Interpretation of $R^D$ (Vectors Space, Dimensions and Functions)
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I'm aware that it's a trivial question, but I want to make sure that I'm understanding correctly what I'm studying, therefore I would like to ask you, can you tell me what does $R^D$ "say" in these three different cases? Does it always say the same thing or the meaning is different?
1) Let "G" be a finite-dimensional vector space of real functions in $R^D$.
2) $x in R^D$
3) {$x_1, x_2, ..., x_m$} $subseteq R^D$
functions vectors dimension-theory
asked Dec 20 '18 at 19:12
Tommaso BendinelliTommaso Bendinelli
14610
$endgroup$
add a comment |
$begingroup$
I'm aware that it's a trivial question, but I want to make sure that I'm understanding correctly what I'm studying, therefore I would like to ask you, can you tell me what does $R^D$ "say" in these three different cases? Does it always say the same thing or the meaning is different?
1) Let "G" be a finite-dimensional vector space of real functions in $R^D$.
2) $x in R^D$
3) {$x_1, x_2, ..., x_m$} $subseteq R^D$
functions vectors dimension-theory
asked Dec 20 '18 at 19:12
Tommaso BendinelliTommaso Bendinelli
14610
$endgroup$
add a comment |
$begingroup$
I'm aware that it's a trivial question, but I want to make sure that I'm understanding correctly what I'm studying, therefore I would like to ask you, can you tell me what does $R^D$ "say" in these three different cases? Does it always say the same thing or the meaning is different?
1) Let "G" be a finite-dimensional vector space of real functions in $R^D$.
2) $x in R^D$
3) {$x_1, x_2, ..., x_m$} $subseteq R^D$
functions vectors dimension-theory
asked Dec 20 '18 at 19:12
Tommaso BendinelliTommaso Bendinelli
14610
$endgroup$
I'm aware that it's a trivial question, but I want to make sure that I'm understanding correctly what I'm studying, therefore I would like to ask you, can you tell me what does $R^D$ "say" in these three different cases? Does it always say the same thing or the meaning is different?
1) Let "G" be a finite-dimensional vector space of real functions in $R^D$.
2) $x in R^D$
3) {$x_1, x_2, ..., x_m$} $subseteq R^D$
functions vectors dimension-theory
functions vectors dimension-theory
asked Dec 20 '18 at 19:12
Tommaso BendinelliTommaso Bendinelli
14610
asked Dec 20 '18 at 19:12
Tommaso BendinelliTommaso Bendinelli
14610
asked Dec 20 '18 at 19:12
Tommaso BendinelliTommaso Bendinelli
14610
asked Dec 20 '18 at 19:12
Tommaso BendinelliTommaso Bendinelli
14610
asked Dec 20 '18 at 19:12
Tommaso BendinelliTommaso Bendinelli
14610
14610
add a comment |
add a comment |
1 Answer
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These are all different, though you may have intended $2$ and $3$ to be the same if you've mixed up your vector and set notation.
For item $1$ the elements of $G$ are all functions. "in" is a little bit unclear, but I think most people would interpret it as meaning that each element $fin G$ is a mapping $f:{mathbb R}^D rightarrow {mathbb K}$ where ${mathbb K}$ is either ${mathbb R}$ or ${mathbb C}$. It could however mean functions taking values in ${mathbb R}^D$. Note that there aren't all that many (mathematically speaking) finite dimensional function spaces: ${mathbb R}^n$ and $({mathbb R}^n)^*$ for each $n$ are most of them (if you throw away enough topological structure you can find a few more, but they're not really interesting then).
For item $2$, ${mathbb R}^D = {mathbb R} times {mathbb R} times cdots {mathbb R}$ is the cartesian product of $D$ copies of ${mathbb R}$, and a typical element $xin {mathbb R}^D$ therefore can be represented as a vector of $D$ real numbers $(x_1, x_2, ldots , x_D)$ As an example, consider the three orthogonal unit vectors in ${mathbb R}^3: (1,0,0), (0,1,0)$ and $(0,0,1)$
For item $3$ you have a set of $m$ elements that is a subset of ${mathbb R}^D$. As an example here, consider ${x in {mathbb R}^3 | |x|leq 1 }$ where the norm is the $max$ norm. Then this set contains all three vectors from the previous example, as well as many more.
answered Dec 20 '18 at 20:38
postmortespostmortes
2,29531422
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$begingroup$
These are all different, though you may have intended $2$ and $3$ to be the same if you've mixed up your vector and set notation.
For item $1$ the elements of $G$ are all functions. "in" is a little bit unclear, but I think most people would interpret it as meaning that each element $fin G$ is a mapping $f:{mathbb R}^D rightarrow {mathbb K}$ where ${mathbb K}$ is either ${mathbb R}$ or ${mathbb C}$. It could however mean functions taking values in ${mathbb R}^D$. Note that there aren't all that many (mathematically speaking) finite dimensional function spaces: ${mathbb R}^n$ and $({mathbb R}^n)^*$ for each $n$ are most of them (if you throw away enough topological structure you can find a few more, but they're not really interesting then).
For item $2$, ${mathbb R}^D = {mathbb R} times {mathbb R} times cdots {mathbb R}$ is the cartesian product of $D$ copies of ${mathbb R}$, and a typical element $xin {mathbb R}^D$ therefore can be represented as a vector of $D$ real numbers $(x_1, x_2, ldots , x_D)$ As an example, consider the three orthogonal unit vectors in ${mathbb R}^3: (1,0,0), (0,1,0)$ and $(0,0,1)$
For item $3$ you have a set of $m$ elements that is a subset of ${mathbb R}^D$. As an example here, consider ${x in {mathbb R}^3 | |x|leq 1 }$ where the norm is the $max$ norm. Then this set contains all three vectors from the previous example, as well as many more.
answered Dec 20 '18 at 20:38
postmortespostmortes
2,29531422
$endgroup$
add a comment |
$begingroup$
These are all different, though you may have intended $2$ and $3$ to be the same if you've mixed up your vector and set notation.
For item $1$ the elements of $G$ are all functions. "in" is a little bit unclear, but I think most people would interpret it as meaning that each element $fin G$ is a mapping $f:{mathbb R}^D rightarrow {mathbb K}$ where ${mathbb K}$ is either ${mathbb R}$ or ${mathbb C}$. It could however mean functions taking values in ${mathbb R}^D$. Note that there aren't all that many (mathematically speaking) finite dimensional function spaces: ${mathbb R}^n$ and $({mathbb R}^n)^*$ for each $n$ are most of them (if you throw away enough topological structure you can find a few more, but they're not really interesting then).
For item $2$, ${mathbb R}^D = {mathbb R} times {mathbb R} times cdots {mathbb R}$ is the cartesian product of $D$ copies of ${mathbb R}$, and a typical element $xin {mathbb R}^D$ therefore can be represented as a vector of $D$ real numbers $(x_1, x_2, ldots , x_D)$ As an example, consider the three orthogonal unit vectors in ${mathbb R}^3: (1,0,0), (0,1,0)$ and $(0,0,1)$
For item $3$ you have a set of $m$ elements that is a subset of ${mathbb R}^D$. As an example here, consider ${x in {mathbb R}^3 | |x|leq 1 }$ where the norm is the $max$ norm. Then this set contains all three vectors from the previous example, as well as many more.
answered Dec 20 '18 at 20:38
postmortespostmortes
2,29531422
$endgroup$
add a comment |
$begingroup$
These are all different, though you may have intended $2$ and $3$ to be the same if you've mixed up your vector and set notation.
For item $1$ the elements of $G$ are all functions. "in" is a little bit unclear, but I think most people would interpret it as meaning that each element $fin G$ is a mapping $f:{mathbb R}^D rightarrow {mathbb K}$ where ${mathbb K}$ is either ${mathbb R}$ or ${mathbb C}$. It could however mean functions taking values in ${mathbb R}^D$. Note that there aren't all that many (mathematically speaking) finite dimensional function spaces: ${mathbb R}^n$ and $({mathbb R}^n)^*$ for each $n$ are most of them (if you throw away enough topological structure you can find a few more, but they're not really interesting then).
For item $2$, ${mathbb R}^D = {mathbb R} times {mathbb R} times cdots {mathbb R}$ is the cartesian product of $D$ copies of ${mathbb R}$, and a typical element $xin {mathbb R}^D$ therefore can be represented as a vector of $D$ real numbers $(x_1, x_2, ldots , x_D)$ As an example, consider the three orthogonal unit vectors in ${mathbb R}^3: (1,0,0), (0,1,0)$ and $(0,0,1)$
For item $3$ you have a set of $m$ elements that is a subset of ${mathbb R}^D$. As an example here, consider ${x in {mathbb R}^3 | |x|leq 1 }$ where the norm is the $max$ norm. Then this set contains all three vectors from the previous example, as well as many more.
answered Dec 20 '18 at 20:38
postmortespostmortes
2,29531422
$endgroup$
These are all different, though you may have intended $2$ and $3$ to be the same if you've mixed up your vector and set notation.
For item $1$ the elements of $G$ are all functions. "in" is a little bit unclear, but I think most people would interpret it as meaning that each element $fin G$ is a mapping $f:{mathbb R}^D rightarrow {mathbb K}$ where ${mathbb K}$ is either ${mathbb R}$ or ${mathbb C}$. It could however mean functions taking values in ${mathbb R}^D$. Note that there aren't all that many (mathematically speaking) finite dimensional function spaces: ${mathbb R}^n$ and $({mathbb R}^n)^*$ for each $n$ are most of them (if you throw away enough topological structure you can find a few more, but they're not really interesting then).
For item $2$, ${mathbb R}^D = {mathbb R} times {mathbb R} times cdots {mathbb R}$ is the cartesian product of $D$ copies of ${mathbb R}$, and a typical element $xin {mathbb R}^D$ therefore can be represented as a vector of $D$ real numbers $(x_1, x_2, ldots , x_D)$ As an example, consider the three orthogonal unit vectors in ${mathbb R}^3: (1,0,0), (0,1,0)$ and $(0,0,1)$
For item $3$ you have a set of $m$ elements that is a subset of ${mathbb R}^D$. As an example here, consider ${x in {mathbb R}^3 | |x|leq 1 }$ where the norm is the $max$ norm. Then this set contains all three vectors from the previous example, as well as many more.
answered Dec 20 '18 at 20:38
postmortespostmortes
2,29531422
answered Dec 20 '18 at 20:38
postmortespostmortes
2,29531422
answered Dec 20 '18 at 20:38
postmortespostmortes
2,29531422
answered Dec 20 '18 at 20:38
postmortespostmortes
2,29531422
2,29531422
add a comment |
add a comment |
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