How Can I Derive Sumations with Multiple Indices and Powers?

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The discussion focuses on deriving a complex summation involving multiple indices and powers of z. The initial summations for P(z) and Q(z) are provided, along with a more intricate double summation that needs simplification. The key to solving the problem lies in switching the order of summation and applying a change of variables. Although one participant suggests that the solution is trivial, others express confusion about the process and the reasoning behind it. The conversation highlights the challenges in manipulating indices and understanding the implications of changing the order of summation.
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I have a hard time trying to figure it out how to derive the following sumations:

We know that:
P(z) = Sumation from j=0 to Infinity of [(Pj * (z^j)]
and
Q(z) = Sumation from j=0 to Infinity of [(Kj * ((z^j)]

Where j is a subindex for P and K, and a power for z.


Sumation from j=0 to Infinity of [ Sumation from i=0 to j+1 of [(z^j) * Pi * Kj-i+1] ]

Where i and j are a subindices for P and K, and a power for z.



I know that I have to play with the indices, but I have no clue. Any advice??
 
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I may be missing something, but I can't see what your question is?
 
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Yes, you are right.
From the information above, it has to be simplified to:
[ (1/z) * P(z) * Q(z) ] - [ (1/z) * Po * Q(z) ]

The author of the book does not give much more information about how he derived this equation, but he says is trivial.
 
The basic trick is to switch the order of summation. Then you will have two parts. The main part has i=1,inf with j=i-1,inf. The other part has i=0 with j=0,inf. For the main part let n=j-i+1 (replacing j), then n=0,inf. Put it all together and you should get the required answer.
 
Sorry

I am sorry, but I still do not get it.
 
Do you understand switching the order of summation?
Do you see the result of switching?
Do you understand the change of variables (n=)?
 

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