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Hard Sequence problemby aznluster
Tags: sequence 
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#1
Dec2010, 12:15 AM

P: 5

Find [tex]\sqrt{1+\sqrt{1+2\sqrt{1+3\sqrt{1+...}}}}[/tex]



#2
Dec2010, 04:39 AM

P: 32

It comes out to be 3. Why? I'm trying to find out.



#3
Dec2010, 05:01 AM

P: 32

Ok so this is basically one of Ramanajan's identities:
[tex] x + n + a = \sqrt{ax + (n+a)^{2} + x \sqrt{a(x +n) + (n+a)^{2} + (x + n) \sqrt{...}}}}[/tex] 


#4
Dec2010, 11:14 PM

P: 5

Hard Sequence problem
Wouldn't that add up to 2? Or maybe I'm doing something wrong. Is there a proof of that identity somewhere?



#5
Dec2010, 11:45 PM

P: 3,014

Let
[tex] x_{n} = \sqrt{1 + n \sqrt{1 + (n +1) \sqrt{1 + \ldots}}} [/tex] Then, we have the backwards recursive formula: [tex] x_{n} = \sqrt{1 + n x_{n + 1}} [/tex] We want to find [itex]x_{1}[/itex]. Let us see what is the limit [itex]x \equiv \lim_{n \rightarrow \infty}{x_{n}}[/itex]? If we want a finite limit, we must have: [tex] x_{n} \sim \frac{1}{n}, \; n \rightarrow \infty [/tex] but, this is impossible since all [itex]x_{n} > 0[/itex]. Therefore, the limit is infinite. Let us assume that is asymptotically a power law: [tex] x_{n} \sim A \, n^{\alpha}, \; n \rightarrow \infty, \; \alpha > 0 [/tex] Then, we have: [tex] A n^{\alpha} \sim \sqrt{1 + n A (n + 1)^{\alpha}} \sim \sqrt{A} \, n^{\frac{1 + \alpha}{2}} [/tex] which means: [tex] \alpha = \frac{1 + \alpha}{2} \Rightarrow \alpha = 1 [/tex] and [tex] A = \sqrt{A} \Rightarrow A = 0 \vee A = 1 [/tex] So, we may conclude: [tex] x_{n} = n (1 + y_{n}), \; y_{n} \rightarrow 0, \; n \rightarrow \infty [/tex] The backwards recursive relation for [itex]y_{n}[/itex] is: [tex] n (1 + y_{n}) = \sqrt{1 + n \cdot n (1 + y_{n + 1})} [/tex] [tex] y_{n} = \sqrt{1 + y_{n + 1} + \frac{1}{n}}  1 [/tex] [tex] y_{n + 1} = y_{n} (2 + y_{n})  \frac{1}{n} [/tex] Define a discrete ztransform: [tex] Y(z) \equiv \sum_{n = 1}^{\infty}{y_{n} z^{n}} [/tex] As we can see, [itex]y_{1}[/itex] is the coefficient in the Laurent series of [itex]Y(z)[/itex] around [itex]z = 0[/itex]. According to the residue theorem: [tex] y_{1} = \mathrm{Res}(Y(z), z = 0) = \frac{1}{2 \pi i} \, \oint_{C_{z = 0}}{Y(z) \, dz} [/tex] We need to find a functional equation for [itex]Y(z)[/itex] from the above (nonlinear) recursive relation. 


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