A Fibonacci convolution












2















A Fibonacci convolution. Recall that $$F(x)=sum_{n=0}^infty F_n x^n =frac{x}{1-x-x^2} =frac{1}{sqrt{5}} left(frac{1}{1-Phi x} -frac{1}{1-bar{Phi}x}right).$$
(a) Prove that $displaystyle sum_{n=0}^infty F_{n+1} x^n =frac{1}{1-x-x^2}$.



(b) Prove that $displaystyle sum_{n=0}^infty (2F_{n+1} -F_n)x^n =frac{2-x}{1-x-x^2} =sum_{n=0}^infty (Phi^n +bar{Phi}^n) x^n$.



(c) Prove that $displaystyle 5F(x)^2 =sum_{n=0}^infty binom{n+1}{1} Phi^n x^n -2sum_{n=0}^infty F_{n+1} x^n +sum_{n=0}^infty binom{n+1}{1} bar{Phi}^n x^n$.



(d) Prove that $boldsymbol{displaystyle sum_{k=0}^n F_k F_{n-k} =frac{2n F_{n+1} -(n+1)F_n}{5}}$.




I've gotten through (a)-(c) but I don't know how to start (d).






combinatorics generating-functions convolution fibonacci-numbers







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  • 1




    Start by noting that the convolution is the generating function for $F(x)^2$.
    – rogerl
    Nov 27 '18 at 21:58










  • That, or bash it by strong induction on $n$ (as usual for Fibonacci-number identities, going down by $1$ and $2$ in the induction step).
    – darij grinberg
    Nov 28 '18 at 3:30


















2















A Fibonacci convolution. Recall that $$F(x)=sum_{n=0}^infty F_n x^n =frac{x}{1-x-x^2} =frac{1}{sqrt{5}} left(frac{1}{1-Phi x} -frac{1}{1-bar{Phi}x}right).$$
(a) Prove that $displaystyle sum_{n=0}^infty F_{n+1} x^n =frac{1}{1-x-x^2}$.



(b) Prove that $displaystyle sum_{n=0}^infty (2F_{n+1} -F_n)x^n =frac{2-x}{1-x-x^2} =sum_{n=0}^infty (Phi^n +bar{Phi}^n) x^n$.



(c) Prove that $displaystyle 5F(x)^2 =sum_{n=0}^infty binom{n+1}{1} Phi^n x^n -2sum_{n=0}^infty F_{n+1} x^n +sum_{n=0}^infty binom{n+1}{1} bar{Phi}^n x^n$.



(d) Prove that $boldsymbol{displaystyle sum_{k=0}^n F_k F_{n-k} =frac{2n F_{n+1} -(n+1)F_n}{5}}$.




I've gotten through (a)-(c) but I don't know how to start (d).






combinatorics generating-functions convolution fibonacci-numbers







share|cite|improve this question




















  • 1




    Start by noting that the convolution is the generating function for $F(x)^2$.
    – rogerl
    Nov 27 '18 at 21:58










  • That, or bash it by strong induction on $n$ (as usual for Fibonacci-number identities, going down by $1$ and $2$ in the induction step).
    – darij grinberg
    Nov 28 '18 at 3:30
















2












2








2


1






A Fibonacci convolution. Recall that $$F(x)=sum_{n=0}^infty F_n x^n =frac{x}{1-x-x^2} =frac{1}{sqrt{5}} left(frac{1}{1-Phi x} -frac{1}{1-bar{Phi}x}right).$$
(a) Prove that $displaystyle sum_{n=0}^infty F_{n+1} x^n =frac{1}{1-x-x^2}$.



(b) Prove that $displaystyle sum_{n=0}^infty (2F_{n+1} -F_n)x^n =frac{2-x}{1-x-x^2} =sum_{n=0}^infty (Phi^n +bar{Phi}^n) x^n$.



(c) Prove that $displaystyle 5F(x)^2 =sum_{n=0}^infty binom{n+1}{1} Phi^n x^n -2sum_{n=0}^infty F_{n+1} x^n +sum_{n=0}^infty binom{n+1}{1} bar{Phi}^n x^n$.



(d) Prove that $boldsymbol{displaystyle sum_{k=0}^n F_k F_{n-k} =frac{2n F_{n+1} -(n+1)F_n}{5}}$.




I've gotten through (a)-(c) but I don't know how to start (d).






combinatorics generating-functions convolution fibonacci-numbers







share|cite|improve this question
















A Fibonacci convolution. Recall that $$F(x)=sum_{n=0}^infty F_n x^n =frac{x}{1-x-x^2} =frac{1}{sqrt{5}} left(frac{1}{1-Phi x} -frac{1}{1-bar{Phi}x}right).$$
(a) Prove that $displaystyle sum_{n=0}^infty F_{n+1} x^n =frac{1}{1-x-x^2}$.



(b) Prove that $displaystyle sum_{n=0}^infty (2F_{n+1} -F_n)x^n =frac{2-x}{1-x-x^2} =sum_{n=0}^infty (Phi^n +bar{Phi}^n) x^n$.



(c) Prove that $displaystyle 5F(x)^2 =sum_{n=0}^infty binom{n+1}{1} Phi^n x^n -2sum_{n=0}^infty F_{n+1} x^n +sum_{n=0}^infty binom{n+1}{1} bar{Phi}^n x^n$.



(d) Prove that $boldsymbol{displaystyle sum_{k=0}^n F_k F_{n-k} =frac{2n F_{n+1} -(n+1)F_n}{5}}$.




I've gotten through (a)-(c) but I don't know how to start (d).






combinatorics generating-functions convolution fibonacci-numbers




combinatorics generating-functions convolution fibonacci-numbers






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edited Nov 28 '18 at 3:29









Rócherz

2,7762721




2,7762721










asked Nov 27 '18 at 21:50









H.B

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  • 1




    Start by noting that the convolution is the generating function for $F(x)^2$.
    – rogerl
    Nov 27 '18 at 21:58










  • That, or bash it by strong induction on $n$ (as usual for Fibonacci-number identities, going down by $1$ and $2$ in the induction step).
    – darij grinberg
    Nov 28 '18 at 3:30
















  • 1




    Start by noting that the convolution is the generating function for $F(x)^2$.
    – rogerl
    Nov 27 '18 at 21:58










  • That, or bash it by strong induction on $n$ (as usual for Fibonacci-number identities, going down by $1$ and $2$ in the induction step).
    – darij grinberg
    Nov 28 '18 at 3:30










1




1




Start by noting that the convolution is the generating function for $F(x)^2$.
– rogerl
Nov 27 '18 at 21:58




Start by noting that the convolution is the generating function for $F(x)^2$.
– rogerl
Nov 27 '18 at 21:58












That, or bash it by strong induction on $n$ (as usual for Fibonacci-number identities, going down by $1$ and $2$ in the induction step).
– darij grinberg
Nov 28 '18 at 3:30






That, or bash it by strong induction on $n$ (as usual for Fibonacci-number identities, going down by $1$ and $2$ in the induction step).
– darij grinberg
Nov 28 '18 at 3:30












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Since you have already mastered a.) to c.) we can conveniently use the results to prove d.).




We obtain
begin{align*}
color{blue}{5F(x)^2}&=sum_{n=0}^{infty}(n+1)left(Phi^n+bar{Phi}^nright)x^n-2sum_{n=0}^infty F_{n+1}x^ntag{1}\
&=sum_{n=0}^infty(n+1)left(2F_{n+1}-F_nright)x^n-2sum_{n=0}^infty F_{n+1}x^ntag{2}\
&,,color{blue}{=sum_{n=0}^inftyleft(2nF_{n+1}-(n+1)F_nright)x^n}
end{align*}
and d.) follows.




Comment:




  • In (1) we apply c.) by using $binom{n+1}{1}=n+1$ and collecting the first and the last sum.


  • In (2) we use from b.) the identity $2F_{n+1}-F_n=Phi^n+bar{Phi}^n$.







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    Since you have already mastered a.) to c.) we can conveniently use the results to prove d.).




    We obtain
    begin{align*}
    color{blue}{5F(x)^2}&=sum_{n=0}^{infty}(n+1)left(Phi^n+bar{Phi}^nright)x^n-2sum_{n=0}^infty F_{n+1}x^ntag{1}\
    &=sum_{n=0}^infty(n+1)left(2F_{n+1}-F_nright)x^n-2sum_{n=0}^infty F_{n+1}x^ntag{2}\
    &,,color{blue}{=sum_{n=0}^inftyleft(2nF_{n+1}-(n+1)F_nright)x^n}
    end{align*}
    and d.) follows.




    Comment:




    • In (1) we apply c.) by using $binom{n+1}{1}=n+1$ and collecting the first and the last sum.


    • In (2) we use from b.) the identity $2F_{n+1}-F_n=Phi^n+bar{Phi}^n$.







    share|cite|improve this answer


























      2














      Since you have already mastered a.) to c.) we can conveniently use the results to prove d.).




      We obtain
      begin{align*}
      color{blue}{5F(x)^2}&=sum_{n=0}^{infty}(n+1)left(Phi^n+bar{Phi}^nright)x^n-2sum_{n=0}^infty F_{n+1}x^ntag{1}\
      &=sum_{n=0}^infty(n+1)left(2F_{n+1}-F_nright)x^n-2sum_{n=0}^infty F_{n+1}x^ntag{2}\
      &,,color{blue}{=sum_{n=0}^inftyleft(2nF_{n+1}-(n+1)F_nright)x^n}
      end{align*}
      and d.) follows.




      Comment:




      • In (1) we apply c.) by using $binom{n+1}{1}=n+1$ and collecting the first and the last sum.


      • In (2) we use from b.) the identity $2F_{n+1}-F_n=Phi^n+bar{Phi}^n$.







      share|cite|improve this answer
























        2












        2








        2






        Since you have already mastered a.) to c.) we can conveniently use the results to prove d.).




        We obtain
        begin{align*}
        color{blue}{5F(x)^2}&=sum_{n=0}^{infty}(n+1)left(Phi^n+bar{Phi}^nright)x^n-2sum_{n=0}^infty F_{n+1}x^ntag{1}\
        &=sum_{n=0}^infty(n+1)left(2F_{n+1}-F_nright)x^n-2sum_{n=0}^infty F_{n+1}x^ntag{2}\
        &,,color{blue}{=sum_{n=0}^inftyleft(2nF_{n+1}-(n+1)F_nright)x^n}
        end{align*}
        and d.) follows.




        Comment:




        • In (1) we apply c.) by using $binom{n+1}{1}=n+1$ and collecting the first and the last sum.


        • In (2) we use from b.) the identity $2F_{n+1}-F_n=Phi^n+bar{Phi}^n$.







        share|cite|improve this answer












        Since you have already mastered a.) to c.) we can conveniently use the results to prove d.).




        We obtain
        begin{align*}
        color{blue}{5F(x)^2}&=sum_{n=0}^{infty}(n+1)left(Phi^n+bar{Phi}^nright)x^n-2sum_{n=0}^infty F_{n+1}x^ntag{1}\
        &=sum_{n=0}^infty(n+1)left(2F_{n+1}-F_nright)x^n-2sum_{n=0}^infty F_{n+1}x^ntag{2}\
        &,,color{blue}{=sum_{n=0}^inftyleft(2nF_{n+1}-(n+1)F_nright)x^n}
        end{align*}
        and d.) follows.




        Comment:




        • In (1) we apply c.) by using $binom{n+1}{1}=n+1$ and collecting the first and the last sum.


        • In (2) we use from b.) the identity $2F_{n+1}-F_n=Phi^n+bar{Phi}^n$.








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 21:37









        Markus Scheuer

        60.1k455143




        60.1k455143






























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