Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Is this number known in other contexts?
#1
Computing the alternating infinite series



for x=1 I came to a certain number for b, where a turning point occurs.
For b>0.15001 I have positive values for , with decreasing b possibly increasing to e, for 0.145 < b <=0.150001... negative values, possibly diverging, and for b<0.145 again positive values. The occuring terms of the series are diverging with alternate signs in all cases, so my computations may not be well approximated. Anyway - is a number b in the near of 0.15 known as significant in any other context in tetration?
Gottfried Helms, Kassel
Reply
#2
I'm not sure I follow, because for bases in that range, the series should be converging (e^-e < b < 1). Maybe I'm just misunderstanding what you're attempting to do?

For bases less than e^-e, the series are divergent, but the terms in the series converge exponentially on a simple alternating series, such that a summation for the divergent series should be possible with advanced methods.
~ Jay Daniel Fox
Reply
#3
jaydfox Wrote:I'm not sure I follow, because for bases in that range, the series should be converging (e^-e < b < 1). Maybe I'm just misunderstanding what you're attempting to do?

For bases less than e^-e, the series are divergent, but the terms in the series converge exponentially on a simple alternating series, such that a summation for the divergent series should be possible with advanced methods.

Hmm, that's true... May be it is an artifact of the numerical computation of the truncated matrix. What I'm doing is, according to the rules of a geometric series for the scalar base x s(x) = 1/(1-x) and for the alternating geometric series, which occurs if x=-y this is then 1/(1+y).
To compute the above alternating sum I build

Bb = tetration-matrix/Bell-matrix to base b
Mb = (matid(n) + Bb)^-1

and then use the second column of Mb as terms for a powerseries in x

s(b,x) = sum r=0..inf mb[r,1]*x^r

This gives perfect results for the usual convergent cases (can be verified by Euler-summation of the series s(b,x) = x - b^x + b^b^x - ... + ... )

Moreover, this approach has the nice property, that using x=0 instead of x=1, I find the negative of the result already in the very first term:
This is justified by the observation that

s(b,1) = 1 - b^1 + b^b^1 - b^b^b^1 + .... = 1 - b + b^b - b^b^b + ....
s(b,0) = 0 - b^0 + b^b^0 - b^b^b^0 + .... = 0 - 1 + b - b^b - ....

and s(b,1) = - s(b,0)

But it is also
s(b,0) = sum r=0..inf mb[r,1]*0^r
s(b,0) = mb[0,1]*0^0 = mb[0,1]

and then simply

s(b,1) = - mb[0,1]

which avoids any divergent summation at all.


For some b decreasing from 0.155 to 0.15001 I get increasing values for s(b,1), say 2.0,2.1,2.2,... 2.6... , (by inspecting -mb[0,1] only) then if I approach 0.15 suddenly negative values occur, and if I decrease b to, say 1.4, I have positive values for s(b,1) again (by the according negative value in mb[0,1]). So I suspect, there is some singularity in that region, may be something like b contains reciprocal powers of e or e +- 1 .

If it is an artifact of the finite truncation this effect should change by change of dimension. I'll check & report that later...

Gottfried
Gottfried Helms, Kassel
Reply
#4
Aha, the singularity seems to occur .
[update] this hope didn't improve to knowledge... ;-( [/update]

By Eigenanalysis the eigenvalues of Bb are diag(1,u,u^2,...),
where u = log(t) and t is the fixpoint of b.

Now the h(b)-function gives a fixpoint for b=0.15.It gives [update]
t0=H(0.150,0) \\ = 0.436708722499
u0=log(t0) \\ = -0.828488845027
b = t0^(1/t0) =0.15
[/update]

Using Branch 1:
t1=H(0.150,1) \\ = -0.467424941871 - 2.38182350872*I
u1=log(t1) + 2*Pi*I \\ = 0.886761198657 + 4.51860497886*I
exp(u1/t1) \\ = 0.150


The eigenvalues of Mb should be diag(1/(1+1),1/(1+u),1/(1+u^2),... )
and I don't see a singularity occuring here. This, however does not use t but its logarithm u - but I'm possibly nearer to the core of the problem here.

Gottfried
Gottfried Helms, Kassel
Reply
#5
I don't know for sure, but one thing I can say, is that 0.14 is close to and it might just be that it is so close that it starts to "appear to diverge" even though it converges mathematically.

Andrew Robbins
Reply


Possibly Related Threads...
Thread Author Replies Views Last Post
  computing teh last digits without computing the number deepinlife 3 6,026 02/24/2009, 09:09 AM
Last Post: deepinlife



Users browsing this thread: 1 Guest(s)