Mixing
Doubt about the role of “equivalence class of functions” in this Least Square example.
$begingroup$
I am dealing with a proof from De Boor (1972) about least square approximations using splines.
Suppose we have a set of data and we want to estimate the least square approximation.
Let $ $ $ be a finite dimensional linear space of functions, we seek a best approximation from $$$ to 'g' with respect to the norm $||cdot||_2 $.
This is we want to find $f^*in$$ such that:
$forall f in $$ $$||'g'-f^*||_2=min_{fin$}||'g'-f||_2$$
The author then says:
We have put g in quotes here because we do not knot g.
Instead we intend to use the approximate value $g_i$ whenever the value $g(x_i)$ is called for. In effect 'g' is an equivalence class of functions.
I am not familiar with the concept of "equivalence class of functions" and don't really understand what are the consequences of 'g' being such.
Obviously we ignore the real function 'g' due to the presence of some error term, but I am lost with the mathematical terminology.
Any hint or suggestion will be greatly appreciated.
functional-analysis least-squares regression-analysis
asked Jan 15 at 14:26
Ramiro ScorolliRamiro Scorolli
644114
$endgroup$
add a comment |
$begingroup$
I am dealing with a proof from De Boor (1972) about least square approximations using splines.
Suppose we have a set of data and we want to estimate the least square approximation.
Let $ $ $ be a finite dimensional linear space of functions, we seek a best approximation from $$$ to 'g' with respect to the norm $||cdot||_2 $.
This is we want to find $f^*in$$ such that:
$forall f in $$ $$||'g'-f^*||_2=min_{fin$}||'g'-f||_2$$
The author then says:
We have put g in quotes here because we do not knot g.
Instead we intend to use the approximate value $g_i$ whenever the value $g(x_i)$ is called for. In effect 'g' is an equivalence class of functions.
I am not familiar with the concept of "equivalence class of functions" and don't really understand what are the consequences of 'g' being such.
Obviously we ignore the real function 'g' due to the presence of some error term, but I am lost with the mathematical terminology.
Any hint or suggestion will be greatly appreciated.
functional-analysis least-squares regression-analysis
asked Jan 15 at 14:26
Ramiro ScorolliRamiro Scorolli
644114
$endgroup$
add a comment |
$begingroup$
I am dealing with a proof from De Boor (1972) about least square approximations using splines.
Suppose we have a set of data and we want to estimate the least square approximation.
Let $ $ $ be a finite dimensional linear space of functions, we seek a best approximation from $$$ to 'g' with respect to the norm $||cdot||_2 $.
This is we want to find $f^*in$$ such that:
$forall f in $$ $$||'g'-f^*||_2=min_{fin$}||'g'-f||_2$$
The author then says:
We have put g in quotes here because we do not knot g.
Instead we intend to use the approximate value $g_i$ whenever the value $g(x_i)$ is called for. In effect 'g' is an equivalence class of functions.
I am not familiar with the concept of "equivalence class of functions" and don't really understand what are the consequences of 'g' being such.
Obviously we ignore the real function 'g' due to the presence of some error term, but I am lost with the mathematical terminology.
Any hint or suggestion will be greatly appreciated.
functional-analysis least-squares regression-analysis
asked Jan 15 at 14:26
Ramiro ScorolliRamiro Scorolli
644114
$endgroup$
I am dealing with a proof from De Boor (1972) about least square approximations using splines.
Suppose we have a set of data and we want to estimate the least square approximation.
Let $ $ $ be a finite dimensional linear space of functions, we seek a best approximation from $$$ to 'g' with respect to the norm $||cdot||_2 $.
This is we want to find $f^*in$$ such that:
$forall f in $$ $$||'g'-f^*||_2=min_{fin$}||'g'-f||_2$$
The author then says:
We have put g in quotes here because we do not knot g.
Instead we intend to use the approximate value $g_i$ whenever the value $g(x_i)$ is called for. In effect 'g' is an equivalence class of functions.
I am not familiar with the concept of "equivalence class of functions" and don't really understand what are the consequences of 'g' being such.
Obviously we ignore the real function 'g' due to the presence of some error term, but I am lost with the mathematical terminology.
Any hint or suggestion will be greatly appreciated.
functional-analysis least-squares regression-analysis
functional-analysis least-squares regression-analysis
asked Jan 15 at 14:26
Ramiro ScorolliRamiro Scorolli
644114
asked Jan 15 at 14:26
Ramiro ScorolliRamiro Scorolli
644114
asked Jan 15 at 14:26
Ramiro ScorolliRamiro Scorolli
644114
asked Jan 15 at 14:26
Ramiro ScorolliRamiro Scorolli
644114
asked Jan 15 at 14:26
Ramiro ScorolliRamiro Scorolli
644114
644114
add a comment |
add a comment |
1 Answer
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I think what the author means is simply the usual "equivalence" of functions under the $L^2$ norm, that is we say $f,gin L^2$ are equivalent if $Vert f-g Vert_2 = 0$, the equivalence class of a function is simply the set of functions with are equivalent to it. Note that $Vert f-g Vert_2 = 0$ iff $int vert f(x)-g(x) vert dx = 0$, this may happen without the functions $f,g$ being truely equal, meaning $f(x) = g(x)$ for all $x$, for example, if I take a Reimann integrable function $f:[0,1]rightarrow mathbb{R}$ and change it at one point, it is still equivalent to $f$.
This is relevant in this case since to approximate a function $g$, it doesn't matter which of the functions of the equivalence class $'g'$ you arppoximate.
answered Jan 15 at 16:39
pitariverpitariver
354112
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1 Answer
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1 Answer
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$begingroup$
I think what the author means is simply the usual "equivalence" of functions under the $L^2$ norm, that is we say $f,gin L^2$ are equivalent if $Vert f-g Vert_2 = 0$, the equivalence class of a function is simply the set of functions with are equivalent to it. Note that $Vert f-g Vert_2 = 0$ iff $int vert f(x)-g(x) vert dx = 0$, this may happen without the functions $f,g$ being truely equal, meaning $f(x) = g(x)$ for all $x$, for example, if I take a Reimann integrable function $f:[0,1]rightarrow mathbb{R}$ and change it at one point, it is still equivalent to $f$.
This is relevant in this case since to approximate a function $g$, it doesn't matter which of the functions of the equivalence class $'g'$ you arppoximate.
answered Jan 15 at 16:39
pitariverpitariver
354112
$endgroup$
add a comment |
$begingroup$
I think what the author means is simply the usual "equivalence" of functions under the $L^2$ norm, that is we say $f,gin L^2$ are equivalent if $Vert f-g Vert_2 = 0$, the equivalence class of a function is simply the set of functions with are equivalent to it. Note that $Vert f-g Vert_2 = 0$ iff $int vert f(x)-g(x) vert dx = 0$, this may happen without the functions $f,g$ being truely equal, meaning $f(x) = g(x)$ for all $x$, for example, if I take a Reimann integrable function $f:[0,1]rightarrow mathbb{R}$ and change it at one point, it is still equivalent to $f$.
This is relevant in this case since to approximate a function $g$, it doesn't matter which of the functions of the equivalence class $'g'$ you arppoximate.
answered Jan 15 at 16:39
pitariverpitariver
354112
$endgroup$
add a comment |
$begingroup$
I think what the author means is simply the usual "equivalence" of functions under the $L^2$ norm, that is we say $f,gin L^2$ are equivalent if $Vert f-g Vert_2 = 0$, the equivalence class of a function is simply the set of functions with are equivalent to it. Note that $Vert f-g Vert_2 = 0$ iff $int vert f(x)-g(x) vert dx = 0$, this may happen without the functions $f,g$ being truely equal, meaning $f(x) = g(x)$ for all $x$, for example, if I take a Reimann integrable function $f:[0,1]rightarrow mathbb{R}$ and change it at one point, it is still equivalent to $f$.
This is relevant in this case since to approximate a function $g$, it doesn't matter which of the functions of the equivalence class $'g'$ you arppoximate.
answered Jan 15 at 16:39
pitariverpitariver
354112
$endgroup$
I think what the author means is simply the usual "equivalence" of functions under the $L^2$ norm, that is we say $f,gin L^2$ are equivalent if $Vert f-g Vert_2 = 0$, the equivalence class of a function is simply the set of functions with are equivalent to it. Note that $Vert f-g Vert_2 = 0$ iff $int vert f(x)-g(x) vert dx = 0$, this may happen without the functions $f,g$ being truely equal, meaning $f(x) = g(x)$ for all $x$, for example, if I take a Reimann integrable function $f:[0,1]rightarrow mathbb{R}$ and change it at one point, it is still equivalent to $f$.
This is relevant in this case since to approximate a function $g$, it doesn't matter which of the functions of the equivalence class $'g'$ you arppoximate.
answered Jan 15 at 16:39
pitariverpitariver
354112
answered Jan 15 at 16:39
pitariverpitariver
354112
answered Jan 15 at 16:39
pitariverpitariver
354112
answered Jan 15 at 16:39
pitariverpitariver
354112
354112
add a comment |
add a comment |
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