_n}{T_{n+1}}$? – Adam Staples Mar 05 '15 at 05:43
  • @AdamStaples The recurrence depends on $x$. You can't make it disappear. – Yuval Filmus Mar 05 '15 at 05:45
  • That is true. Maybe I misunderstood the question. I don't have the problem in front of me in order to know how to change it to the correct problem. maybe it's the limit of the sequence? It said I should have something like (135...(2n-1))/(...) but I cannot recall exactly. – Adam Staples Mar 05 '15 at 05:50
  • @science Is there a more general result or a way of deriving this result? – DanielV Mar 05 '15 at 06:08
  • @DanielV: Yes there are. – science Mar 05 '15 at 06:09
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    @DanielV Notice the similarity of the recursion formula with the addition formula $\coth(2x) = \frac{\coth^2(x)+1}{2\coth(x)}$. If $a_n = \coth(2^n\alpha)$ then the identity above implies that $a_{n+1} = \frac{a_n^2 + 1}{2a_n}$. This type of procedure can be applied to other (trigonometric) addition formulas to get recursion formulas (or reversed engineered to find solutions). For example since $\sin(2x) = 2\sin(x)\cos(x)$ we have that $a_n = \sin(2^n \alpha)$ solves $a_{n+1} = 2a_n\sqrt{1-a_n^2}$. – Winther Mar 05 '15 at 06:22
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    So not really a general approach, but an interesting set of singular cases from trig ID. Thanks~ – DanielV Mar 05 '15 at 07:59