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

44.7k8117272




44.7k8117272










asked Dec 1 '18 at 7:32









TegernakoTegernako

897




897












  • $begingroup$
    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









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    2 Answers
    2






    active

    oldest

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    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

    94.3k1063134




    94.3k1063134












    • $begingroup$
      @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

        154k22123226




        154k22123226






























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            Ida-Boy-Ed-Garten

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            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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