Root sequence question

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Short root sequence question

For the recursive sequence
[tex] R_n = x + \sqrt {x - \sqrt {R_{n - 2} } } [/tex]
[tex] R_0 = x = k^2 - k + 1 [/tex]

[tex] \forall k \in \mathbb{N},\;k > 1 [/tex] why does [tex] \mathop {\lim }\limits_{n \to \infty } R_n = k^2 [/tex] ??
 
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  • #2
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Perhaps if I redefine the whole problem and open it to the reals ([tex] \mathbb{R} [/tex]),
it will become clearer and easier to solve :smile: .

First off, the sequence shall be defined recursively as
[tex] R_n = \sqrt {x + \sqrt {x - \sqrt {R_{n - 2} } } } [/tex]
(just takin' the root of what's in my original post! (It's a root sequence anyway))

So:
If we start with [tex] R_0 [/tex], we will follow with [tex] \left\{ {R_2 ,R_4 , \ldots ,R_n } \right\} [/tex]
If we start with [tex] R_1 [/tex], we will follow with [tex] \left\{ {R_3 ,R_5 , \ldots ,R_n } \right\} [/tex]

Such that:
[tex] R_0 \in \left\{ {x|x > \sqrt {x + \sqrt {x - \sqrt x } } > 1 ,x \in \mathbb{R}} \right\} [/tex] (numerically, x > 1.311 for [tex] R_0 [/tex])
[tex] R_1 \in \left\{ {x|x > \sqrt {x + \sqrt x } > 1 ,x \in \mathbb{R}} \right\} [/tex] (numerically, x > 1.755 for [tex] R_1 [/tex])

We will find that:
[tex] \mathop {\lim }\limits_{n \to \infty } R_n = \frac{{1 + \sqrt {4x - 3} }}{2} [/tex]
(doesn't matter which [tex] R_0 [/tex] or [tex] R_1 [/tex] you choose...
just as long as it's inside the appropriate interval!:biggrin:)
------------------------------------------------------------------
Now, why exactly does:
[tex] \mathop {\lim }\limits_{n \to \infty } R_n = \frac{{1 + \sqrt {4x - 3} }}{2} [/tex] ?
 
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  • #3
CRGreathouse
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bomba923 said:
For the recursive sequence
[tex] R_n = x + \sqrt {x - \sqrt {R_{n - 2} } } [/tex]
[tex] R_0 = x = k^2 - k + 1 [/tex]

[tex] \forall k \in \mathbb{N},\;k > 1 [/tex] why does [tex] \mathop {\lim }\limits_{n \to \infty } R_n = k^2 [/tex] ??
If the limit exists, you should just be able to set [tex]R_n=R_{n-2}[/tex] and solve. This gives

[tex]((R-k^2+k-1)^2-k^2+k-1)^2-R=0[/tex]

which has solutions

[tex](R-k^2)(R-(k^2-2k+1))(R^2+(-2k^2+2k-3)R+(k^4-2k^3+3k^2-2k+1))[/tex]

Why this is defined for all [tex]k\in\mathbb{Z}^+[/tex] and why it has the one particular solution out of the four possible ones I don't know offhand.
 
  • #4
Hurkyl
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[tex]
\lim_{n \rightarrow \infty} R_n =
\lim_{n \rightarrow \infty} \sqrt{x + \sqrt{x - \sqrt{R_{n-2}}} }
= \sqrt{ \lim_{n \rightarrow \infty}x + \sqrt{x - \sqrt{R_{n-2}}} }
= \cdots
[/tex]
 
  • #5
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Hurkyl said:
[tex]
\lim_{n \rightarrow \infty} R_n =
\lim_{n \rightarrow \infty} \sqrt{x + \sqrt{x - \sqrt{R_{n-2}}} }
= \sqrt{ \lim_{n \rightarrow \infty}x + \sqrt{x - \sqrt{R_{n-2}}} }
= \cdots
[/tex]
But how does that simplify [tex] \mathop {\lim }\limits_{n \to \infty } R_n [/tex] down to '"[tex] \frac{{1 + \sqrt {4x - 3} }}{2} [/tex]"' ?
 
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  • #6
Hurkyl
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Continue through the dots. :tongue2: It might help you visualize to use L to denote [itex]\lim_{x \rightarrow \infty} R_n[/itex].
 
  • #7
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Oh--ic now
!
 
  • #8
saltydog
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You mind explaining this a little further?

I realize the limit of an infinitely nested radical:

[tex]\sqrt{x+\sqrt{x+\sqrt{x+...}}}[/tex]

is:

[tex]\frac{1+\sqrt{1+4x}}{2}[/tex]

but don't see how to get the 3 in there.
 
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  • #9
Hurkyl
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I think your problem is that you're considering a different sequence. :smile:
 
  • #10
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saltydog said:
You mind explaining this a little further?

I realize the limit of an infinitely nested radical:

[tex]\sqrt{x+\sqrt{x+\sqrt{x+...}}}[/tex]

is:

[tex]\frac{1+\sqrt{1+4x}}{2}[/tex]

but don't see how to get the 3 in there.
My sequence alternates the signs of the square roots as I embed them, as
[tex]\sqrt{x+\sqrt{x-\sqrt{x+...}}}[/tex]
 
  • #11
saltydog
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Hello Bomba. Thanks for responding, you too Hurkyl. I know you're just telling me the minimum for my own good but you know what, I couldn't even get past the definition and I'd like to understand it. This is what I have:

If I start from [itex]R_0[/itex] and you say calculate R2, R4, R6,. . ., and the recursion relation is always in reference to R(n-2), then it's just like doing it for each N and set up the relation for R(n-1). Is that not correct? That is, if I start with [itex]R_0[/itex], then it seems to me, just ignore the odd members in the sequence. Perhaps that is not correct though. When I experiment with it though in Mathematica, say for R0=1.9, it doesn't seem to converge to the value you specify. This is how I set it up in Mathematica. Perhaps you can correct my interpretation:

[tex]f[x,xm1]=\sqrt{x+\sqrt{x-\sqrt{xm1}}}[/tex];

[tex]xstart=1.9[/tex]

[tex]ntotal=alargenumber[/tex]

[tex]valist=Table[{0},{ntotal}][/tex]

[tex]valist[[1]]=f[xstart,xstart][/tex]

[tex]For[n=2,n<ntotal,n++,[/tex]

[tex]valist[[n]]=f[xstart,valist[[n-1]]];[/tex]

When I do that, it seems to converge to 1.639 . . .

But:

[tex]\frac{1+\sqrt{4xstart-3}}{2}=1.572...[/tex]

Would you (or Hurkyl or anyone else) tell me where my problem is?

Thanks,
Salty

Edit: corrected to reflect relation of even members in terms of R0
 
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  • #12
Hurkyl
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Hrm. The problem is that I had assumed that what was posted as "the answer" was correct. 1.639... is, in fact, the correct limit. (Assuming that my calculuator is sufficiently smart and would tell me if there were additional real solutions to the fixed point equation)
 

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