Z transformation with $k$ from non-zero

Question

we known that $Z$ transformation of $f_k$ is defined as $$F(z)=\sum_{k=0}^{\infty}f_k z^k$$

My problem is if $k$ starts from $m$, where $m >0$, then $\sum_{k=m}^{\infty}f_{k+m} z^{k+m}$ is still equals $F(z)$, Is it correct?

Answer 1

Is $\sum_{k=m}^{\infty}f_{k+m} z^{k+m}$ still equal to $F(z)=\sum_{k=0}^{\infty}f_k z^k$ ?

In general, no.

You have $$ \begin{align} \sum_{k=m}^{\infty}f_{k+m} z^{k+m}&=\sum_{p=2m}^{\infty}f_{p} z^{p} \quad (p:=k+m)\\\\ &=\sum_{p=0}^{\infty}f_{p} z^{p}-\sum_{p=0}^{2m-1}f_{p} z^{p}\\\\ &=F(z)-\sum_{p=0}^{2m-1}f_{p} z^{p}. \end{align} $$

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But you have

$$ \begin{align} \sum_{k=m}^{\infty}f_{k-m} z^{k-m}&=\sum_{p=0}^{\infty}f_{p} z^{p} \quad (p:=k-m)\\\\ &=F(z). \end{align} $$