Finding the closed form of a generating function












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$begingroup$


Given that $k$ is a positive integer and $f(x)$ is the generating function of the sequence $(b_0,b_1,b_2,...)$ where $b_n = {n choose k};, forall ;n$, show that: $$f(x)=frac{x^k}{(1-x)^{k+1}}$$
I tried writing a few terms of $f(x)$:$$f(x)=x^k+{k+1 choose k}x^{k+1}+{k+2 choose k}x^{k+2}+...$$$$;;;;;;;;;=x^kleft(1+{k+1 choose k}x^1+{k+2 choose k}x^2+...right)$$$$=x^kcdot h(x);;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; $$
Where $h(x)$ is defined as the expression in parenthesis, then I tried some manipulation, for example I computed $h(x)-xh(x)$, since: ${k+n+1 choose k}-{k+n choose k}={k+n choose k-1}$ we have:



$h(x)-xh(x)=1+{k choose k-1}x+{k+1 choose k-1}x^2+...$



But I'm stuck, not sure how I should procede






generating-functions







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    2












    $begingroup$


    Given that $k$ is a positive integer and $f(x)$ is the generating function of the sequence $(b_0,b_1,b_2,...)$ where $b_n = {n choose k};, forall ;n$, show that: $$f(x)=frac{x^k}{(1-x)^{k+1}}$$
    I tried writing a few terms of $f(x)$:$$f(x)=x^k+{k+1 choose k}x^{k+1}+{k+2 choose k}x^{k+2}+...$$$$;;;;;;;;;=x^kleft(1+{k+1 choose k}x^1+{k+2 choose k}x^2+...right)$$$$=x^kcdot h(x);;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; $$
    Where $h(x)$ is defined as the expression in parenthesis, then I tried some manipulation, for example I computed $h(x)-xh(x)$, since: ${k+n+1 choose k}-{k+n choose k}={k+n choose k-1}$ we have:



    $h(x)-xh(x)=1+{k choose k-1}x+{k+1 choose k-1}x^2+...$



    But I'm stuck, not sure how I should procede






    generating-functions







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      0



      $begingroup$


      Given that $k$ is a positive integer and $f(x)$ is the generating function of the sequence $(b_0,b_1,b_2,...)$ where $b_n = {n choose k};, forall ;n$, show that: $$f(x)=frac{x^k}{(1-x)^{k+1}}$$
      I tried writing a few terms of $f(x)$:$$f(x)=x^k+{k+1 choose k}x^{k+1}+{k+2 choose k}x^{k+2}+...$$$$;;;;;;;;;=x^kleft(1+{k+1 choose k}x^1+{k+2 choose k}x^2+...right)$$$$=x^kcdot h(x);;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; $$
      Where $h(x)$ is defined as the expression in parenthesis, then I tried some manipulation, for example I computed $h(x)-xh(x)$, since: ${k+n+1 choose k}-{k+n choose k}={k+n choose k-1}$ we have:



      $h(x)-xh(x)=1+{k choose k-1}x+{k+1 choose k-1}x^2+...$



      But I'm stuck, not sure how I should procede






      generating-functions







      share|cite|improve this question









      $endgroup$




      Given that $k$ is a positive integer and $f(x)$ is the generating function of the sequence $(b_0,b_1,b_2,...)$ where $b_n = {n choose k};, forall ;n$, show that: $$f(x)=frac{x^k}{(1-x)^{k+1}}$$
      I tried writing a few terms of $f(x)$:$$f(x)=x^k+{k+1 choose k}x^{k+1}+{k+2 choose k}x^{k+2}+...$$$$;;;;;;;;;=x^kleft(1+{k+1 choose k}x^1+{k+2 choose k}x^2+...right)$$$$=x^kcdot h(x);;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; $$
      Where $h(x)$ is defined as the expression in parenthesis, then I tried some manipulation, for example I computed $h(x)-xh(x)$, since: ${k+n+1 choose k}-{k+n choose k}={k+n choose k-1}$ we have:



      $h(x)-xh(x)=1+{k choose k-1}x+{k+1 choose k-1}x^2+...$



      But I'm stuck, not sure how I should procede






      generating-functions




      generating-functions






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      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 25 at 15:54









      Spasoje DurovicSpasoje Durovic

      38210




      38210






















          2 Answers
          2






          active

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          $begingroup$

          Here is a variation based upon the coefficient of operator $[x^n]$ which denotes the coefficient of $x^n$ of a series.




          We obtain
          begin{align*}
          color{blue}{b_n=[x^n]f(x)}&=[x^n]frac{x^k}{(1-x)^{k+1}}\
          &=[x^{n-k}](1-x)^{-k-1}tag{1}\
          &=[x^{n-k}]sum_{j=0}^infty binom{-k-1}{j}(-x)^jtag{2}\
          &=[x^{n-k}]sum_{j=0}^infty binom{k+j}{j}x^jtag{3}\
          &=binom{n}{n-k}tag{4}\
          &,,color{blue}{=binom{n}{k}}tag{5}
          end{align*}



          and the claim follows.




          Comment:




          • In (1) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.


          • In (2) we apply the binomial series expansion.


          • In (3) we use the binomial identity $binom{-p}{q}=binom{p+q-1}{q}(-1)^q$.


          • In (4) we select the coefficient of $x^{n-k}$.


          • In (5) we use the binomial identity $binom{p}{q}=binom{p}{p-q}$.







          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            HINT:



            Inductive hypothesis
            begin{eqnarray*}
            f_k(x)= sum_{n=k}^{infty} binom{n}{k} x^n = frac{x^k}{(1-x)^{k+1}}
            end{eqnarray*}
            and use
            begin{eqnarray*}
            binom{n}{k} + binom{n}{k+1}= binom{n+1}{k+1}.
            end{eqnarray*}



            More detail available on request.



            Further Hint :



            begin{eqnarray*}
            f_{k+1}(x) &=& sum_{n=k+1}^{infty} binom{n}{k+1} x^n = sum_{n=k}^{infty} binom{n+1}{k+1} x^{n+1} \
            &=& sum_{n=k}^{infty} binom{n}{k} x^{n+1} + sum_{n=k}^{infty} binom{n}{k+1} x^{n+1} \ cdots
            end{eqnarray*}






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              From this I got $f_{k+1}(x)=frac{x^k}{(1-x)^{k+2}}$ so $f_{k+1}(x)=frac{1}{(1-x)}f_k(x)$ I never used induction on generating functions so I'm not sure what I'm supposed to do or if I did this correctly
              $endgroup$
              – Spasoje Durovic
              Jan 25 at 16:39













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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Here is a variation based upon the coefficient of operator $[x^n]$ which denotes the coefficient of $x^n$ of a series.




            We obtain
            begin{align*}
            color{blue}{b_n=[x^n]f(x)}&=[x^n]frac{x^k}{(1-x)^{k+1}}\
            &=[x^{n-k}](1-x)^{-k-1}tag{1}\
            &=[x^{n-k}]sum_{j=0}^infty binom{-k-1}{j}(-x)^jtag{2}\
            &=[x^{n-k}]sum_{j=0}^infty binom{k+j}{j}x^jtag{3}\
            &=binom{n}{n-k}tag{4}\
            &,,color{blue}{=binom{n}{k}}tag{5}
            end{align*}



            and the claim follows.




            Comment:




            • In (1) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.


            • In (2) we apply the binomial series expansion.


            • In (3) we use the binomial identity $binom{-p}{q}=binom{p+q-1}{q}(-1)^q$.


            • In (4) we select the coefficient of $x^{n-k}$.


            • In (5) we use the binomial identity $binom{p}{q}=binom{p}{p-q}$.







            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Here is a variation based upon the coefficient of operator $[x^n]$ which denotes the coefficient of $x^n$ of a series.




              We obtain
              begin{align*}
              color{blue}{b_n=[x^n]f(x)}&=[x^n]frac{x^k}{(1-x)^{k+1}}\
              &=[x^{n-k}](1-x)^{-k-1}tag{1}\
              &=[x^{n-k}]sum_{j=0}^infty binom{-k-1}{j}(-x)^jtag{2}\
              &=[x^{n-k}]sum_{j=0}^infty binom{k+j}{j}x^jtag{3}\
              &=binom{n}{n-k}tag{4}\
              &,,color{blue}{=binom{n}{k}}tag{5}
              end{align*}



              and the claim follows.




              Comment:




              • In (1) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.


              • In (2) we apply the binomial series expansion.


              • In (3) we use the binomial identity $binom{-p}{q}=binom{p+q-1}{q}(-1)^q$.


              • In (4) we select the coefficient of $x^{n-k}$.


              • In (5) we use the binomial identity $binom{p}{q}=binom{p}{p-q}$.







              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Here is a variation based upon the coefficient of operator $[x^n]$ which denotes the coefficient of $x^n$ of a series.




                We obtain
                begin{align*}
                color{blue}{b_n=[x^n]f(x)}&=[x^n]frac{x^k}{(1-x)^{k+1}}\
                &=[x^{n-k}](1-x)^{-k-1}tag{1}\
                &=[x^{n-k}]sum_{j=0}^infty binom{-k-1}{j}(-x)^jtag{2}\
                &=[x^{n-k}]sum_{j=0}^infty binom{k+j}{j}x^jtag{3}\
                &=binom{n}{n-k}tag{4}\
                &,,color{blue}{=binom{n}{k}}tag{5}
                end{align*}



                and the claim follows.




                Comment:




                • In (1) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.


                • In (2) we apply the binomial series expansion.


                • In (3) we use the binomial identity $binom{-p}{q}=binom{p+q-1}{q}(-1)^q$.


                • In (4) we select the coefficient of $x^{n-k}$.


                • In (5) we use the binomial identity $binom{p}{q}=binom{p}{p-q}$.







                share|cite|improve this answer









                $endgroup$



                Here is a variation based upon the coefficient of operator $[x^n]$ which denotes the coefficient of $x^n$ of a series.




                We obtain
                begin{align*}
                color{blue}{b_n=[x^n]f(x)}&=[x^n]frac{x^k}{(1-x)^{k+1}}\
                &=[x^{n-k}](1-x)^{-k-1}tag{1}\
                &=[x^{n-k}]sum_{j=0}^infty binom{-k-1}{j}(-x)^jtag{2}\
                &=[x^{n-k}]sum_{j=0}^infty binom{k+j}{j}x^jtag{3}\
                &=binom{n}{n-k}tag{4}\
                &,,color{blue}{=binom{n}{k}}tag{5}
                end{align*}



                and the claim follows.




                Comment:




                • In (1) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.


                • In (2) we apply the binomial series expansion.


                • In (3) we use the binomial identity $binom{-p}{q}=binom{p+q-1}{q}(-1)^q$.


                • In (4) we select the coefficient of $x^{n-k}$.


                • In (5) we use the binomial identity $binom{p}{q}=binom{p}{p-q}$.








                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 26 at 18:52









                Markus ScheuerMarkus Scheuer

                62.6k459149




                62.6k459149























                    1












                    $begingroup$

                    HINT:



                    Inductive hypothesis
                    begin{eqnarray*}
                    f_k(x)= sum_{n=k}^{infty} binom{n}{k} x^n = frac{x^k}{(1-x)^{k+1}}
                    end{eqnarray*}
                    and use
                    begin{eqnarray*}
                    binom{n}{k} + binom{n}{k+1}= binom{n+1}{k+1}.
                    end{eqnarray*}



                    More detail available on request.



                    Further Hint :



                    begin{eqnarray*}
                    f_{k+1}(x) &=& sum_{n=k+1}^{infty} binom{n}{k+1} x^n = sum_{n=k}^{infty} binom{n+1}{k+1} x^{n+1} \
                    &=& sum_{n=k}^{infty} binom{n}{k} x^{n+1} + sum_{n=k}^{infty} binom{n}{k+1} x^{n+1} \ cdots
                    end{eqnarray*}






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      From this I got $f_{k+1}(x)=frac{x^k}{(1-x)^{k+2}}$ so $f_{k+1}(x)=frac{1}{(1-x)}f_k(x)$ I never used induction on generating functions so I'm not sure what I'm supposed to do or if I did this correctly
                      $endgroup$
                      – Spasoje Durovic
                      Jan 25 at 16:39


















                    1












                    $begingroup$

                    HINT:



                    Inductive hypothesis
                    begin{eqnarray*}
                    f_k(x)= sum_{n=k}^{infty} binom{n}{k} x^n = frac{x^k}{(1-x)^{k+1}}
                    end{eqnarray*}
                    and use
                    begin{eqnarray*}
                    binom{n}{k} + binom{n}{k+1}= binom{n+1}{k+1}.
                    end{eqnarray*}



                    More detail available on request.



                    Further Hint :



                    begin{eqnarray*}
                    f_{k+1}(x) &=& sum_{n=k+1}^{infty} binom{n}{k+1} x^n = sum_{n=k}^{infty} binom{n+1}{k+1} x^{n+1} \
                    &=& sum_{n=k}^{infty} binom{n}{k} x^{n+1} + sum_{n=k}^{infty} binom{n}{k+1} x^{n+1} \ cdots
                    end{eqnarray*}






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      From this I got $f_{k+1}(x)=frac{x^k}{(1-x)^{k+2}}$ so $f_{k+1}(x)=frac{1}{(1-x)}f_k(x)$ I never used induction on generating functions so I'm not sure what I'm supposed to do or if I did this correctly
                      $endgroup$
                      – Spasoje Durovic
                      Jan 25 at 16:39
















                    1












                    1








                    1





                    $begingroup$

                    HINT:



                    Inductive hypothesis
                    begin{eqnarray*}
                    f_k(x)= sum_{n=k}^{infty} binom{n}{k} x^n = frac{x^k}{(1-x)^{k+1}}
                    end{eqnarray*}
                    and use
                    begin{eqnarray*}
                    binom{n}{k} + binom{n}{k+1}= binom{n+1}{k+1}.
                    end{eqnarray*}



                    More detail available on request.



                    Further Hint :



                    begin{eqnarray*}
                    f_{k+1}(x) &=& sum_{n=k+1}^{infty} binom{n}{k+1} x^n = sum_{n=k}^{infty} binom{n+1}{k+1} x^{n+1} \
                    &=& sum_{n=k}^{infty} binom{n}{k} x^{n+1} + sum_{n=k}^{infty} binom{n}{k+1} x^{n+1} \ cdots
                    end{eqnarray*}






                    share|cite|improve this answer











                    $endgroup$



                    HINT:



                    Inductive hypothesis
                    begin{eqnarray*}
                    f_k(x)= sum_{n=k}^{infty} binom{n}{k} x^n = frac{x^k}{(1-x)^{k+1}}
                    end{eqnarray*}
                    and use
                    begin{eqnarray*}
                    binom{n}{k} + binom{n}{k+1}= binom{n+1}{k+1}.
                    end{eqnarray*}



                    More detail available on request.



                    Further Hint :



                    begin{eqnarray*}
                    f_{k+1}(x) &=& sum_{n=k+1}^{infty} binom{n}{k+1} x^n = sum_{n=k}^{infty} binom{n+1}{k+1} x^{n+1} \
                    &=& sum_{n=k}^{infty} binom{n}{k} x^{n+1} + sum_{n=k}^{infty} binom{n}{k+1} x^{n+1} \ cdots
                    end{eqnarray*}







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 25 at 16:52

























                    answered Jan 25 at 16:08









                    Donald SplutterwitDonald Splutterwit

                    22.9k21446




                    22.9k21446












                    • $begingroup$
                      From this I got $f_{k+1}(x)=frac{x^k}{(1-x)^{k+2}}$ so $f_{k+1}(x)=frac{1}{(1-x)}f_k(x)$ I never used induction on generating functions so I'm not sure what I'm supposed to do or if I did this correctly
                      $endgroup$
                      – Spasoje Durovic
                      Jan 25 at 16:39




















                    • $begingroup$
                      From this I got $f_{k+1}(x)=frac{x^k}{(1-x)^{k+2}}$ so $f_{k+1}(x)=frac{1}{(1-x)}f_k(x)$ I never used induction on generating functions so I'm not sure what I'm supposed to do or if I did this correctly
                      $endgroup$
                      – Spasoje Durovic
                      Jan 25 at 16:39


















                    $begingroup$
                    From this I got $f_{k+1}(x)=frac{x^k}{(1-x)^{k+2}}$ so $f_{k+1}(x)=frac{1}{(1-x)}f_k(x)$ I never used induction on generating functions so I'm not sure what I'm supposed to do or if I did this correctly
                    $endgroup$
                    – Spasoje Durovic
                    Jan 25 at 16:39






                    $begingroup$
                    From this I got $f_{k+1}(x)=frac{x^k}{(1-x)^{k+2}}$ so $f_{k+1}(x)=frac{1}{(1-x)}f_k(x)$ I never used induction on generating functions so I'm not sure what I'm supposed to do or if I did this correctly
                    $endgroup$
                    – Spasoje Durovic
                    Jan 25 at 16:39




















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