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.

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.

1

Hans Dolhaine

Hans Dolhaine, retired

Posted 8 years ago

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]