## 3 Neural Network-Based Q-Analogies of the Numerical Scheme
The authors declare no conflicts of interest. ## Appendix A

Proof of Theorem 1.: Let \(\zeta\) be a simple root of \(f\), and \(x^{[\sigma]}=\zeta+\vartheta^{[\sigma]}\) and \(z^{[\sigma]}=\zeta+\vartheta^{[\sigma+1]}\). By q-Taylor's series expansion of \(\left(\partial_{q}^{[1]}f\right)(x^{[\sigma]})\) around \(x^{[\sigma]}=\zeta\), and taking \(f(\zeta)=0\), we get

\[f(x^{[\sigma]}) = \left(\partial_{q}^{[1]}f\right)(\zeta)\left(\left[\vartheta^{[ \sigma]}\right]^{1};q\right)+\left(\partial_{q}^{[2]}f\right)(\zeta)\left( \left[\vartheta^{[\sigma]}\right]^{2};q\right)+\] (A1) \[\left(\partial_{q}^{[3]}f\right)(\zeta)\left(\left[\vartheta^{[ \sigma]}\right]^{3};q\right)+\ldots\]

\[f(x^{[\sigma]})=\left(\left[\vartheta^{[\sigma]}\right]^{1};q\right)+\frac{ \left(\partial_{q}^{[2]}f\right)(\zeta)}{\left(\partial_{q}^{[1]}f\right)( \zeta)}\left(\left[\vartheta^{[\sigma]}\right]^{2};q\right)+\frac{\left( \partial_{q}^{[3]}f\right)}{\left(\partial_{q}^{[1]}f\right)(\zeta)}(\zeta) \left(\left[\vartheta^{[\sigma]}\right]^{3};q\right)+\ldots\] (A2)

and

\[\left(\partial_{q}^{[1]}f\right)(x^{[\sigma]})=\left(\partial_{q}^{[1]}f\right) (\zeta)\left\{ \begin{array}{c}1+2\frac{\left(\partial_{q}^{[2]}f\right)( \zeta)}{[2!](\partial_{q}^{[1]}f)(\zeta)}\left(\left[\vartheta^{[\sigma]} \right];q\right)\\ +3\frac{\left(\partial_{q}^{[3]}f\right)(\zeta)}{[3!](\left\lfloor\vartheta_{ d}f\right\rangle)(\zeta)}\left(\left[\vartheta^{[\sigma]}\right]^{2};q\right)+ \ldots\end{array} \right\}.\] (A3)

When (A1) is divided by (A3), we have:

\[\frac{f(x^{[\sigma]})}{\left(\partial_{q}^{[1]}f\right)(x^{[ \sigma]})} = \left(\vartheta^{[\sigma]};q\right)-\left(\frac{\left(\partial_ {q}^{[2]}f\right)(\zeta)}{[2!](\partial_{q}f)(\zeta)}\right)\left(\left[ \vartheta^{[\sigma]}\right]^{2};q\right)-\] (A4) \[\left( \begin{array}{c}2\left(\frac{\left(\partial_{q}^{[3]}f \right)(\zeta)}{[3!](\left\left\lfloor\zeta_{d}f\right\rangle)(\zeta)} \right)-\\ 2\left(\frac{\left(\partial_{q}^{[2]}f\right)(\zeta)}{[2!](\left\lfloor\zeta_ {d}f\right\rangle)}\right)^{2}\end{array} \right)\left(\left[\vartheta^{[ \sigma]}\right]^{3};q\right)+\ldots\]\[1+\alpha_{2}f(x^{[\sigma]}) = 1+\alpha_{2}\Big{(}\theta^{[\sigma]};q\Big{)}+\alpha_{2}\frac{ \Big{(}\bar{\cup}_{q}^{[2]}f\Big{)}(\zeta)}{\Big{(}\bar{\cup}_{q}^{[1]}f\Big{)}( \zeta)}\Big{(}\Big{[}\theta^{[\sigma]}\Big{]}^{2};q\Big{)}\] (A5) \[+\alpha_{2}\frac{\Big{(}\bar{\cup}_{q}^{[3]}f\Big{)}(\zeta)}{\Big{(} \bar{\cup}_{q}^{[1]}f\Big{)}(\zeta)}\Big{(}\Big{[}\theta^{[\sigma]}\Big{]}^{3}; q\Big{)}+\ldots\]

\[\frac{\alpha_{1}f(x^{[\sigma]})}{1+\alpha_{2}f(x^{[\sigma]})}=\alpha_{1}\Big{(} \theta^{[\sigma]};q\Big{)}+\Bigg{(}\alpha\frac{\Big{(}\bar{\cup}_{q}^{[2]}f \Big{)}(\zeta)}{\Big{(}\bar{\cup}_{q}^{[1]}f\Big{)}(\zeta)}-\alpha_{1}\alpha_{ 2}\Bigg{)}\Big{(}\Big{[}\theta^{[\sigma]}\Big{]}^{2};q\Big{)}+O\bigg{(}\Big{[} \theta^{[\sigma]}\Big{]}^{3};q\Big{)}.\] (A6)

By using this expression in the first-step of (13), we have:

\[1-\frac{\alpha_{1}f(x^{[\sigma]})}{1+\alpha_{2}f(x^{[\sigma]})}=1-\alpha_{1} \Big{(}\theta^{[\sigma]};q\Big{)}-\Bigg{(} \begin{array}{c}\alpha\frac{ \Big{(}\bar{\cup}_{q}^{[3]}f\Big{)}(\zeta)}{\Big{(}\bar{\cup}_{q}^{[1]}f\Big{)} (\zeta)}-\\ \alpha_{1}\alpha_{2}\end{array} \Bigg{)}\Big{(}\Big{[}\theta^{[\sigma]}\Big{]} ^{2};q\Big{)}+O\Big{(}\Big{[}\theta^{[\sigma]}\Big{]}^{3};q\Big{)}.\] (A7)

We can take the inverse of (A7):

\[\Bigg{[}1-\frac{\alpha_{1}f(x^{[\sigma]})}{1+\alpha_{2}f(x^{[\sigma]})}\Bigg{]} ^{-1}=1+\alpha_{1}\Big{(}\theta^{[\sigma]};q\Big{)}+\Bigg{(} \begin{array}{c }\alpha_{1}\frac{\Big{(}\bar{\cup}_{q}^{[2]}f\Big{)}(\zeta)}{\Big{(}\bar{\cup }_{q}^{[1]}f\Big{)}(\zeta)}\\ -\alpha_{1}\alpha_{2}\end{array}