3.7. An eigenfunction of the summation operation

Because the exponential function is an eigenfunction of the finite differences derivative and according to Eq. (46), we can say that

$$\begin{aligned} \sum\_{j=n}^{m} 2\chi(\boldsymbol{\upsilon}, \boldsymbol{\Delta}) \boldsymbol{\upsilon} \, \boldsymbol{\varepsilon}^{\eta \eta\_{j}} &= \sum\_{j=n}^{m} 2\chi(\boldsymbol{\upsilon}, \boldsymbol{\Delta}) (\mathbf{D} \boldsymbol{\varepsilon}^{\eta \eta})\_{j} = \sum\_{j=n}^{m} (\boldsymbol{\varepsilon}^{\eta \eta\_{j+1}} - \boldsymbol{\varepsilon}^{\eta \eta\_{j-1}}) \\ &= \boldsymbol{\varepsilon}^{\eta \eta\_{m+1}} + \boldsymbol{\varepsilon}^{\eta \eta\_{m}} - \boldsymbol{\varepsilon}^{\eta \eta\_{n}} - \boldsymbol{\varepsilon}^{\eta \eta\_{n-1}}, \end{aligned} \tag{47}$$

in agreement with the corresponding continuous variable equality Ð <sup>x</sup> <sup>a</sup> dx v evx <sup>¼</sup> <sup>e</sup>vx � <sup>e</sup>va. However, here, we have to deal with two values at each boundary.
