Iterated exponentiation

[Guest bgjyd834]

Is there any way to find out what the following becomes?

 

 

Thanks in advance!

[Xittenn]

What is 0.690347126114964 ?????

 

0.69034712611496429 @ n = 10e10 :|

 

well calculating seems to have been a very bad idea ... double precision and the library used wasn't good enough and I noticed that when I started at n=odd vs n=even I get two different numbers ... I'll have to look at this again in the morning if nobody comes up with a better or more viable solution ...

[Shadow]

Here is a plot of the sequence as n goes from 0 to fifty (blue) as well as the average of consecutive values (red).

 

 

My computational resources are limited, but to the best of my knowledge the sequence converges (super, super slowly) to 0.6743563626906477538..., which is just some random number according to WolframAlpha. Hope this helps.

Edited April 27, 2011 by Shadow

[imatfaal]

Isn't it simply 1/2^(1/n!)

 

edit there is a power of 1/2 in the exponent that I dont want in the above answer

 

so it would be 1/2^(2/n!)

Edited April 27, 2011 by imatfaal

[Shadow]

You're confusing tetration with repeated exponentiation.

Edited April 27, 2011 by Shadow

[imatfaal]

Shadow / Xitten Could you explain what calc you are running as I am totally lost. Surely this is just nested roots or power tower. Cube root of a half is about ..79, fourth root of .79ish is .94ish ; it's gonna converge on one

 

Shadow Just read your link on tetration - digesting and will revert . To aid could you elucidate on the calc you did so that I can follow it. Thanks

 

Last edit - So in the question the power of 1/4 is acting on the 1/3 not on the whole below it. I think I get it (ie what you are saying the question is). Thanks

I would still love to see the bones of your calc - cos I still think the original question reads as one half, raised to the power of a third, the answer of which is in turn raised to the power of one fourth, etc. And as you multiply exponents like this I would just multiply out to say that exponent is 1/(3*4*5...n) ie 2/n!

Edited April 27, 2011 by imatfaal

[Xittenn]

I was just using C++, I was installing SAGE to do a better job(I hadn't installed it yet) but Shadow has posted. I am curious about Shadows projected figure and I still plan to double check this in SAGE and maybe a few other places.

 

n = 50

 

x = 1/n

 

for( i = n - 1; n > 1; n-- ) x = pow( 1/i, x );

 

 

I went for large numbers to start which was a little foolish; I was half asleep dammit !

 

> 9000

 

I don't get why the value bounces like that :/

[Shadow]

You're going in the wrong direction; when evaluating the power tower, one must first evaluate [math]{\frac{1}{3}}^{\frac{1}{4}} = 0.7598...[/math] and then [math]\frac{1}{2}^{0.7598...} = 0.5905...[/math] etc.

 

I'm short on time right now, I'll post more details on how I calculated the result later.

Edited April 27, 2011 by Shadow

[Xittenn]

You're going in the wrong direction; when evaluating the power tower, one must first evaluate [math]{\frac{1}{3}}^{\frac{1}{4}} = 0.7598...[/math] and then [math]\frac{1}{2}^0.7598... = 0.5905...[/math] etc.

 

I'm short on time right now, I'll post more details on how I calculated the result later.

 

 

note that n = 50 and i = n - 1 and i-- <===== the right way .... my numbers were actually the same as yours .... I don't understand why an odd n gives ~.65 and an even n gives ~.69 but I can see how this could converge on the avg. I am sure when I think a little more on it, it will become more clear to me.

 

 

n = 50

 

x = 1/n

 

for( i = n - 1; n > 1; n-- ) x = pow( 1/i, x );

 

I accidentally put a bunch of n's where i's should have been :/

 

i > 1 and i--

Edited April 27, 2011 by Xittenn

[Shadow]

I know, I was answering imatfaal's post

 

Also, with a little more time on my hands I was able to look at you're code, and your algorithm is wrong. What you are calculating is [math](((x^y)^z)^w)^q=x^{y \cdot z \cdot w \cdot w \cdot q} \neq x^{y^{z^{w^q}}} [/math], which is what the OP is trying to find.

 

I used Mathematica, specifically the function Fold, which is defined as

Fold[f, x, {a, b, c, d}] = f[f[f[f[x, a], b], c], d]

(The length of the list can be arbitrary)

 

I then defined f:

f[x, y] = y^x

made x = 1/n and generated the list containing the second function arguments (In the function definition above they are a, b, c, d) using the function Table as follows:

Table[1/n, {n-1, 2, -1}] 

which generated the list

{1/(n-1), 1/(n-2), 1/(n-3)...1/4, 1/3, 1/2}.

The final code for the function which gave the numeric value of the tetration operation was:

T[n] = Fold[f, 1/n, Table[1/n, {n-1, 2, -1}]]

I then plotted the following two sequences:

Table[T[n], {n, 3, 50}] (blue)

Table[(T[n]+T[n+1])/2, {n, 3, 50}] (red)

Et voilà. In essence, I was building the power tower backwards, ie. starting from the top and building down. The number I posted is equal to (T[101]+T[102])/2.

Edited April 28, 2011 by Shadow

[Xittenn]

I know, I was answering imatfaal's post

 

Also, with a little more time on my hands I was able to look at you're code, and you're algorithm is wrong. What you are calculating is [math](((x^y)^z)^w)^q=x^{y \cdot z \cdot w \cdot w \cdot q} \neq x^{y^{z^{w^q}}} [/math], which is what the OP is trying to find.

 

 

No my algorithm is fine :/ .... but thanks for your solution ... I will be trying in Sage when I finish doing some computer updates ....

 

I am still conflicted about how the plot is achieved where odd/even n starting points are affecting the outcome ...

[Xittenn]

I don't see any proof positive of there being convergence .....

 

http://www.wolframalpha.com/input/?i=plot%20%281%2Fn%29^%281%2F%28n%2B1%29%29%20n%3D2%20to%2050

 

http://www.wolframalpha.com/input/?i=plot+%281%2Fn%29^%281%2F%28n%2B1%29%29^%281%2F%28n%2B2%29%29+n%3D2+to+50

 

http://www.wolframalpha.com/input/?i=plot+%281%2Fn%29^%281%2F%28n%2B1%29%29^%281%2F%28n%2B2%29%29^%281%2F%28n%2B3%29%29+n%3D2+to+50

 

The red average is just that an average of two plots. There is the set of exponential terms which sum to an odd value of total terms. There is the set of exponential terms which sum to an positive value of total terms. I think there is actually evidence that suggests that each converge on their own values ..... I am not sure how to prove this indefinitely in either direction though :/

 

There is no continuous function that represents this problem, as far as I can tell!

[Shadow]

I am still conflicted about how the plot is achieved where odd/even n starting points are affecting the outcome ...

 

The plot was generated by plotting the points [n, T[n]], with n from 3 to 50. The first few point would then be:

 

n: 3 -> [3, [math]\frac{1}{2}^{\frac{1}{3}}[/math]] = [3, 0.7937...]

n: 4 -> [4, [math]\frac{1}{2}^{\frac{1}{3}^{\frac{1}{4}}}[/math]] = [4, 0.590563...]

n: 5 -> [5, [math]\frac{1}{2}^{\frac{1}{3}^{\frac{1}{4}^{\frac{1}{5}}}}[/math]] = [5, 0.73973...]

n: 6 -> [6, [math]\frac{1}{2}^{\frac{1}{3}^{\frac{1}{4}^{\frac{1}{5}^{\frac{1}{6}}}}}[/math] = [6, 0.62266...]

.

.

.

 

It is clear that for even n, the values increase with increasing n, while for odd n, the values decrease with increasing n. I did make one assumption which I hadn't realized at the time, and that is that they both converge to the same number (such as [math]a_n = (-\frac{1}{2})^x[/math] and that might not be true. I actually wouldn't be surprised if this were the case, because it does seem to converge awfully slowly. I'll have a look later and see if I can figure it out.

 

The red average is just that an average of two plots.

 

Yes, since that would be the approximate value of the limit if both the "even branch" and the "odd branch" converged to the same number. As I said, I unknowingly made an incorrect assumption.

Edited April 28, 2011 by Shadow

[imatfaal]

Shadow - thanks for taking the time to explain. Hopefully OP is still around to get benefit as well!

 

 

Appreciated and plusrepped.

 

Matthew

[Xittenn]

It is clear that for even n, the values increase with increasing n, while for odd n, the values decrease with increasing n.

 

Yes, since that would be the approximate value of the limit if both the "even branch" and the "odd branch" converged to the same number. As I said, I unknowingly made an incorrect assumption.

 

Note that given the generalized case it is in fact insufficient to say where n is odd or even but in fact is the parity of the sum total of the number of terms performing an operation that delineates the effect.

 

[math] where \; n \geq 1 [/math]

 

[math] \lim_{m\rightarrow\infty} (\frac{1}{n})^{(\frac{1}{n+1})^{.^{.^{.^{(\frac{1}{n + m})}}}}} [/math]

 

I wonder what [math] n = 1 [/math] says about the convergence of [math] n > 1 [/math] in the generalization?

Edited April 28, 2011 by Xittenn

[Xittenn]

Finch, Steven R. Mathematical constants, Volume 94. Pg 449 describes this as the hyperpower analogue of the harmonic series. He goes on to say that it is in fact divergent; though, it does converge in its odd/even parts.

 

[math] \lim_{n\rightarrow\infty} H_{2n} = 0.65836559922 . . . < 0.6903471261 . . . = \lim_{n\rightarrow\infty} H_{2n+1} [/math]

 

He does not give proof and I am finding it rather difficult to find one for any similar series.

 

Are the same types of proofs, such as the Ratio or Alternating Series proofs applicable to Hyperpower in some way? The Ratio test seems rather easily applied . . .

 

[math] where \: n \geq 1 [/math]

 

[math] \lim_{m\rightarrow\infty} \Xi_{m=0} \left ( \frac{1}{n+m} \right ) [/math]

 

The Number Theory book I have dances around a lot of this but never actually dips into!

 

o.o

Edited April 29, 2011 by Xittenn