What Property Connects the Steps in a Geometric Series Calculation?

[nacho-man]

Hi, just wanted to know what property/?? was used to get from the first red-box to the second one.

It looks like it has to do with the geometric sum, but my series/sequence is verrrrrry rusty.

any help appreciated, I'd like to find the general property so I can apply this rule to different examples. I can see that the bounds of the sum has altered the outcome also.

IF anyone could explain this to me, or link me to some relevant theory I would be ecstatic.many thanks in advance.

(see attached image)edit: my brain is fried, there are tables which give these direct results! awesome.
just wondering though, is there a way to solve for z-transforms without having to consult a table?
Although this last question is probably reserved for a different sub-board.

thanks anyway.!

 

[Dethrone]

That is the sum of a geometric series, consider:

We'll represent a geometric series by $S_n=\sum ar^{n-1}$

$$S_n=a+ar+ar^2+...+ar^{n-1}$$

Multiply both sides by $r$:

$$rS_n=ar+ar^2+...+ar^{n-1}+ar^n$$

Subtract from $S_n$ so that most of the terms cancel:

$$S_n - rS_n=a-ar^n$$
$$(1-r)S_n=a(1-r^n)$$
$$S_n = \frac{a(1-r^n)}{1-r}$$

$a$ is the first term, which in your case, is $1$.
$r$ is the common ratio which is what we're multiplying "$n$" times: $az^{-1}$

Applying $S_n$, we get from the first red box to the second :D I don't think I'll be able to answer your other questions though, I've only just started learning series.

 

[I like Serena]

edit: my brain is fried, there are tables which give these direct results! awesome.
just wondering though, is there a way to solve for z-transforms without having to consult a table?

Hi nacho,

Evaluating a z-transform without tables is exactly what is done here.
It depends on evaluating a series.

Tables or math software make your life much easier for complicated z-transforms though.

 

[nacho-man]

wonderful, thanks heaps guys.
A revision session is long overdue haha!