Computing the solution of a recurrence relation











up vote
0
down vote

favorite












Let $y_k=dfrac{16}{3}y_{k-1}-dfrac{5}{3}y_{k-2}$



with $y_0=1$ and $y_1=dfrac{1}{2}$ for $k geq 2, k in mathbb{N}$.



To find the solution $lbrace y_k rbrace_{k in mathbb{N_0}}$ I used:



Firstly, I computed it for $k=2$:



$y_2=dfrac{16}{3}dfrac{1}{2}-dfrac{5}{3}1=1$



What has to be done next? Induction isn't helpful here.



I don't see how can it be found for $lbrace y_k rbrace_{k in mathbb{N_0}}$.






recurrence-relations







share|cite|improve this question




















  • 3




    There's a whole method for solving linear recurrences. Try opening your notebook
    – Jakobian
    15 hours ago















up vote
0
down vote

favorite












Let $y_k=dfrac{16}{3}y_{k-1}-dfrac{5}{3}y_{k-2}$



with $y_0=1$ and $y_1=dfrac{1}{2}$ for $k geq 2, k in mathbb{N}$.



To find the solution $lbrace y_k rbrace_{k in mathbb{N_0}}$ I used:



Firstly, I computed it for $k=2$:



$y_2=dfrac{16}{3}dfrac{1}{2}-dfrac{5}{3}1=1$



What has to be done next? Induction isn't helpful here.



I don't see how can it be found for $lbrace y_k rbrace_{k in mathbb{N_0}}$.






recurrence-relations







share|cite|improve this question




















  • 3




    There's a whole method for solving linear recurrences. Try opening your notebook
    – Jakobian
    15 hours ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $y_k=dfrac{16}{3}y_{k-1}-dfrac{5}{3}y_{k-2}$



with $y_0=1$ and $y_1=dfrac{1}{2}$ for $k geq 2, k in mathbb{N}$.



To find the solution $lbrace y_k rbrace_{k in mathbb{N_0}}$ I used:



Firstly, I computed it for $k=2$:



$y_2=dfrac{16}{3}dfrac{1}{2}-dfrac{5}{3}1=1$



What has to be done next? Induction isn't helpful here.



I don't see how can it be found for $lbrace y_k rbrace_{k in mathbb{N_0}}$.






recurrence-relations







share|cite|improve this question















Let $y_k=dfrac{16}{3}y_{k-1}-dfrac{5}{3}y_{k-2}$



with $y_0=1$ and $y_1=dfrac{1}{2}$ for $k geq 2, k in mathbb{N}$.



To find the solution $lbrace y_k rbrace_{k in mathbb{N_0}}$ I used:



Firstly, I computed it for $k=2$:



$y_2=dfrac{16}{3}dfrac{1}{2}-dfrac{5}{3}1=1$



What has to be done next? Induction isn't helpful here.



I don't see how can it be found for $lbrace y_k rbrace_{k in mathbb{N_0}}$.






recurrence-relations




recurrence-relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 14 hours ago









Yadati Kiran

348110




348110










asked 15 hours ago









Olsgur

323




323








  • 3




    There's a whole method for solving linear recurrences. Try opening your notebook
    – Jakobian
    15 hours ago














  • 3




    There's a whole method for solving linear recurrences. Try opening your notebook
    – Jakobian
    15 hours ago








3




3




There's a whole method for solving linear recurrences. Try opening your notebook
– Jakobian
15 hours ago




There's a whole method for solving linear recurrences. Try opening your notebook
– Jakobian
15 hours ago










1 Answer
1






active

oldest

votes

















up vote
-1
down vote













$$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



$$[1,1/2,1,9/2,67/3,2009/18,15067/27,452009/162,3390067/243,101702009/1458,762765067/2187]$$



Solution details:



Char. equation is
$${{t}^{2}}-frac{16 t}{3}+frac{5}{3}=0$$
$$t_1=5,quad t_2=frac13$$
Then solution is
$$y_k=A,5^k+B,left(frac13right)^k$$
From initial conditions
$$A,5^0+B,left(frac13right)^0=1,\
A,5^1+B,left(frac13right)^1=frac12$$
$Rightarrow$
$$A=frac{1}{28},quad B=frac{27}{28}$$
$Rightarrow$
$$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



see for example https://www.youtube.com/watch?v=aHw7hAAjbD0






share|cite|improve this answer























    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














     

    draft saved


    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004137%2fcomputing-the-solution-of-a-recurrence-relation%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    -1
    down vote













    $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



    $$[1,1/2,1,9/2,67/3,2009/18,15067/27,452009/162,3390067/243,101702009/1458,762765067/2187]$$



    Solution details:



    Char. equation is
    $${{t}^{2}}-frac{16 t}{3}+frac{5}{3}=0$$
    $$t_1=5,quad t_2=frac13$$
    Then solution is
    $$y_k=A,5^k+B,left(frac13right)^k$$
    From initial conditions
    $$A,5^0+B,left(frac13right)^0=1,\
    A,5^1+B,left(frac13right)^1=frac12$$
    $Rightarrow$
    $$A=frac{1}{28},quad B=frac{27}{28}$$
    $Rightarrow$
    $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



    see for example https://www.youtube.com/watch?v=aHw7hAAjbD0






    share|cite|improve this answer



























      up vote
      -1
      down vote













      $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



      $$[1,1/2,1,9/2,67/3,2009/18,15067/27,452009/162,3390067/243,101702009/1458,762765067/2187]$$



      Solution details:



      Char. equation is
      $${{t}^{2}}-frac{16 t}{3}+frac{5}{3}=0$$
      $$t_1=5,quad t_2=frac13$$
      Then solution is
      $$y_k=A,5^k+B,left(frac13right)^k$$
      From initial conditions
      $$A,5^0+B,left(frac13right)^0=1,\
      A,5^1+B,left(frac13right)^1=frac12$$
      $Rightarrow$
      $$A=frac{1}{28},quad B=frac{27}{28}$$
      $Rightarrow$
      $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



      see for example https://www.youtube.com/watch?v=aHw7hAAjbD0






      share|cite|improve this answer

























        up vote
        -1
        down vote










        up vote
        -1
        down vote









        $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



        $$[1,1/2,1,9/2,67/3,2009/18,15067/27,452009/162,3390067/243,101702009/1458,762765067/2187]$$



        Solution details:



        Char. equation is
        $${{t}^{2}}-frac{16 t}{3}+frac{5}{3}=0$$
        $$t_1=5,quad t_2=frac13$$
        Then solution is
        $$y_k=A,5^k+B,left(frac13right)^k$$
        From initial conditions
        $$A,5^0+B,left(frac13right)^0=1,\
        A,5^1+B,left(frac13right)^1=frac12$$
        $Rightarrow$
        $$A=frac{1}{28},quad B=frac{27}{28}$$
        $Rightarrow$
        $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



        see for example https://www.youtube.com/watch?v=aHw7hAAjbD0






        share|cite|improve this answer














        $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



        $$[1,1/2,1,9/2,67/3,2009/18,15067/27,452009/162,3390067/243,101702009/1458,762765067/2187]$$



        Solution details:



        Char. equation is
        $${{t}^{2}}-frac{16 t}{3}+frac{5}{3}=0$$
        $$t_1=5,quad t_2=frac13$$
        Then solution is
        $$y_k=A,5^k+B,left(frac13right)^k$$
        From initial conditions
        $$A,5^0+B,left(frac13right)^0=1,\
        A,5^1+B,left(frac13right)^1=frac12$$
        $Rightarrow$
        $$A=frac{1}{28},quad B=frac{27}{28}$$
        $Rightarrow$
        $$y_k=frac{{{5}^{k}}}{28}+frac{{{3}^{3-k}}}{28}$$



        see for example https://www.youtube.com/watch?v=aHw7hAAjbD0







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 4 hours ago

























        answered 15 hours ago









        Aleksas Domarkas

        7085




        7085






























             

            draft saved


            draft discarded



















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004137%2fcomputing-the-solution-of-a-recurrence-relation%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

            Image
            28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
            Read more

            Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

            Image
            up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
            Read more

            A Topological Invariant for $pi_3(U(n))$

            Image
            3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
            Read more