Linear dependence of three functions: $f(x) = sin(x)$, $g(x) = cos(x)$ and $h(x)=x$
up vote
3
down vote
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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
edited Dec 1 at 14:53
Martin Sleziak
44.6k7115269
asked Dec 1 at 7:32
Tegernako
596
add a comment |
up vote
3
down vote
favorite
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
edited Dec 1 at 14:53
Martin Sleziak
44.6k7115269
asked Dec 1 at 7:32
Tegernako
596
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
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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
edited Dec 1 at 14:53
Martin Sleziak
44.6k7115269
asked Dec 1 at 7:32
Tegernako
596
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
edited Dec 1 at 14:53
Martin Sleziak
44.6k7115269
asked Dec 1 at 7:32
Tegernako
596
edited Dec 1 at 14:53
Martin Sleziak
44.6k7115269
asked Dec 1 at 7:32
Tegernako
596
edited Dec 1 at 14:53
Martin Sleziak
44.6k7115269
edited Dec 1 at 14:53
Martin Sleziak
44.6k7115269
edited Dec 1 at 14:53
Martin Sleziak
44.6k7115269
44.6k7115269
asked Dec 1 at 7:32
Tegernako
596
asked Dec 1 at 7:32
Tegernako
596
asked Dec 1 at 7:32
Tegernako
596
596
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
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?
answered Dec 1 at 7:52
Robert Z
92.2k1058129
@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
|
show 1 more comment
up vote
2
down vote
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.
answered Dec 1 at 7:56
José Carlos Santos
146k22117217
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
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?
answered Dec 1 at 7:52
Robert Z
92.2k1058129
@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
|
show 1 more comment
up vote
7
down vote
accepted
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?
answered Dec 1 at 7:52
Robert Z
92.2k1058129
@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
|
show 1 more comment
up vote
7
down vote
accepted
up vote
7
down vote
accepted
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?
answered Dec 1 at 7:52
Robert Z
92.2k1058129
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?
answered Dec 1 at 7:52
Robert Z
92.2k1058129
edited Dec 1 at 10:44
answered Dec 1 at 7:52
Robert Z
92.2k1058129
answered Dec 1 at 7:52
Robert Z
92.2k1058129
answered Dec 1 at 7:52
Robert Z
92.2k1058129
92.2k1058129
@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
|
show 1 more comment
@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
|
show 1 more comment
up vote
2
down vote
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.
answered Dec 1 at 7:56
José Carlos Santos
146k22117217
add a comment |
up vote
2
down vote
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.
answered Dec 1 at 7:56
José Carlos Santos
146k22117217
add a comment |
up vote
2
down vote
up vote
2
down vote
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.
answered Dec 1 at 7:56
José Carlos Santos
146k22117217
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.
answered Dec 1 at 7:56
José Carlos Santos
146k22117217
answered Dec 1 at 7:56
José Carlos Santos
146k22117217
answered Dec 1 at 7:56
José Carlos Santos
146k22117217
answered Dec 1 at 7:56
José Carlos Santos
146k22117217
146k22117217
add a comment |
add a comment |
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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