Solve for the coefficient of an even generating function












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$begingroup$


Using a generating function, find the number of ways to select 10 candies from a huge pile of red, blue, and green lollipops if the selection has an even number of blue lollipops.



I started, but I don't understand how to continue



$(1 + x + x^2 + x^3 ...)^2 cdot (1 + x^2 + x^4 + x^6 ...)$



$=(frac{1}{1-x})^2 cdot frac{1}{1-x^2}$



$=big(1 + binom{1 + 2 - 1}{1}x + binom{2 + 2 - 1}{2}x^2 ...big) cdot ???$



How do you find coefficient to $x^{10}$?



Also, our textbook tells us $(1 + x^2 + x^4 + x^6 ...)$ = $frac{1}{1-x^2}$ how?






combinatorics generating-functions







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    1












    $begingroup$


    Using a generating function, find the number of ways to select 10 candies from a huge pile of red, blue, and green lollipops if the selection has an even number of blue lollipops.



    I started, but I don't understand how to continue



    $(1 + x + x^2 + x^3 ...)^2 cdot (1 + x^2 + x^4 + x^6 ...)$



    $=(frac{1}{1-x})^2 cdot frac{1}{1-x^2}$



    $=big(1 + binom{1 + 2 - 1}{1}x + binom{2 + 2 - 1}{2}x^2 ...big) cdot ???$



    How do you find coefficient to $x^{10}$?



    Also, our textbook tells us $(1 + x^2 + x^4 + x^6 ...)$ = $frac{1}{1-x^2}$ how?






    combinatorics generating-functions







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Using a generating function, find the number of ways to select 10 candies from a huge pile of red, blue, and green lollipops if the selection has an even number of blue lollipops.



      I started, but I don't understand how to continue



      $(1 + x + x^2 + x^3 ...)^2 cdot (1 + x^2 + x^4 + x^6 ...)$



      $=(frac{1}{1-x})^2 cdot frac{1}{1-x^2}$



      $=big(1 + binom{1 + 2 - 1}{1}x + binom{2 + 2 - 1}{2}x^2 ...big) cdot ???$



      How do you find coefficient to $x^{10}$?



      Also, our textbook tells us $(1 + x^2 + x^4 + x^6 ...)$ = $frac{1}{1-x^2}$ how?






      combinatorics generating-functions







      share|cite|improve this question











      $endgroup$




      Using a generating function, find the number of ways to select 10 candies from a huge pile of red, blue, and green lollipops if the selection has an even number of blue lollipops.



      I started, but I don't understand how to continue



      $(1 + x + x^2 + x^3 ...)^2 cdot (1 + x^2 + x^4 + x^6 ...)$



      $=(frac{1}{1-x})^2 cdot frac{1}{1-x^2}$



      $=big(1 + binom{1 + 2 - 1}{1}x + binom{2 + 2 - 1}{2}x^2 ...big) cdot ???$



      How do you find coefficient to $x^{10}$?



      Also, our textbook tells us $(1 + x^2 + x^4 + x^6 ...)$ = $frac{1}{1-x^2}$ how?






      combinatorics generating-functions




      combinatorics generating-functions






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      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 27 '18 at 0:06







      Math Newbie

















      asked Oct 31 '18 at 23:29









      Math NewbieMath Newbie

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          Notice that $left(frac{1}{1-x}right)^2=frac{d}{dx}left(frac{1}{1-x}right)$, and hence $left(frac{1}{1-x}right)^2=1+2x+3x^2+dots$ You can then find the $x^{10}$ coefficient just by multiplying out.



          The equality stated in the textbooks holds for the same reason that $1+x+x^2+dots=frac{1}{1-x}$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            The original problem is $(1 + x + x^2 + x^3 ...)^2 cdot (1 + x^2 + x^4 + x^6 ...)$



            $frac{1}{1-x} = (1 + x + x^2 + x^3 ...)$



            Which can be shown by,



            $1 = (1 + x + x^2 + x^3 ...)(1-x)$



            $1 = (1 + x + x^2 + x^3 ...) - x cdot (1 + x + x^2 + x^3 ...)$



            $1 =1$



            To show, $frac{1}{1-x^2} = (1 + x^2 + x^4 + x^6 ...)$



            we substitute $y = x^2$



            $frac{1}{1-y} = (1 + y + y^2 + y^3 ...)$ which is equivalent due to $frac{1}{1-x} = (1 + x + x^2 + x^3 ...)$ shown above.



            To solve the equation,



            $=(frac{1}{1-x})^2 cdot frac{1}{1-x^2}$



            $=frac{1}{(1-x)^2} cdot frac{1}{1-x^2}$



            $=big(1 + binom{1 + 2 - 1}{1}x + binom{2 + 2 - 1}{2}x^2 ...big) cdot (1 + x^2 + x^4 + x^6 ...)$



            Thus the coefficient of $x^{10}$ is the coefficient of



            $=1 cdot x^{10} + binom{3}{1}x^2 cdot x^{8} + binom{5}{1}x^4 cdot x^{6} + binom{7}{1}x^6 cdot x^{4} + binom{9}{1}x^8 cdot x^{2} + binom{11}{1}x^{10} cdot 1$






            share|cite|improve this answer









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              2 Answers
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              $begingroup$

              Notice that $left(frac{1}{1-x}right)^2=frac{d}{dx}left(frac{1}{1-x}right)$, and hence $left(frac{1}{1-x}right)^2=1+2x+3x^2+dots$ You can then find the $x^{10}$ coefficient just by multiplying out.



              The equality stated in the textbooks holds for the same reason that $1+x+x^2+dots=frac{1}{1-x}$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Notice that $left(frac{1}{1-x}right)^2=frac{d}{dx}left(frac{1}{1-x}right)$, and hence $left(frac{1}{1-x}right)^2=1+2x+3x^2+dots$ You can then find the $x^{10}$ coefficient just by multiplying out.



                The equality stated in the textbooks holds for the same reason that $1+x+x^2+dots=frac{1}{1-x}$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Notice that $left(frac{1}{1-x}right)^2=frac{d}{dx}left(frac{1}{1-x}right)$, and hence $left(frac{1}{1-x}right)^2=1+2x+3x^2+dots$ You can then find the $x^{10}$ coefficient just by multiplying out.



                  The equality stated in the textbooks holds for the same reason that $1+x+x^2+dots=frac{1}{1-x}$.






                  share|cite|improve this answer









                  $endgroup$



                  Notice that $left(frac{1}{1-x}right)^2=frac{d}{dx}left(frac{1}{1-x}right)$, and hence $left(frac{1}{1-x}right)^2=1+2x+3x^2+dots$ You can then find the $x^{10}$ coefficient just by multiplying out.



                  The equality stated in the textbooks holds for the same reason that $1+x+x^2+dots=frac{1}{1-x}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 1 '18 at 1:03









                  Harry PetytHarry Petyt

                  1435




                  1435























                      1












                      $begingroup$

                      The original problem is $(1 + x + x^2 + x^3 ...)^2 cdot (1 + x^2 + x^4 + x^6 ...)$



                      $frac{1}{1-x} = (1 + x + x^2 + x^3 ...)$



                      Which can be shown by,



                      $1 = (1 + x + x^2 + x^3 ...)(1-x)$



                      $1 = (1 + x + x^2 + x^3 ...) - x cdot (1 + x + x^2 + x^3 ...)$



                      $1 =1$



                      To show, $frac{1}{1-x^2} = (1 + x^2 + x^4 + x^6 ...)$



                      we substitute $y = x^2$



                      $frac{1}{1-y} = (1 + y + y^2 + y^3 ...)$ which is equivalent due to $frac{1}{1-x} = (1 + x + x^2 + x^3 ...)$ shown above.



                      To solve the equation,



                      $=(frac{1}{1-x})^2 cdot frac{1}{1-x^2}$



                      $=frac{1}{(1-x)^2} cdot frac{1}{1-x^2}$



                      $=big(1 + binom{1 + 2 - 1}{1}x + binom{2 + 2 - 1}{2}x^2 ...big) cdot (1 + x^2 + x^4 + x^6 ...)$



                      Thus the coefficient of $x^{10}$ is the coefficient of



                      $=1 cdot x^{10} + binom{3}{1}x^2 cdot x^{8} + binom{5}{1}x^4 cdot x^{6} + binom{7}{1}x^6 cdot x^{4} + binom{9}{1}x^8 cdot x^{2} + binom{11}{1}x^{10} cdot 1$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        The original problem is $(1 + x + x^2 + x^3 ...)^2 cdot (1 + x^2 + x^4 + x^6 ...)$



                        $frac{1}{1-x} = (1 + x + x^2 + x^3 ...)$



                        Which can be shown by,



                        $1 = (1 + x + x^2 + x^3 ...)(1-x)$



                        $1 = (1 + x + x^2 + x^3 ...) - x cdot (1 + x + x^2 + x^3 ...)$



                        $1 =1$



                        To show, $frac{1}{1-x^2} = (1 + x^2 + x^4 + x^6 ...)$



                        we substitute $y = x^2$



                        $frac{1}{1-y} = (1 + y + y^2 + y^3 ...)$ which is equivalent due to $frac{1}{1-x} = (1 + x + x^2 + x^3 ...)$ shown above.



                        To solve the equation,



                        $=(frac{1}{1-x})^2 cdot frac{1}{1-x^2}$



                        $=frac{1}{(1-x)^2} cdot frac{1}{1-x^2}$



                        $=big(1 + binom{1 + 2 - 1}{1}x + binom{2 + 2 - 1}{2}x^2 ...big) cdot (1 + x^2 + x^4 + x^6 ...)$



                        Thus the coefficient of $x^{10}$ is the coefficient of



                        $=1 cdot x^{10} + binom{3}{1}x^2 cdot x^{8} + binom{5}{1}x^4 cdot x^{6} + binom{7}{1}x^6 cdot x^{4} + binom{9}{1}x^8 cdot x^{2} + binom{11}{1}x^{10} cdot 1$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          The original problem is $(1 + x + x^2 + x^3 ...)^2 cdot (1 + x^2 + x^4 + x^6 ...)$



                          $frac{1}{1-x} = (1 + x + x^2 + x^3 ...)$



                          Which can be shown by,



                          $1 = (1 + x + x^2 + x^3 ...)(1-x)$



                          $1 = (1 + x + x^2 + x^3 ...) - x cdot (1 + x + x^2 + x^3 ...)$



                          $1 =1$



                          To show, $frac{1}{1-x^2} = (1 + x^2 + x^4 + x^6 ...)$



                          we substitute $y = x^2$



                          $frac{1}{1-y} = (1 + y + y^2 + y^3 ...)$ which is equivalent due to $frac{1}{1-x} = (1 + x + x^2 + x^3 ...)$ shown above.



                          To solve the equation,



                          $=(frac{1}{1-x})^2 cdot frac{1}{1-x^2}$



                          $=frac{1}{(1-x)^2} cdot frac{1}{1-x^2}$



                          $=big(1 + binom{1 + 2 - 1}{1}x + binom{2 + 2 - 1}{2}x^2 ...big) cdot (1 + x^2 + x^4 + x^6 ...)$



                          Thus the coefficient of $x^{10}$ is the coefficient of



                          $=1 cdot x^{10} + binom{3}{1}x^2 cdot x^{8} + binom{5}{1}x^4 cdot x^{6} + binom{7}{1}x^6 cdot x^{4} + binom{9}{1}x^8 cdot x^{2} + binom{11}{1}x^{10} cdot 1$






                          share|cite|improve this answer









                          $endgroup$



                          The original problem is $(1 + x + x^2 + x^3 ...)^2 cdot (1 + x^2 + x^4 + x^6 ...)$



                          $frac{1}{1-x} = (1 + x + x^2 + x^3 ...)$



                          Which can be shown by,



                          $1 = (1 + x + x^2 + x^3 ...)(1-x)$



                          $1 = (1 + x + x^2 + x^3 ...) - x cdot (1 + x + x^2 + x^3 ...)$



                          $1 =1$



                          To show, $frac{1}{1-x^2} = (1 + x^2 + x^4 + x^6 ...)$



                          we substitute $y = x^2$



                          $frac{1}{1-y} = (1 + y + y^2 + y^3 ...)$ which is equivalent due to $frac{1}{1-x} = (1 + x + x^2 + x^3 ...)$ shown above.



                          To solve the equation,



                          $=(frac{1}{1-x})^2 cdot frac{1}{1-x^2}$



                          $=frac{1}{(1-x)^2} cdot frac{1}{1-x^2}$



                          $=big(1 + binom{1 + 2 - 1}{1}x + binom{2 + 2 - 1}{2}x^2 ...big) cdot (1 + x^2 + x^4 + x^6 ...)$



                          Thus the coefficient of $x^{10}$ is the coefficient of



                          $=1 cdot x^{10} + binom{3}{1}x^2 cdot x^{8} + binom{5}{1}x^4 cdot x^{6} + binom{7}{1}x^6 cdot x^{4} + binom{9}{1}x^8 cdot x^{2} + binom{11}{1}x^{10} cdot 1$







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                          answered Dec 27 '18 at 0:22









                          Math NewbieMath Newbie

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