# Approximation by Bicomplex Favard-Szasz-Mirakjan Operators
## 3 Construction of Bicomplex Favard-Szasz-Mirakjan Operators
Therefore, we have  
\[+\bigg{(}e^{-n(t_{1}+i_{1}t_{2})}\sum_{j=0}^{\infty}\frac{(n(t_{1} +i_{1}t_{2}))^{j}}{j!}\bigg{(}\frac{j}{n}\bigg{)}^{k}\bigg{)}e_{2}.\]4 \[+\bigg{(}e^{-n(t_{1}+i_{1}t_{2})}\sum_{j=0}^{\infty}\frac{(n(t_{1} +i_{1}t_{2}))^{j}}{j!}\bigg{(}\frac{j}{n}\bigg{)}^{k}\bigg{)}e_{2}.\]5  
and\[+\bigg{(}e^{-n(t_{1}+i_{1}t_{2})}\sum_{j=0}^{\infty}\frac{(n(t_{1} +i_{1}t_{2}))^{j}}{j!}\bigg{(}\frac{j}{n}\bigg{)}^{k}\bigg{)}e_{2}.\]6 \[+\bigg{(}e^{-n(t_{1}+i_{1}t_{2})}\sum_{j=0}^{\infty}\frac{(n(t_{1} +i_{1}t_{2}))^{j}}{j!}\bigg{(}\frac{j}{n}\bigg{)}^{k}\bigg{)}e_{2}.\]7  
Now, let \(n\geq 1\) be an arbitrary constant. Employing the mathematical induction on \(k\), we prove that this recurrence leads to the following inequalities:  
\[+\bigg{(}e^{-n(t_{1}+i_{1}t_{2})}\sum_{j=0}^{\infty}\frac{(n(t_{1} +i_{1}t_{2}))^{j}}{j!}\bigg{(}\frac{j}{n}\bigg{)}^{k}\bigg{)}e_{2}.\]8  
and  
\[+\bigg{(}e^{-n(t_{1}+i_{1}t_{2})}\sum_{j=0}^{\infty}\frac{(n(t_{1} +i_{1}t_{2}))^{j}}{j!}\bigg{(}\frac{j}{n}\bigg{)}^{k}\bigg{)}e_{2}.\]9  
for all \(k\geq 2\), \(n\geq 1\). Indeed, given the first inequality above for \(k=2\) and \(n\geq 1\), the left-hand sides emerge as \(\frac{r_{1}}{n}\) and \(\frac{r_{2}}{n}\), while the right-hand sides are \(\frac{3r_{1}}{n}\) and \(\frac{3r_{2}}{n}\). Assuming that the first inequality holds for \(k\), we proceed to establish the result for \(k+1\). Thus, we can express  
\[D_{t_{1}+i_{2}t_{2}}(T_{n,k})(t_{1}+i_{2}t_{2}) =D_{t_{1}-i_{1}t_{2}}\bigg{[}\bigg{(}e^{-n(t_{1}-i_{1}t_{2})}\sum _{j=0}^{\infty}\frac{(n(t_{1}-i_{1}t_{2}))^{j}}{j!}\bigg{(}\frac{j}{n}\bigg{)} ^{k}\bigg{)}e_{1}\bigg{]}\]0  
It remains to prove that  
\[D_{t_{1}+i_{2}t_{2}}(T_{n,k})(t_{1}+i_{2}t_{2}) =D_{t_{1}-i_{1}t_{2}}\bigg{[}\bigg{(}e^{-n(t_{1}-i_{1}t_{2})}\sum _{j=0}^{\infty}\frac{(n(t_{1}-i_{1}t_{2}))^{j}}{j!}\bigg{(}\frac{j}{n}\bigg{)} ^{k}\bigg{)}e_{1}\bigg{]}\]1  
or we can write  
\[D_{t_{1}+i_{2}t_{2}}(T_{n,k})(t_{1}+i_{2}t_{2}) =D_{t_{1}-i_{1}t_{2}}\bigg{[}\bigg{(}e^{-n(t_{1}-i_{1}t_{2})}\sum _{j=0}^{\infty}\frac{(n(t_{1}-i_{1}t_{2}))^{j}}{j!}\bigg{(}\frac{j}{n}\bigg{)} ^{k}\bigg{)}e_{1}\bigg{]}\]2  
It is easy to see that this last inequality holds true for all \(k\geq 2\), \(n\geq 1\). We can use a similar method for \(\left\|T_{n,k}-e_{k}\right\|_{r_{2}}\). This immediately implies the estimates for both \(\left\|T_{n,k}-e_{k}\right\|_{r_{1}}\) and \(\left\|T_{n,k}-e_{k}\right\|_{r_{2}}\) for all \(k=0,1,\ldots\) and \(n\in\mathbb{N}\). Note that from this hypothesis, \(g(t_{1}+i_{2}t_{2})=\sum_{k=0}^{\infty}(c_{k}+i_{2}d_{k})(t_{1}+i_{2}t_{2})^{k}\) is absolutely and uniformly convergent in \(|t_{1}-i_{1}t_{2}|<r_{1}\) and \(|t_{1}+i_{1}t_{2}|<r_{2}\), respectively. Then, for all \(1\leq r_{1}<\frac{1}{S_{1}}\), \(1\leq r_{2}<\frac{1}{S_{2}}\) and \(S_{1}\in(\frac{1}{R_{1}^{2}},1)\), \(S_{2}\in(\frac{1}{R_{2}^{2}},1)\), the power series on the right hand side can be differentiated twice, i.e.,  
\[D_{t_{1}+i_{2}t_{2}}(T_{n,k})(t_{1}+i_{2}t_{2}) =D_{t_{1}-i_{1}t_{2}}\bigg{[}\bigg{(}e^{-n(t_{1}-i_{1}t_{2})}\sum _{j=0}^{\infty}\frac{(n(t_{1}-i_{1}t_{2}))^{j}}{j!}\bigg{(}\frac{j}{n}\bigg{)} ^{k}\bigg{)}e_{1}\bigg{]}\]3 \[D_{t_{1}+i_{2}t_{2}}(T_{n,k})(t_{1}+i_{2}t_{2}) =D_{t_{1}-i_{1}t_{2}}\bigg{[}\bigg{(}e^{-n(t_{1}-i_{1}t_{2})}\sum _{j=0}^{\infty}\frac{(n(t_{1}-i_{1}t_{2}))^{j}}{j!}\bigg{(}\frac{j}{n}\bigg{)} ^{k}\bigg{)}e_{1}\bigg{]}\]4  
where \(|c_{k}-i_{1}d_{k}|\leq M\frac{S_{1}^{k}}{kr}\), \(|c_{k}+i_{1}d_{k}|\leq M\frac{S_{1}^{k}}{kr}\), and  
\[D_{t_{1}+i_{2}t_{2}}(T_{n,k})(t_{1}+i_{2}t_{2}) =D_{t_{1}-i_{1}t_{2}}\bigg{[}\bigg{(}e^{-n(t_{1}-i_{1}t_{2})}\sum _{j=0}^{\infty}\frac{(n(t_{1}-i_{1}t_{2}))^{j}}{j!}\bigg{(}\frac{j}{n}\bigg{)} ^{k}\bigg{)}e_{1}\bigg{]}\]5  
by taking into account the series \(\sum\limits_{k=2}^{\infty}(r_{1}S_{1})^{k+1}\), \(\sum\limits_{k=2}^{\infty}(r_{2}S_{2})^{k+1}\), meaning their derivatives \(\sum\limits_{k=2}^{\infty}(k+1)(r_{1}S_{1})^{k}\), \(\sum\limits_{k=2}^{\infty}(k+1)(r_{2}S_{2})^{k}\) are uniformly and absolutely convergent.