## 3 Reconstruction Algorithms
### Algorithm
Then_

\[q_{-}^{(n)}=\min_{1\leq i\leq I^{(n)}}\{q_{i}^{(n)}\}\,,\quad\text{A}^{(n)}= \sum_{i=1}^{I^{(n)}-1}a_{i}^{(n)}\,, \tag{6}\]5

_as \(p\to\infty\), and \(\text{A}^{(n)}\to 0\) as \(n\to 0\)._

Proof. : Let \(k\geq 1\) and \(n=\varphi(p)\in\left[m_{k},n_{k}\right]\), where \(p\in\left[p_{k}+1,p_{k+1}\right]\). Let \(\text{A}^{(n)}\) be given by Equation (6), \(a_{+}^{(n)}\) be given y Equation (7), and \(i_{+}^{(n)}\) be given by Equation (8). At iteration \(n+1\) one has:

\[x_{i}^{(n+1)}= \begin{cases}x_{i}^{(n)}&\text{for }1\leq i\leq i_{+}^{(n)},\\ \frac{1}{2}\!\left(x_{i_{+}^{(n)}}^{(n)}+x_{i_{+}^{(n)}+1}^{(n)}\right)&\text{ for }i=i_{+}^{(n)}+1,\\ x_{i-1}^{(n)}&\text{for }i_{+}^{(n)}+2\leq i\leq I^{(n+1)}.\end{cases} \]

Also \(y_{i}^{(n+1)}\leq y_{i+1}^{(n+1)}\) for \(1\leq i\leq I^{(n+1)}-1\). One may check that \(a_{+}^{(n)}=2a_{i_{+}^{(n)}}^{(n+1)}+2a_{i_{+}^{(n)}+1}^{(n+1)}\) (see Figure 2) and therefore:

\[\text{A}^{(n+1)}=\text{A}^{(n)}-a_{+}^{(n)}+a_{i_{+}^{(n)}}^{(n+1)}+a_{i_{+}^ {(n)}+1}^{(n+1)}=\text{A}^{(n)}-\frac{1}{2}a_{+}^{(n)}\,. \tag{13}\]

Besides \(\text{A}^{(n)}\leq(I^{(n)}-1)a_{+}^{(n)}\) so that one has:

\[\text{A}^{(n+1)} \leq\text{A}^{(n)}-\frac{\text{A}^{(n)}}{2(I^{(n)}-1)}\] \[\leq\text{A}^{(n)}\!\left(\frac{2(I^{(n)}-1)-1}{2(I^{(n)}-1)} \right),\]\[\mathrm{A}^{(\varphi(p)+1)}\leq\mathrm{A}^{(\varphi(p))}\left(\frac{2(I^{(\varphi(p ))}-1)-1}{2(I^{(\varphi(p))}-1)}\right). \tag{14}\]

At this stage two situations arise:

* either \(p\in[p_{k}+1,p_{k+1}-1]\), in which case \(\varphi(p)+1=\varphi(p+1)\);
* or \(p=p_{k+1}\), in which case by our algorithm \(\mathrm{A}^{(n)}\) is kept constant from \(n=n_{k}+1\) to \(n=m_{k+1}\); that is \(\mathrm{A}^{(n_{k}+1)}=\mathrm{A}^{(m_{k+1})}\), or: \[\mathrm{A}^{(\varphi(p_{k+1})+1)}=\mathrm{A}^{(\varphi(p_{k+1}+1))}\,.\] The choice of \(k\) being arbitrary, one concludes that Equation (14) also reads \(\forall p\in\mathbb{N}\): \[\mathrm{A}^{(\varphi(p+1))} \leq\mathrm{A}^{(\varphi(p))}\left(\frac{2(I^{(\varphi(p))}-1)- 1}{2(I^{(\varphi(p))}-1)}\right)\] \[\leq\mathrm{A}^{(\varphi(p))}\left(\frac{2(I^{(\varphi(0))}+p-1) -1}{2(I^{(\varphi(0))}+p-1)}\right).\] Thus: \[\mathrm{A}^{(\varphi(p))} \leq\mathrm{A}^{(\varphi(1))}\prod_{i=1}^{p-1}\left(\frac{2(I^{( \varphi(0))}+i-1)-1}{2(I^{(\varphi(0))}+i-1)}\right)\] \[\leq\mathrm{A}^{(\varphi(1))}\prod_{i=1}^{p-1}\left(\frac{1+ \frac{\alpha}{i}}{1+\frac{\beta}{i}}\right),\] letting \(\alpha=I^{(\varphi(0))}-\frac{3}{2}\) and \(\beta=I^{(\varphi(0))}-1\). However,

\[\sum_{i=1}^{p}\log\Bigl{(}1+\frac{\alpha}{i}\Bigr{)}=\alpha\sum_{i=1}^{p} \frac{1}{i}+C_{p}^{\prime\prime}\]

where \(\lim_{p\to\infty}C_{p}^{\prime\prime}=C^{\prime\prime}\), and

\[\text{A}^{(n+1)} \leq\text{A}^{(n)}-\frac{\text{A}^{(n)}}{2(I^{(n)}-1)}\]0

where \(\gamma\) is the Euler constant and \(\lim_{p\to\infty}e_{p}^{\prime}=0\). Consequently:

\[\text{A}^{(n+1)} \leq\text{A}^{(n)}-\frac{\text{A}^{(n)}}{2(I^{(n)}-1)}\]1 \[\text{A}^{(n+1)} \leq\text{A}^{(n)}-\frac{\text{A}^{(n)}}{2(I^{(n)}-1)}\]2 \[\text{A}^{(n+1)} \leq\text{A}^{(n)}-\frac{\text{A}^{(n)}}{2(I^{(n)}-1)}\]3

since \(\alpha-\beta=-\frac{1}{2}\); again \(C_{p}\) and \(C_{p}^{\prime}\) are sequences with constant limits \(\lim_{p\to\infty}C_{p}=C\) and \(\lim_{p\to\infty}C_{p}^{\prime}=C^{\prime}\). Therefore,

\[\text{A}^{(n+1)} \leq\text{A}^{(n)}-\frac{\text{A}^{(n)}}{2(I^{(n)}-1)}\]4where \(\mathcal{C}\) is a constant, and \(\lim_{p\to\infty}\epsilon_{p}=0\). One also concludes that \(\mathrm{A}^{(n)}\), which is either kept constant or equal to \(\mathrm{A}^{(\varphi(p))}\), converges to 0 as \(n\to\infty\). Hence the claimed results hold. Proposition 3: \(E^{-}\) _is infinite._

Proof: Let us assume that \(E^{-}\) is finite: \(\exists N\) such that \(\forall n\geq N\), \(n\in E^{+}\). Therefore we are in the situation \(\mathrm{WA}^{(n)}\geq\mathcal{E}>0\), and \(\varphi(n)\) has the form \(\varphi(n)=n-n_{0}\), \(n\geq N\) for some \(n_{0}\in\mathbb{N}\).