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












-1














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 differential-equations vector-spaces







share|cite|improve this question









New contributor




Raluca lok is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2




    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.
    – Nick Peterson
    Jan 4 at 17:03
















-1














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 differential-equations vector-spaces







share|cite|improve this question









New contributor




Raluca lok is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2




    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.
    – Nick Peterson
    Jan 4 at 17:03














-1












-1








-1


1





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 differential-equations vector-spaces







share|cite|improve this question









New contributor




Raluca lok is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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 differential-equations vector-spaces




linear-algebra differential-equations vector-spaces






share|cite|improve this question









New contributor




Raluca lok is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Raluca lok is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 16:51









Davide Giraudo

125k16150261




125k16150261






New contributor




Raluca lok is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Jan 4 at 16:48









Raluca lokRaluca lok

21




21




New contributor




Raluca lok is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Raluca lok is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Raluca lok is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 2




    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.
    – Nick Peterson
    Jan 4 at 17:03














  • 2




    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.
    – Nick Peterson
    Jan 4 at 17:03








2




2




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.
– Nick Peterson
Jan 4 at 17:03




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.
– Nick Peterson
Jan 4 at 17:03










1 Answer
1






active

oldest

votes


















1














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























    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
    });


    }
    });






    Raluca lok is a new contributor. Be nice, and check out our Code of Conduct.










    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














    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




























      1














      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


























        1












        1








        1






        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














        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

        44.2k22963




        44.2k22963






















            Raluca lok is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            Raluca lok is a new contributor. Be nice, and check out our Code of Conduct.













            Raluca lok is a new contributor. Be nice, and check out our Code of Conduct.












            Raluca lok is a new contributor. Be nice, and check out our Code of Conduct.
















            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%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

            1300-talet

            Image
            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .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 ...
            Read more

            1300-talet

            Image
            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .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 ...
            Read more

            Display a custom attribute below product name in the front-end Magento 1.9.3.8

            Image
            0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
            Read more