Linear dependence of three functions: $f(x) = sin(x)$, $g(x) = cos(x)$ and $h(x)=x$












3












$begingroup$



Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.
The following functions in the vector space are defined as follows:
$$f(x) = sin(x),qquad g(x) = cos(x),qquad h(x) = x.$$
Is the triple $(f,g,h)$ linearly dependent?




I have a feeling it is, because $h(x)$ is the identity function, so $h(sin(x)) = sin(x)$, and I know the definitions, I can't figure out how to put my explanation (if it is correct) properly.



Thanks in advance.






linear-algebra functions vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Linear dependent means there are coefficients $a,b,c$ (not all equal to zero) such that $asin(x)+bcos(x)+cx=0$ for all $xinmathbb{R}$
    $endgroup$
    – Fakemistake
    Dec 1 '18 at 7:38












  • $begingroup$
    $h(x)$ is not the identity function when the operation is function addition, $f(x) = 0$ is. Function composition does not form a vector space, so linear dependence is as defined in the above comment.
    $endgroup$
    – Anthony Ter
    Dec 1 '18 at 7:40


















3












$begingroup$



Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.
The following functions in the vector space are defined as follows:
$$f(x) = sin(x),qquad g(x) = cos(x),qquad h(x) = x.$$
Is the triple $(f,g,h)$ linearly dependent?




I have a feeling it is, because $h(x)$ is the identity function, so $h(sin(x)) = sin(x)$, and I know the definitions, I can't figure out how to put my explanation (if it is correct) properly.



Thanks in advance.






linear-algebra functions vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Linear dependent means there are coefficients $a,b,c$ (not all equal to zero) such that $asin(x)+bcos(x)+cx=0$ for all $xinmathbb{R}$
    $endgroup$
    – Fakemistake
    Dec 1 '18 at 7:38












  • $begingroup$
    $h(x)$ is not the identity function when the operation is function addition, $f(x) = 0$ is. Function composition does not form a vector space, so linear dependence is as defined in the above comment.
    $endgroup$
    – Anthony Ter
    Dec 1 '18 at 7:40
















3












3








3





$begingroup$



Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.
The following functions in the vector space are defined as follows:
$$f(x) = sin(x),qquad g(x) = cos(x),qquad h(x) = x.$$
Is the triple $(f,g,h)$ linearly dependent?




I have a feeling it is, because $h(x)$ is the identity function, so $h(sin(x)) = sin(x)$, and I know the definitions, I can't figure out how to put my explanation (if it is correct) properly.



Thanks in advance.






linear-algebra functions vector-spaces







share|cite|improve this question











$endgroup$





Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.
The following functions in the vector space are defined as follows:
$$f(x) = sin(x),qquad g(x) = cos(x),qquad h(x) = x.$$
Is the triple $(f,g,h)$ linearly dependent?




I have a feeling it is, because $h(x)$ is the identity function, so $h(sin(x)) = sin(x)$, and I know the definitions, I can't figure out how to put my explanation (if it is correct) properly.



Thanks in advance.






linear-algebra functions vector-spaces




linear-algebra functions vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 14:53









Martin Sleziak

44.8k10118272




44.8k10118272










asked Dec 1 '18 at 7:32









TegernakoTegernako

908




908












  • $begingroup$
    Linear dependent means there are coefficients $a,b,c$ (not all equal to zero) such that $asin(x)+bcos(x)+cx=0$ for all $xinmathbb{R}$
    $endgroup$
    – Fakemistake
    Dec 1 '18 at 7:38












  • $begingroup$
    $h(x)$ is not the identity function when the operation is function addition, $f(x) = 0$ is. Function composition does not form a vector space, so linear dependence is as defined in the above comment.
    $endgroup$
    – Anthony Ter
    Dec 1 '18 at 7:40




















  • $begingroup$
    Linear dependent means there are coefficients $a,b,c$ (not all equal to zero) such that $asin(x)+bcos(x)+cx=0$ for all $xinmathbb{R}$
    $endgroup$
    – Fakemistake
    Dec 1 '18 at 7:38












  • $begingroup$
    $h(x)$ is not the identity function when the operation is function addition, $f(x) = 0$ is. Function composition does not form a vector space, so linear dependence is as defined in the above comment.
    $endgroup$
    – Anthony Ter
    Dec 1 '18 at 7:40


















$begingroup$
Linear dependent means there are coefficients $a,b,c$ (not all equal to zero) such that $asin(x)+bcos(x)+cx=0$ for all $xinmathbb{R}$
$endgroup$
– Fakemistake
Dec 1 '18 at 7:38






$begingroup$
Linear dependent means there are coefficients $a,b,c$ (not all equal to zero) such that $asin(x)+bcos(x)+cx=0$ for all $xinmathbb{R}$
$endgroup$
– Fakemistake
Dec 1 '18 at 7:38














$begingroup$
$h(x)$ is not the identity function when the operation is function addition, $f(x) = 0$ is. Function composition does not form a vector space, so linear dependence is as defined in the above comment.
$endgroup$
– Anthony Ter
Dec 1 '18 at 7:40






$begingroup$
$h(x)$ is not the identity function when the operation is function addition, $f(x) = 0$ is. Function composition does not form a vector space, so linear dependence is as defined in the above comment.
$endgroup$
– Anthony Ter
Dec 1 '18 at 7:40












2 Answers
2






active

oldest

votes


















7












$begingroup$

Your feeling is not correct. Recall that, by definition, if such functions are linearly dependent then there are real numbers $A,B,C$ not all zero such that
$$Asin(x)+Bcos(x)+Cx=0 quad forall xin mathbb{R}.$$
Now by letting $x=0,pi/2,pi$, we have three linear equations
$$begin{cases}
Acdot 0+Bcdot 1+Ccdot 0=0qquad &(x=0)\
Acdot 1+Bcdot 0+Ccdot frac{pi}{2}=0qquad &(x=frac{pi}{2})\
Acdot 0+Bcdot (-1)+Ccdot pi=0qquad &(x=pi)\
end{cases}$$
What may we conclude about $A,B,C$? Can you take it from here?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @caverac What do you mean? OP is interested in the vector space $mathbb{R}^mathbb{R}$, so the domain for $x$ is the whole real line.
    $endgroup$
    – Robert Z
    Dec 1 '18 at 10:14












  • $begingroup$
    You mean "your feeling is incorrect", right?
    $endgroup$
    – Pedro A
    Dec 1 '18 at 10:26










  • $begingroup$
    @PedroA Yes, you are right, thanks for pointing out!
    $endgroup$
    – Robert Z
    Dec 1 '18 at 10:45












  • $begingroup$
    Yes, but I realize issue is with my approach
    $endgroup$
    – caverac
    Dec 1 '18 at 12:01










  • $begingroup$
    @RobertZ that we ultimately get that A=B=C=0 for all cases?
    $endgroup$
    – Tegernako
    Dec 1 '18 at 12:49





















2












$begingroup$

Asserting that ${f,g,h}$ is linearly independent means that$$(forall a,b,cinmathbb{R}):af+bg+ch=0implies a=b=c=0.$$If $af+bg+ch=0$, then $af(0)+bg(0)+ch(0)=0$, which means that $b=0$. You still have $af+ch=0$. But then $afleft(fracpi2right)+cfracpi2=0$, which means that $a+cfracpi2=0$, and $af(pi)+ch(pi)=0$, which means that $cpi=0$. So, $a=c=0$ too.






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%2f3021094%2flinear-dependence-of-three-functions-fx-sinx-gx-cosx-and-h%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









    7












    $begingroup$

    Your feeling is not correct. Recall that, by definition, if such functions are linearly dependent then there are real numbers $A,B,C$ not all zero such that
    $$Asin(x)+Bcos(x)+Cx=0 quad forall xin mathbb{R}.$$
    Now by letting $x=0,pi/2,pi$, we have three linear equations
    $$begin{cases}
    Acdot 0+Bcdot 1+Ccdot 0=0qquad &(x=0)\
    Acdot 1+Bcdot 0+Ccdot frac{pi}{2}=0qquad &(x=frac{pi}{2})\
    Acdot 0+Bcdot (-1)+Ccdot pi=0qquad &(x=pi)\
    end{cases}$$
    What may we conclude about $A,B,C$? Can you take it from here?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @caverac What do you mean? OP is interested in the vector space $mathbb{R}^mathbb{R}$, so the domain for $x$ is the whole real line.
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:14












    • $begingroup$
      You mean "your feeling is incorrect", right?
      $endgroup$
      – Pedro A
      Dec 1 '18 at 10:26










    • $begingroup$
      @PedroA Yes, you are right, thanks for pointing out!
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:45












    • $begingroup$
      Yes, but I realize issue is with my approach
      $endgroup$
      – caverac
      Dec 1 '18 at 12:01










    • $begingroup$
      @RobertZ that we ultimately get that A=B=C=0 for all cases?
      $endgroup$
      – Tegernako
      Dec 1 '18 at 12:49


















    7












    $begingroup$

    Your feeling is not correct. Recall that, by definition, if such functions are linearly dependent then there are real numbers $A,B,C$ not all zero such that
    $$Asin(x)+Bcos(x)+Cx=0 quad forall xin mathbb{R}.$$
    Now by letting $x=0,pi/2,pi$, we have three linear equations
    $$begin{cases}
    Acdot 0+Bcdot 1+Ccdot 0=0qquad &(x=0)\
    Acdot 1+Bcdot 0+Ccdot frac{pi}{2}=0qquad &(x=frac{pi}{2})\
    Acdot 0+Bcdot (-1)+Ccdot pi=0qquad &(x=pi)\
    end{cases}$$
    What may we conclude about $A,B,C$? Can you take it from here?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @caverac What do you mean? OP is interested in the vector space $mathbb{R}^mathbb{R}$, so the domain for $x$ is the whole real line.
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:14












    • $begingroup$
      You mean "your feeling is incorrect", right?
      $endgroup$
      – Pedro A
      Dec 1 '18 at 10:26










    • $begingroup$
      @PedroA Yes, you are right, thanks for pointing out!
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:45












    • $begingroup$
      Yes, but I realize issue is with my approach
      $endgroup$
      – caverac
      Dec 1 '18 at 12:01










    • $begingroup$
      @RobertZ that we ultimately get that A=B=C=0 for all cases?
      $endgroup$
      – Tegernako
      Dec 1 '18 at 12:49
















    7












    7








    7





    $begingroup$

    Your feeling is not correct. Recall that, by definition, if such functions are linearly dependent then there are real numbers $A,B,C$ not all zero such that
    $$Asin(x)+Bcos(x)+Cx=0 quad forall xin mathbb{R}.$$
    Now by letting $x=0,pi/2,pi$, we have three linear equations
    $$begin{cases}
    Acdot 0+Bcdot 1+Ccdot 0=0qquad &(x=0)\
    Acdot 1+Bcdot 0+Ccdot frac{pi}{2}=0qquad &(x=frac{pi}{2})\
    Acdot 0+Bcdot (-1)+Ccdot pi=0qquad &(x=pi)\
    end{cases}$$
    What may we conclude about $A,B,C$? Can you take it from here?






    share|cite|improve this answer











    $endgroup$



    Your feeling is not correct. Recall that, by definition, if such functions are linearly dependent then there are real numbers $A,B,C$ not all zero such that
    $$Asin(x)+Bcos(x)+Cx=0 quad forall xin mathbb{R}.$$
    Now by letting $x=0,pi/2,pi$, we have three linear equations
    $$begin{cases}
    Acdot 0+Bcdot 1+Ccdot 0=0qquad &(x=0)\
    Acdot 1+Bcdot 0+Ccdot frac{pi}{2}=0qquad &(x=frac{pi}{2})\
    Acdot 0+Bcdot (-1)+Ccdot pi=0qquad &(x=pi)\
    end{cases}$$
    What may we conclude about $A,B,C$? Can you take it from here?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 1 '18 at 10:44

























    answered Dec 1 '18 at 7:52









    Robert ZRobert Z

    97.4k1066137




    97.4k1066137












    • $begingroup$
      @caverac What do you mean? OP is interested in the vector space $mathbb{R}^mathbb{R}$, so the domain for $x$ is the whole real line.
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:14












    • $begingroup$
      You mean "your feeling is incorrect", right?
      $endgroup$
      – Pedro A
      Dec 1 '18 at 10:26










    • $begingroup$
      @PedroA Yes, you are right, thanks for pointing out!
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:45












    • $begingroup$
      Yes, but I realize issue is with my approach
      $endgroup$
      – caverac
      Dec 1 '18 at 12:01










    • $begingroup$
      @RobertZ that we ultimately get that A=B=C=0 for all cases?
      $endgroup$
      – Tegernako
      Dec 1 '18 at 12:49




















    • $begingroup$
      @caverac What do you mean? OP is interested in the vector space $mathbb{R}^mathbb{R}$, so the domain for $x$ is the whole real line.
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:14












    • $begingroup$
      You mean "your feeling is incorrect", right?
      $endgroup$
      – Pedro A
      Dec 1 '18 at 10:26










    • $begingroup$
      @PedroA Yes, you are right, thanks for pointing out!
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:45












    • $begingroup$
      Yes, but I realize issue is with my approach
      $endgroup$
      – caverac
      Dec 1 '18 at 12:01










    • $begingroup$
      @RobertZ that we ultimately get that A=B=C=0 for all cases?
      $endgroup$
      – Tegernako
      Dec 1 '18 at 12:49


















    $begingroup$
    @caverac What do you mean? OP is interested in the vector space $mathbb{R}^mathbb{R}$, so the domain for $x$ is the whole real line.
    $endgroup$
    – Robert Z
    Dec 1 '18 at 10:14






    $begingroup$
    @caverac What do you mean? OP is interested in the vector space $mathbb{R}^mathbb{R}$, so the domain for $x$ is the whole real line.
    $endgroup$
    – Robert Z
    Dec 1 '18 at 10:14














    $begingroup$
    You mean "your feeling is incorrect", right?
    $endgroup$
    – Pedro A
    Dec 1 '18 at 10:26




    $begingroup$
    You mean "your feeling is incorrect", right?
    $endgroup$
    – Pedro A
    Dec 1 '18 at 10:26












    $begingroup$
    @PedroA Yes, you are right, thanks for pointing out!
    $endgroup$
    – Robert Z
    Dec 1 '18 at 10:45






    $begingroup$
    @PedroA Yes, you are right, thanks for pointing out!
    $endgroup$
    – Robert Z
    Dec 1 '18 at 10:45














    $begingroup$
    Yes, but I realize issue is with my approach
    $endgroup$
    – caverac
    Dec 1 '18 at 12:01




    $begingroup$
    Yes, but I realize issue is with my approach
    $endgroup$
    – caverac
    Dec 1 '18 at 12:01












    $begingroup$
    @RobertZ that we ultimately get that A=B=C=0 for all cases?
    $endgroup$
    – Tegernako
    Dec 1 '18 at 12:49






    $begingroup$
    @RobertZ that we ultimately get that A=B=C=0 for all cases?
    $endgroup$
    – Tegernako
    Dec 1 '18 at 12:49













    2












    $begingroup$

    Asserting that ${f,g,h}$ is linearly independent means that$$(forall a,b,cinmathbb{R}):af+bg+ch=0implies a=b=c=0.$$If $af+bg+ch=0$, then $af(0)+bg(0)+ch(0)=0$, which means that $b=0$. You still have $af+ch=0$. But then $afleft(fracpi2right)+cfracpi2=0$, which means that $a+cfracpi2=0$, and $af(pi)+ch(pi)=0$, which means that $cpi=0$. So, $a=c=0$ too.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Asserting that ${f,g,h}$ is linearly independent means that$$(forall a,b,cinmathbb{R}):af+bg+ch=0implies a=b=c=0.$$If $af+bg+ch=0$, then $af(0)+bg(0)+ch(0)=0$, which means that $b=0$. You still have $af+ch=0$. But then $afleft(fracpi2right)+cfracpi2=0$, which means that $a+cfracpi2=0$, and $af(pi)+ch(pi)=0$, which means that $cpi=0$. So, $a=c=0$ too.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Asserting that ${f,g,h}$ is linearly independent means that$$(forall a,b,cinmathbb{R}):af+bg+ch=0implies a=b=c=0.$$If $af+bg+ch=0$, then $af(0)+bg(0)+ch(0)=0$, which means that $b=0$. You still have $af+ch=0$. But then $afleft(fracpi2right)+cfracpi2=0$, which means that $a+cfracpi2=0$, and $af(pi)+ch(pi)=0$, which means that $cpi=0$. So, $a=c=0$ too.






        share|cite|improve this answer









        $endgroup$



        Asserting that ${f,g,h}$ is linearly independent means that$$(forall a,b,cinmathbb{R}):af+bg+ch=0implies a=b=c=0.$$If $af+bg+ch=0$, then $af(0)+bg(0)+ch(0)=0$, which means that $b=0$. You still have $af+ch=0$. But then $afleft(fracpi2right)+cfracpi2=0$, which means that $a+cfracpi2=0$, and $af(pi)+ch(pi)=0$, which means that $cpi=0$. So, $a=c=0$ too.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 '18 at 7:56









        José Carlos SantosJosé Carlos Santos

        161k22127232




        161k22127232






























            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%2f3021094%2flinear-dependence-of-three-functions-fx-sinx-gx-cosx-and-h%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

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
            Read more