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Infinite Pentation (and x-srt-x)
#11
nuninho1980 Wrote:yes. but I don't get value "x" of a^^x if x<1 using your code "tetration and slog". Sad

Don't be sad, that's to be expected. My code only applies when (a>1), which would require that (x>1) for this function.
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#12
So this is my attempt at a proof, and there are a bunch of inequalities, which always confuse me Tongue, so let me know if there are any mistakes.

Let as usual. I will use k (instead of n) to avoid confusion with .

Lemma 1 (Knoebel). iff for all .

Lemma 2. for all real .
Proof.
The base- tetrational function is continuous and monotonic?

Lemma 3. for all positive real and real .
Proof.
The function for all , thus . If , then , so .

Lemma 4 (lower bound). For all integer , .
Proof.
Since , then obviously . Together with lemma (2), this implies that . Substituting (hypothesis), this can be written as which implies by lemma (1).

Lemma 5 (decreasing). For all integer and , .
Proof.
We have for all integer by definition. It follows that
by lemma (3). Thus which implies by lemma (1).

Theorem.
Proof.
for all by lemma (4) and lemma (5).
... In the limit, the squeeze theorem and completeness should guarantee that the limit exists and converges to .

Is this right?

Andrew Robbins
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#13
So I tried using regular and natural tetration to fill out the rest of the graph. I almost ignored using bases where because these bases would have 2 fixed points: (1) the upper fixed point would overlap the "blue" region which I already calculated with interpolation (but a good area for comparison), and (2) the lower fixed point would overlap the "green" region which I already calculated with natural iteration (also a good area for comparison).

The only part I used regular tetration for was the "red" region, which corresponds to bases where . Using pure regular iteration from the primary fixed point of the base-a tetrational is real-valued only for integers, and complex-valued everywhere else. So I fudged a little bit to make the graph, I solve the equation and these points are what you see in the "red" region of the graph. I'm pretty sure these are wrong, and that the graph would have to be complex here (if tetrationals for bases less than 1 form complex outputs for real inputs), so it's more of a heuristic than a continuation of the blue curve.

   

Andrew Robbins


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#14
andydude Wrote: for all by lemma (4) and lemma (5).
... In the limit, the squeeze theorem and completeness should guarantee that the limit exists and converges to .

Is this right?

I dont think so. When the sequence is decreasing and bounded it has a limit. But this limit could be probably bigger than . Or did I overlook something?
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#15
(04/13/2009, 05:01 PM)bo198214 Wrote:
andydude Wrote: for all by lemma (4) and lemma (5).
... In the limit, the squeeze theorem and completeness should guarantee that the limit exists and converges to .

Is this right?

I dont think so. When the sequence is decreasing and bounded it has a limit. But this limit could be probably bigger than . Or did I overlook something?

for all

so the limit must be eta when lim k = oo because the rule is < and not =< , i think bo has overlooked that !?!

or did i miss something ?
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#16
(05/03/2009, 10:42 PM)tommy1729 Wrote: for all

so the limit must be eta when lim k = oo because the rule is < and not =< , i think bo has overlooked that !?!

or did i miss something ?

yes, take a similar example:


by your argumentation .
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#17
(05/03/2009, 10:51 PM)bo198214 Wrote:
(05/03/2009, 10:42 PM)tommy1729 Wrote: for all

so the limit must be eta when lim k = oo because the rule is < and not =< , i think bo has overlooked that !?!

or did i miss something ?

yes, take a similar example:


by your argumentation .

thats not the same , the limit value cannot possibly be smaller than eta , so eta is the ultimate boundary.

and because any infinite tetration of real base (eta + q) = oo

our resulting limit must be between eta < " limit "< eta + q

taking lim q = 0+ => " limit " = eta.
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#18
(05/03/2009, 10:57 PM)tommy1729 Wrote: thats not the same , the limit value cannot possibly be smaller than eta ,
Of course not the limit in the above case is 0. It is *greater* than -1.
Thats what I already said in my original objection: the limit can possibly be *greater* than .

Quote:and because any infinite tetration of real base (eta + q) = oo

proof?
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#19
(05/03/2009, 10:57 PM)tommy1729 Wrote: and because any infinite tetration of real base (eta + q) = oo

I thought this was about infinite pentation, not infinite tetration.
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#20
(05/04/2009, 08:20 PM)BenStandeven Wrote:
(05/03/2009, 10:57 PM)tommy1729 Wrote: and because any infinite tetration of real base (eta + q) = oo

I thought this was about infinite pentation, not infinite tetration.

Well depends how you interpret it. Surely the title "infinite pentation" means

which is the same as infinitely repeated tetration, which tommy refers to.
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