5
$\begingroup$

$x_0=2,\ x_n=x_{n-1}+\log \left(x_{n-1}\right)\quad$ has a series expansion about $1.$

Since $x_0=2,$ $(x_0-1)^k=1,$ so the recurrence can be written, up to the first $5$ terms as

\begin{align} 1+\frac{773788}{604800}\ 2^n-\frac{247200}{604800}\ 2^{2 n}+\frac{118825}{604800}\ 2^{3 n}-\frac{54300}{604800}\ 2^{4 n}+\frac{13687}{604800}\ 2^{5 n}-\dots \end{align}

found with

Evaluate[Plus @@ (((FindSequenceFunction[With[{nn = #}, (Rest@(List @@ 
Series[Nest[# + Log[#] &, x, #], {x, 1, 20}] & /@ Range[0, 20]))
[[All, 3, nn + 1]]]] & /@ Range[0, 5] /. # -> n)[[All, 1]]))]

and then combining like terms. The problem is finding the limits of the coefficients of $2^{k n}.$ Theoretically, the recurrence relation should be rewriteable as $$1+\sum _{k=1}^{\infty } (-1)^{k+1} c_k 2^{k n}$$

but I don't know how to go about finding a general solution for $c_k.$

Added

Building on @AntonioVargas' comment below, the recurrence relation seems very close numerically to $$\sum _{k=1}^{\infty } (-1)^{k+1} \left(n c_k \log (n)-\frac{1}{2} n \log \left(c_k\right)\right)\approx \pi n/2 +n \log n$$

(based on the first $9$ terms of the series expansion).

Numerically, an approximation of the coefficients can be found with something like

Abs[With[{aa = SplitBy[SortBy[Flatten@Join[{1, 2^n}, Drop[List @@@ 
Expand[Evaluate[(((FindSequenceFunction[With[{nn = #}, (Rest@(List @@ 
Series[Nest[# + Log[#] &, x, #], {x, 1, 20(*LARGER THAN number of terms*)}] 
& /@ Range[0,20(*LARGER THAN number of terms*)]))[[All, 3, nn + 1]]]] & /@ 
Range[0, 9(*number of terms*)] /. # -> n)[[All, 1]]))]], 2]],  
Coefficient[PowerExpand[Log2[#]], n] &], Coefficient[PowerExpand[Log2[#]], 
n] &]}, (a /. Solve[Plus @@ (aa[[#]]) == a 2^( (# - 1) n), a][[1]] // N) & 
/@ Range[2, Length@aa]]]
$\endgroup$
  • 3
    $\begingroup$ I doubt it. Any random recurrence you write is unlikely to have a closed-form solution. Only those with a very special form will have a simple formula. $\endgroup$ – Jair Taylor May 14 '15 at 17:33
  • 2
    $\begingroup$ @martin Hmm, don't know. I suppose it's possible there's some brilliant Ramanujan-esque series solution for the sequence :) $\endgroup$ – Jair Taylor May 14 '15 at 17:59
  • 3
    $\begingroup$ @JairTaylor any Ramanujans out there?!!! $\endgroup$ – martin May 14 '15 at 18:01
  • 1
    $\begingroup$ I'm actually experimenting with Faà di Bruno's formula because with finite n it can be viewed as a composition series of the series produced by $log(x)+x$ $\endgroup$ – Histograms May 15 '15 at 8:19
  • 1
    $\begingroup$ I understand that you are asking for coefficients $c^n_k$ such that, for every $n\geqslant0$, if $x_0=x$ is close enough to $1$ and if $x_i=x_{i-1}+\ln x_{î-1}$ for every $1\leqslant i\leqslant n$, then $x_n=\sum\limits_kc^n_k\cdot(x-1)^k$. Right? $\endgroup$ – Did May 17 '15 at 10:11

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