The below image is from my book. I am confused how the following is true: $$\mathcal{F}(y(x), \Delta_\delta y(x), \Delta^2_\delta y(x), \cdots; x) = \mathcal{F}(y(x), y(x+\delta), y(x + 2\delta), \cdots; x)$$ This is stated in line $(1.14)$ (bottom right).
So, how is $\Delta_\delta y(x)$ equivalent to $y(x + \delta)$?
By definition of the difference operator, $\Delta_\delta y(x) = \frac{y(x+\delta) - y(x)}{\delta}$ so...what is going on?
No! Note the bar on $\bar{\mathcal{F}}$. You rewrite every appearance of $\Delta_\delta y(x)=\frac{y(x+\delta)-y(x)}{\delta}$, $\Delta_\delta^2 y(x)=\frac{y(x+2\delta)-2y(x+\delta)+y(x)}{\delta^2}$, etc., so an equation of the form $$ \mathcal{F}(y(x),\Delta_\delta y(x), \Delta_\delta^2 y(x),\dots;x)=0 $$ becomes an equation involving only $x, y(x), y(x+\delta), y(x+2\delta),\dots$ which you denote by $$ \bar{\mathcal{F}}(y(x), y(x+\delta), y(x+2\delta),\dots;x)=0. $$