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












3















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
























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












  • $h(x)$ 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.
    – Anthony Ter
    Dec 1 at 7:40


















3















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
























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












  • $h(x)$ 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.
    – Anthony Ter
    Dec 1 at 7:40
















3












3








3








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
















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 at 14:53









Martin Sleziak

44.7k7115270




44.7k7115270










asked Dec 1 at 7:32









Tegernako

756




756












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












  • $h(x)$ 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.
    – Anthony Ter
    Dec 1 at 7:40




















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












  • $h(x)$ 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.
    – Anthony Ter
    Dec 1 at 7:40


















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






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














$h(x)$ 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.
– Anthony Ter
Dec 1 at 7:40






$h(x)$ 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.
– Anthony Ter
Dec 1 at 7:40












2 Answers
2






active

oldest

votes


















7














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























  • @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.
    – Robert Z
    Dec 1 at 10:14












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










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












  • Yes, but I realize issue is with my approach
    – caverac
    Dec 1 at 12:01










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





















2














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





















    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














    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























    • @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.
      – Robert Z
      Dec 1 at 10:14












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










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












    • Yes, but I realize issue is with my approach
      – caverac
      Dec 1 at 12:01










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


















    7














    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























    • @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.
      – Robert Z
      Dec 1 at 10:14












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










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












    • Yes, but I realize issue is with my approach
      – caverac
      Dec 1 at 12:01










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
















    7












    7








    7






    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














    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 at 10:44

























    answered Dec 1 at 7:52









    Robert Z

    93.2k1061132




    93.2k1061132












    • @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.
      – Robert Z
      Dec 1 at 10:14












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










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












    • Yes, but I realize issue is with my approach
      – caverac
      Dec 1 at 12:01










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




















    • @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.
      – Robert Z
      Dec 1 at 10:14












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










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












    • Yes, but I realize issue is with my approach
      – caverac
      Dec 1 at 12:01










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


















    @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.
    – Robert Z
    Dec 1 at 10:14






    @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.
    – Robert Z
    Dec 1 at 10:14














    You mean "your feeling is incorrect", right?
    – Pedro A
    Dec 1 at 10:26




    You mean "your feeling is incorrect", right?
    – Pedro A
    Dec 1 at 10:26












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






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














    Yes, but I realize issue is with my approach
    – caverac
    Dec 1 at 12:01




    Yes, but I realize issue is with my approach
    – caverac
    Dec 1 at 12:01












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






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













    2














    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


























      2














      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
























        2












        2








        2






        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












        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 at 7:56









        José Carlos Santos

        150k22120221




        150k22120221






























            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%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

            AnyDesk - Fatal Program Failure

            Image
            up vote 0 down vote favorite Good day all. I have a window that pops up saying Fatal Program Failure (AnyDesk). This is what I have done so far: Uninstalled AnyDesk and reinstalled it. ( The window still appeared ) I disabled AnyDesk start up in Msconfig and the window seems to have disapeared. Lastly I made sure that AnyDesk worked by remoting into it, and it worked. Today the same window appeared. How do I get rid of this window? remote-desktop share | improve this question edited 10 hours ago Tomasz Jakub Rup 552 3 13 ...
            Read more

            How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

            Image
            up vote 0 down vote favorite I need help with an issue we're facing in a project where we are moving from a Windows-based POS environment to one running on CentOS7. The problem consists of a number of factors, please find the details below: The supplier of our POS application will only deliver an application that is hard-coded to run in a 1024:768 4:3 format. The hardware we are using is a POS unit from HP that has a built-in 16:9 touch-screen. The employees using this solution is not supposed to see nor have access to anything but the POS application. Our sales organization demands that touch is an available feature when going live with the new POS environment. Going further into this, allow me to explain what we have encountered on the way trying to achieve the goal of having a touch-screen: We've simply change...
            Read more

            QoS: MAC-Priority for clients behind a repeater

            Image
            1 1 I've setup a wireless network using a Linksys-Router (DD-WRT) as an AP and a TP-Link repeater for range extension. To manage bandwith for specific users, I've setup QoS-Rules on the AP (DD-WRT) using MAC-Priority: However the AP shows only the Repeater's MAC and the MACs of users inside the AP-Range. The MAC addresses of users behind the Repeater are not listed : The purple MAC is not listed in the AP-Clientlist. Its related to a device connected via the TP-Link repeater. Q: How can I setup MAC-Priority for repeater-clients? Do I simply add the purple MAC in AP-MAC priority list although it's not listed as a client? I want to setup specific priority instead of giving one priority for all clients behind the repeater by adding the repeater's MAC in the MAC priority list in the AP as ...
            Read more