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Hard Sequence problem

by aznluster
Tags: sequence
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aznluster
#1
Dec20-10, 12:15 AM
P: 5
Find [tex]\sqrt{1+\sqrt{1+2\sqrt{1+3\sqrt{1+...}}}}[/tex]
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QuantumJG
#2
Dec20-10, 04:39 AM
P: 32
It comes out to be 3. Why? I'm trying to find out.
QuantumJG
#3
Dec20-10, 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]

aznluster
#4
Dec20-10, 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?
Dickfore
#5
Dec20-10, 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 z-transform:

[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 (non-linear) recursive relation.


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