Formula for product of two polynomials












1












$begingroup$


I have found such a formula
$$
left(sum_{k=0}^n a_k z^k right)cdot left(sum_{j=0}^m b_j z^j right)=sum_{k=0}^{n+m} left( sum_{j=k-min(n,m)}^{min(k,max(n,m))}a_j b_{k-j} right)z^k,
$$
but seems it's too complicated. Is there any other formulas?






combinatorics polynomials







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$endgroup$








  • 1




    $begingroup$
    The inner sum corresponds to a convolution of the coefficients. You may simplify it by going in the frequency domain (Fourier) for example (or $z$ transform), but it is unlikely it is what you are looking for.
    $endgroup$
    – Damien
    Dec 20 '18 at 19:18










  • $begingroup$
    @Damien Can you explane it as an asnwer?
    $endgroup$
    – Leox
    Dec 20 '18 at 19:28






  • 1




    $begingroup$
    I have no time now, sorry. You can find information on convolution here for example: en.wikipedia.org/wiki/Convolution. It is a well known operation, widely studied. Used for example to represent filtering
    $endgroup$
    – Damien
    Dec 20 '18 at 19:41










  • $begingroup$
    Not sure why the answer in there is not correct but may help: math.stackexchange.com/questions/1937543/…
    $endgroup$
    – NoChance
    Dec 20 '18 at 20:22


















1












$begingroup$


I have found such a formula
$$
left(sum_{k=0}^n a_k z^k right)cdot left(sum_{j=0}^m b_j z^j right)=sum_{k=0}^{n+m} left( sum_{j=k-min(n,m)}^{min(k,max(n,m))}a_j b_{k-j} right)z^k,
$$
but seems it's too complicated. Is there any other formulas?






combinatorics polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The inner sum corresponds to a convolution of the coefficients. You may simplify it by going in the frequency domain (Fourier) for example (or $z$ transform), but it is unlikely it is what you are looking for.
    $endgroup$
    – Damien
    Dec 20 '18 at 19:18










  • $begingroup$
    @Damien Can you explane it as an asnwer?
    $endgroup$
    – Leox
    Dec 20 '18 at 19:28






  • 1




    $begingroup$
    I have no time now, sorry. You can find information on convolution here for example: en.wikipedia.org/wiki/Convolution. It is a well known operation, widely studied. Used for example to represent filtering
    $endgroup$
    – Damien
    Dec 20 '18 at 19:41










  • $begingroup$
    Not sure why the answer in there is not correct but may help: math.stackexchange.com/questions/1937543/…
    $endgroup$
    – NoChance
    Dec 20 '18 at 20:22
















1












1








1





$begingroup$


I have found such a formula
$$
left(sum_{k=0}^n a_k z^k right)cdot left(sum_{j=0}^m b_j z^j right)=sum_{k=0}^{n+m} left( sum_{j=k-min(n,m)}^{min(k,max(n,m))}a_j b_{k-j} right)z^k,
$$
but seems it's too complicated. Is there any other formulas?






combinatorics polynomials







share|cite|improve this question











$endgroup$




I have found such a formula
$$
left(sum_{k=0}^n a_k z^k right)cdot left(sum_{j=0}^m b_j z^j right)=sum_{k=0}^{n+m} left( sum_{j=k-min(n,m)}^{min(k,max(n,m))}a_j b_{k-j} right)z^k,
$$
but seems it's too complicated. Is there any other formulas?






combinatorics polynomials




combinatorics polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 19:16







Leox

















asked Dec 20 '18 at 19:09









LeoxLeox

5,3481424




5,3481424








  • 1




    $begingroup$
    The inner sum corresponds to a convolution of the coefficients. You may simplify it by going in the frequency domain (Fourier) for example (or $z$ transform), but it is unlikely it is what you are looking for.
    $endgroup$
    – Damien
    Dec 20 '18 at 19:18










  • $begingroup$
    @Damien Can you explane it as an asnwer?
    $endgroup$
    – Leox
    Dec 20 '18 at 19:28






  • 1




    $begingroup$
    I have no time now, sorry. You can find information on convolution here for example: en.wikipedia.org/wiki/Convolution. It is a well known operation, widely studied. Used for example to represent filtering
    $endgroup$
    – Damien
    Dec 20 '18 at 19:41










  • $begingroup$
    Not sure why the answer in there is not correct but may help: math.stackexchange.com/questions/1937543/…
    $endgroup$
    – NoChance
    Dec 20 '18 at 20:22
















  • 1




    $begingroup$
    The inner sum corresponds to a convolution of the coefficients. You may simplify it by going in the frequency domain (Fourier) for example (or $z$ transform), but it is unlikely it is what you are looking for.
    $endgroup$
    – Damien
    Dec 20 '18 at 19:18










  • $begingroup$
    @Damien Can you explane it as an asnwer?
    $endgroup$
    – Leox
    Dec 20 '18 at 19:28






  • 1




    $begingroup$
    I have no time now, sorry. You can find information on convolution here for example: en.wikipedia.org/wiki/Convolution. It is a well known operation, widely studied. Used for example to represent filtering
    $endgroup$
    – Damien
    Dec 20 '18 at 19:41










  • $begingroup$
    Not sure why the answer in there is not correct but may help: math.stackexchange.com/questions/1937543/…
    $endgroup$
    – NoChance
    Dec 20 '18 at 20:22










1




1




$begingroup$
The inner sum corresponds to a convolution of the coefficients. You may simplify it by going in the frequency domain (Fourier) for example (or $z$ transform), but it is unlikely it is what you are looking for.
$endgroup$
– Damien
Dec 20 '18 at 19:18




$begingroup$
The inner sum corresponds to a convolution of the coefficients. You may simplify it by going in the frequency domain (Fourier) for example (or $z$ transform), but it is unlikely it is what you are looking for.
$endgroup$
– Damien
Dec 20 '18 at 19:18












$begingroup$
@Damien Can you explane it as an asnwer?
$endgroup$
– Leox
Dec 20 '18 at 19:28




$begingroup$
@Damien Can you explane it as an asnwer?
$endgroup$
– Leox
Dec 20 '18 at 19:28




1




1




$begingroup$
I have no time now, sorry. You can find information on convolution here for example: en.wikipedia.org/wiki/Convolution. It is a well known operation, widely studied. Used for example to represent filtering
$endgroup$
– Damien
Dec 20 '18 at 19:41




$begingroup$
I have no time now, sorry. You can find information on convolution here for example: en.wikipedia.org/wiki/Convolution. It is a well known operation, widely studied. Used for example to represent filtering
$endgroup$
– Damien
Dec 20 '18 at 19:41












$begingroup$
Not sure why the answer in there is not correct but may help: math.stackexchange.com/questions/1937543/…
$endgroup$
– NoChance
Dec 20 '18 at 20:22






$begingroup$
Not sure why the answer in there is not correct but may help: math.stackexchange.com/questions/1937543/…
$endgroup$
– NoChance
Dec 20 '18 at 20:22












1 Answer
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$begingroup$

You may simplify a little by omitting the extremes of summation.



Define $a_i=0$ if $inotin{0,dots,n}$ and $b_j=0$ if $jnotin{1,dots,m}$.
Let $c_k$ be the coefficient of $z^j$ in the product. Then the formula is saying that
$$
c_k = sum_{substack{i,jinmathbb N\i+j=k}} a_ib_j
= sum_{iinmathbb N}a_i b_{k-i} .
$$
This is as simple as it gets.



This way you can extend the formula to Laurent series.






share|cite|improve this answer









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    $begingroup$

    You may simplify a little by omitting the extremes of summation.



    Define $a_i=0$ if $inotin{0,dots,n}$ and $b_j=0$ if $jnotin{1,dots,m}$.
    Let $c_k$ be the coefficient of $z^j$ in the product. Then the formula is saying that
    $$
    c_k = sum_{substack{i,jinmathbb N\i+j=k}} a_ib_j
    = sum_{iinmathbb N}a_i b_{k-i} .
    $$
    This is as simple as it gets.



    This way you can extend the formula to Laurent series.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You may simplify a little by omitting the extremes of summation.



      Define $a_i=0$ if $inotin{0,dots,n}$ and $b_j=0$ if $jnotin{1,dots,m}$.
      Let $c_k$ be the coefficient of $z^j$ in the product. Then the formula is saying that
      $$
      c_k = sum_{substack{i,jinmathbb N\i+j=k}} a_ib_j
      = sum_{iinmathbb N}a_i b_{k-i} .
      $$
      This is as simple as it gets.



      This way you can extend the formula to Laurent series.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You may simplify a little by omitting the extremes of summation.



        Define $a_i=0$ if $inotin{0,dots,n}$ and $b_j=0$ if $jnotin{1,dots,m}$.
        Let $c_k$ be the coefficient of $z^j$ in the product. Then the formula is saying that
        $$
        c_k = sum_{substack{i,jinmathbb N\i+j=k}} a_ib_j
        = sum_{iinmathbb N}a_i b_{k-i} .
        $$
        This is as simple as it gets.



        This way you can extend the formula to Laurent series.






        share|cite|improve this answer









        $endgroup$



        You may simplify a little by omitting the extremes of summation.



        Define $a_i=0$ if $inotin{0,dots,n}$ and $b_j=0$ if $jnotin{1,dots,m}$.
        Let $c_k$ be the coefficient of $z^j$ in the product. Then the formula is saying that
        $$
        c_k = sum_{substack{i,jinmathbb N\i+j=k}} a_ib_j
        = sum_{iinmathbb N}a_i b_{k-i} .
        $$
        This is as simple as it gets.



        This way you can extend the formula to Laurent series.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 20 '18 at 19:28









        FedericoFederico

        5,124514




        5,124514






























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