Integrating minimum of two periodic functions












0












$begingroup$


I am trying to see what is the simplest way to calculate the definite integral of a function of the form



$$f(x) = min(sin(x), sin(2x))$$



I can of course check when $sin(x)<sin(2x)$ and combine the definite integrals for each section. I can imagine how I would do this when $x$ is in units of $2pi$, as I can just multiply the result for the first period. But let's say $x=6.5pi$, would the correct approach be to have the simplified calculation for the first $6pi$ and then for the remaining section separately?






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Visualisation: $f$ is periodic with period $2pi$.
    $endgroup$
    – TheSimpliFire
    Dec 27 '18 at 8:34
















0












$begingroup$


I am trying to see what is the simplest way to calculate the definite integral of a function of the form



$$f(x) = min(sin(x), sin(2x))$$



I can of course check when $sin(x)<sin(2x)$ and combine the definite integrals for each section. I can imagine how I would do this when $x$ is in units of $2pi$, as I can just multiply the result for the first period. But let's say $x=6.5pi$, would the correct approach be to have the simplified calculation for the first $6pi$ and then for the remaining section separately?






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Visualisation: $f$ is periodic with period $2pi$.
    $endgroup$
    – TheSimpliFire
    Dec 27 '18 at 8:34














0












0








0


1



$begingroup$


I am trying to see what is the simplest way to calculate the definite integral of a function of the form



$$f(x) = min(sin(x), sin(2x))$$



I can of course check when $sin(x)<sin(2x)$ and combine the definite integrals for each section. I can imagine how I would do this when $x$ is in units of $2pi$, as I can just multiply the result for the first period. But let's say $x=6.5pi$, would the correct approach be to have the simplified calculation for the first $6pi$ and then for the remaining section separately?






calculus







share|cite|improve this question











$endgroup$




I am trying to see what is the simplest way to calculate the definite integral of a function of the form



$$f(x) = min(sin(x), sin(2x))$$



I can of course check when $sin(x)<sin(2x)$ and combine the definite integrals for each section. I can imagine how I would do this when $x$ is in units of $2pi$, as I can just multiply the result for the first period. But let's say $x=6.5pi$, would the correct approach be to have the simplified calculation for the first $6pi$ and then for the remaining section separately?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 8:39









Robert Z

98.6k1068139




98.6k1068139










asked Dec 27 '18 at 8:28









CaharpukaCaharpuka

1012




1012












  • $begingroup$
    Visualisation: $f$ is periodic with period $2pi$.
    $endgroup$
    – TheSimpliFire
    Dec 27 '18 at 8:34


















  • $begingroup$
    Visualisation: $f$ is periodic with period $2pi$.
    $endgroup$
    – TheSimpliFire
    Dec 27 '18 at 8:34
















$begingroup$
Visualisation: $f$ is periodic with period $2pi$.
$endgroup$
– TheSimpliFire
Dec 27 '18 at 8:34




$begingroup$
Visualisation: $f$ is periodic with period $2pi$.
$endgroup$
– TheSimpliFire
Dec 27 '18 at 8:34










2 Answers
2






active

oldest

votes


















2












$begingroup$

Hint. Use the fact that
$$
begin{align}min(sin(x),sin(2x))&=frac{1}{2}left(sin(x)+sin(2x)-|sin(x)-sin(2x)|right).
end{align}$$
Hence if we integrate over a period,
$$begin{align}
int_0^{2pi}min(sin(x),sin(2x)),dx&=
-frac{1}{2}int_0^{2pi}|sin(x)-sin(2x)|,dx\
&=-int_0^{pi}|sin(x)-sin(2x)|,dx\
&=int_0^{pi/3}(sin(x)-sin(2x)),dx\
&quad-int_{pi/3}^pi(sin(x)-sin(2x)),dx
end{align}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That the absolute value of sin(x) can be integrated more simply? I thought this was just passing the problem on to somewhere else, but you're right this way it simplifies the integration because I can use sgn() instead.
    $endgroup$
    – Caharpuka
    Dec 27 '18 at 8:41



















0












$begingroup$

$$f(x)=sin2x-sin x=2sindfrac x2cosdfrac{3x}2$$



As the $f(x+2pi)=f(x)$



we can safely focus on $0le xle2pi$ where $sindfrac x2not<0$



$f(x)$ will be $>0$



if $cosdfrac{3x}2>0$



$(i)0<dfrac{3x}2<dfracpi2$ or $(ii)dfrac{3pi}2<dfrac{3x}2<2pi$



$(i)0<x<dfracpi3$ or $(ii)pi<x<dfrac{4pi}3$



Consequently, $displaystyleint_0^{2pi}$min$(sin x,sin2x) dx$



$$=int_0^{pi/3}sin x dx+int_{pi/3}^pisin2x dx+int_pi^{4pi/3}sin x dx+int_{4pi/3}^{2pi }sin2x dx=?$$






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%2f3053708%2fintegrating-minimum-of-two-periodic-functions%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Hint. Use the fact that
    $$
    begin{align}min(sin(x),sin(2x))&=frac{1}{2}left(sin(x)+sin(2x)-|sin(x)-sin(2x)|right).
    end{align}$$
    Hence if we integrate over a period,
    $$begin{align}
    int_0^{2pi}min(sin(x),sin(2x)),dx&=
    -frac{1}{2}int_0^{2pi}|sin(x)-sin(2x)|,dx\
    &=-int_0^{pi}|sin(x)-sin(2x)|,dx\
    &=int_0^{pi/3}(sin(x)-sin(2x)),dx\
    &quad-int_{pi/3}^pi(sin(x)-sin(2x)),dx
    end{align}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      That the absolute value of sin(x) can be integrated more simply? I thought this was just passing the problem on to somewhere else, but you're right this way it simplifies the integration because I can use sgn() instead.
      $endgroup$
      – Caharpuka
      Dec 27 '18 at 8:41
















    2












    $begingroup$

    Hint. Use the fact that
    $$
    begin{align}min(sin(x),sin(2x))&=frac{1}{2}left(sin(x)+sin(2x)-|sin(x)-sin(2x)|right).
    end{align}$$
    Hence if we integrate over a period,
    $$begin{align}
    int_0^{2pi}min(sin(x),sin(2x)),dx&=
    -frac{1}{2}int_0^{2pi}|sin(x)-sin(2x)|,dx\
    &=-int_0^{pi}|sin(x)-sin(2x)|,dx\
    &=int_0^{pi/3}(sin(x)-sin(2x)),dx\
    &quad-int_{pi/3}^pi(sin(x)-sin(2x)),dx
    end{align}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      That the absolute value of sin(x) can be integrated more simply? I thought this was just passing the problem on to somewhere else, but you're right this way it simplifies the integration because I can use sgn() instead.
      $endgroup$
      – Caharpuka
      Dec 27 '18 at 8:41














    2












    2








    2





    $begingroup$

    Hint. Use the fact that
    $$
    begin{align}min(sin(x),sin(2x))&=frac{1}{2}left(sin(x)+sin(2x)-|sin(x)-sin(2x)|right).
    end{align}$$
    Hence if we integrate over a period,
    $$begin{align}
    int_0^{2pi}min(sin(x),sin(2x)),dx&=
    -frac{1}{2}int_0^{2pi}|sin(x)-sin(2x)|,dx\
    &=-int_0^{pi}|sin(x)-sin(2x)|,dx\
    &=int_0^{pi/3}(sin(x)-sin(2x)),dx\
    &quad-int_{pi/3}^pi(sin(x)-sin(2x)),dx
    end{align}$$






    share|cite|improve this answer











    $endgroup$



    Hint. Use the fact that
    $$
    begin{align}min(sin(x),sin(2x))&=frac{1}{2}left(sin(x)+sin(2x)-|sin(x)-sin(2x)|right).
    end{align}$$
    Hence if we integrate over a period,
    $$begin{align}
    int_0^{2pi}min(sin(x),sin(2x)),dx&=
    -frac{1}{2}int_0^{2pi}|sin(x)-sin(2x)|,dx\
    &=-int_0^{pi}|sin(x)-sin(2x)|,dx\
    &=int_0^{pi/3}(sin(x)-sin(2x)),dx\
    &quad-int_{pi/3}^pi(sin(x)-sin(2x)),dx
    end{align}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 27 '18 at 8:50

























    answered Dec 27 '18 at 8:35









    Robert ZRobert Z

    98.6k1068139




    98.6k1068139












    • $begingroup$
      That the absolute value of sin(x) can be integrated more simply? I thought this was just passing the problem on to somewhere else, but you're right this way it simplifies the integration because I can use sgn() instead.
      $endgroup$
      – Caharpuka
      Dec 27 '18 at 8:41


















    • $begingroup$
      That the absolute value of sin(x) can be integrated more simply? I thought this was just passing the problem on to somewhere else, but you're right this way it simplifies the integration because I can use sgn() instead.
      $endgroup$
      – Caharpuka
      Dec 27 '18 at 8:41
















    $begingroup$
    That the absolute value of sin(x) can be integrated more simply? I thought this was just passing the problem on to somewhere else, but you're right this way it simplifies the integration because I can use sgn() instead.
    $endgroup$
    – Caharpuka
    Dec 27 '18 at 8:41




    $begingroup$
    That the absolute value of sin(x) can be integrated more simply? I thought this was just passing the problem on to somewhere else, but you're right this way it simplifies the integration because I can use sgn() instead.
    $endgroup$
    – Caharpuka
    Dec 27 '18 at 8:41











    0












    $begingroup$

    $$f(x)=sin2x-sin x=2sindfrac x2cosdfrac{3x}2$$



    As the $f(x+2pi)=f(x)$



    we can safely focus on $0le xle2pi$ where $sindfrac x2not<0$



    $f(x)$ will be $>0$



    if $cosdfrac{3x}2>0$



    $(i)0<dfrac{3x}2<dfracpi2$ or $(ii)dfrac{3pi}2<dfrac{3x}2<2pi$



    $(i)0<x<dfracpi3$ or $(ii)pi<x<dfrac{4pi}3$



    Consequently, $displaystyleint_0^{2pi}$min$(sin x,sin2x) dx$



    $$=int_0^{pi/3}sin x dx+int_{pi/3}^pisin2x dx+int_pi^{4pi/3}sin x dx+int_{4pi/3}^{2pi }sin2x dx=?$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $$f(x)=sin2x-sin x=2sindfrac x2cosdfrac{3x}2$$



      As the $f(x+2pi)=f(x)$



      we can safely focus on $0le xle2pi$ where $sindfrac x2not<0$



      $f(x)$ will be $>0$



      if $cosdfrac{3x}2>0$



      $(i)0<dfrac{3x}2<dfracpi2$ or $(ii)dfrac{3pi}2<dfrac{3x}2<2pi$



      $(i)0<x<dfracpi3$ or $(ii)pi<x<dfrac{4pi}3$



      Consequently, $displaystyleint_0^{2pi}$min$(sin x,sin2x) dx$



      $$=int_0^{pi/3}sin x dx+int_{pi/3}^pisin2x dx+int_pi^{4pi/3}sin x dx+int_{4pi/3}^{2pi }sin2x dx=?$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $$f(x)=sin2x-sin x=2sindfrac x2cosdfrac{3x}2$$



        As the $f(x+2pi)=f(x)$



        we can safely focus on $0le xle2pi$ where $sindfrac x2not<0$



        $f(x)$ will be $>0$



        if $cosdfrac{3x}2>0$



        $(i)0<dfrac{3x}2<dfracpi2$ or $(ii)dfrac{3pi}2<dfrac{3x}2<2pi$



        $(i)0<x<dfracpi3$ or $(ii)pi<x<dfrac{4pi}3$



        Consequently, $displaystyleint_0^{2pi}$min$(sin x,sin2x) dx$



        $$=int_0^{pi/3}sin x dx+int_{pi/3}^pisin2x dx+int_pi^{4pi/3}sin x dx+int_{4pi/3}^{2pi }sin2x dx=?$$






        share|cite|improve this answer









        $endgroup$



        $$f(x)=sin2x-sin x=2sindfrac x2cosdfrac{3x}2$$



        As the $f(x+2pi)=f(x)$



        we can safely focus on $0le xle2pi$ where $sindfrac x2not<0$



        $f(x)$ will be $>0$



        if $cosdfrac{3x}2>0$



        $(i)0<dfrac{3x}2<dfracpi2$ or $(ii)dfrac{3pi}2<dfrac{3x}2<2pi$



        $(i)0<x<dfracpi3$ or $(ii)pi<x<dfrac{4pi}3$



        Consequently, $displaystyleint_0^{2pi}$min$(sin x,sin2x) dx$



        $$=int_0^{pi/3}sin x dx+int_{pi/3}^pisin2x dx+int_pi^{4pi/3}sin x dx+int_{4pi/3}^{2pi }sin2x dx=?$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 12:08









        lab bhattacharjeelab bhattacharjee

        226k15157275




        226k15157275






























            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%2f3053708%2fintegrating-minimum-of-two-periodic-functions%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