# The Euler-Type Universal Numerical Integrator (E-TUNI) with Backward Integration
## 5 The Euler-Type Universal Numerical Integrator (E-TUNI)
### General E-TUNI Algorithm with Forward and Backward Integrations  
This section describes first-order Euler integrators with forward and backward integration. They are divided into two classes: (i) the conventional Euler integrator, which works with instantaneous derivative functions (low numerical precision and variable step), and (ii) the E-TUNI type integrator, which works with mean derivative functions (high numerical precision and fixed step). Therefore, it is important to mathematically relate the mean derivatives functions with the instantaneous derivatives functions (see again Figure 2). Thus, it is easily seen that if \(\dot{y}=f(y)\) then,  
\[{}^{k}\!\!y^{i}_{i}=f({}^{k}\!\!y^{i})=\underset{\Delta t\to 0}{lim} tan_{\Delta t}{}^{k}\!\!y^{i}_{j}=\underset{\Delta t\to 0}{lim}\frac{{}^{k+1}\!\!y^{i}_{j}-{}^{k}\! \!y^{i}_{j}}{\Delta t} \tag{8}\] \[for\quad j=1,2,\ldots,n\]  
or  
\[\underset{\Delta t\to 0}{lim} tan_{\Delta t}{}^{k}\!\!y^{i}_{j}=tan^{k}\!\!\Theta^{i}_{j} \tag{9}\] \[for\quad j=1,2,\ldots,n\]  
Figure 2 graphically illustrates the convergence of mean derivatives functions to instantaneous derivatives functions when \(\Delta t\) tends to zero. Based on this, it is possible to enunciate two basic types of first-order numerical integrators or Euler integrators. An Euler integrator that uses instantaneous derivatives functions, given by \(tan^{k}\Theta^{i}_{j}=f_{j}({}^{k}\!\!y^{i})\) for \(j=1,2,\ldots,n\) and an Euler integrator that uses mean derivatives functions, given by \(tan^{k}_{\Delta t}y^{i}_{j}=\frac{{}^{k+1}\!\!y^{i}_{j}-{}^{k}\!\!y^{i}_{j}}{ \Delta t}=\frac{{}^{k+1}\!\!y^{i}_{j}-{}^{k}\!\!y^{i}_{j}}{\Delta t}=g_{j}({}^{k }\!\!y^{i}_{j}+{}^{k}\!\!y^{i}_{j},\Delta t)\). Additionally, like \(f_{j}^{+}(k^{\prime}y^{\prime})=f_{j}^{-}(k^{\prime}y^{i})=f_{j}(k^{\prime}y^{i})\), the Euler integrator with instantaneous derivatives functions has a symmetric solution between intervals \([t_{k-1},t_{k}]\) and \([t_{k},t_{k+1}]\), thus implying low numerical precision, for the outputs \(k^{-1}y^{i}_{j}\) and \(k^{+1}y^{i}_{j}\) for \(j=1,2,\ldots,n\). To say that a solution is symmetric at the point \(k\) means to say that \(|k^{\prime}y^{i}_{j}-k^{-1}y^{i}_{j}|=|k^{+1}y^{i}_{j}-k^{\prime}y^{i}_{j}|\) to \(j=1,2,\ldots,n\). This Euler integrator symmetry property only occurs when instantaneous derivative functions are used. However, in [PERSON]'s integrator, designed with the mean derivatives functions, this symmetry of the solution around the point \(t_{k}\) is not verified (see Figure 8 again). This is important because in [PERSON]'s integrator, designed with mean derivatives functions, the backward solution does not necessarily decrease when the forward solution grows. In [32], several difference operators can be applied to a function \(k^{\prime}y^{l}_{j}\) for \(j=1,2,\ldots,n\). However, the two operators that interest us are the forward difference and the backward difference, given below, respectively:  
\[\Delta^{k}y^{i}_{j}=^{k+1}y^{i}_{j}-^{k}y^{i}_{j}\quad for\quad j=1,2,\ldots,n \tag{14}\]  
\[\
abla^{k}y^{i}_{j}=^{k}y^{i}_{j}-^{k-1}y^{i}_{j}\quad for\quad j=1,2,\ldots,n \tag{15}\]  
Thus, using the finite difference operators, there are two different ways to represent the secant at the point \(k^{\prime}y^{i}_{j}\) and with integration step \(\Delta t\):  
\[tan^{+}_{\Delta t}k^{i}_{j} =\frac{\Delta^{k}y^{i}_{j}}{\Delta t}=\frac{k^{+1}y^{i}_{j}-^{k} y^{i}_{j}}{\Delta t} \tag{16}\] \[for\quad j=1,2,\ldots,n\]  
\[tan^{-}_{\Delta t}k^{i}_{j} =\frac{\
abla^{k}y^{i}_{j}}{\Delta t}=\frac{k^{y}_{j}-^{k-1}y^{i} _{j}}{\Delta t} \tag{17}\] \[for\quad j=1,2,\ldots,n\]  
Comparing Equation (16) with the result of Theorem 4, one can obtain the following recursive and exact equation to represent the solution of autonomous non-linear ordinary differential equations system for a forward integration process (i.e., with high numerical accuracy):  
\[ \begin{gathered}{}^{k+1}y^{i}_{j}=tan^{+}_{\Delta t}k^{i}_{j} \cdot\Delta t+^{k}y^{i}_{j}\\ for\quad j=1,2,\ldots,n\end{gathered}  \tag{18}\]  
or  
\[