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Attached a snapshot of problem and solution at the bottom. Need to find the inverse of the given function.
I perfectly understand how they got the right-sided sequence by differentiating below :

$$\dfrac{1}{1-z^{-1}} = \sum\limits_{n=0}^{\infty}z^{-n}$$

I understand why the right-sided sequence starts at $n=2$. No issues with this.

However I'm terribly confused about the left sided sequence. I know it should be a ramp. But why is it starting at $n=2$ ? I'm not able to see this. Also why the terms are positive ? I'm asking this because in all previous example problems I noticed that the left sided sequence always has opposite signed terms : For example, if $a^nu[n]$ is the right-sided sequence, then the left-sided sequence would be $\color{red}{-}a^nu[-n-1]$. But in the present problem, both left-sided and right-sided sequences have positive terms. I'm trying to understand why this is the case, but it seems too much for me. Help appreciated...

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AgentS
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1 Answers1

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Looks I've figured this out! Starting with $$\dfrac{1}{1-z} = \sum\limits_{n=-\infty}^{\infty} u[n]z^n$$

Since we want a left-sided sequence, plugin $n=-m$ above and get $$\dfrac{1}{1-z} = \sum\limits_{m=-\infty}^{\infty} u[-m]z^{-m}$$

which is same as $$\dfrac{1}{1-z} = \sum\limits_{m=-\infty}^{0} z^{-m}$$

Differentiating both sides $$\dfrac{1}{(1-z)^2} = \sum\limits_{m=-\infty}^{0} (-m)z^{-m-1} $$

$$\dfrac{1}{z(1-z)^2} = \sum\limits_{m=-\infty}^{0} (-m)z^{-m-2} $$

Clearly the right hand side expression is a shifted version of the ramp and its terms are positive because $-m$ is positive when $m$ is negative. Wow! A bit lengthy I guess... I'd still love to see if there are other more smart ways to work this. Thanks :)

AgentS
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