# Derivation of nabla exponential T=hZ

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Here we are working in the time scale $\mathbb{T}=h\mathbb{Z}$. Rearrange the equation $y^{\nabla}(t)=p(t)y(t);y(s)=1$ to get the equation $y(t)=\dfrac{y(t-h)}{1-hp(t-h)}$. From this it is clear that $$y(s+h)=\dfrac{y(s)}{1-hp(s+h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+h}{h}} \dfrac{1}{1-hp(hk)},$$ $$y(s+2h)=\dfrac{y(s+h)}{1-hp(s+2h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+2h}{h}} \dfrac{1}{1-hp(hk)},$$ $$\vdots$$ $$y(s+nh)=\dfrac{y(s+(n-1)h)}{1-hp(s+nh)} = \displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+nh}{h}} \dfrac{1}{1-hp(hk)},$$ so that if $t=s+nh$ then $y(t)=\displaystyle\prod_{k=\frac{s}{h}+1}^{\frac{t}{h}} \dfrac{1}{1-hp(hk)}.$ Now rearrange the IVP to get $y(t-h)=(1-hp(t))y(t),$ from which we see $$y(s-h)=(1-hp(s))y(s)=1-hp(s)=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s}{h}} 1-hp(hk),$$ $$y(s-2h)=(1-hp(s-h))y(s-h)=(1-hp(s-h))(1-hp(s))=\displaystyle\prod_{k=\frac{s-2h}{h}+1}^{\frac{s}{h}} 1-hp(hk),$$ $$\vdots$$ $$y(s-nh)=(1-hp(s-nh))y(s-(n-1)h)=\displaystyle\prod_{k=\frac{s-nh}{h}+1}^{\frac{s}{h}} 1-hp(hk),$$ so that if $t=s-nh$ then $y(t)=\displaystyle\prod_{k=\frac{t}{h}+1}^{\frac{s}{h}} 1-hp(hk).$ Hence we have derived $$\hat{e}_p(t,s)=\left\{ \begin{array}{ll} \displaystyle\prod_{k=\frac{t}{h}+1}^{\frac{s}{h}} (1-hp(hk)) &; t \lt s \\ 1 &; t=s \\ \displaystyle\prod_{k=\frac{s}{h}+1}^{\frac{t}{h}} \dfrac{1}{1-hp(hk)} &; t \gt s. \end{array} \right.$$