Find the sup of an integral succession











up vote
0
down vote

favorite












Let
$$
a_n=int^pi_0e^{-nx}sin(n^2x)dx
$$



find $sup{a_n:ninmathbb N}$





I found myself the solution.



Integrating twice by parts we have
$$
a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
$$

then
$$
a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
$$










share|cite|improve this question




















  • 1




    Did you compute $a_n$ ? If you did, what I hope and wish, include your result in the post.
    – Claude Leibovici
    Nov 22 at 11:23

















up vote
0
down vote

favorite












Let
$$
a_n=int^pi_0e^{-nx}sin(n^2x)dx
$$



find $sup{a_n:ninmathbb N}$





I found myself the solution.



Integrating twice by parts we have
$$
a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
$$

then
$$
a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
$$










share|cite|improve this question




















  • 1




    Did you compute $a_n$ ? If you did, what I hope and wish, include your result in the post.
    – Claude Leibovici
    Nov 22 at 11:23















up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let
$$
a_n=int^pi_0e^{-nx}sin(n^2x)dx
$$



find $sup{a_n:ninmathbb N}$





I found myself the solution.



Integrating twice by parts we have
$$
a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
$$

then
$$
a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
$$










share|cite|improve this question















Let
$$
a_n=int^pi_0e^{-nx}sin(n^2x)dx
$$



find $sup{a_n:ninmathbb N}$





I found myself the solution.



Integrating twice by parts we have
$$
a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
$$

then
$$
a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
$$







calculus integration supremum-and-infimum






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 3:13









user302797

19.7k92252




19.7k92252










asked Nov 22 at 11:07









P De Donato

3567




3567








  • 1




    Did you compute $a_n$ ? If you did, what I hope and wish, include your result in the post.
    – Claude Leibovici
    Nov 22 at 11:23
















  • 1




    Did you compute $a_n$ ? If you did, what I hope and wish, include your result in the post.
    – Claude Leibovici
    Nov 22 at 11:23










1




1




Did you compute $a_n$ ? If you did, what I hope and wish, include your result in the post.
– Claude Leibovici
Nov 22 at 11:23






Did you compute $a_n$ ? If you did, what I hope and wish, include your result in the post.
– Claude Leibovici
Nov 22 at 11:23












1 Answer
1






active

oldest

votes

















up vote
0
down vote



accepted










I found myself the solution.



Integrating twice by parts we have
$$
a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
$$

then
$$
a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
$$






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%2f3009008%2ffind-the-sup-of-an-integral-succession%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
    0
    down vote



    accepted










    I found myself the solution.



    Integrating twice by parts we have
    $$
    a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
    a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
    a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
    a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
    $$

    then
    $$
    a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
    $$






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      I found myself the solution.



      Integrating twice by parts we have
      $$
      a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
      a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
      a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
      a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
      $$

      then
      $$
      a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
      $$






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        I found myself the solution.



        Integrating twice by parts we have
        $$
        a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
        a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
        a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
        a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
        $$

        then
        $$
        a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
        $$






        share|cite|improve this answer












        I found myself the solution.



        Integrating twice by parts we have
        $$
        a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
        a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
        a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
        a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
        $$

        then
        $$
        a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 21:29









        P De Donato

        3567




        3567






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3009008%2ffind-the-sup-of-an-integral-succession%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)

            Quarter-circle Tiles

            Mont Emei