What does “∏” mean?

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In the following formula:
$$P_i(z)= prod_j=0^k-1(z+2^jx_i) mod 2^k.$$



  • What does "∏" mean?





notation definition







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  • 1




    Capital Pi notation
    – mwt
    1 hour ago






  • 1




    Yes product as others have stated, it's the poduct equivalent of stating that $Sigma$ means sum.
    – Kevin
    1 hour ago














up vote
2
down vote

favorite
1












In the following formula:
$$P_i(z)= prod_j=0^k-1(z+2^jx_i) mod 2^k.$$



  • What does "∏" mean?





notation definition







share|cite|improve this question



















  • 1




    Capital Pi notation
    – mwt
    1 hour ago






  • 1




    Yes product as others have stated, it's the poduct equivalent of stating that $Sigma$ means sum.
    – Kevin
    1 hour ago












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





In the following formula:
$$P_i(z)= prod_j=0^k-1(z+2^jx_i) mod 2^k.$$



  • What does "∏" mean?





notation definition







share|cite|improve this question















In the following formula:
$$P_i(z)= prod_j=0^k-1(z+2^jx_i) mod 2^k.$$



  • What does "∏" mean?





notation definition




notation definition






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









MRobinson

88315




88315










asked 1 hour ago









R1-

1318




1318







  • 1




    Capital Pi notation
    – mwt
    1 hour ago






  • 1




    Yes product as others have stated, it's the poduct equivalent of stating that $Sigma$ means sum.
    – Kevin
    1 hour ago












  • 1




    Capital Pi notation
    – mwt
    1 hour ago






  • 1




    Yes product as others have stated, it's the poduct equivalent of stating that $Sigma$ means sum.
    – Kevin
    1 hour ago







1




1




Capital Pi notation
– mwt
1 hour ago




Capital Pi notation
– mwt
1 hour ago




1




1




Yes product as others have stated, it's the poduct equivalent of stating that $Sigma$ means sum.
– Kevin
1 hour ago




Yes product as others have stated, it's the poduct equivalent of stating that $Sigma$ means sum.
– Kevin
1 hour ago










3 Answers
3






active

oldest

votes

















up vote
4
down vote



accepted










For comparison (if you are familiar with the sigma-notation):



$$sum_i=0^n a_i=a_0+a_1+a_2+dots+a_n$$
$$prod_i=0^na_i=a_0times a_1times a_2timesdotstimes a_n$$



The only difference is that you multiply instead of adding. So $$boxedprod_j=0^k-1(z+2^jx_i)=(z+2^0x_i)times (z+2^1x_i)times(z+2^2x_i)timesdotstimes(z+2^k-1x_i)\=(z+x_i)(z+2x_i)(z+4x_i)timesdotstimes (z+2^k-1x_i)$$






share|cite|improve this answer





























    up vote
    1
    down vote













    The Product of.



    So multiply together all the $z + 2^jx_i$ for each $j=0$ to $k-1$.






    share|cite|improve this answer



























      up vote
      1
      down vote













      The symbol you expressed as $$Pi$$ is known as the product of an expression



      in the formula, it says (in words), "P sub i of z = "the product" when j = 0 and goes to k - 1 of z + 2 to the j separated by x sub i where the whole formula is modded by 2 to the k"






      share|cite|improve this answer








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        Your Answer




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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        4
        down vote



        accepted










        For comparison (if you are familiar with the sigma-notation):



        $$sum_i=0^n a_i=a_0+a_1+a_2+dots+a_n$$
        $$prod_i=0^na_i=a_0times a_1times a_2timesdotstimes a_n$$



        The only difference is that you multiply instead of adding. So $$boxedprod_j=0^k-1(z+2^jx_i)=(z+2^0x_i)times (z+2^1x_i)times(z+2^2x_i)timesdotstimes(z+2^k-1x_i)\=(z+x_i)(z+2x_i)(z+4x_i)timesdotstimes (z+2^k-1x_i)$$






        share|cite|improve this answer


























          up vote
          4
          down vote



          accepted










          For comparison (if you are familiar with the sigma-notation):



          $$sum_i=0^n a_i=a_0+a_1+a_2+dots+a_n$$
          $$prod_i=0^na_i=a_0times a_1times a_2timesdotstimes a_n$$



          The only difference is that you multiply instead of adding. So $$boxedprod_j=0^k-1(z+2^jx_i)=(z+2^0x_i)times (z+2^1x_i)times(z+2^2x_i)timesdotstimes(z+2^k-1x_i)\=(z+x_i)(z+2x_i)(z+4x_i)timesdotstimes (z+2^k-1x_i)$$






          share|cite|improve this answer
























            up vote
            4
            down vote



            accepted







            up vote
            4
            down vote



            accepted






            For comparison (if you are familiar with the sigma-notation):



            $$sum_i=0^n a_i=a_0+a_1+a_2+dots+a_n$$
            $$prod_i=0^na_i=a_0times a_1times a_2timesdotstimes a_n$$



            The only difference is that you multiply instead of adding. So $$boxedprod_j=0^k-1(z+2^jx_i)=(z+2^0x_i)times (z+2^1x_i)times(z+2^2x_i)timesdotstimes(z+2^k-1x_i)\=(z+x_i)(z+2x_i)(z+4x_i)timesdotstimes (z+2^k-1x_i)$$






            share|cite|improve this answer














            For comparison (if you are familiar with the sigma-notation):



            $$sum_i=0^n a_i=a_0+a_1+a_2+dots+a_n$$
            $$prod_i=0^na_i=a_0times a_1times a_2timesdotstimes a_n$$



            The only difference is that you multiply instead of adding. So $$boxedprod_j=0^k-1(z+2^jx_i)=(z+2^0x_i)times (z+2^1x_i)times(z+2^2x_i)timesdotstimes(z+2^k-1x_i)\=(z+x_i)(z+2x_i)(z+4x_i)timesdotstimes (z+2^k-1x_i)$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 1 hour ago









            lioness99a

            3,6382527




            3,6382527










            answered 1 hour ago









            cansomeonehelpmeout

            5,7083830




            5,7083830




















                up vote
                1
                down vote













                The Product of.



                So multiply together all the $z + 2^jx_i$ for each $j=0$ to $k-1$.






                share|cite|improve this answer
























                  up vote
                  1
                  down vote













                  The Product of.



                  So multiply together all the $z + 2^jx_i$ for each $j=0$ to $k-1$.






                  share|cite|improve this answer






















                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    The Product of.



                    So multiply together all the $z + 2^jx_i$ for each $j=0$ to $k-1$.






                    share|cite|improve this answer












                    The Product of.



                    So multiply together all the $z + 2^jx_i$ for each $j=0$ to $k-1$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    MRobinson

                    88315




                    88315




















                        up vote
                        1
                        down vote













                        The symbol you expressed as $$Pi$$ is known as the product of an expression



                        in the formula, it says (in words), "P sub i of z = "the product" when j = 0 and goes to k - 1 of z + 2 to the j separated by x sub i where the whole formula is modded by 2 to the k"






                        share|cite|improve this answer








                        New contributor




                        jonathan nicholson jon123276 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.





















                          up vote
                          1
                          down vote













                          The symbol you expressed as $$Pi$$ is known as the product of an expression



                          in the formula, it says (in words), "P sub i of z = "the product" when j = 0 and goes to k - 1 of z + 2 to the j separated by x sub i where the whole formula is modded by 2 to the k"






                          share|cite|improve this answer








                          New contributor




                          jonathan nicholson jon123276 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.



















                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            The symbol you expressed as $$Pi$$ is known as the product of an expression



                            in the formula, it says (in words), "P sub i of z = "the product" when j = 0 and goes to k - 1 of z + 2 to the j separated by x sub i where the whole formula is modded by 2 to the k"






                            share|cite|improve this answer








                            New contributor




                            jonathan nicholson jon123276 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            The symbol you expressed as $$Pi$$ is known as the product of an expression



                            in the formula, it says (in words), "P sub i of z = "the product" when j = 0 and goes to k - 1 of z + 2 to the j separated by x sub i where the whole formula is modded by 2 to the k"







                            share|cite|improve this answer








                            New contributor




                            jonathan nicholson jon123276 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            share|cite|improve this answer



                            share|cite|improve this answer






                            New contributor




                            jonathan nicholson jon123276 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            answered 1 hour ago









                            jonathan nicholson jon123276

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                            311




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                            New contributor





                            jonathan nicholson jon123276 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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