# Identify $\sum\limits_{n=0}^\infty \left(x_n-L\right)$ where $x_{n+1}=\sqrt[3]{1+x_n}$ and $L^3=L+1$

Let $x>2$ and let $f(x) = \sqrt[3] {1 + x}$.

Let $f^n(x)$ be the $n$ th iterate of $f(x)$.

Let $L$ be the positive fixpoint of $\sqrt[3] { 1 + x}$: the plastic constant. See http://mathworld.wolfram.com/PlasticConstant.html

Consider for $x>2$ :

$$Q(x) = x - L + f^1(x) - L + f^2(x) - L + ...$$

Is there a closed form for $Q(x)$ ?

Special values Maybe ? ( like $f(e) = \pi$ or something ).

This is similar to Telescoping exercise with iterations?

I tried this

$$Q(x) - Q(f(x)) = x - L$$

So

$$f(x) = \sqrt[3] { 1 + x }= Q^{-1} ( Q(x) - x + L )$$

But then i got stuck.

I assume $Q(x)$ is analytic.

• For any convergent iteration you can define $$Q(x)=\sum_{n=0}^\infty \left(f^n(x)-f^{\infty}(x)\right)$$ what's so special about this one? – polfosol Jan 10 '17 at 6:32
• I mean, why should it have a closed form? – polfosol Jan 10 '17 at 6:34
• See the links for similar Sums with closed form solutions. – mick Jan 10 '17 at 12:28
• mick Speaking of which... when are you going to clean the mess on this other page and, possibly, post a solution there, so that the thing does not stay hanging? – Did Jan 23 '17 at 12:50

I would write things in the following way. Starting with your definitions $f(x)= \sqrt[3]{1+x}$ and for the fixpoint $L = \lim_{h \to \infty } x_h$ .

Let's now introduce the shorthand notations $x' = x+ L$ and $x^" = x-L$ .

Also I would explicitely introduce the function $$g(x) = f(x+L)-L$$ as a much common usage in the discussion of functional iteration.

A notationally short notation is now $$g(x) = f(x')^" \qquad \qquad \text{and } \qquad f(x) = g(x^")' \qquad \text{ .}$$ With this let us furtherly denote the iterates of the $g()$ function with the common index-notation
$$x_h:=g^{°h}(x) \qquad .$$ Note btw., that the power series for $g(x)$ has no constant term and we could generate power series for fractional iterates and so on.

Iterates of $f(x)$ are now expressible by iterates of $g(x)$: $$f^{°h} (x) = g^{°h} (x-L)+L = g^{°h} (x^")' = (x^"_h)' \qquad .$$

Since $L$ is an attracting fixpoint for $f(x)$ the iteration of $g(x)$ (at least for a certain interval for $x$) converges to zero and a iteration-series based on $g(x)$ might also be convergent - however at this point this is not yet sure!

Then I'd make the iteration-series depending on $g(x)$ such that $$Q_g(x) = \sum_{h=0}^\infty x_h$$ and your definition for $Q(x)$ would be $$Q(x) = Q_g(x^")$$

With that bit longish preliminaries it is easy to create a formal power series for $Q_g(x)$ ; I like especially the simple way using Carlemanmatrices and Neumann-series with that Carlemanmatrices.

Let $G$ denote the Carlemanmatrix for function $g(x)$, let $V(x)$ denote a row-vector of consecutive powers of $x$ such that $V(x)=[1,x,x^2,x^3,...]$ Then with the little matrix-algebra $$\begin{array} {rll} V(x_0) \cdot G &= [ 1, x_1, (x_1)^2, (x_1)^3 ,...] &= V(x_1)) \\ V(x_1) \cdot G &= [ 1, x_2, (x_2)^2 , (x_2)^3,...] &= V(x_2) \\ ... \end{array}$$ and $$V(x) \cdot ( I + G + G^2 + ... +G^h) = \sum_{k=0}^h V(x_k)$$ The idea is now, that the above partial geometric series of $G$ above should possibly be expressed by the difference of two full geometric series. However, one exemplar of the full geometric series provides already the full iteration series of your question; and the shortcut for geometric series (here with a matrix argument, this is called "Neumann-series") gets then involved: $$V(x) \cdot ( I + G + G^2 + ...) = V(x) \cdot (I-G)^{-1} = \sum_{k=0}^\infty x_k$$

For finite truncations of the (infinite) Carlemanmatrix that reciprocal term $(I-G)^{-1}$ cannot be done because the top-left entry and thus the whole first column would become zero. But in the case of this function $g(x)$ we can have a workaround, in that we insert a value $1$ in the top-left of $I-G$, (but which I don't want to discuss deeper here).

This suggests to give that reciprocal the matrix-name $Q_g$. (Note, that by this operation $Q_g$ is not of the Carleman type!) We can evaluate $$V(x) \cdot Q_g = [1, \sum_{h=0}^\infty x_h , \sum_{h=0}^\infty x_h^2, \sum_{h=0}^\infty x_h^3, ... ]$$ Of course we are only interested in the second result, so we can write - using the second column of $Q_g$ only: $$V(x) \cdot Q_g [,1] = \sum_{h=0}^\infty x_h = Q_g(x)$$ and your searched result is then given by the power series evaluation $$Q(x) = V(x^") \cdot Q_g [,1]$$ The formal power series which occurs by the rhs dot-product begins as $$Q_g(x) = \small{ 1.2344868 x - 0.034880742 x^2 + 0.0084538027 x^3 - 0.0024444375 x^4 \\ + 0.00077551090 x^5 - 0.00026057755 x^6 + 0.000091048921 x^7 \\ - 0.000032729496 x^8 + 0.000012021537 x^9 - 0.0000044908285 x^{10} \\ + 0.0000017006482 x^{11} - 0.00000065129523 x^{12} + 0.00000025178258 x^{13} \\ - 0.000000098117179 x^{14} + 0.000000038499197 x^{15} + O(x^{16}) }$$

Conclusion: Other than in your earlier question, where in the $Q$-matrix of that function we found polynomial expressions leading to the exact result $Q(x) = 1/8 - x^2$ (see there) this $Q$-matrix here does not show such polynomial expression, so I rather think there are no such (simple) closed forms. But - the columns in $Q_g$ and $Q$ provide coefficients for power series so any characteristic in the result might be possible; I could not yet give a hint for the question of closed forms by this analysis only. (Perhaps soemone else can - at least we have now a power series for the function of $Q(x)$)

for reference: here the top-left of matrix $Q_g$

     1               0               0               0               0               0
0       1.2344868               0               0               0               0
0    -0.034880742       1.0374302               0               0               0
0    0.0084538027    -0.010808060       1.0069005               0               0
0   -0.0024444375    0.0033709884   -0.0029721978       1.0013034               0
0   0.00077551090   -0.0011184147    0.0011374829  -0.00074777195       1.0002473
0  -0.00026057755   0.00038742656  -0.00042842516   0.00033959599  -0.00017732328
0  0.000091048921  -0.00013845479   0.00016195039  -0.00014353963  0.000093231469