Prove that a vector x from L2[a,b] is in the subspace generated by f1,f2,f3












-1












$begingroup$


Let $f_1(t)=e^t$, $f_2(t)=e^{it}$ and $f_3(t)=e^{-it}$. Prove that a vector $x$ from L2[a,b] is in the subspace generated by $f_1,f_2,f_3$ if and only if $x$ verifies the differential equation $x'''-x''+x'-x=0$.






linear-algebra ordinary-differential-equations vector-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Welcome to Math.SE! As written, this question isn't likely to get a lot of attention. But, if you update it with information about what you've tried and where you're getting stuck, I'm sure somebody will be happy to give you some pointers.
    $endgroup$
    – Nick Peterson
    Jan 4 at 17:03
















-1












$begingroup$


Let $f_1(t)=e^t$, $f_2(t)=e^{it}$ and $f_3(t)=e^{-it}$. Prove that a vector $x$ from L2[a,b] is in the subspace generated by $f_1,f_2,f_3$ if and only if $x$ verifies the differential equation $x'''-x''+x'-x=0$.






linear-algebra ordinary-differential-equations vector-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Welcome to Math.SE! As written, this question isn't likely to get a lot of attention. But, if you update it with information about what you've tried and where you're getting stuck, I'm sure somebody will be happy to give you some pointers.
    $endgroup$
    – Nick Peterson
    Jan 4 at 17:03














-1












-1








-1


1



$begingroup$


Let $f_1(t)=e^t$, $f_2(t)=e^{it}$ and $f_3(t)=e^{-it}$. Prove that a vector $x$ from L2[a,b] is in the subspace generated by $f_1,f_2,f_3$ if and only if $x$ verifies the differential equation $x'''-x''+x'-x=0$.






linear-algebra ordinary-differential-equations vector-spaces







share|cite|improve this question











$endgroup$




Let $f_1(t)=e^t$, $f_2(t)=e^{it}$ and $f_3(t)=e^{-it}$. Prove that a vector $x$ from L2[a,b] is in the subspace generated by $f_1,f_2,f_3$ if and only if $x$ verifies the differential equation $x'''-x''+x'-x=0$.






linear-algebra ordinary-differential-equations vector-spaces




linear-algebra ordinary-differential-equations vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 16:51









Davide Giraudo

127k17154268




127k17154268










asked Jan 4 at 16:48









Raluca lokRaluca lok

21




21








  • 2




    $begingroup$
    Welcome to Math.SE! As written, this question isn't likely to get a lot of attention. But, if you update it with information about what you've tried and where you're getting stuck, I'm sure somebody will be happy to give you some pointers.
    $endgroup$
    – Nick Peterson
    Jan 4 at 17:03














  • 2




    $begingroup$
    Welcome to Math.SE! As written, this question isn't likely to get a lot of attention. But, if you update it with information about what you've tried and where you're getting stuck, I'm sure somebody will be happy to give you some pointers.
    $endgroup$
    – Nick Peterson
    Jan 4 at 17:03








2




2




$begingroup$
Welcome to Math.SE! As written, this question isn't likely to get a lot of attention. But, if you update it with information about what you've tried and where you're getting stuck, I'm sure somebody will be happy to give you some pointers.
$endgroup$
– Nick Peterson
Jan 4 at 17:03




$begingroup$
Welcome to Math.SE! As written, this question isn't likely to get a lot of attention. But, if you update it with information about what you've tried and where you're getting stuck, I'm sure somebody will be happy to give you some pointers.
$endgroup$
– Nick Peterson
Jan 4 at 17:03










1 Answer
1






active

oldest

votes


















1












$begingroup$

To begin, we note that each of the functions



$f_1(t) = e^t, ; f_2(t) = e^{it}, f_3(t) = e^{-it} in L_2[a, b] tag 1$



satisfy the differential equation



$x''' - x'' + x' - x = 0, tag 2$



as is easily verified by simply substituting the $f_i(t)$, $1 le i le 3$ into (2) and performing the indicated operations. Then if



$f(t) = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) in L_2[a, b], tag 3$



$f(t)$ also satisfies (2) by linearity; since the functions (3) are precisely those in the subspace



$V = langle f_1(t), f_2(t), f_3(t) rangle subset L_2[a, b], tag 4$



we see that every element of this subspace $V$ obeys (2).



To go the other way, we find the solutions to the given differential equation (2). If we make the tentative hypothesis that there exist solutions of the form



$x = e^{rt}, tag 5$



so that



$x' = re^{rt}, ; x'' = r^2 e^{rt}, ; x''' = r^3 e^{rt}, tag 6$



we find that



$r^3 e^{rt} - r^2 e^{rt} + re^{rt} - e^{rt} = x''' - x'' + x' - x = 0; tag 7$



we may divide out $e^{rt}$:



$r^3 - r^2 + r - 1 = 0; tag 8$



we may factor the cubic on the left:



$r^3 - r^2 + r - 1 = (r^2 + 1)(r - 1); tag 9$



thus,



$(r^2 + 1)(r - 1) = 0, tag{10}$



whence



$r = 1, ; r = pm i; tag{11}$



we thus find three functions



$f_1(t)= e^t, ; f_2(t) = e^{it}, ; f_3(t) = e^{-it}, tag{12}$



all of which satisfy (2); taking it as known these $f_i(t)$ are linearly independent, we infer they span the solution space of (2), since it is of order (3); thus the set of solutions to (2) all lie in $V$.






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%2f3061821%2fprove-that-a-vector-x-from-l2a-b-is-in-the-subspace-generated-by-f1-f2-f3%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









    1












    $begingroup$

    To begin, we note that each of the functions



    $f_1(t) = e^t, ; f_2(t) = e^{it}, f_3(t) = e^{-it} in L_2[a, b] tag 1$



    satisfy the differential equation



    $x''' - x'' + x' - x = 0, tag 2$



    as is easily verified by simply substituting the $f_i(t)$, $1 le i le 3$ into (2) and performing the indicated operations. Then if



    $f(t) = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) in L_2[a, b], tag 3$



    $f(t)$ also satisfies (2) by linearity; since the functions (3) are precisely those in the subspace



    $V = langle f_1(t), f_2(t), f_3(t) rangle subset L_2[a, b], tag 4$



    we see that every element of this subspace $V$ obeys (2).



    To go the other way, we find the solutions to the given differential equation (2). If we make the tentative hypothesis that there exist solutions of the form



    $x = e^{rt}, tag 5$



    so that



    $x' = re^{rt}, ; x'' = r^2 e^{rt}, ; x''' = r^3 e^{rt}, tag 6$



    we find that



    $r^3 e^{rt} - r^2 e^{rt} + re^{rt} - e^{rt} = x''' - x'' + x' - x = 0; tag 7$



    we may divide out $e^{rt}$:



    $r^3 - r^2 + r - 1 = 0; tag 8$



    we may factor the cubic on the left:



    $r^3 - r^2 + r - 1 = (r^2 + 1)(r - 1); tag 9$



    thus,



    $(r^2 + 1)(r - 1) = 0, tag{10}$



    whence



    $r = 1, ; r = pm i; tag{11}$



    we thus find three functions



    $f_1(t)= e^t, ; f_2(t) = e^{it}, ; f_3(t) = e^{-it}, tag{12}$



    all of which satisfy (2); taking it as known these $f_i(t)$ are linearly independent, we infer they span the solution space of (2), since it is of order (3); thus the set of solutions to (2) all lie in $V$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      To begin, we note that each of the functions



      $f_1(t) = e^t, ; f_2(t) = e^{it}, f_3(t) = e^{-it} in L_2[a, b] tag 1$



      satisfy the differential equation



      $x''' - x'' + x' - x = 0, tag 2$



      as is easily verified by simply substituting the $f_i(t)$, $1 le i le 3$ into (2) and performing the indicated operations. Then if



      $f(t) = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) in L_2[a, b], tag 3$



      $f(t)$ also satisfies (2) by linearity; since the functions (3) are precisely those in the subspace



      $V = langle f_1(t), f_2(t), f_3(t) rangle subset L_2[a, b], tag 4$



      we see that every element of this subspace $V$ obeys (2).



      To go the other way, we find the solutions to the given differential equation (2). If we make the tentative hypothesis that there exist solutions of the form



      $x = e^{rt}, tag 5$



      so that



      $x' = re^{rt}, ; x'' = r^2 e^{rt}, ; x''' = r^3 e^{rt}, tag 6$



      we find that



      $r^3 e^{rt} - r^2 e^{rt} + re^{rt} - e^{rt} = x''' - x'' + x' - x = 0; tag 7$



      we may divide out $e^{rt}$:



      $r^3 - r^2 + r - 1 = 0; tag 8$



      we may factor the cubic on the left:



      $r^3 - r^2 + r - 1 = (r^2 + 1)(r - 1); tag 9$



      thus,



      $(r^2 + 1)(r - 1) = 0, tag{10}$



      whence



      $r = 1, ; r = pm i; tag{11}$



      we thus find three functions



      $f_1(t)= e^t, ; f_2(t) = e^{it}, ; f_3(t) = e^{-it}, tag{12}$



      all of which satisfy (2); taking it as known these $f_i(t)$ are linearly independent, we infer they span the solution space of (2), since it is of order (3); thus the set of solutions to (2) all lie in $V$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        To begin, we note that each of the functions



        $f_1(t) = e^t, ; f_2(t) = e^{it}, f_3(t) = e^{-it} in L_2[a, b] tag 1$



        satisfy the differential equation



        $x''' - x'' + x' - x = 0, tag 2$



        as is easily verified by simply substituting the $f_i(t)$, $1 le i le 3$ into (2) and performing the indicated operations. Then if



        $f(t) = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) in L_2[a, b], tag 3$



        $f(t)$ also satisfies (2) by linearity; since the functions (3) are precisely those in the subspace



        $V = langle f_1(t), f_2(t), f_3(t) rangle subset L_2[a, b], tag 4$



        we see that every element of this subspace $V$ obeys (2).



        To go the other way, we find the solutions to the given differential equation (2). If we make the tentative hypothesis that there exist solutions of the form



        $x = e^{rt}, tag 5$



        so that



        $x' = re^{rt}, ; x'' = r^2 e^{rt}, ; x''' = r^3 e^{rt}, tag 6$



        we find that



        $r^3 e^{rt} - r^2 e^{rt} + re^{rt} - e^{rt} = x''' - x'' + x' - x = 0; tag 7$



        we may divide out $e^{rt}$:



        $r^3 - r^2 + r - 1 = 0; tag 8$



        we may factor the cubic on the left:



        $r^3 - r^2 + r - 1 = (r^2 + 1)(r - 1); tag 9$



        thus,



        $(r^2 + 1)(r - 1) = 0, tag{10}$



        whence



        $r = 1, ; r = pm i; tag{11}$



        we thus find three functions



        $f_1(t)= e^t, ; f_2(t) = e^{it}, ; f_3(t) = e^{-it}, tag{12}$



        all of which satisfy (2); taking it as known these $f_i(t)$ are linearly independent, we infer they span the solution space of (2), since it is of order (3); thus the set of solutions to (2) all lie in $V$.






        share|cite|improve this answer











        $endgroup$



        To begin, we note that each of the functions



        $f_1(t) = e^t, ; f_2(t) = e^{it}, f_3(t) = e^{-it} in L_2[a, b] tag 1$



        satisfy the differential equation



        $x''' - x'' + x' - x = 0, tag 2$



        as is easily verified by simply substituting the $f_i(t)$, $1 le i le 3$ into (2) and performing the indicated operations. Then if



        $f(t) = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) in L_2[a, b], tag 3$



        $f(t)$ also satisfies (2) by linearity; since the functions (3) are precisely those in the subspace



        $V = langle f_1(t), f_2(t), f_3(t) rangle subset L_2[a, b], tag 4$



        we see that every element of this subspace $V$ obeys (2).



        To go the other way, we find the solutions to the given differential equation (2). If we make the tentative hypothesis that there exist solutions of the form



        $x = e^{rt}, tag 5$



        so that



        $x' = re^{rt}, ; x'' = r^2 e^{rt}, ; x''' = r^3 e^{rt}, tag 6$



        we find that



        $r^3 e^{rt} - r^2 e^{rt} + re^{rt} - e^{rt} = x''' - x'' + x' - x = 0; tag 7$



        we may divide out $e^{rt}$:



        $r^3 - r^2 + r - 1 = 0; tag 8$



        we may factor the cubic on the left:



        $r^3 - r^2 + r - 1 = (r^2 + 1)(r - 1); tag 9$



        thus,



        $(r^2 + 1)(r - 1) = 0, tag{10}$



        whence



        $r = 1, ; r = pm i; tag{11}$



        we thus find three functions



        $f_1(t)= e^t, ; f_2(t) = e^{it}, ; f_3(t) = e^{-it}, tag{12}$



        all of which satisfy (2); taking it as known these $f_i(t)$ are linearly independent, we infer they span the solution space of (2), since it is of order (3); thus the set of solutions to (2) all lie in $V$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 5 at 0:27

























        answered Jan 4 at 17:56









        Robert LewisRobert Lewis

        48.1k23067




        48.1k23067






























            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%2f3061821%2fprove-that-a-vector-x-from-l2a-b-is-in-the-subspace-generated-by-f1-f2-f3%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

            Bressuire

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
            Read more

            Gyllenstierna

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
            Read more