2
$\begingroup$

$x_{1}[k]$ is a function with an argument $k$

$$f(z)=\sum_{k=-\infty }^{+\infty} \left( x_{1}[k] \left( \sum_{n=-\infty }^{+\infty} \frac{x_{2}[n-k]}{z ^{n} } \right) \right)$$

$$p:=n-k$$

$$n:-\infty\rightarrow +\infty$$

$$ \left(\therefore ~~p:-\infty\rightarrow +\infty\right)$$

$$f(z)=\sum_{k=-\infty }^{+\infty} \left( x_{1}[k] \left( \sum_{p=-\infty }^{+\infty} \frac{x_{2}[p]}{z ^{p+k} } \right) \right)$$

What I can't get currently is the range of $p~$ stated above.

As endpoints of the range of $k$ are not infinite things ,I can get $\left(p:-\infty\rightarrow +\infty\right) ~\text{as for instance }~\left(k:-10^{9}\rightarrow +10^{9}\right)$

However the actual endpoints of range of $k$ are infinite values so as $k=\infty~$ ,

$$p=n-(\infty)$$

is held and as $n=-\infty$, I can get $p=-\infty-\infty=-\infty$ however as $n=\infty$,

$p=\infty-\infty=0?$

So how do I interpret like as above formula arose?

I thought below.

As $k=\infty$ and $n=\infty$ , $~~~n-k$ seems takes 0 however $n$ has the priority(which means that the value of n was determined later the determination of value of k) so $\infty_{\text{k}}<\infty_{\text{n}}$ holds so actually the below equation can be held.

$$n-k=\infty_{n}-\infty_{k}=\infty$$

Is this thought correct?

New contributor
mechatronics enthusiast is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$

Your Answer

mechatronics enthusiast is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.