Mixing
What does a “linear space of all continuous functions from R to R” mean?
$begingroup$
So far, I have learned that linear spaces are defined by vectors or tensors and the span of linearly independent vectors produces a vector space.
But, check this question out:
Let $C(mathbb{R})$ be the linear space of all continuous functions from $mathbb{R}$ to $mathbb{R}$. Let $S$ be the set of differentiable functions $u(x)$ that satisfy the differential equation $u' = 2xu + c$
for all real $x$. For which value(s) of the real constant $c$ is this set a linear subspace of $C(mathbb{R})$?
Here, what does the phrase "linear space of all continuous functions" mean?
linear-algebra abstract-algebra statistics discrete-mathematics
edited Jan 22 at 9:20
Yuval Gat
572213
asked Jan 22 at 9:00
Chirag RajuChirag Raju
1
$endgroup$
add a comment |
$begingroup$
So far, I have learned that linear spaces are defined by vectors or tensors and the span of linearly independent vectors produces a vector space.
But, check this question out:
Let $C(mathbb{R})$ be the linear space of all continuous functions from $mathbb{R}$ to $mathbb{R}$. Let $S$ be the set of differentiable functions $u(x)$ that satisfy the differential equation $u' = 2xu + c$
for all real $x$. For which value(s) of the real constant $c$ is this set a linear subspace of $C(mathbb{R})$?
Here, what does the phrase "linear space of all continuous functions" mean?
linear-algebra abstract-algebra statistics discrete-mathematics
edited Jan 22 at 9:20
Yuval Gat
572213
asked Jan 22 at 9:00
Chirag RajuChirag Raju
1
$endgroup$
add a comment |
$begingroup$
So far, I have learned that linear spaces are defined by vectors or tensors and the span of linearly independent vectors produces a vector space.
But, check this question out:
Let $C(mathbb{R})$ be the linear space of all continuous functions from $mathbb{R}$ to $mathbb{R}$. Let $S$ be the set of differentiable functions $u(x)$ that satisfy the differential equation $u' = 2xu + c$
for all real $x$. For which value(s) of the real constant $c$ is this set a linear subspace of $C(mathbb{R})$?
Here, what does the phrase "linear space of all continuous functions" mean?
linear-algebra abstract-algebra statistics discrete-mathematics
edited Jan 22 at 9:20
Yuval Gat
572213
asked Jan 22 at 9:00
Chirag RajuChirag Raju
1
$endgroup$
So far, I have learned that linear spaces are defined by vectors or tensors and the span of linearly independent vectors produces a vector space.
But, check this question out:
Let $C(mathbb{R})$ be the linear space of all continuous functions from $mathbb{R}$ to $mathbb{R}$. Let $S$ be the set of differentiable functions $u(x)$ that satisfy the differential equation $u' = 2xu + c$
for all real $x$. For which value(s) of the real constant $c$ is this set a linear subspace of $C(mathbb{R})$?
Here, what does the phrase "linear space of all continuous functions" mean?
linear-algebra abstract-algebra statistics discrete-mathematics
linear-algebra abstract-algebra statistics discrete-mathematics
edited Jan 22 at 9:20
Yuval Gat
572213
asked Jan 22 at 9:00
Chirag RajuChirag Raju
1
edited Jan 22 at 9:20
Yuval Gat
572213
asked Jan 22 at 9:00
Chirag RajuChirag Raju
1
edited Jan 22 at 9:20
Yuval Gat
572213
edited Jan 22 at 9:20
Yuval Gat
572213
edited Jan 22 at 9:20
Yuval Gat
572213
572213
asked Jan 22 at 9:00
Chirag RajuChirag Raju
1
asked Jan 22 at 9:00
Chirag RajuChirag Raju
1
asked Jan 22 at 9:00
Chirag RajuChirag Raju
1
1
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Addition and scalar multiplication are the usual operations on functions. A set of real functions on $mathbb R$ functions is a vector space of it is closed under addition and scalar multiplication. In this case the zero function belongs to the space only if $c=0$. And if $c=0$ the set of solutions of the DE are closed under addition and scalar multiplication so it is a vector space. [ $u'=2xu,v'=2xv$ imply $(u+v)'=2x(u+v)$ and $(au)'=2x(au)$ ].
answered Jan 22 at 9:04
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
$endgroup$
$begingroup$
How did we determine c=0?
$endgroup$
– Chirag Raju
Jan 22 at 9:14
1
$begingroup$
@ChiragRaju The zero element is the function $u$ defined by $u(x)=0$ for all $x$. If this belongs to the given set then the equation $u'=2xu+c$ give $0=0+c$. Hence $c=0$.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:18
add a comment |
$begingroup$
Here "linear space" is used synonymous for "vector space", you consider the set of all continuous functions $Bbb RtoBbb R$ with the addition of functions defined by
$$(f+g)(x) = f(x) + g(x)$$
and scalar multiples of functions defined by
$$(lambda cdot f)(x) = lambdacdot f(x).$$
Now in this vector space you have to figure out for each $cinmathbb R$ if the set of all differentiable $ucolon Bbb RtoBbb R$ satisfying the differential equation $u'=2xu+c$ is a subspace, that is, is it non-empty and closed under addition of functions and taking scalar multiples of functions?
answered Jan 22 at 9:04
ChristophChristoph
12.5k1642
$endgroup$
$begingroup$
Thank you!! could you tell me how to proceed with the problem? after integrating by parts the LHS : u'- 2xu, we get a c* exp(-x^2) on the RHS. How do I proceed from there?
$endgroup$
– Chirag Raju
Jan 22 at 9:12
$begingroup$
You don't actually need to solve the differential equation. You only need to check if the solutions satisfy the property of being a linear subspace. For example every linear subspace contains the zero vector which already tells you a lot about what values of $c$ are possible, see Kavi's answer.
$endgroup$
– Christoph
Jan 22 at 11:01
add a comment |
$begingroup$
Note that you do not need to solve the d.e. to answer this question.
Just suppose that you have two functions $f$ and $g$ which you know are in $S$. So you know that
$f'=2xf+c\g'=2xg+c$
Then if $S$ is a linear subspace you need $lambda f$ to be in $S$ for any $lambda in mathbb{R}$. So you need to have
$(lambda f)' = 2x lambda f + c$
but $(lambda f)' = lambda f' = lambda(2xf+c)=2xlambda f + lambda c$, so this can only work for any $lambda$ if $c=0$.
Similarly, $f+g$ will only be in $S$ if $c=0$.
answered Jan 22 at 10:24
gandalf61gandalf61
8,936725
$endgroup$
add a comment |
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3 Answers
3
active
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votes
3 Answers
3
active
oldest
votes
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oldest
votes
$begingroup$
Addition and scalar multiplication are the usual operations on functions. A set of real functions on $mathbb R$ functions is a vector space of it is closed under addition and scalar multiplication. In this case the zero function belongs to the space only if $c=0$. And if $c=0$ the set of solutions of the DE are closed under addition and scalar multiplication so it is a vector space. [ $u'=2xu,v'=2xv$ imply $(u+v)'=2x(u+v)$ and $(au)'=2x(au)$ ].
answered Jan 22 at 9:04
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
$endgroup$
$begingroup$
How did we determine c=0?
$endgroup$
– Chirag Raju
Jan 22 at 9:14
1
$begingroup$
@ChiragRaju The zero element is the function $u$ defined by $u(x)=0$ for all $x$. If this belongs to the given set then the equation $u'=2xu+c$ give $0=0+c$. Hence $c=0$.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:18
add a comment |
$begingroup$
Addition and scalar multiplication are the usual operations on functions. A set of real functions on $mathbb R$ functions is a vector space of it is closed under addition and scalar multiplication. In this case the zero function belongs to the space only if $c=0$. And if $c=0$ the set of solutions of the DE are closed under addition and scalar multiplication so it is a vector space. [ $u'=2xu,v'=2xv$ imply $(u+v)'=2x(u+v)$ and $(au)'=2x(au)$ ].
answered Jan 22 at 9:04
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
$endgroup$
$begingroup$
How did we determine c=0?
$endgroup$
– Chirag Raju
Jan 22 at 9:14
1
$begingroup$
@ChiragRaju The zero element is the function $u$ defined by $u(x)=0$ for all $x$. If this belongs to the given set then the equation $u'=2xu+c$ give $0=0+c$. Hence $c=0$.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:18
add a comment |
$begingroup$
Addition and scalar multiplication are the usual operations on functions. A set of real functions on $mathbb R$ functions is a vector space of it is closed under addition and scalar multiplication. In this case the zero function belongs to the space only if $c=0$. And if $c=0$ the set of solutions of the DE are closed under addition and scalar multiplication so it is a vector space. [ $u'=2xu,v'=2xv$ imply $(u+v)'=2x(u+v)$ and $(au)'=2x(au)$ ].
answered Jan 22 at 9:04
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
$endgroup$
Addition and scalar multiplication are the usual operations on functions. A set of real functions on $mathbb R$ functions is a vector space of it is closed under addition and scalar multiplication. In this case the zero function belongs to the space only if $c=0$. And if $c=0$ the set of solutions of the DE are closed under addition and scalar multiplication so it is a vector space. [ $u'=2xu,v'=2xv$ imply $(u+v)'=2x(u+v)$ and $(au)'=2x(au)$ ].
answered Jan 22 at 9:04
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
answered Jan 22 at 9:04
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
answered Jan 22 at 9:04
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
answered Jan 22 at 9:04
Kavi Rama MurthyKavi Rama Murthy
65.8k42867
65.8k42867
$begingroup$
How did we determine c=0?
$endgroup$
– Chirag Raju
Jan 22 at 9:14
1
$begingroup$
@ChiragRaju The zero element is the function $u$ defined by $u(x)=0$ for all $x$. If this belongs to the given set then the equation $u'=2xu+c$ give $0=0+c$. Hence $c=0$.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:18
add a comment |
$begingroup$
How did we determine c=0?
$endgroup$
– Chirag Raju
Jan 22 at 9:14
1
$begingroup$
@ChiragRaju The zero element is the function $u$ defined by $u(x)=0$ for all $x$. If this belongs to the given set then the equation $u'=2xu+c$ give $0=0+c$. Hence $c=0$.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:18
$begingroup$
How did we determine c=0?
$endgroup$
– Chirag Raju
Jan 22 at 9:14
$begingroup$
How did we determine c=0?
$endgroup$
– Chirag Raju
Jan 22 at 9:14
1
1
$begingroup$
@ChiragRaju The zero element is the function $u$ defined by $u(x)=0$ for all $x$. If this belongs to the given set then the equation $u'=2xu+c$ give $0=0+c$. Hence $c=0$.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:18
$begingroup$
@ChiragRaju The zero element is the function $u$ defined by $u(x)=0$ for all $x$. If this belongs to the given set then the equation $u'=2xu+c$ give $0=0+c$. Hence $c=0$.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 9:18
add a comment |
$begingroup$
Here "linear space" is used synonymous for "vector space", you consider the set of all continuous functions $Bbb RtoBbb R$ with the addition of functions defined by
$$(f+g)(x) = f(x) + g(x)$$
and scalar multiples of functions defined by
$$(lambda cdot f)(x) = lambdacdot f(x).$$
Now in this vector space you have to figure out for each $cinmathbb R$ if the set of all differentiable $ucolon Bbb RtoBbb R$ satisfying the differential equation $u'=2xu+c$ is a subspace, that is, is it non-empty and closed under addition of functions and taking scalar multiples of functions?
answered Jan 22 at 9:04
ChristophChristoph
12.5k1642
$endgroup$
$begingroup$
Thank you!! could you tell me how to proceed with the problem? after integrating by parts the LHS : u'- 2xu, we get a c* exp(-x^2) on the RHS. How do I proceed from there?
$endgroup$
– Chirag Raju
Jan 22 at 9:12
$begingroup$
You don't actually need to solve the differential equation. You only need to check if the solutions satisfy the property of being a linear subspace. For example every linear subspace contains the zero vector which already tells you a lot about what values of $c$ are possible, see Kavi's answer.
$endgroup$
– Christoph
Jan 22 at 11:01
add a comment |
$begingroup$
Here "linear space" is used synonymous for "vector space", you consider the set of all continuous functions $Bbb RtoBbb R$ with the addition of functions defined by
$$(f+g)(x) = f(x) + g(x)$$
and scalar multiples of functions defined by
$$(lambda cdot f)(x) = lambdacdot f(x).$$
Now in this vector space you have to figure out for each $cinmathbb R$ if the set of all differentiable $ucolon Bbb RtoBbb R$ satisfying the differential equation $u'=2xu+c$ is a subspace, that is, is it non-empty and closed under addition of functions and taking scalar multiples of functions?
answered Jan 22 at 9:04
ChristophChristoph
12.5k1642
$endgroup$
$begingroup$
Thank you!! could you tell me how to proceed with the problem? after integrating by parts the LHS : u'- 2xu, we get a c* exp(-x^2) on the RHS. How do I proceed from there?
$endgroup$
– Chirag Raju
Jan 22 at 9:12
$begingroup$
You don't actually need to solve the differential equation. You only need to check if the solutions satisfy the property of being a linear subspace. For example every linear subspace contains the zero vector which already tells you a lot about what values of $c$ are possible, see Kavi's answer.
$endgroup$
– Christoph
Jan 22 at 11:01
add a comment |
$begingroup$
Here "linear space" is used synonymous for "vector space", you consider the set of all continuous functions $Bbb RtoBbb R$ with the addition of functions defined by
$$(f+g)(x) = f(x) + g(x)$$
and scalar multiples of functions defined by
$$(lambda cdot f)(x) = lambdacdot f(x).$$
Now in this vector space you have to figure out for each $cinmathbb R$ if the set of all differentiable $ucolon Bbb RtoBbb R$ satisfying the differential equation $u'=2xu+c$ is a subspace, that is, is it non-empty and closed under addition of functions and taking scalar multiples of functions?
answered Jan 22 at 9:04
ChristophChristoph
12.5k1642
$endgroup$
Here "linear space" is used synonymous for "vector space", you consider the set of all continuous functions $Bbb RtoBbb R$ with the addition of functions defined by
$$(f+g)(x) = f(x) + g(x)$$
and scalar multiples of functions defined by
$$(lambda cdot f)(x) = lambdacdot f(x).$$
Now in this vector space you have to figure out for each $cinmathbb R$ if the set of all differentiable $ucolon Bbb RtoBbb R$ satisfying the differential equation $u'=2xu+c$ is a subspace, that is, is it non-empty and closed under addition of functions and taking scalar multiples of functions?
answered Jan 22 at 9:04
ChristophChristoph
12.5k1642
answered Jan 22 at 9:04
ChristophChristoph
12.5k1642
answered Jan 22 at 9:04
ChristophChristoph
12.5k1642
answered Jan 22 at 9:04
ChristophChristoph
12.5k1642
12.5k1642
$begingroup$
Thank you!! could you tell me how to proceed with the problem? after integrating by parts the LHS : u'- 2xu, we get a c* exp(-x^2) on the RHS. How do I proceed from there?
$endgroup$
– Chirag Raju
Jan 22 at 9:12
$begingroup$
You don't actually need to solve the differential equation. You only need to check if the solutions satisfy the property of being a linear subspace. For example every linear subspace contains the zero vector which already tells you a lot about what values of $c$ are possible, see Kavi's answer.
$endgroup$
– Christoph
Jan 22 at 11:01
add a comment |
$begingroup$
Thank you!! could you tell me how to proceed with the problem? after integrating by parts the LHS : u'- 2xu, we get a c* exp(-x^2) on the RHS. How do I proceed from there?
$endgroup$
– Chirag Raju
Jan 22 at 9:12
$begingroup$
You don't actually need to solve the differential equation. You only need to check if the solutions satisfy the property of being a linear subspace. For example every linear subspace contains the zero vector which already tells you a lot about what values of $c$ are possible, see Kavi's answer.
$endgroup$
– Christoph
Jan 22 at 11:01
$begingroup$
Thank you!! could you tell me how to proceed with the problem? after integrating by parts the LHS : u'- 2xu, we get a c* exp(-x^2) on the RHS. How do I proceed from there?
$endgroup$
– Chirag Raju
Jan 22 at 9:12
$begingroup$
Thank you!! could you tell me how to proceed with the problem? after integrating by parts the LHS : u'- 2xu, we get a c* exp(-x^2) on the RHS. How do I proceed from there?
$endgroup$
– Chirag Raju
Jan 22 at 9:12
$begingroup$
You don't actually need to solve the differential equation. You only need to check if the solutions satisfy the property of being a linear subspace. For example every linear subspace contains the zero vector which already tells you a lot about what values of $c$ are possible, see Kavi's answer.
$endgroup$
– Christoph
Jan 22 at 11:01
$begingroup$
You don't actually need to solve the differential equation. You only need to check if the solutions satisfy the property of being a linear subspace. For example every linear subspace contains the zero vector which already tells you a lot about what values of $c$ are possible, see Kavi's answer.
$endgroup$
– Christoph
Jan 22 at 11:01
add a comment |
$begingroup$
Note that you do not need to solve the d.e. to answer this question.
Just suppose that you have two functions $f$ and $g$ which you know are in $S$. So you know that
$f'=2xf+c\g'=2xg+c$
Then if $S$ is a linear subspace you need $lambda f$ to be in $S$ for any $lambda in mathbb{R}$. So you need to have
$(lambda f)' = 2x lambda f + c$
but $(lambda f)' = lambda f' = lambda(2xf+c)=2xlambda f + lambda c$, so this can only work for any $lambda$ if $c=0$.
Similarly, $f+g$ will only be in $S$ if $c=0$.
answered Jan 22 at 10:24
gandalf61gandalf61
8,936725
$endgroup$
add a comment |
$begingroup$
Note that you do not need to solve the d.e. to answer this question.
Just suppose that you have two functions $f$ and $g$ which you know are in $S$. So you know that
$f'=2xf+c\g'=2xg+c$
Then if $S$ is a linear subspace you need $lambda f$ to be in $S$ for any $lambda in mathbb{R}$. So you need to have
$(lambda f)' = 2x lambda f + c$
but $(lambda f)' = lambda f' = lambda(2xf+c)=2xlambda f + lambda c$, so this can only work for any $lambda$ if $c=0$.
Similarly, $f+g$ will only be in $S$ if $c=0$.
answered Jan 22 at 10:24
gandalf61gandalf61
8,936725
$endgroup$
add a comment |
$begingroup$
Note that you do not need to solve the d.e. to answer this question.
Just suppose that you have two functions $f$ and $g$ which you know are in $S$. So you know that
$f'=2xf+c\g'=2xg+c$
Then if $S$ is a linear subspace you need $lambda f$ to be in $S$ for any $lambda in mathbb{R}$. So you need to have
$(lambda f)' = 2x lambda f + c$
but $(lambda f)' = lambda f' = lambda(2xf+c)=2xlambda f + lambda c$, so this can only work for any $lambda$ if $c=0$.
Similarly, $f+g$ will only be in $S$ if $c=0$.
answered Jan 22 at 10:24
gandalf61gandalf61
8,936725
$endgroup$
Note that you do not need to solve the d.e. to answer this question.
Just suppose that you have two functions $f$ and $g$ which you know are in $S$. So you know that
$f'=2xf+c\g'=2xg+c$
Then if $S$ is a linear subspace you need $lambda f$ to be in $S$ for any $lambda in mathbb{R}$. So you need to have
$(lambda f)' = 2x lambda f + c$
but $(lambda f)' = lambda f' = lambda(2xf+c)=2xlambda f + lambda c$, so this can only work for any $lambda$ if $c=0$.
Similarly, $f+g$ will only be in $S$ if $c=0$.
answered Jan 22 at 10:24
gandalf61gandalf61
8,936725
answered Jan 22 at 10:24
gandalf61gandalf61
8,936725
answered Jan 22 at 10:24
gandalf61gandalf61
8,936725
answered Jan 22 at 10:24
gandalf61gandalf61
8,936725
8,936725
add a comment |
add a comment |
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