/sci/ - Science & Math

>>14711048
Did anyone get a numerical answer last thread? I was at work and forgot about this

>>14711048
Thanks, forgot about this. Maybe I'll think some more over the weekend.

The sequence converges to 0.23385218...
The recursion enters a fixed point cycle with each new numerator converging to 0.71191079...

log(0.71191079) / log(0.23385218) = 0.23385218

>>14711048
Basically, I got to:
f(x/2)*log(f(x)) = log(f(x+1))
expanded in powers of 1/x, f(x) is roughly:
1 + (-1/2)/x + (3/8)/x^2 + (-7/24)/x^3 + (89/384)/x^4 + (-809/4320)/x^5 + (7073/46080)/x^6
Solving X^X = f(1) should give the answer.
I've just been trying to prove that the coefficients are alternating and decreasing.
If you let f(x) = e^g(x) then differentiate, you get
g'(x/2)g(x)g(x+1)/2 + g'(x)g(x+1) = g(x)g'(x+1).
This should make generating coefficients easier.

>>14711381
How did you get f(x/2)log f(x) = log f(x+1) ? What is f(x) in this context?

>>14711702
The sequence in question can be described by the the recurrence:
L(0,x)=x
L(1,x) = log(x+1)/log(x) = log(L(0,x+2^0))/log(L(0,x))
...
L(n+1,x) = log(L(n,x+2^n))/log(L(n,x))
Plugging in x=2 corresponds to OP's sequence.
Define a new sequence F(n,x) = L(n,x*2^n).
F(n+1,x/2) = log(F(n,x+1))/log(F(n,x))

It can be shown with a little work that F(n,x) converges to some f(x) near x = +infinity as n tends to infinity.
To prove this you would need to expand each L(n,x) around x = infinity and compute recurrences for each coefficient.
Using strong induction you can show the coefficient of 1/x^k is asymptotic to b(k)*2^(nk) as n tends to infinity.
The induction hypothesis is that the coefficients of 1/x^k in L(n,x) are asymptotic to b(k)*2^(nk) as n tends to infinity for all k<K.

>>14711825
>Expand in the neighborhood of infinity
Oh tucky you out here trying to solve this /sci/ head scratchers too? No wonder we need infinity hats.

>>14711048
Whatever number this sequence converges to,
we're claiming it and naming it after someone.

>>14712061
Anon’s theorem

>>14712115
>>14712061
Hmm...not bad. I'd accept it.

>>14711048
This looks like new math. If this number converges that means there's probably a whole host of sequences out there like this that converge.

>>14711825
I see, very clever. I guess once you have that and take limits it becomes a problem of solving [math] f(x)^{f(x/2)} = f(x+1) [/math]. There's probably a way of calculating the exact coefficients of a power series of f.

>>14712584
Are you able to find a value of f?

>>14712584
If we can't compute a value of f the power series is meaningless. I don't fully understand but it sounds like >>14711825 got an asymptotic expansion which wouldn't help.

>>14712614
>got an asymptotic expansion which wouldn't help
I plugged in x=1 to
>1 + (-1/2)/x + (3/8)/x^2 + (-7/24)/x^3 + (89/384)/x^4 + (-809/4320)/x^5 + (7073/46080)/x^6
and did a 3rd order shanks transformation on the partial sums to get
>385035044785342820194769960414379643/540846770058945445942960933552987392
which is roughly
0.71191151745 which agrees with >>14711064
I think the asymptotic series converges at 1.

>>14711064
I ran a few more terms piecewise out to 2^32+1 and tried some generalizations.

0.2338521816856... is an attractor for all input values that don't cause the sequence to hit a landmine at -1, 0, or 1. Even in cases where the terms go complex, the imaginary component converges to 0.

For example, consider f(x,A) where x is the number used to seed the sequence (as in >>14711825) and A is a constant (linear) difference between terms. x and A can be positive or negative. OP would be f(2,1) giving the input values {2,3,4,5...} and f(1/2,-2) would give the inputs {1/2, -3/2, -7/2, -11/2...}. All such f(x,A) that don't hit -1, 0, or 1 converge to the real number 0.2338521816856...

This generalizes further to nonlinear differences and we get the same result for f(x;b,c) where the input values are stretched exponentially or polynomially by b^n or n^c at the nth input. Iow, OP would be f(1;0,1) and f(1;2,3) would give {4,13,36,81...}. These also all converge to the real number 0.2338521816856...

Heuristically, this works because any constant tower of exponents is eventually dwarfed by the infinite nest of logs.

>>14713249
At this point this problem is probably solvable. Not just it, but the generalizations of it.

1=0!
2=1! 2
8=2! 2^2
24 = 3! 2^2
384 = 4! 4^2
4320 = 5! 6^2
46080 = 6! 8^2

>>14714977
Its annoying the first few terms break the pattern. I don't immesiately see a clearcut pattern for the numerator, but its possible there's cancellation happening between the numerator and denominator.

>>14714977
Its annoying the first few terms break the pattern. I didn't notice a pattern for the numerator, but its possible there's cancellation going on. I highly believe there's a pattern tho.

>>14713249
>1,3,7,89,809 are all prime numbers

Is this one of those rare threads where something actually happens here?
Like the Haruhi thing?
Posting for posterity just in case.

>>14715137
The question is which scientist isn't afraid of widely being known as a shitposter enough to publish?

bump

>>14714545
Can the attracting number 0.233852... be written
as some combination of familiar constants (e, pi, etc.)? Or is it a brand new number entirely of its own?

>>14718151
The number is actually [math]\,\dfrac{e^{\pi}}{99}\,[/math]. I have discovered a truly marvelous proof of this but I'm afraid the character limit on 4chan is too small to contain it.

>>14718151
The number is actually [math]\dfrac{e^{\pi}}{99}[/math]. I have discovered a marvelous proof of this but I'm afraid the character limit on 4chan is too small to contain it

>>14713249
we really need the next term in the expansion

>>14711048
The first terms of the sequence are:
1.58496250072116
0.324063081651386
0.358631620258398
0.324694803240390
0.294193751040354
0.273035374807878
0.259263565405933
0.250439104086099
0.244774786342303
0.241108249061561
0.238710187920748
0.237125803359093
0.236069607230701
0.235360251486783
0.234880984171790
0.234555659838973
0.234334041529749
0.234182662833368
0.234079053853288
0.234008034485765
0.233959300492526
where I used 100 precision digits.
It seems that L(2n-1,1) is decreasing (see >>14711825). It should be possible to prove this using induction, which would imply that the sequence is convergent.
Also, numerical experiments suggest that for k >=1 and l in {0, ..., 2^k-1}
L(n, 1+l*2^(n-k)) / ( log(L(n, 1+l*2^(n-k))) * L(n, 1+l*2^(n-k)) ) converges to 2l+1 when n, k -> infty,
which should say something intereseting about the limit function f(x) in >>14711825.

If no one steps in to solves it I will. Just too busy to spend time working this out right. Hopefully thread stays alive or the contributions get archived.

>>14719068
*right now

>>14718987
how did you get so many nestings?
my jank python script only gets me 9
at 10 I get a math domain error for some reason
>>14719068
xholic was so comfy

>>14719476
nvmd
chugging along now

>>14718151
I was also curious about that originally. That the convergence turned out to be a logarithmic attractor (not proven, just my experimental analysis) suggests it probably can't.

For example, a simpler such attractor is the solution to w = log(w). If you pick any number (that isn't zero or a nonnegative integer tetration of e) and keep logging it forever, the resulting sequence always converges to 0.3181315... + 1.3372357...i. This number is essentially Lambert's W(-1) but with the sign on the real component flipped from - to +.

I'm now more curious as to whether the attractor might be arithmetically expressible in terms of W() (or derivatives thereof).

>>14711048
This is the OP of this thread.
Here is the link to the previous thread where you
can source the work done thus far.
>>14691513

Highlights:
>some initial joke constants...maybe the converging number is expressible by known constants or is a new standalone number
>[math]f^{1}_k=k+1[/math], [math]f^{n+1}_k={{\log(f^{n}_{2k})}\over{\log(f^{n}_{2k-1})}}[/math] is the concise recursive setup of problem but [math]L(n,x)[/math] notation is favored for asymptotic analysis
>some Mathematica code, some Python code...
>complex numbers in the sequence?--don't worry, it's unlikely
>and much more...

pic unrelated

>>14719817
Here's the continuation of the data in >>14693064 to term 32.

1.58496250072115618145373894394782
0.32406308165138598033097465319887
0.35863162025839799602578777565449
0.32469480324038951431940962164433
0.29419375104035399647662845574938
0.27303537480787765688581680617525
0.25926356540593274173168470875008
0.25043910408609927865935129074951
0.24477478634230313500332513456255
0.24110824906156070776857083427248
0.23871018792074832206114543098676
0.23712580335909256059458670248218
0.23606960723070077440265280383401
0.23536025148678325240355301870987
0.23488098417179024078794173861603
0.23455565983897277554473464053969
0.23433404152974945114634426854941
0.23418266283336785796390976873517
0.23407905385328760279763197139090
0.23400803448576490451466745621378
0.23395930049252607655564862945559
0.23392583205553177602961378132442
0.23390283391766757463334520733887
0.23388702383064308808147551866682
0.23387615183327968056001026961937
0.23386867391312655786113575993118
0.23386352966670702402036271730424
0.23385999040541319660666064252962
0.23385755517930682655212369500577
0.23385587949742962162487440805138
0.23385472641025172724276465070717
0.23385393291243149954538999433500

>>14719794
>>14718151
If you can throw in a 'x' in the sequence
somewhere like the definition of the W function,
[math]W(xe^x)=x[/math], then you may have it.

>>14719533
>python 2
explain yourself

>>14719817
>>14711825
Here there is another recursion for the sequence.
Let Q(k, 0) = log(k+1) and
Q(k, n+1) = -log(Q(2k+1,n) - Q(2k,n)).
This has the advantage of having a difference and a single logarithm and not a quotient of logarithms as the one in >>14711825.

Note that, with the notation of >>14711825, we have that
L(k, n) = exp(Q(k, n)).
Hence, we are looking for
Lim exp(Q(1, n)).

>>14718201
Fuck off Fermat.

bump

>>14711381
How did you generate the coefficients of the series? When I tried to differentiate various forms of f(x/2) log f(x) = log f(x+1) I got cancellation on the 1/x coefficient. Expanding the LHS and RHS into infinite series also gave me cancellation of the 1/x coefficient.

bump b4 bed

>>14712061
4chan's number

>>14722587
>muh vanity thread

>>14721830
Don't differentiate.
Just expand the logs around 1 and expand the 1/(x+1)^k = (1/x^k)/(1+1/x)^k in powers of 1/x.
You'll need to use the coefficient of 1/x^(k+1) to compute the coefficient of 1/x^k.
(1 + 2A/x + 4B/x^2 ...) * (A/x + B/x^2 - (A^2)/(2x^2) ...) = (A/x + (B-A)/x^2 - (A^2)/(2x^2) ...)
Looking at [1/x^2] gives B - (A^2)/2 + 2A^2 = B - A - (A^2)/2
which gives 2A^2 = -A
which gives A = -1/2

>>14723558
Anon's Constant.

>>14712061
No one has proven convergence yet.

just move from the inside out you nitwits

>>14724062
Somewhat obnoxious: I replaced f(x) with g(x) = e^f(1/x). Then got a closed "recursive" formula for the coefficients of g using

e^g(2x) g(x) = g(x/(1+x)).

But the terms aren't cooperating. There is cancellation on one term, then no more cancellation on the other terms.

Expanding log f(1+x) in closed form doable, but less simple.

>>14724062
Do you have any idea WHY the cancellation on the terms would be so wonky like this?

>>14724062
Its really bizarre the coefficients are like that. Do you know any philosophical reason why it should be that way?

I got a recursive formula for the coefficients that is manageable. Gonna compute a lot of terms, look for a pattern, and try to induct on the pattern.

bump

I'm able to reduce this to one of two forms. Either

[math]\sum_{k=1}^\infty \left[f(x/2) [f(x) - 1]^k - [f(x+1) - 1]^k\right] = 0 [/math]

or

[math]\sum_{m=0}^\infty \left[f(x/2) [f(x)]^m - [f(x+1)]^m\right] = 0 [/math]

this should help writing the series. I'm too lazy.

>>14726560
Wow!

>>14726560
Both are wrong.
Expanding log(f(x)) should look like:
log(f(x)) = log(1 + f(x) - 1) = (f(x)-1) - (1/2)*(f(x)-1)^2 + (1/3)*(f(x)-1)^3 ...
Idk what you are doing.

>>14719847
my best rational approximation is 76/325, which seems too arbitrary

2/(e*pi) is close, but it seems to be converging down to something lower, so I doubt that's it

>>14727310
Guessing the constant is a long shot IMO. I think the best chance is finding a pattern in the coefficients which look very fishy. See >>14714977.

>>14724062
Found a neat formula for the coefficients of g(x) = log f(1/x) = B_0 x + B_1 x^2 + ...

[eqn] B_n P(0) + B_{n-1} P(1) + ... + B_0 P(n) = B_n {n \choose 0} - B_{n-1} {n \choose 1} + ... + (-1)^n B_0 {n \choose n}
[/eqn]

where P(k) are polynomials in B_0,...,B_{k-1}. But computing terms for x^6 and higher is being a chore.

>>14727831
Here's the first few terms incase anyone notices something. B0 = -1/2, B1 = 1/4, B2 = -7/48, B3 = 3/32, B4 = 2233/34560, B5= 649/13824

The numbers are fishy, but suspect there's cancellation going on between the numerator and denominator making it harder to find any pattern.

Probably won't be able to spend more time on this for a while. Already behind in my job.

First few polynomials are
P(0)= 1,
P(1) = 2B0
P(2) = 2 B0^2 +4 B1
P(3) = 4 B0^3/3 + 8 B0 B1 + 8 B2

P(4) and higher is longer, but they're precisely defined.

>>14727831
>>14711381
I don't think that f has enough information to determine the limit we are looking for.
Nevertheless, I programmed some mathematica code to compute the coefficients:

deg = 13;
f[x_] := 1 + Total[Table[c[k]/x^k, {k, 1, deg}]];
expr = Series[
fun[x/2]*Log[f[x]] - Log[f[1 + x]],
{x, Infinity, deg}];
eqs = Table[SeriesCoefficient[expr, i] == 0, {i, deg}];
coeffs = Solve[eqs, Table[c[i], {i, deg - 1}]][[1]]

It gives:
1 + 3/(8 x^2) - 7/(24 x^3) + 89/(384 x^4) - 809/( 4320 x^5) + 7073/(46080 x^6) - 3576131/(28062720 x^7) + 57672625/( 538804224 x^8) - 12196850723/(134219980800 x^9) + 260258179571/( 3340586188800 x^10) - 3766996323960469/( 55891347524812800 x^11) + 157728806913261211/(2682784681191014400 x^12)

Y'all a bunch of losers. I guess that's why you're posting here.

The series approaches zero.
>But muh heckin 10 billion terms
>Muh heckin 3rd order Shankerino
Yeah, and? These are nested logs. You can put 10^50 terms in and see only a 10% reduction in the sequence. Logs grow very, very, very slowly. Nested logs even more so.

>>14729224
>The series approaches zero.
An interesting conclusion with a reasonable argument, and yet your post contains no proof.

>>14729236
Log(2n)/log(n) -> 1

>>14729243
And what does log(log(2n+2)/log(n+1))/log(log(2n)/log(n)) approach?

>>14729224
brainlet redditor

>>14728966
Why would f not be sufficient? Don't we need f(1)?

>28062720 is randomly divisible by 29
OOF. That hurts.

Sorry OP, >>14727831 may be about the best you'll get. It "defines" the limit, but not much else.

P(m) are 2^m times the ordinary Bell Polynomals. There's a lot of identities for the Bell Polynomials, but I didn't see an applicable one.

Almost time to abandon the infinite series computation and hope there's some clever trick elsewhere.

>>14730062
Maybe the series of f at infinity does not converge at 1.
>Sorry OP, >>14727831 may be about the best you'll get. It "defines" the limit, but not much else.
Finding the limit is too ambitious. It's more reasonable to try to prove first that the limit exists, and then try to find oiut if it is positive or zero.

>>14730305
I'd be shocked if that wasn't doable.

>>14729737
Lhopital

>>14731832
...psychiatrique

>>14730305
>It's more reasonable to try to prove first that the limit exists
Proving the limit exists amounts to proving the coefficients alternate and are o(1).
It might be possible to use induction (assume coefficients are bounded by C/n^a for n<N) then use the recurrence to prove the assumption holds for N.

>>14732684
Nigga I told you already it goes to zero.

>>14732684
Its easy to prove they alternate using the recursion with Bell Polynomials. Decreasing not so sure

>>14732686
Bruh.

bump

>>14715137
It could very well be.

bump

>>14719533
go back the fuck to r*ddit and stay there.

>>14736153
>>>/r9k/

>>14736156
go back to r*ddit newfaggot.

>>14727831
I went one step further.
Starting with f(x/2)*log(f(x) = log(f(x+1))
first I defined g(x) satisfying f(x) = e^g(x)
resulting in g(x)*e^g(x/2) = g(x+1).
Logging gives g(x/2) = log(g(x+1)) - log(g(x)).
Then I defined h(x) satisfying g(x) = (-1/(2x))*e^h(x)
which gives (-1/x)*e^h(x/2) = h(x+1) - h(x) - log(1+1/x).
Multiplying by -x then differentiating gives
(1/2)*h'(x/2)*e^h(x/2) = (1/x)*e^h(x/2) -x*(h'(x+1) - h'(x)) - 1/(x+1).
This gives [1 - x*h'(x/2)/2]*[h(x+1) - h(x) - log(1+1/x)] = -x*[h'(x+1) - h'(x)] - (1/x)/(1+1/x)
You get a more manageable recurrence with this.
[math](1+\sum\limits_{k=1}^{\infty}{k h_k 2^k \over x^k})
(\sum\limits_{m=1}^{\infty}{1 \over x^m}({(-1)^m \over m}+\sum\limits_{k=1}^{m-1}{m-1 \choose k-1}(-1)^{m-k}h_k))= \\
(\sum\limits_{m=1}^{\infty}{1 \over x^m}((-1)^m +\sum\limits_{k=1}^{m-1}{m \choose k}(-1)^{m-k}kh_k))[/math]

>>14736207
desmos link?

>>14736207
Taking the coefficient of 1/x^(N+1) should give a recurrence for the Nth term.
[math]Let\ \tilde{h_k}=\ (-1)^{k}h_k.\\
{1 \over N+1}+\sum\limits_{k=1}^{N}{N \choose k-1}\tilde{h_k}\ +
\sum\limits_{k=1}^{N}k \tilde{h_k}2^k({1 \over N+1-k}+\sum\limits_{n=1}^{N-k}{N-k \choose n-1}\tilde{h_n})=\\
1+\sum\limits_{k=1}^{N}{N+1 \choose k}k \tilde{h_k}.[/math]

>>14736304
>>14736207
Did you check this?

Its easy to generate the first few coefficients of h with wolfram by plugging in the f,g coefficients we got.

You can further simplify using
[eqn]{N+1 \choose k+1} = \frac{N+1}{k+1} {N \choose k} [/eqn]
however summations of N choose k over k do not have a closed form unless its from 0 to N.(Summing over N does)

>>14736304
If there aren't mistakes it MAY be possible to massage this.

>>14711063
Y

>>14736361
I got ~h_1 = 1/2, ~h_2 = 1/6
these give the first three of >>14727846 correctly
It seems to be working.

>>14736391
~h_1 = 1/2, ~h_2 = 1/6, ~h_3 = 1/12, ~h_4 = 217/4320
These are agreeing with >>14727846
The series appears to be alternating and bounded above by 1/n^2.

>>14736487
I proved by induction |h_k| < 1/k^2. Should be enough to say h converges.

Its not difficult. Solve >>14736304 for ~h_N, apply this formula >>14736361 to bound

[eqn]
\sum {M \choose k-1} h_k \leq 2^M/M
[/eqn]
then take the absolute value of both sides and use triangle inequality.

>>14736487
There may be an annoying factor of 2, but I'm like 99% the induction argument is there with a little tweaking.

>>14736487
>>14736837
There may be an annoying factor of 2 or something, but I'm like 99% sure the induction argument is there with minor tweaking.

This should also be sufficient to show the nested logs converge to a positive number

>>14736837
>>14736854
>>14736875
Hopefully, an anon will write the full proof with all the missing steps described through the posts in this thread.

>>14736998
The most annoying part isn't writing those steps. Its that OP's question probably has significant generalizations which aren't far away.

>>14736998
OP's question seems to take an interval and assign it a ratio. Then the next ratio is given by finding what power you need to raise the previous intervals ratio to get the next intervals ratio.

As you can see such a concept is hardly limited by OP's singular example

Well the rest is up to /sci/. Too kuch for me

bump

>>14738032
I'm not OP but I like this thread, sorry I've been away. We can generalize it from a logarithmic attractor to an eigensequence. The only other generalized "convergence" I've found is an adherence by oscillation to Grandi's series. I'll post some stuff probably next fortnight when I have a break and some time to write it up. In any case, OP obviously converges (and not to 0) by some basic log identities. Think of it as a long tail off to 1 and the head gets up to 4.27620556195698... or so (which is 1/(what I wrote before). Identical output, just a clearer way to see it.

from math import log
nums = []
nums2 = []
r = 9

for i in range(2**r):
nums.append(log(i+2))

for i in range(r):
for i in range(int(len(nums)/2)):
i *= 2
nums2.append(
log(
nums[i+1] / nums[i]
)
)
nums = nums2[::]
nums2 = []

print(f"Answer = {nums[0]}")

>>14738296
4chan fucked up the indentation. But this python script calculates what you want to know.

With r = 9 the answer is -1.3795.....

Increasing r increases the size of ur pattern or whatever OP is talking about

bump

>>14738065
Brother coming in clutch.

bump

I don't understand the interest in this, but just represent it as a series and take the limit as the nested number inside log approaches infinity.
This isn't particularly interesting and is just basic freshman stuff.

do your own homework dude.

>>14740884
>I don't understand
nice abstract

>>14740884
The series is implicitly defined.

>>14711048

>>14741473
retard

I fucking hate women

>>14711048
try it for finite cyclic groups

>>14731832
hospital

>>14742288
you're gonna need one for your post-op

bump

>>14741552
no u
why do retards always say try X when they clearly haven't tried X themselves
also op what is the purpose of this jpg

bump

>>14743648
Someone wanted to know how the sequence of
logs continues if OP's pic was one of the terms
in the sequence, and what it converges to if at all.

bump

>>14732686
Hmm...

Looks like >>14738065 may not deliver.
Hopefully someone picks up the torch and writes up the problem. Its more or less solved in these posts.
>>14736207
>>14736304
>>14736837
>>14711825

The past 2 weeks have been fun. See you. Think its time to leave 4chan again.

>>14749038
Has it been a fortnight yet? In any case, the way I have it framed is much more generalized, doesn't involve that recursion, and doesn't involve extracting coefficients from any gf or egf. So if you claim to see a solution in those posts then why don't you simply write up the solution you claim to see instead of pointing at the posts over and over again and hoping someone else will connect whatever dots you think need to be connected?

bump

Man y'all be wasting time. I fuckin told you. It goes to zero. Once one of you finally figures it out formally, you're all gonna feel so stupid.

>>14750315
>It goes to zero
nope

>>14718201
if only that were true...

bump

bump

>>14749217
quick abstract until after next weekend hopefully
> [math]f_{0}[/math](x) = x+1
[math]f_{0}[/math] can be any function from N to C; this is OP [math]f_{0}[/math]
> [math]f_{n}[/math](x) = log [math]f_{n-1}[/math](2x) / log [math]f_{n-1}[/math](2x-1)
[math]f_{n}[/math] is the nth transformation of [math]f_{0}[/math]
> [math]\phi_{1}[/math](n) = [math]f_{n}[/math](1)
the nth element of [math]\phi_{1}[/math] is the first element of the nth transformation of [math]f_{0}[/math]; [math]\phi_{1}[/math] is OP sequence
> [math]\phi_{k}[/math](n) = [math]f_{n}[/math](k)
e.g. [math]\phi_{2}[/math] is the sequence of new numerators
> [math]\Phi[/math](x) = [math]\lim_{x\rightarrow\infty}[/math](x)
if [math]f_{0}[/math] converges for all [math]\phi_{k}[/math] (as it does for OP [math]f_{0}[/math]) then [math]\Phi[/math] is an eigensequence
>>14750315
OP can't go to 0 because
> [math]\phi_{1}[/math] converges to 0 iff [math]\phi_{2}[/math] converges to 1
> [math]\phi_{k}[/math] is majorized by (4k-3)/(4k-1)
also
> [math]\Phi[/math](2) < 1/e of course gives no real solution to [math]\Phi[/math](1)
the generalization has a bunch of interesting inflection points, jumps, and other discontinuities; it's all quite enjoyable to explore

>>14755394
wow that looks like shit

>>14755394
wow that looks like shit; see instead picrel

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>>14721830
imagine that happened? holy shit

this is already looking good

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>>14711048
0

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>>14711048
pretty

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>>14763934
It's not zero

>>14711048
I guess you could say it's a big log.

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so we don't actually have a full fleshed out proof it converges at the moment right?

>>14770190
We have an abstract from >>14755394 of what
to expect, the proof may still be forming.

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>>14768732
yeah

Answer : 2

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>>14774966

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>>14755394
Where are you!? Where's the result you're going to
give?

>>14711048
It doesn't converge to any number, I didn't read the previous thread now it is deleted so I assume that it is a nested tree based on powers of two to infinity

Always will there be an expression like that

[math]\cfrac{\log\left(\cfrac{\log\left(n+1\right)}{\log\left(n\right)}\right)}{\log\left(\cfrac{\log\left(k+1\right)}{\log\left(k\right)}\right)}\rightarrow k+2=n[/math]

Taking the limit:

[math]\lim_{k\rightarrow\infty}\left(\lim_{n\rightarrow\infty}\cfrac{\log\left(\cfrac{\log\left(n+1\right)}{\log\left(n\right)}\right)}{\log\left(\cfrac{\log\left(k+1\right)}{\log\left(k\right)}\right)}\right)[/math]

[math]\lim_{k\rightarrow\infty}\left(\lim_{n\rightarrow\infty}\cfrac{\log\left(\log\left(\cfrac{1}{n}+1\right)\right)}{\log\left(\log\left(\cfrac{1}{k}+1\right)\right)}\right)[/math]

[math]\cfrac{\log\left(\log\left(1\right)\right)}{\log\left(\log\left(1\right)\right)}=\cfrac{\log\left(0\right)}{\log\left(0\right)}=\text{Indeterminate}[/math]

So the entire three will collapse in an indetermination when higher values are reached

>>14778495
Some anons posted data about the number and according to the data the value always gets down with every iteration. That means that either it converges to zero or some number greater than zero.

>>14778543
>That means that either it converges to zero or some number greater than zero
yeah but is an infinitelly recusive function and you need to re evaluate the log whit that zero like values and you will get a indetermination

I realize that when numerically evaluating on 2**16 and the result was 'nan'

>>14778495
I think you can still have access to the deleted thread through the warosu site here:

>>/sci/thread/S14691513#p14691513

>anons computed the answer in 2 different ways and got the same result to 10 decimal places
>/sci/ keeps insisting it goes to 0 or is undefined

>>14778827
shitty python scripts aren't mathematical proofs

>>14778960
>he thinks people got the decimal place to 10 digits in 2 different ways by chance

>>14778967
Well you can compute zeros Riemann zeta function and get many of real values of 1/2 print it and claim 1M dollar prize

Its recursive function is undetermined when goes to infinity

>>14778827
anon, find the limit of 1/log^(10)(x) when x goes to infinity. where here i mean to take the log ten times, i.e. log(log(log(log...))))

use the same precision as in this thread, and higher. hell, go up to a googolplex, and tell me what the value is.

>>14778988
>>14779042
Argue whatever you want. Two different methods converged to 10 decimal places. >>14711064 >>14713249
You aren't squirming out of that. Besides its already been proved.

>>14749217 and I are just too lazy to write it up.

>>14779070
you don't understand my post. if you decided to actually investigate my new question 1/log^(10)(x) as x goes to infinity, you'll realize with your silly computer script you could put x as high as googolplex and still get the same 10-digit precision as 2^(32) or whatever other precision was used ITT. this is because of how slowly the logarithm function grows.

using two different approaches to achieve 10-digit precision isn't a proof that it tends to a non-zero value. not even close. it's meaningless, and all it shows is that the two approaches have similar degrees of precision. it says nothing about how good that precision actually is.

>>14779079
The series used to approximate the answer is alternating and decreasing.

>>14779092
not a proof that it converges to a nonzero value.

>>14779102
Yes, it does.

Besides even if it wasn't alternating, the first term is larger than the next infinity.

>>14779110
have fun wasting more time on a problem with either no actual answer or a trivial answer of 0, kek.

>>14711048
Previous thread:

>>/sci/thread/S14691513#p14691513

>>14779118
Brutal brah. Lay off the impenetrable burns

>>14778588
Well, just like you could evaluate [math]\dfrac{x}{2x}[/math] where [math]x[/math] approaches zero, you could also evaluate OP's number, even though you would get an indetermination when [math]x=0[/math]

>>14719847
I got similar results in the first terms, but it suddenly goes to complex, whats wrong? numeric precision?

[0, 2.0, 0.0],
[0, 2.0, 0.0],
[1, 1.5849625007211561815, 0.0],
[2, 0.32406308165138598046, 0.0],
[3, 0.35863162025839799622, -0.0],
[4, 0.32469480324038952303, -0.0],
[4, 0.32469480324038952303, -0.0],
[5, 0.29419375104035465232, -0.0],
[6, 0.27303537480777983842, -0.0],
[7, 0.25926356541716866796, -0.0],
[8, 0.2504391040059657708, -0.0],
[8, 0.2504391040059657708, -0.0],
[9, 0.24478050996779064533, -0.0],
[10, 0.23908398924107082895, -0.0],
[11, -1.4492189760715020163, 1.2126122545785537823],
[12, -0.09824800619768097299, -0.2309594769460118942],
[12, -0.09824800619768097299, -0.2309594769460118942],
[13, 0.14160099095382852404, -0.17354802823944018362],
[14, 0.2844927486499665074, 0.7074435925309152299],
[15, 0.09518837766519388301, 0.20162304660561336568],
[16, 0.07476756431804393172, -0.43125454039700780352]

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>>14779451
numerical pricision
see>>14719476
>>14719533

>>14779451
>numeric precision?
Yep, you can try multiplying numerator and denominator by powers of 10 to keep your calculations stable. It looks like yours already went off the rails at row 11, since most others have found that slow convergence to about .23X or .24ish

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*bumps intensity*

*bumps intensify*

>172 / 27 / 47 / 1
of 172 posts, over 100 from the same IP, in a month old thread.
desperate and crinj

>>14784548
It’s good content though

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>>14785006
I agree, nothing wrong with good threads with
plenty of chances at some activity or quality content.

It’s amazing how this sequence seems to be converging to [math]\frac{e^\pi}{10^2}[/math], combining four of the most fundamental constants in mathematics…

>>14787315
If that seems to be the case, that'll be quite
interesting. This was a question from the first
thread whether the converging constant can be
resolved as a combination of known constants
or is a brand new new one for this sequence.

>>14787315
I thought it only agreed on the first few digits.
e^pi/10^2 = 0.23140692632
>>14711064
gave 0.23385218

>>14787996
So close, but not there yet.

>>14732203
HAHAHA merci j'ai ri

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please do not let this thread die i need answers

>>14793686
There will be an answer, the effort needs to be
concentrated for just a bit. And if not this thread,
then in the next. The past thread has the info
you need here: >>14779123 .

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>>14711048
I feel like there's context missing. What are the conditions of the task at hand?

>>14797993
What OP was trying to do is to extrapolate the
sequence defined by the recursive form on
>>14719817 (#2). This form is rather tough but
others have settled for some asymptotic analysis.

The nesting of logs grows without bound as well
as the natural numbers from deep within. The
question is where does the sequence converge.
A very good advance was made here: >>14755394
and the previous thread about this is found here: >>14779123

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https://wiki.scholastia.xyz/stem/math/labs encrypted autism space if you want it

0

Bumping quality thread

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>>14711048
It's 42

Please don't scream. What are the implications for this type of math, just theory fun or can you do stuff with this logic?

We should name 0.23385218... the bump constant.

>>14711048
>(((log)))
FUCKING JEWS!!

Whatever you guys try to workout with this, it is undefined as x goes to infinity, other than that I don't know what else one could find with this useless looking expression.

If it was me I would call this constant "Thrembo"

hint to the post above me:

Limit arithmetic, I know we (I) already did that in calculus classes but anyone can do it if he tries

>>14805661
underrrated post

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bumping for a resolution to this sequence problem

The solution is e/pi

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morning bump squad checking in

>>14810354
No, it's not.

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To summarize:
It's easy to see the sequence is bounded and cauchy, so it converges and therefore from the results of the numerical simulations we can be confident in the first few digits, which are .23385218...

There may not be anything more to say about it than this.

>>14711048
This is the OP of this thread.
After much time, the main point of the question
has been reached. If the sequence is extended
in the way described in >>14719817, the answer
converges to .23385218... as mentioned by >>14815428
and others.

Although the answer is computed, a cogent
proof remains to be seen (>>14755394). Another interesting
issue would be if this number can be made with
existing constants (e, pi, etc.) or is a brand
new constant of its own (and perhaps be named by us).
Any input is welcomed up to another thread, if
needed. For now, the matter is somewhat settled.

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>>14802247
It's not zero

>>14822313
And you know this because...?

>>14822334
>>14822313

From the table at >>14718987, the sequence of
values is plateauing around 0.23 even with anon's
100 digit precision. The nature of logs is that they
approach values very slowly but the sequence
surprisingly converges to that number.

>>14822421
>plateauing
Define this in mathematical terms?

>>14822421
>>14823150
By plateauing, I mean that each subsequent term
of the sequence has numbers in the decimals that
stay the same; a sign of convergence.

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>>14816301
The constant already exists. It's zero.

>>14822421
Wrong. Logs converge slowly, and nested logs even more slowly. You can throw a billion terms at this, and only get a change fifty decimal points out. Your table doesn't prove shit except that it's a decreasing function.

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This shit is numerically so unstable, and it's impossible to prove