## 3 Problem Formulation and a First Solution
From this equation, we get  
\[D^{\alpha}[x(t)-x_{0}\epsilon(t)]+x(t)=\frac{1}{q}x\bigg{(}\frac{t}{q}\bigg{)} \quad t\geq 0,\ q\geq\]3  
or  
\[D^{\alpha}[x(t)-x_{0}\epsilon(t)]+x(t)=\frac{1}{q}x\bigg{(}\frac{t}{q}\bigg{)} \quad t\geq 0,\ q\geq\]4  
With \(n=1\), we obtain  
\[D^{\alpha}[x(t)-x_{0}\epsilon(t)]+x(t)=\frac{1}{q}x\bigg{(}\frac{t}{q}\bigg{)} \quad t\geq 0,\ q\geq\]5  
and, using (21), we are led to  
\[D^{\alpha}[x(t)-x_{0}\epsilon(t)]+x(t)=\frac{1}{q}x\bigg{(}\frac{t}{q}\bigg{)} \quad t\geq 0,\ q\geq\]6 \[D^{\alpha}[x(t)-x_{0}\epsilon(t)]+x(t)=\frac{1}{q}x\bigg{(}\frac{t}{q}\bigg{)} \quad t\geq 0,\ q\geq\]7  
However,  
\[D^{\alpha}[x(t)-x_{0}\epsilon(t)]+x(t)=\frac{1}{q}x\bigg{(}\frac{t}{q}\bigg{)} \quad t\geq 0,\ q\geq\]8  
which leads to  
\[D^{\alpha}[x(t)-x_{0}\epsilon(t)]+x(t)=\frac{1}{q}x\bigg{(}\frac{t}{q}\bigg{)} \quad t\geq 0,\ q\geq\]9 \[x(t)=\sum_{n=0}^{\infty}C_{n}\frac{t^{n\alpha}}{\Gamma(n\alpha+1)},\quad t>0, \tag{13}\]0  
Continuing from (23),  
\[x(t)=\sum_{n=0}^{\infty}C_{n}\frac{t^{n\alpha}}{\Gamma(n\alpha+1)},\quad t>0, \tag{13}\]1  
which gives  
\[x(t)=\sum_{n=0}^{\infty}C_{n}\frac{t^{n\alpha}}{\Gamma(n\alpha+1)},\quad t>0, \tag{13}\]2 \[x(t)=\sum_{n=0}^{\infty}C_{n}\frac{t^{n\alpha}}{\Gamma(n\alpha+1)},\quad t>0, \tag{13}\]3  
where \(B_{n}\) are real parameters to be computed. Proof. : Begin by inserting (24) into (21) to get  
\[x(t)=\sum_{n=0}^{\infty}C_{n}\frac{t^{n\alpha}}{\Gamma(n\alpha+1)},\quad t>0, \tag{13}\]4  
and  
\[x(t)=\sum_{n=0}^{\infty}C_{n}\frac{t^{n\alpha}}{\Gamma(n\alpha+1)},\quad t>0, \tag{13}\]5  
On the other hand,  
\[x(t)=\sum_{n=0}^{\infty}C_{n}\frac{t^{n\alpha}}{\Gamma(n\alpha+1)},\quad t>0, \tag{13}\]6  
that leads to  
\[x(t)=\sum_{n=0}^{\infty}C_{n}\frac{t^{n\alpha}}{\Gamma(n\alpha+1)},\quad t>0, \tag{13}\]7  
from where we obtain  
\[x(t)=\sum_{n=0}^{\infty}C_{n}\frac{t^{n\alpha}}{\Gamma(n\alpha+1)},\quad t>0, \tag{13}\]8  
that leads to (19). The relation (20) is obtained from the initial value theorem [48]. To finish, we must study the convergence of (18). First, we note that each function  
\[x(t)=\sum_{n=0}^{\infty}C_{n}\frac{t^{n\alpha}}{\Gamma(n\alpha+1)},\quad t>0, \tag{13}\]9  
has a pseudo-pole [49], since the zero of the denominator is in the second Riemann surface in the complex plane, due to the fact that \(q\) is a positive parameter. Therefore, it does not have a pole. Consequently, the inverse Laplace transform, \(E_{\alpha}(-q^{-n\alpha}t^{\alpha})\), is bounded [46; 49]. In fact, it is a continuous function for \(t>0\)[46] with initial and final values \(1\) and \(0\), respectively. Therefore, it has a maximum and minimum. This implies absolute and uniform convergence of (18), since the \(|B_{n}|\) sequence decreases exponentially to \(0\). **Remark 2**. : _The recursion (25) leaves undetermined the value of \(B_{0}\). Therefore, we start with a \"seed\" \(\hat{B}_{0}=1\) to obtain a sequence \(\hat{B}_{n},\ n=1,2,\cdots\) that has to be normalised so that the corresponding sum equals \(x_{0}\). Let \(S=\sum_{n=0}^{\infty}\hat{B}_{n}\). Define a normalising factor \(F=\frac{x_{0}}{S}\). The normalised coefficients are given by \(B_{n}=\hat{B}_{n}\cdot F\)._  
The exponential decreasing character of the series coefficients (19) explains why the series in (18) has a fast convergence, as we illustrate in the following. In Figure 2 we depict some plots illustrating the variation of the \(B_{n}\) sequence for several values of \(\alpha\) and \(q\). As seen, above a given \(n_{th}\), the sequence becomes negligible, implying that the series solutions (18), or (24), have a finite number of meaningful terms, so that they can be truncated, disregarding the terms for \(n>n_{th}\). This means that the solution of (3) can be approximated by a finite linear combination of Mittag-Leffler functions. These considerations may lead us to conclude that the solutions stated in (13) and in (18) are different and could lead to a loss of the unicity of the solution of (3), which is impossible, since it is a linear equation.