I'm reading some lecture notes my lecture posted online on Z Transforms and here's what one of the notes says:
We may differentiate a sequence $\space {x_n} = {a^n} \space$ with respect to $\space a \space$ to get $\space {{dx_n}\over da} = na^{n-1} \space$.
Here's where he lost me completely:
Then:
$$\space (na^{n-1}) = {d\over da} Z(a^n) = {d\over da} {z\over {z-a}} = {z\over {(z-a)^2}} \space$$
How is $ \space (na^{n-1}) = {d\over da} Z(a^n) \space$?
Isn't that assuming that:
$$a^{n} = Z(a^{n})$$