Why is $F_n = r^n$ a solution of the difference equation if $r$ satisfies $r^2-r-1=0$?












1












$begingroup$


The following is from p.4 of https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch3.pdf




The terms in the Fibonacci sequence are uniquely determined by the linear
difference equation



$$F_n − F_{n−1} − F_{n−2} = 0, n ≥ 3$$



with the initial conditions $F_1 = 1, F_2 = 1$.



We see that $F_n = r^n$ is a solution of the difference equation if $r$ satisfies
$r^2 − r − 1 = 0$ which gives $r = phi$ or −$dfrac{1}{phi}$ where $phi = dfrac{1 + sqrt{5}}{2}approx 1.61803$.




I'm unable to see why $F_n = r^n$ is a solution of the difference equation if $r$ satisfies $r^2-r-1=0$.






recurrence-relations fibonacci-numbers golden-ratio







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$endgroup$








  • 3




    $begingroup$
    Do you mean $F_n$ not $F^n$? If so, plugging in $F_n=r^n$, we get $r^n-r^{n-1}-r^{n-2}=r^{n-2}(r^2-r-1)=0$ and clearly $rne0$.
    $endgroup$
    – TheSimpliFire
    Dec 28 '18 at 19:12












  • $begingroup$
    See the answer at this duplicate.
    $endgroup$
    – Dietrich Burde
    Dec 28 '18 at 19:15










  • $begingroup$
    @TheSimpliFire: Yes, I meant $F_n$.
    $endgroup$
    – K.M
    Dec 28 '18 at 19:28
















1












$begingroup$


The following is from p.4 of https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch3.pdf




The terms in the Fibonacci sequence are uniquely determined by the linear
difference equation



$$F_n − F_{n−1} − F_{n−2} = 0, n ≥ 3$$



with the initial conditions $F_1 = 1, F_2 = 1$.



We see that $F_n = r^n$ is a solution of the difference equation if $r$ satisfies
$r^2 − r − 1 = 0$ which gives $r = phi$ or −$dfrac{1}{phi}$ where $phi = dfrac{1 + sqrt{5}}{2}approx 1.61803$.




I'm unable to see why $F_n = r^n$ is a solution of the difference equation if $r$ satisfies $r^2-r-1=0$.






recurrence-relations fibonacci-numbers golden-ratio







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Do you mean $F_n$ not $F^n$? If so, plugging in $F_n=r^n$, we get $r^n-r^{n-1}-r^{n-2}=r^{n-2}(r^2-r-1)=0$ and clearly $rne0$.
    $endgroup$
    – TheSimpliFire
    Dec 28 '18 at 19:12












  • $begingroup$
    See the answer at this duplicate.
    $endgroup$
    – Dietrich Burde
    Dec 28 '18 at 19:15










  • $begingroup$
    @TheSimpliFire: Yes, I meant $F_n$.
    $endgroup$
    – K.M
    Dec 28 '18 at 19:28














1












1








1


0



$begingroup$


The following is from p.4 of https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch3.pdf




The terms in the Fibonacci sequence are uniquely determined by the linear
difference equation



$$F_n − F_{n−1} − F_{n−2} = 0, n ≥ 3$$



with the initial conditions $F_1 = 1, F_2 = 1$.



We see that $F_n = r^n$ is a solution of the difference equation if $r$ satisfies
$r^2 − r − 1 = 0$ which gives $r = phi$ or −$dfrac{1}{phi}$ where $phi = dfrac{1 + sqrt{5}}{2}approx 1.61803$.




I'm unable to see why $F_n = r^n$ is a solution of the difference equation if $r$ satisfies $r^2-r-1=0$.






recurrence-relations fibonacci-numbers golden-ratio







share|cite|improve this question











$endgroup$




The following is from p.4 of https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch3.pdf




The terms in the Fibonacci sequence are uniquely determined by the linear
difference equation



$$F_n − F_{n−1} − F_{n−2} = 0, n ≥ 3$$



with the initial conditions $F_1 = 1, F_2 = 1$.



We see that $F_n = r^n$ is a solution of the difference equation if $r$ satisfies
$r^2 − r − 1 = 0$ which gives $r = phi$ or −$dfrac{1}{phi}$ where $phi = dfrac{1 + sqrt{5}}{2}approx 1.61803$.




I'm unable to see why $F_n = r^n$ is a solution of the difference equation if $r$ satisfies $r^2-r-1=0$.






recurrence-relations fibonacci-numbers golden-ratio




recurrence-relations fibonacci-numbers golden-ratio






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 19:31









TheSimpliFire

12.5k62460




12.5k62460










asked Dec 28 '18 at 19:10









K.MK.M

700412




700412








  • 3




    $begingroup$
    Do you mean $F_n$ not $F^n$? If so, plugging in $F_n=r^n$, we get $r^n-r^{n-1}-r^{n-2}=r^{n-2}(r^2-r-1)=0$ and clearly $rne0$.
    $endgroup$
    – TheSimpliFire
    Dec 28 '18 at 19:12












  • $begingroup$
    See the answer at this duplicate.
    $endgroup$
    – Dietrich Burde
    Dec 28 '18 at 19:15










  • $begingroup$
    @TheSimpliFire: Yes, I meant $F_n$.
    $endgroup$
    – K.M
    Dec 28 '18 at 19:28














  • 3




    $begingroup$
    Do you mean $F_n$ not $F^n$? If so, plugging in $F_n=r^n$, we get $r^n-r^{n-1}-r^{n-2}=r^{n-2}(r^2-r-1)=0$ and clearly $rne0$.
    $endgroup$
    – TheSimpliFire
    Dec 28 '18 at 19:12












  • $begingroup$
    See the answer at this duplicate.
    $endgroup$
    – Dietrich Burde
    Dec 28 '18 at 19:15










  • $begingroup$
    @TheSimpliFire: Yes, I meant $F_n$.
    $endgroup$
    – K.M
    Dec 28 '18 at 19:28








3




3




$begingroup$
Do you mean $F_n$ not $F^n$? If so, plugging in $F_n=r^n$, we get $r^n-r^{n-1}-r^{n-2}=r^{n-2}(r^2-r-1)=0$ and clearly $rne0$.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 19:12






$begingroup$
Do you mean $F_n$ not $F^n$? If so, plugging in $F_n=r^n$, we get $r^n-r^{n-1}-r^{n-2}=r^{n-2}(r^2-r-1)=0$ and clearly $rne0$.
$endgroup$
– TheSimpliFire
Dec 28 '18 at 19:12














$begingroup$
See the answer at this duplicate.
$endgroup$
– Dietrich Burde
Dec 28 '18 at 19:15




$begingroup$
See the answer at this duplicate.
$endgroup$
– Dietrich Burde
Dec 28 '18 at 19:15












$begingroup$
@TheSimpliFire: Yes, I meant $F_n$.
$endgroup$
– K.M
Dec 28 '18 at 19:28




$begingroup$
@TheSimpliFire: Yes, I meant $F_n$.
$endgroup$
– K.M
Dec 28 '18 at 19:28










2 Answers
2






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oldest

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1












$begingroup$

Hint: Multiply $r^2 − r − 1 = 0$ by $r^{n-2}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Multiplying $r^2 -r-1=0$ by $r^{n-2}$, I get that $r^{n}-r^{n-1}-r^{n-2}=0$, but I'm not sure how I would get that $F_{n}=r^{n}$ without first assuming that $F_n=r^{n}$ and then plugging this into $r^{n}-r^{n-1}-r^{n-2}=0$
    $endgroup$
    – K.M
    Dec 29 '18 at 21:14



















0












$begingroup$

Usually, the complete solution to a difference equation with two distinct roots in the characteristic equation is in the form $$A{r_1}^n + B{r_2}^n$$.



In your case, the distinct roots for $r_1$ and $r_2$ are $phi = dfrac {1+sqrt{5}}{2}$ and $-dfrac {1}{phi} = dfrac {1-sqrt{5}}{2}$. Thus the complete solution is $$F_n = dfrac {1}{sqrt {5}} left (phi^n - dfrac {1}{phi}^n right)$$






share|cite|improve this answer











$endgroup$













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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









    1












    $begingroup$

    Hint: Multiply $r^2 − r − 1 = 0$ by $r^{n-2}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Multiplying $r^2 -r-1=0$ by $r^{n-2}$, I get that $r^{n}-r^{n-1}-r^{n-2}=0$, but I'm not sure how I would get that $F_{n}=r^{n}$ without first assuming that $F_n=r^{n}$ and then plugging this into $r^{n}-r^{n-1}-r^{n-2}=0$
      $endgroup$
      – K.M
      Dec 29 '18 at 21:14
















    1












    $begingroup$

    Hint: Multiply $r^2 − r − 1 = 0$ by $r^{n-2}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Multiplying $r^2 -r-1=0$ by $r^{n-2}$, I get that $r^{n}-r^{n-1}-r^{n-2}=0$, but I'm not sure how I would get that $F_{n}=r^{n}$ without first assuming that $F_n=r^{n}$ and then plugging this into $r^{n}-r^{n-1}-r^{n-2}=0$
      $endgroup$
      – K.M
      Dec 29 '18 at 21:14














    1












    1








    1





    $begingroup$

    Hint: Multiply $r^2 − r − 1 = 0$ by $r^{n-2}$.






    share|cite|improve this answer









    $endgroup$



    Hint: Multiply $r^2 − r − 1 = 0$ by $r^{n-2}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 28 '18 at 19:13









    lhflhf

    165k10171396




    165k10171396












    • $begingroup$
      Multiplying $r^2 -r-1=0$ by $r^{n-2}$, I get that $r^{n}-r^{n-1}-r^{n-2}=0$, but I'm not sure how I would get that $F_{n}=r^{n}$ without first assuming that $F_n=r^{n}$ and then plugging this into $r^{n}-r^{n-1}-r^{n-2}=0$
      $endgroup$
      – K.M
      Dec 29 '18 at 21:14


















    • $begingroup$
      Multiplying $r^2 -r-1=0$ by $r^{n-2}$, I get that $r^{n}-r^{n-1}-r^{n-2}=0$, but I'm not sure how I would get that $F_{n}=r^{n}$ without first assuming that $F_n=r^{n}$ and then plugging this into $r^{n}-r^{n-1}-r^{n-2}=0$
      $endgroup$
      – K.M
      Dec 29 '18 at 21:14
















    $begingroup$
    Multiplying $r^2 -r-1=0$ by $r^{n-2}$, I get that $r^{n}-r^{n-1}-r^{n-2}=0$, but I'm not sure how I would get that $F_{n}=r^{n}$ without first assuming that $F_n=r^{n}$ and then plugging this into $r^{n}-r^{n-1}-r^{n-2}=0$
    $endgroup$
    – K.M
    Dec 29 '18 at 21:14




    $begingroup$
    Multiplying $r^2 -r-1=0$ by $r^{n-2}$, I get that $r^{n}-r^{n-1}-r^{n-2}=0$, but I'm not sure how I would get that $F_{n}=r^{n}$ without first assuming that $F_n=r^{n}$ and then plugging this into $r^{n}-r^{n-1}-r^{n-2}=0$
    $endgroup$
    – K.M
    Dec 29 '18 at 21:14











    0












    $begingroup$

    Usually, the complete solution to a difference equation with two distinct roots in the characteristic equation is in the form $$A{r_1}^n + B{r_2}^n$$.



    In your case, the distinct roots for $r_1$ and $r_2$ are $phi = dfrac {1+sqrt{5}}{2}$ and $-dfrac {1}{phi} = dfrac {1-sqrt{5}}{2}$. Thus the complete solution is $$F_n = dfrac {1}{sqrt {5}} left (phi^n - dfrac {1}{phi}^n right)$$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Usually, the complete solution to a difference equation with two distinct roots in the characteristic equation is in the form $$A{r_1}^n + B{r_2}^n$$.



      In your case, the distinct roots for $r_1$ and $r_2$ are $phi = dfrac {1+sqrt{5}}{2}$ and $-dfrac {1}{phi} = dfrac {1-sqrt{5}}{2}$. Thus the complete solution is $$F_n = dfrac {1}{sqrt {5}} left (phi^n - dfrac {1}{phi}^n right)$$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Usually, the complete solution to a difference equation with two distinct roots in the characteristic equation is in the form $$A{r_1}^n + B{r_2}^n$$.



        In your case, the distinct roots for $r_1$ and $r_2$ are $phi = dfrac {1+sqrt{5}}{2}$ and $-dfrac {1}{phi} = dfrac {1-sqrt{5}}{2}$. Thus the complete solution is $$F_n = dfrac {1}{sqrt {5}} left (phi^n - dfrac {1}{phi}^n right)$$






        share|cite|improve this answer











        $endgroup$



        Usually, the complete solution to a difference equation with two distinct roots in the characteristic equation is in the form $$A{r_1}^n + B{r_2}^n$$.



        In your case, the distinct roots for $r_1$ and $r_2$ are $phi = dfrac {1+sqrt{5}}{2}$ and $-dfrac {1}{phi} = dfrac {1-sqrt{5}}{2}$. Thus the complete solution is $$F_n = dfrac {1}{sqrt {5}} left (phi^n - dfrac {1}{phi}^n right)$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 28 '18 at 19:48

























        answered Dec 28 '18 at 19:42









        bjcolby15bjcolby15

        1,47411016




        1,47411016






























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