# [✓] 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.
11 months ago
5 Replies
 Hans Dolhaine 1 Vote 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] 
11 months ago
 Hans Dolhaine 1 Vote 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:
 Hans Dolhaine 2 Votes 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.