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

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

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2

-- AI

P.S -> .dettimbus steg egassem taht os dedda texT

-- AI

P.S -> .dettimbus steg egassem taht os dedda texT

- #3

AKG

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For all natural n:

[tex]n = \sqrt{1 + (n-1)\sqrt{1 + n\sqrt{1 + (n + 1)\sqrt{\dots }}}}[/tex]

For n = 1, we see the right side is:

[tex]\sqrt{1 + 0\sqrt{\dots }} = \sqrt{1} = 1[/tex]

**Note that for now, we're assuming that what we're looking at is even a number, and doesn't go to infinity**. Next, assume that:

[tex]k = \sqrt{1 + (k-1)\sqrt{1 + k\sqrt{1 + (k + 1)\sqrt{\dots }}}}[/tex]

for some natural k. Then:

[tex]k+1 = \frac{(k - 1)(k + 1)}{(k - 1)}[/tex]

From this, TenaliRaman's answer follows. It remains to prove, however, that the stuff really is a number and doesn't "diverge".**Any ideas as to how to prove that?**

[tex]n = \sqrt{1 + (n-1)\sqrt{1 + n\sqrt{1 + (n + 1)\sqrt{\dots }}}}[/tex]

For n = 1, we see the right side is:

[tex]\sqrt{1 + 0\sqrt{\dots }} = \sqrt{1} = 1[/tex]

[tex]k = \sqrt{1 + (k-1)\sqrt{1 + k\sqrt{1 + (k + 1)\sqrt{\dots }}}}[/tex]

for some natural k. Then:

[tex]k+1 = \frac{(k - 1)(k + 1)}{(k - 1)}[/tex]

[tex]= \frac{k^2 - 1}{k - 1}[/tex]

[tex]= \frac{\left (\sqrt{1 + (k-1)\sqrt{1 + k\sqrt{1 + (k + 1)\sqrt{\dots }}}}\right )^2 - 1}{k - 1}[/tex]

[tex]= \frac{(k - 1)\sqrt{1 + k\sqrt{1 + (k + 1)\sqrt{\dots }}}}{k - 1}[/tex]

[tex]= \sqrt{1 + k\sqrt{1 + (k + 1)\sqrt{\dots }}}[/tex]

as required.[tex]= \frac{\left (\sqrt{1 + (k-1)\sqrt{1 + k\sqrt{1 + (k + 1)\sqrt{\dots }}}}\right )^2 - 1}{k - 1}[/tex]

[tex]= \frac{(k - 1)\sqrt{1 + k\sqrt{1 + (k + 1)\sqrt{\dots }}}}{k - 1}[/tex]

[tex]= \sqrt{1 + k\sqrt{1 + (k + 1)\sqrt{\dots }}}[/tex]

From this, TenaliRaman's answer follows. It remains to prove, however, that the stuff really is a number and doesn't "diverge".

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Let f(x) = x+1

f(x)

= sqrt((x+1)^2)

= sqrt(x^2 + 2x + 1)

= sqrt(1 + x (x+2))

= sqrt(1 + xf(x+1))

Recursively applying the above gives,

f(x) = sqrt(1+x*sqrt(1+(x+1)*sqrt(1+(x+2)*sqrt(......))) = x + 1

place x = 1 and et voila!

-- AI

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ahhh, i see... how did you see that?

- #7

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Its one of those *aha* moments. Personally, these sort of "functional equations" are good fun to play around with (i have played with this one, quite a while back). Just remember the trick i used above and try to come up with some more radicals just like those.EvLer said:ahhh, i see... how did you see that?

For example,

f(x) = x + a = sqrt(x^2+2ax+a^2) = sqrt(a^2+x*f(x+a))

Recursively apply the above and get a new oh_so_cool radical expression, fit in a value for a and set some value for x and give it to your friends to torture them for a while.

Note in all of the above, its important to analyse that the infinite recursion is valid and that the infinite recursion actually converges. This aspect ofcourse i have pushed under the rug for now, since its a puzzle after all :tongue2: .

There is a paper by T Vijayaraghavan (IIRC) which discussed convergence issues of such radicals, which could probably be the starting point of such analysis.

-- AI

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have a link? or a reference?TenaliRaman said:There is a paper by T Vijayaraghavan (IIRC) which discussed convergence issues of such radicals, which could probably be the starting point of such analysis.

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AKG

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How would you check that the recursion is valid? EDIT: actually, since each step in the recursion produces the same number, why would the infinite recursion be any different? If it is different, what we have to do to check convergence?TenaliRaman said:Note in all of the above, its important to analyse that the infinite recursion is valid and that the infinite recursion actually converges.

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When i said recursion being valid, i meant, making sure that the final number doesnt change by infinite application of recursion.AKG said:How would you check that the recursion is valid?

umm, let me try and give an example,actually, since each step in the recursion produces the same number, why would the infinite recursion be any different?

Let f(x) = x

f(x) = x = (x+1) - 1

f(x) = f(x+1)-1

.....-3-2-1 = f(x) = x ,for any x

I guess this example is pretty lame but i hope it drives home the point.

I am not sure whether there is a convergence check (the paper i mentioned above, discusses convergence of infinite radicals IIRC and not exactly recursion, and to be honest, when i had read that paper, i couldnt understand a penny out of it, :uhh: cmon i am just 21 and an engineer at that so give me a break ) .If it is different, what we have to do to check convergence?

However, i guess one could model the convergence condition as that in the convergence of a series,

Let R(k,f(x)) denote the value of recursive function f(x) at depth k.

Then a possible convergence condition would be,

R(1,f(x)) = R(2,f(x)) = R(k,f(x)) = R(k+1,f(x))

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If i had, i would have definitely linked to it in the original post. I dont even remember where i had read it, it was quite a long time back. Hell i was a toddler in maths when i had read that, i was just able to quote that paper because i had read that in reference to Srinivasa Ramanujan, who btw was the brains behind such infinite radicals.EvLer said:have a link? or a reference?

-- AI

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Hurkyl

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Why not apply theLet R(k,f(x)) denote the value of recursive function f(x) at depth k.

Then a possible convergence condition would be,

R(1,f(x)) = R(2,f(x)) = R(k,f(x)) = R(k+1,f(x))

I.E. does this limit exist?

[tex]

\lim_{k \rightarrow \infty} R(k, f)

[/tex]

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Indeed!Hurkyl said:Why not apply theusualcondition for convergence of a sequence? :tongue2:

I.E. does this limit exist?

[tex]

\lim_{k \rightarrow \infty} R(k, f)

[/tex]

Damn, my post looks stupid now!

-- AI