Proof of $frac{q_n}{q_{n-1}} = [a_n,a_{n-1},a_{n-2},…,a_2,a_1]$?












1












$begingroup$


Proof of continued fractions axiom.



Let $c=[a_0,a_1,a_2,dots,a_n,dots] = a_0 + cfrac{1}{a_1 + cfrac{1}{a_2 + ddots}}$ be a continued fraction which could be finite or infinite.



By $frac{p_n}{q_n}$ we denote the $n$-th convergent of the continued fraction, i.e.
$$frac{p_n}{q_n}=[a_0,a_1,a_2,dots,a_n].$$



How to prove the formula formula: $$frac{q_n}{q_{n-1}} = [a_n,a_{n-1},a_{n-2},...,a_2,a_1]?$$



This formula is stated in the book Olds C.D. Continued fractions (Math.Assoc.Am., Yale, 1963) as a part of Problem 7 on page 26, together with a related formula
$$frac{p_n}{p_{n-1}} = [a_n,a_{n-1},dots,a_1,a_0].$$



It is also stated as formula (31) in Wolfram Mathworld article on continued fractions.



visual formula - https://s31.postimg.org/6l4ehkd8r/pic.png






fractions continued-fractions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I have tried to edit the post. Together with the OPs edits, I do not think the close reason "unclear what you're asking" applies any longer.
    $endgroup$
    – Martin Sleziak
    Jun 19 '16 at 14:45
















1












$begingroup$


Proof of continued fractions axiom.



Let $c=[a_0,a_1,a_2,dots,a_n,dots] = a_0 + cfrac{1}{a_1 + cfrac{1}{a_2 + ddots}}$ be a continued fraction which could be finite or infinite.



By $frac{p_n}{q_n}$ we denote the $n$-th convergent of the continued fraction, i.e.
$$frac{p_n}{q_n}=[a_0,a_1,a_2,dots,a_n].$$



How to prove the formula formula: $$frac{q_n}{q_{n-1}} = [a_n,a_{n-1},a_{n-2},...,a_2,a_1]?$$



This formula is stated in the book Olds C.D. Continued fractions (Math.Assoc.Am., Yale, 1963) as a part of Problem 7 on page 26, together with a related formula
$$frac{p_n}{p_{n-1}} = [a_n,a_{n-1},dots,a_1,a_0].$$



It is also stated as formula (31) in Wolfram Mathworld article on continued fractions.



visual formula - https://s31.postimg.org/6l4ehkd8r/pic.png






fractions continued-fractions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I have tried to edit the post. Together with the OPs edits, I do not think the close reason "unclear what you're asking" applies any longer.
    $endgroup$
    – Martin Sleziak
    Jun 19 '16 at 14:45














1












1








1





$begingroup$


Proof of continued fractions axiom.



Let $c=[a_0,a_1,a_2,dots,a_n,dots] = a_0 + cfrac{1}{a_1 + cfrac{1}{a_2 + ddots}}$ be a continued fraction which could be finite or infinite.



By $frac{p_n}{q_n}$ we denote the $n$-th convergent of the continued fraction, i.e.
$$frac{p_n}{q_n}=[a_0,a_1,a_2,dots,a_n].$$



How to prove the formula formula: $$frac{q_n}{q_{n-1}} = [a_n,a_{n-1},a_{n-2},...,a_2,a_1]?$$



This formula is stated in the book Olds C.D. Continued fractions (Math.Assoc.Am., Yale, 1963) as a part of Problem 7 on page 26, together with a related formula
$$frac{p_n}{p_{n-1}} = [a_n,a_{n-1},dots,a_1,a_0].$$



It is also stated as formula (31) in Wolfram Mathworld article on continued fractions.



visual formula - https://s31.postimg.org/6l4ehkd8r/pic.png






fractions continued-fractions







share|cite|improve this question











$endgroup$




Proof of continued fractions axiom.



Let $c=[a_0,a_1,a_2,dots,a_n,dots] = a_0 + cfrac{1}{a_1 + cfrac{1}{a_2 + ddots}}$ be a continued fraction which could be finite or infinite.



By $frac{p_n}{q_n}$ we denote the $n$-th convergent of the continued fraction, i.e.
$$frac{p_n}{q_n}=[a_0,a_1,a_2,dots,a_n].$$



How to prove the formula formula: $$frac{q_n}{q_{n-1}} = [a_n,a_{n-1},a_{n-2},...,a_2,a_1]?$$



This formula is stated in the book Olds C.D. Continued fractions (Math.Assoc.Am., Yale, 1963) as a part of Problem 7 on page 26, together with a related formula
$$frac{p_n}{p_{n-1}} = [a_n,a_{n-1},dots,a_1,a_0].$$



It is also stated as formula (31) in Wolfram Mathworld article on continued fractions.



visual formula - https://s31.postimg.org/6l4ehkd8r/pic.png






fractions continued-fractions




fractions continued-fractions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 19 '16 at 14:39









Martin Sleziak

44.8k9118272




44.8k9118272










asked Jun 18 '16 at 16:22









qwerryqwerry

314




314








  • 1




    $begingroup$
    I have tried to edit the post. Together with the OPs edits, I do not think the close reason "unclear what you're asking" applies any longer.
    $endgroup$
    – Martin Sleziak
    Jun 19 '16 at 14:45














  • 1




    $begingroup$
    I have tried to edit the post. Together with the OPs edits, I do not think the close reason "unclear what you're asking" applies any longer.
    $endgroup$
    – Martin Sleziak
    Jun 19 '16 at 14:45








1




1




$begingroup$
I have tried to edit the post. Together with the OPs edits, I do not think the close reason "unclear what you're asking" applies any longer.
$endgroup$
– Martin Sleziak
Jun 19 '16 at 14:45




$begingroup$
I have tried to edit the post. Together with the OPs edits, I do not think the close reason "unclear what you're asking" applies any longer.
$endgroup$
– Martin Sleziak
Jun 19 '16 at 14:45










1 Answer
1






active

oldest

votes


















2












$begingroup$

Ok, I guess it is by far simple answer, sorry for posting silly questions :(.



$frac{q_n}{q_{n-1}}$ =
$frac{a_n*q_{n-1}+q_{n-2}}{q_{n-1}}$ =
$frac{a_n*q_{n-1}}{q_{n-1}}$ +
$frac{q_{n-2}}{q_{n-1}}$ =
$a_n$ +
$frac{q_{n-2}}{q_{n-1}}$ =
$a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



So $frac{q_n}{q_{n-1}}$ = $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



and $frac{q_{n-1}}{q_{n-2}}$ = $a_{n-1}$ + $frac{1}{frac{q_{n-2}}{q_{n-3}}}$



and $frac{q_{n-2}}{q_{n-3}}$ = $a_{n-2}$ + $frac{1}{frac{q_{n-3}}{q_{n-4}}}$



$vdots$



So by induction you will have:



$frac{q_n}{q_{n-1}}$ =
$a_n$ + $frac{1}
{a_{n-1} + frac{1}
{a_{n-2} + frac{1}
{frac{q_{n-3}}{
ddots}}}}$ = $[a_{n},a_{n-1},a_{n-2},dots,a_{1}]$






share|cite|improve this answer











$endgroup$













    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',
    autoActivateHeartbeat: false,
    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%2f1831001%2fproof-of-fracq-nq-n-1-a-n-a-n-1-a-n-2-a-2-a-1%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









    2












    $begingroup$

    Ok, I guess it is by far simple answer, sorry for posting silly questions :(.



    $frac{q_n}{q_{n-1}}$ =
    $frac{a_n*q_{n-1}+q_{n-2}}{q_{n-1}}$ =
    $frac{a_n*q_{n-1}}{q_{n-1}}$ +
    $frac{q_{n-2}}{q_{n-1}}$ =
    $a_n$ +
    $frac{q_{n-2}}{q_{n-1}}$ =
    $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



    So $frac{q_n}{q_{n-1}}$ = $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



    and $frac{q_{n-1}}{q_{n-2}}$ = $a_{n-1}$ + $frac{1}{frac{q_{n-2}}{q_{n-3}}}$



    and $frac{q_{n-2}}{q_{n-3}}$ = $a_{n-2}$ + $frac{1}{frac{q_{n-3}}{q_{n-4}}}$



    $vdots$



    So by induction you will have:



    $frac{q_n}{q_{n-1}}$ =
    $a_n$ + $frac{1}
    {a_{n-1} + frac{1}
    {a_{n-2} + frac{1}
    {frac{q_{n-3}}{
    ddots}}}}$     = $[a_{n},a_{n-1},a_{n-2},dots,a_{1}]$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Ok, I guess it is by far simple answer, sorry for posting silly questions :(.



      $frac{q_n}{q_{n-1}}$ =
      $frac{a_n*q_{n-1}+q_{n-2}}{q_{n-1}}$ =
      $frac{a_n*q_{n-1}}{q_{n-1}}$ +
      $frac{q_{n-2}}{q_{n-1}}$ =
      $a_n$ +
      $frac{q_{n-2}}{q_{n-1}}$ =
      $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



      So $frac{q_n}{q_{n-1}}$ = $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



      and $frac{q_{n-1}}{q_{n-2}}$ = $a_{n-1}$ + $frac{1}{frac{q_{n-2}}{q_{n-3}}}$



      and $frac{q_{n-2}}{q_{n-3}}$ = $a_{n-2}$ + $frac{1}{frac{q_{n-3}}{q_{n-4}}}$



      $vdots$



      So by induction you will have:



      $frac{q_n}{q_{n-1}}$ =
      $a_n$ + $frac{1}
      {a_{n-1} + frac{1}
      {a_{n-2} + frac{1}
      {frac{q_{n-3}}{
      ddots}}}}$       = $[a_{n},a_{n-1},a_{n-2},dots,a_{1}]$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Ok, I guess it is by far simple answer, sorry for posting silly questions :(.



        $frac{q_n}{q_{n-1}}$ =
        $frac{a_n*q_{n-1}+q_{n-2}}{q_{n-1}}$ =
        $frac{a_n*q_{n-1}}{q_{n-1}}$ +
        $frac{q_{n-2}}{q_{n-1}}$ =
        $a_n$ +
        $frac{q_{n-2}}{q_{n-1}}$ =
        $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



        So $frac{q_n}{q_{n-1}}$ = $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



        and $frac{q_{n-1}}{q_{n-2}}$ = $a_{n-1}$ + $frac{1}{frac{q_{n-2}}{q_{n-3}}}$



        and $frac{q_{n-2}}{q_{n-3}}$ = $a_{n-2}$ + $frac{1}{frac{q_{n-3}}{q_{n-4}}}$



        $vdots$



        So by induction you will have:



        $frac{q_n}{q_{n-1}}$ =
        $a_n$ + $frac{1}
        {a_{n-1} + frac{1}
        {a_{n-2} + frac{1}
        {frac{q_{n-3}}{
        ddots}}}}$         = $[a_{n},a_{n-1},a_{n-2},dots,a_{1}]$






        share|cite|improve this answer











        $endgroup$



        Ok, I guess it is by far simple answer, sorry for posting silly questions :(.



        $frac{q_n}{q_{n-1}}$ =
        $frac{a_n*q_{n-1}+q_{n-2}}{q_{n-1}}$ =
        $frac{a_n*q_{n-1}}{q_{n-1}}$ +
        $frac{q_{n-2}}{q_{n-1}}$ =
        $a_n$ +
        $frac{q_{n-2}}{q_{n-1}}$ =
        $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



        So $frac{q_n}{q_{n-1}}$ = $a_n$ + $frac{1}{frac{q_{n-1}}{q_{n-2}}}$



        and $frac{q_{n-1}}{q_{n-2}}$ = $a_{n-1}$ + $frac{1}{frac{q_{n-2}}{q_{n-3}}}$



        and $frac{q_{n-2}}{q_{n-3}}$ = $a_{n-2}$ + $frac{1}{frac{q_{n-3}}{q_{n-4}}}$



        $vdots$



        So by induction you will have:



        $frac{q_n}{q_{n-1}}$ =
        $a_n$ + $frac{1}
        {a_{n-1} + frac{1}
        {a_{n-2} + frac{1}
        {frac{q_{n-3}}{
        ddots}}}}$         = $[a_{n},a_{n-1},a_{n-2},dots,a_{1}]$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 11 '18 at 14:28









        Algebear

        616319




        616319










        answered Jun 19 '16 at 21:17









        qwerryqwerry

        314




        314






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1831001%2fproof-of-fracq-nq-n-1-a-n-a-n-1-a-n-2-a-2-a-1%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

            Ellipse (mathématiques)

            Image
            Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
            Read more

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
            Read more