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[✓] Find relationships between coefficients - Equality?

GROUPS:

Hello everyone. Is it possible to find the relationships between the coefficients Subscript[b,m] Subscript[c,m] Subscript[a,n] Subscript[d,n] so that the following expression:

Sum[Subscript[b, m] z^(-m), {m, 0, M}]/
 Sum[Subscript[a, n] z^(-n), {n, 0, N}]

is equal to:

(Subscript[b, 0]/
   Subscript[a, 0]) Product[(1 - Subscript[c, m] z^(-1)), {m, 0, M}]/
  Product[(1 - Subscript[d, n] z^(-1)), {n, 0, N}]

?

Thank you for your time.

POSTED BY: Gennaro Arguzzi
Answer
3 months ago

I consider only one Sum / Product..

Check out the following for differnt M. note the value of sum1 - sum2

M = 5;
sum1 = Collect[Expand[(\!\(
\*UnderoverscriptBox[\(\[Product]\), \(m = 0\), \(M\)]\((1 - 
\*FractionBox[
SubscriptBox[\(c\), \(m\)], \(z\)])\)\)) 1], z]
tt = Table[Subscript[c, j], {j, 0, M}]
ss = Subsets[tt, {2, M + 1}]
sum2 = 1 - (\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(j = 0\), \(M\)]
\*SubscriptBox[\(c\), \(j\)]\))/z + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(j = 2\), \(M + 1\)]\(
FractionBox[\(Plus @@ \((Apply[Times, 
        Select[ss, Length[#] == j &], {1}])\)\), 
SuperscriptBox[\(z\), \(j\)]] 
\*SuperscriptBox[\((\(-1\))\), \(j\)]\)\)

Simplify[sum1 - sum2]
POSTED BY: Hans Dolhaine
Answer
3 months ago

Having gotten the idea about the subsets (which is of course well known in the Expansion of a polynomial given in factorized form) there is a second method to do it. See Notebook for the 1st and 2nd method.

Attachments:
POSTED BY: Hans Dolhaine
Answer
3 months ago

Hi @Hans Dolhaine . Why there aren't equalities in your solution? I expect for example c1c2=c6c4.

POSTED BY: Gennaro Arguzzi
Answer
3 months ago

Why there aren't equalities in your solution? I expect for example c1c2=c6c4.

Hm, I find your manner of addressing me somewhat inpolite. A tiny "thank you" perhaps? I confess, that makes me reluctant to do any further work.

Why aren't there equalities? Because you should do a bit of thinking yourself.

You asked for a relationship between the coefficients of the two representations of a function. And this is implicitly given in my answer.

If you compare in the notebook (for M = mt = 4 for example ) the expression for sum3 ( Second Method )

with that given by you for the numerator

Sum[Subscript[b, m] z^(-m), {m, 0, M}]

you will find

b1 ==  -  b0  ( c0 +c1+c2 +c3 +c4 )
b2 == b0 ( c0 c1 +  c0 c2 + ..... + c3 c4 )

and so on.

Now, given the cj's you can calcultate the b's. But i think it is very hard to calculate the c's if the b's are known.

POSTED BY: Hans Dolhaine
Answer
3 months ago

@Hans Dolhaine . I apologize if that's how it sounded. I am glad for the help you gave me. At the beginning I did not understand your answer and for this I asked you another question. My intention was to say thanks with an up vote after I understand your answer. I apologize again, I am mortified.

POSTED BY: Gennaro Arguzzi
Answer
3 months ago

Group Abstract Group Abstract