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I'm reading a book (Digital Signal Processing by Jonathan Steyn) which defines the infinite sum:

$$r\equiv\sum^\infty_{m=0}z^{-m}x_n$$

This is just the sum of $x_1 + x_2 + x_3 +\dots +x_n$ to infinty. Then it defines the finite difference as $(z^{-1}$ being the previous value):

$$\Delta\equiv(1-z^{-1})$$

Then it claims that they are related such that the one applied to the other equals the identity operator. By referring to the telescoping series $x_1 - x_0 + x_2 - x_1\dots x_n - x_{n-1} = x_n - x_0$, they say you should be able to prove this.

However, this has me stumped. If you take the series $xk = 1,2,3,\dots$ Then you get $x_n - x_0 = x_n - 1$. How can this be the identity operator?

moinudin
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  • Can you please provide the name of the book and author that you're reading for clarification (you can edit this into your original post, to avoid adding a comment below). – Jessie Mar 05 '21 at 18:51
  • Surely you mean $x_1z^{-1}+ x_2z^{-2}+ x_3z^{-3}+ \cdot\cdot\cdot+ x_nz^{-n}+ \cdot\cdot\cdot$. – user247327 Mar 05 '21 at 20:20
  • @user No, because $z^{-m}$ shifts the $n$ in $x_n$ back by $m$, making it $x_{n-m} $. At least in my understanding. Please correct me if you still think I'm wrong. – moinudin Mar 05 '21 at 20:36

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