Link between Z-transform and Taylor series expansion

fatpotato
Homework Statement
Inverting a Z-transform using Taylor series expansion
Relevant Equations
See embedded image in my post
Hello,

I am reading a course on signal processing involving the Z-transform, and I just read something that leaves me confused.

Let ##F(z)## be the given Z-transform of a numerical function ##f[n]## (discrete amplitudes, discrete variable), which has a positive semi-finite support and finite ROC. My textbook says that we should be able to compute the inverse Z-transform of ##F(z)##, so the ##f[n]## values, using the fact that "##f[n]## values are given by the Taylor series expansion of ##z^{-n_0} F(z^{-1})## around ##z=0##", using the following equation:

1654787498356.png

This sounds fascinating, but I don't understand! How are Z-transform and Taylor series expansion linked, and why? I only ever learned about inverting the Z-transform using tables, long division or using the inverse transform definition involving contour integral in the complex plane (which I never had the chance to use).

Any information is welcome!
 
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Look at the definition: <br /> F(z) = \sum_{n=0}^\infty f[n]z^{-n} Setting t = z^{-1} this is <br /> F(t^{-1}) = \sum_{n=0}^\infty f[n]t^{n}. This is a power series in t, so if G(t) = F(t^{-1}) has a convergent Taylor series at 0 this is <br /> \sum_{n=0}^\infty f[n]t^{n} = \sum_{n=0}^\infty \frac{1}{n!} G^{(n)}(0)t^n and hence f[n] = \frac{1}{n!} G^{(n)}(0) = \frac{1}{n!}\left.\frac{d^n}{dt^n}F(t^{-1})\right|_{t=0}.
 
Beautiful, thank you for the derivation!

In a classical engineering program setting, I tend to believe that applying Taylor series would be limited to Calculus and I would have never thought of using it here.

Would you happen to have a suggestion on where I can expand my horizons to come up with such reasonings myself?
 
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