I was supposed to find the Inverse z transform of $$\dfrac{1}{(z-5)^3}$$.
Approach:
I first tried to find out the z transform of $\dfrac{z}{(z-5)^3}$.Partial fraction decomposition tells us:
$$\dfrac{z}{(z-5)^3}= \dfrac{-1}{50}\dfrac{5z}{(z-5)^2} +\dfrac{1}{50} \dfrac{5z(z+5)}{(z-5)^3} \equiv \dfrac{1}{50}Z(-5^nn + 5^nn^2) $$
Therefore, $$ Z^{-1}( \dfrac{z}{(z-5)^3})=\dfrac{5^n(n^2-n)}{50} \equiv Z^{1}(F(z))=f(n)$$
Now, we basically want $Z^{-1}(F(z)/z)$, which, we know from the shifting property is equal to $f(n-1)$, which should mean the answer is $$\dfrac{5^n(n^2-3n+2)}{250}$$
My answer key, and Wolfram alpha both reveal the answer is slightly off. The answer given in my sheet is $f(n-1)$ for $n>0$ and $0$ for $n=0$. (and In my course we always assume that everything is defined only for $n \geq 0$). Wolfram gave the answer as $f(n-1)u(n-1)$, which is basically the same answer.
What exactly is it in my procedure, that causes problems for $n=0$?
From what I did, the answer should be $f(n-1)$, for all $n \geq 0$....
I wasn't careful with the use of the shifting property. The correct formula is (for a one sided z transform, and assuming $f(n)=0$ for $n<0$) :
$$Z^{-1} (Z^{-k}F(z))= f(n-k)$$ for $$n \geq k$$ and $0$ for $$n<k$$
(basically, the role of $n$ is now played by $(n-k)$)
The correct answer now follows from this.