Intuition Wanted: Why Define Integrals Component-Wise
$begingroup$
In our analysis course, we just defined the following:
Let $g := (g_1, ldots, g_n): [a, b] to mathbb{R}^n$, where $g_1, ldots, g_n: [a,b] to mathbb{R}$ are integrable.
Then we call the integral of $g$ over $[a,b]$
begin{equation*}
int_{a}^{b} g(t) dt
:= begin{pmatrix}
int_{a}^{b} g_1(t) dt \ vdots \ int_{a}^{b} g_n(t) dt
end{pmatrix}.
end{equation*}
I came across this definition again at the beginning of measure theory, when we stated:
We ultimately want to integration functions $mathbb{R}^n to mathbb{R}^m$, but because of the above definition we can, without loss of generality, restrict ourselves to the case $m = 1$.
My Question
What is the intuition behind this definition, why does it make sense, if you will, ''on a deeper level''
integration measure-theory intuition big-picture
asked Dec 20 '18 at 21:16
Viktor GlombikViktor Glombik
799527
$endgroup$
add a comment |
$begingroup$
In our analysis course, we just defined the following:
Let $g := (g_1, ldots, g_n): [a, b] to mathbb{R}^n$, where $g_1, ldots, g_n: [a,b] to mathbb{R}$ are integrable.
Then we call the integral of $g$ over $[a,b]$
begin{equation*}
int_{a}^{b} g(t) dt
:= begin{pmatrix}
int_{a}^{b} g_1(t) dt \ vdots \ int_{a}^{b} g_n(t) dt
end{pmatrix}.
end{equation*}
I came across this definition again at the beginning of measure theory, when we stated:
We ultimately want to integration functions $mathbb{R}^n to mathbb{R}^m$, but because of the above definition we can, without loss of generality, restrict ourselves to the case $m = 1$.
My Question
What is the intuition behind this definition, why does it make sense, if you will, ''on a deeper level''
integration measure-theory intuition big-picture
asked Dec 20 '18 at 21:16
Viktor GlombikViktor Glombik
799527
$endgroup$
add a comment |
$begingroup$
In our analysis course, we just defined the following:
Let $g := (g_1, ldots, g_n): [a, b] to mathbb{R}^n$, where $g_1, ldots, g_n: [a,b] to mathbb{R}$ are integrable.
Then we call the integral of $g$ over $[a,b]$
begin{equation*}
int_{a}^{b} g(t) dt
:= begin{pmatrix}
int_{a}^{b} g_1(t) dt \ vdots \ int_{a}^{b} g_n(t) dt
end{pmatrix}.
end{equation*}
I came across this definition again at the beginning of measure theory, when we stated:
We ultimately want to integration functions $mathbb{R}^n to mathbb{R}^m$, but because of the above definition we can, without loss of generality, restrict ourselves to the case $m = 1$.
My Question
What is the intuition behind this definition, why does it make sense, if you will, ''on a deeper level''
integration measure-theory intuition big-picture
asked Dec 20 '18 at 21:16
Viktor GlombikViktor Glombik
799527
$endgroup$
In our analysis course, we just defined the following:
Let $g := (g_1, ldots, g_n): [a, b] to mathbb{R}^n$, where $g_1, ldots, g_n: [a,b] to mathbb{R}$ are integrable.
Then we call the integral of $g$ over $[a,b]$
begin{equation*}
int_{a}^{b} g(t) dt
:= begin{pmatrix}
int_{a}^{b} g_1(t) dt \ vdots \ int_{a}^{b} g_n(t) dt
end{pmatrix}.
end{equation*}
I came across this definition again at the beginning of measure theory, when we stated:
We ultimately want to integration functions $mathbb{R}^n to mathbb{R}^m$, but because of the above definition we can, without loss of generality, restrict ourselves to the case $m = 1$.
My Question
What is the intuition behind this definition, why does it make sense, if you will, ''on a deeper level''
integration measure-theory intuition big-picture
integration measure-theory intuition big-picture
asked Dec 20 '18 at 21:16
Viktor GlombikViktor Glombik
799527
asked Dec 20 '18 at 21:16
Viktor GlombikViktor Glombik
799527
edited Dec 21 '18 at 19:05
Viktor Glombik
asked Dec 20 '18 at 21:16
Viktor GlombikViktor Glombik
799527
asked Dec 20 '18 at 21:16
Viktor GlombikViktor Glombik
799527
asked Dec 20 '18 at 21:16
Viktor GlombikViktor Glombik
799527
799527
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
The Riemann Integral, say, is based on sums. The sums of vectors are defined component-wise. And different norms on $mathbb R$ are topologically equivalent. Therefore, this is the exact thing you would end up with anyway for Riemann integrals if you defined them by analogy instead of component-wise explicitly.
answered Dec 20 '18 at 21:25
Mark S.Mark S.
11.8k22670
$endgroup$
$begingroup$
I think some of the confusion came from me always thinking of the integral as a functional mapping functions to numbers, but here we get a vector, which suddenly isn't intuitive as a measure.
$endgroup$
– Viktor Glombik
Dec 21 '18 at 19:26
1
$begingroup$
@Viktor Depending on your interests, you might be able to bring things down to earth by examining a physical integral that results in a vector, like the Biot-Savart law.
$endgroup$
– Mark S.
Dec 21 '18 at 19:36
$begingroup$
One thing to consider any time you define something component-wise is whether change of basis respects it. In this case, one might want to check that taking the integral and then changing bases results in the same answer as changing bases and then taking the integral.
$endgroup$
– Acccumulation
Dec 21 '18 at 21:32
add a comment |
$begingroup$
Addition, subtraction, scalar multiplication, scalar division, limits, and differentiation all act component-wise on vectors. You would need a good reason to make integration inconsistent with those!
For the "deeper level": However we define integration of vectors, we want it to be:
- Linear, $int(a+b)=int a+int b$
- Covariant with vector space isomorphisms, $int T f=Tint f$.
- Consistent with the usual embedding of $mathbb R$ in $mathbb R^n$, $int (f, 0, 0, ldots)=(int f, 0, 0, ldots)$.
These axioms force it to act component-wise.
answered Dec 20 '18 at 21:38
Chris CulterChris Culter
20.9k43786
$endgroup$
add a comment |
$begingroup$
On a much more general level, considering sets, mapping into a product $f:Ato Btimes C$ is essentially really just two mapping $f_B:Ato B$ and $f_C:Ato C$. Similarly, mapping out of a disjoint union of sets is just two functions. This extends to products of more than two things and in particular to functions to $mathbb R^n$. This phenomenon justifies the reduction you mention. It is a categorical observation valid in many different contexts.
answered Dec 20 '18 at 21:38
Ittay WeissIttay Weiss
63.8k6101183
$endgroup$
add a comment |
$begingroup$
Perhaps I'm missing the point of your quest for intuition, but we wish to define integrals component-wise because each component "should be independent" from one another. Having $mathbb{R}$ as a codomain is really special; when you try to move towards $mathbb{R}^m$ you lose the ability to totally order things in a natural way, and then all of your geometric intuition about signed areas and things being "under" curves flies out the window.
So how could you get around this? Well I'm not going to say there is not some extremely clever and brilliant definition lurking out there, but why not just pull apart your codomain until each piece looks like stuff you are used to? Why reinvent the wheel when it comes to interpretations of what the sums of strange hyeprvolumes are when we can simply iterate what we already know on multiple dimensions? And going back to my earlier comment, what possible other definition can we come up with that is compatible with our intution for the case of $m=1$?
answered Dec 20 '18 at 21:26
DeficientMathDudeDeficientMathDude
1212
$endgroup$
1
$begingroup$
Why reinvent the wheel? Because when you integrate functions with values in infinite-dimensional spaces (e.g., values in a Banach space that is infinite-dimensional) there are no natural coordinates at all. Conceptually, the definition of integration of functions with values in a normed vector space should not involve coordinates in the definition, but instead prove that as a theorem. In introductory courses, using coordinatewise integration as a definition of integration of functions with values in $mathbf R^n$ is concrete. But for later generalizations, this approach is not good.
$endgroup$
– KCd
Dec 22 '18 at 4:14
add a comment |
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The idea of component-wise integrals has a concrete geometrical meaning: to decompose a function curve by projecting it onto planes, then to take the (signed) areas of these projections in their respective planes.
For example, let's take a look at a function v(t)=(x(t),y(t)). In the picture below, the two components of the integral of v are the area of the red curve in its horizontal plane and the area of the yellow curve in its vertical plane.
Another way to define the integral of v would be to take the integral of the norms ||g(t)||, but I guess component-wise integrals have more respect for structure and thus have nicer algebraic properties.
We can also look at a physical interpretation, to make this even more concrete. In the above, interpret t as time, and v(t) as the velocity of an object O moving within some plane, so that x(t) is its horizontal velocity and y(t) is its vertical velocity, at any instant t. Then the integral of x is the horizontal distance traveled by O, and the integral of y is the vertical distance traveled by O, so that the component-wise integral of v is the 2D distance traveled by O. So with component-wise integrals, the integral of velocity remains the distance, even in multidimensional situations.
(Illustration produced thanks to Octave and this script.)
answered Dec 22 '18 at 20:52
FornostFornost
335
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The Riemann Integral, say, is based on sums. The sums of vectors are defined component-wise. And different norms on $mathbb R$ are topologically equivalent. Therefore, this is the exact thing you would end up with anyway for Riemann integrals if you defined them by analogy instead of component-wise explicitly.
answered Dec 20 '18 at 21:25
Mark S.Mark S.
11.8k22670
$endgroup$
$begingroup$
I think some of the confusion came from me always thinking of the integral as a functional mapping functions to numbers, but here we get a vector, which suddenly isn't intuitive as a measure.
$endgroup$
– Viktor Glombik
Dec 21 '18 at 19:26
1
$begingroup$
@Viktor Depending on your interests, you might be able to bring things down to earth by examining a physical integral that results in a vector, like the Biot-Savart law.
$endgroup$
– Mark S.
Dec 21 '18 at 19:36
$begingroup$
One thing to consider any time you define something component-wise is whether change of basis respects it. In this case, one might want to check that taking the integral and then changing bases results in the same answer as changing bases and then taking the integral.
$endgroup$
– Acccumulation
Dec 21 '18 at 21:32
add a comment |
$begingroup$
The Riemann Integral, say, is based on sums. The sums of vectors are defined component-wise. And different norms on $mathbb R$ are topologically equivalent. Therefore, this is the exact thing you would end up with anyway for Riemann integrals if you defined them by analogy instead of component-wise explicitly.
answered Dec 20 '18 at 21:25
Mark S.Mark S.
11.8k22670
$endgroup$
$begingroup$
I think some of the confusion came from me always thinking of the integral as a functional mapping functions to numbers, but here we get a vector, which suddenly isn't intuitive as a measure.
$endgroup$
– Viktor Glombik
Dec 21 '18 at 19:26
1
$begingroup$
@Viktor Depending on your interests, you might be able to bring things down to earth by examining a physical integral that results in a vector, like the Biot-Savart law.
$endgroup$
– Mark S.
Dec 21 '18 at 19:36
$begingroup$
One thing to consider any time you define something component-wise is whether change of basis respects it. In this case, one might want to check that taking the integral and then changing bases results in the same answer as changing bases and then taking the integral.
$endgroup$
– Acccumulation
Dec 21 '18 at 21:32
add a comment |
$begingroup$
The Riemann Integral, say, is based on sums. The sums of vectors are defined component-wise. And different norms on $mathbb R$ are topologically equivalent. Therefore, this is the exact thing you would end up with anyway for Riemann integrals if you defined them by analogy instead of component-wise explicitly.
answered Dec 20 '18 at 21:25
Mark S.Mark S.
11.8k22670
$endgroup$
The Riemann Integral, say, is based on sums. The sums of vectors are defined component-wise. And different norms on $mathbb R$ are topologically equivalent. Therefore, this is the exact thing you would end up with anyway for Riemann integrals if you defined them by analogy instead of component-wise explicitly.
answered Dec 20 '18 at 21:25
Mark S.Mark S.
11.8k22670
answered Dec 20 '18 at 21:25
Mark S.Mark S.
11.8k22670
answered Dec 20 '18 at 21:25
Mark S.Mark S.
11.8k22670
answered Dec 20 '18 at 21:25
Mark S.Mark S.
11.8k22670
11.8k22670
$begingroup$
I think some of the confusion came from me always thinking of the integral as a functional mapping functions to numbers, but here we get a vector, which suddenly isn't intuitive as a measure.
$endgroup$
– Viktor Glombik
Dec 21 '18 at 19:26
1
$begingroup$
@Viktor Depending on your interests, you might be able to bring things down to earth by examining a physical integral that results in a vector, like the Biot-Savart law.
$endgroup$
– Mark S.
Dec 21 '18 at 19:36
$begingroup$
One thing to consider any time you define something component-wise is whether change of basis respects it. In this case, one might want to check that taking the integral and then changing bases results in the same answer as changing bases and then taking the integral.
$endgroup$
– Acccumulation
Dec 21 '18 at 21:32
add a comment |
$begingroup$
I think some of the confusion came from me always thinking of the integral as a functional mapping functions to numbers, but here we get a vector, which suddenly isn't intuitive as a measure.
$endgroup$
– Viktor Glombik
Dec 21 '18 at 19:26
1
$begingroup$
@Viktor Depending on your interests, you might be able to bring things down to earth by examining a physical integral that results in a vector, like the Biot-Savart law.
$endgroup$
– Mark S.
Dec 21 '18 at 19:36
$begingroup$
One thing to consider any time you define something component-wise is whether change of basis respects it. In this case, one might want to check that taking the integral and then changing bases results in the same answer as changing bases and then taking the integral.
$endgroup$
– Acccumulation
Dec 21 '18 at 21:32
$begingroup$
I think some of the confusion came from me always thinking of the integral as a functional mapping functions to numbers, but here we get a vector, which suddenly isn't intuitive as a measure.
$endgroup$
– Viktor Glombik
Dec 21 '18 at 19:26
$begingroup$
I think some of the confusion came from me always thinking of the integral as a functional mapping functions to numbers, but here we get a vector, which suddenly isn't intuitive as a measure.
$endgroup$
– Viktor Glombik
Dec 21 '18 at 19:26
1
1
$begingroup$
@Viktor Depending on your interests, you might be able to bring things down to earth by examining a physical integral that results in a vector, like the Biot-Savart law.
$endgroup$
– Mark S.
Dec 21 '18 at 19:36
$begingroup$
@Viktor Depending on your interests, you might be able to bring things down to earth by examining a physical integral that results in a vector, like the Biot-Savart law.
$endgroup$
– Mark S.
Dec 21 '18 at 19:36
$begingroup$
One thing to consider any time you define something component-wise is whether change of basis respects it. In this case, one might want to check that taking the integral and then changing bases results in the same answer as changing bases and then taking the integral.
$endgroup$
– Acccumulation
Dec 21 '18 at 21:32
$begingroup$
One thing to consider any time you define something component-wise is whether change of basis respects it. In this case, one might want to check that taking the integral and then changing bases results in the same answer as changing bases and then taking the integral.
$endgroup$
– Acccumulation
Dec 21 '18 at 21:32
add a comment |
$begingroup$
Addition, subtraction, scalar multiplication, scalar division, limits, and differentiation all act component-wise on vectors. You would need a good reason to make integration inconsistent with those!
For the "deeper level": However we define integration of vectors, we want it to be:
- Linear, $int(a+b)=int a+int b$
- Covariant with vector space isomorphisms, $int T f=Tint f$.
- Consistent with the usual embedding of $mathbb R$ in $mathbb R^n$, $int (f, 0, 0, ldots)=(int f, 0, 0, ldots)$.
These axioms force it to act component-wise.
answered Dec 20 '18 at 21:38
Chris CulterChris Culter
20.9k43786
$endgroup$
add a comment |
$begingroup$
Addition, subtraction, scalar multiplication, scalar division, limits, and differentiation all act component-wise on vectors. You would need a good reason to make integration inconsistent with those!
For the "deeper level": However we define integration of vectors, we want it to be:
- Linear, $int(a+b)=int a+int b$
- Covariant with vector space isomorphisms, $int T f=Tint f$.
- Consistent with the usual embedding of $mathbb R$ in $mathbb R^n$, $int (f, 0, 0, ldots)=(int f, 0, 0, ldots)$.
These axioms force it to act component-wise.
answered Dec 20 '18 at 21:38
Chris CulterChris Culter
20.9k43786
$endgroup$
add a comment |
$begingroup$
Addition, subtraction, scalar multiplication, scalar division, limits, and differentiation all act component-wise on vectors. You would need a good reason to make integration inconsistent with those!
For the "deeper level": However we define integration of vectors, we want it to be:
- Linear, $int(a+b)=int a+int b$
- Covariant with vector space isomorphisms, $int T f=Tint f$.
- Consistent with the usual embedding of $mathbb R$ in $mathbb R^n$, $int (f, 0, 0, ldots)=(int f, 0, 0, ldots)$.
These axioms force it to act component-wise.
answered Dec 20 '18 at 21:38
Chris CulterChris Culter
20.9k43786
$endgroup$
Addition, subtraction, scalar multiplication, scalar division, limits, and differentiation all act component-wise on vectors. You would need a good reason to make integration inconsistent with those!
For the "deeper level": However we define integration of vectors, we want it to be:
- Linear, $int(a+b)=int a+int b$
- Covariant with vector space isomorphisms, $int T f=Tint f$.
- Consistent with the usual embedding of $mathbb R$ in $mathbb R^n$, $int (f, 0, 0, ldots)=(int f, 0, 0, ldots)$.
These axioms force it to act component-wise.
answered Dec 20 '18 at 21:38
Chris CulterChris Culter
20.9k43786
answered Dec 20 '18 at 21:38
Chris CulterChris Culter
20.9k43786
answered Dec 20 '18 at 21:38
Chris CulterChris Culter
20.9k43786
answered Dec 20 '18 at 21:38
Chris CulterChris Culter
20.9k43786
20.9k43786
add a comment |
add a comment |
$begingroup$
On a much more general level, considering sets, mapping into a product $f:Ato Btimes C$ is essentially really just two mapping $f_B:Ato B$ and $f_C:Ato C$. Similarly, mapping out of a disjoint union of sets is just two functions. This extends to products of more than two things and in particular to functions to $mathbb R^n$. This phenomenon justifies the reduction you mention. It is a categorical observation valid in many different contexts.
answered Dec 20 '18 at 21:38
Ittay WeissIttay Weiss
63.8k6101183
$endgroup$
add a comment |
$begingroup$
On a much more general level, considering sets, mapping into a product $f:Ato Btimes C$ is essentially really just two mapping $f_B:Ato B$ and $f_C:Ato C$. Similarly, mapping out of a disjoint union of sets is just two functions. This extends to products of more than two things and in particular to functions to $mathbb R^n$. This phenomenon justifies the reduction you mention. It is a categorical observation valid in many different contexts.
answered Dec 20 '18 at 21:38
Ittay WeissIttay Weiss
63.8k6101183
$endgroup$
add a comment |
$begingroup$
On a much more general level, considering sets, mapping into a product $f:Ato Btimes C$ is essentially really just two mapping $f_B:Ato B$ and $f_C:Ato C$. Similarly, mapping out of a disjoint union of sets is just two functions. This extends to products of more than two things and in particular to functions to $mathbb R^n$. This phenomenon justifies the reduction you mention. It is a categorical observation valid in many different contexts.
answered Dec 20 '18 at 21:38
Ittay WeissIttay Weiss
63.8k6101183
$endgroup$
On a much more general level, considering sets, mapping into a product $f:Ato Btimes C$ is essentially really just two mapping $f_B:Ato B$ and $f_C:Ato C$. Similarly, mapping out of a disjoint union of sets is just two functions. This extends to products of more than two things and in particular to functions to $mathbb R^n$. This phenomenon justifies the reduction you mention. It is a categorical observation valid in many different contexts.
answered Dec 20 '18 at 21:38
Ittay WeissIttay Weiss
63.8k6101183
answered Dec 20 '18 at 21:38
Ittay WeissIttay Weiss
63.8k6101183
answered Dec 20 '18 at 21:38
Ittay WeissIttay Weiss
63.8k6101183
answered Dec 20 '18 at 21:38
Ittay WeissIttay Weiss
63.8k6101183
63.8k6101183
add a comment |
add a comment |
$begingroup$
Perhaps I'm missing the point of your quest for intuition, but we wish to define integrals component-wise because each component "should be independent" from one another. Having $mathbb{R}$ as a codomain is really special; when you try to move towards $mathbb{R}^m$ you lose the ability to totally order things in a natural way, and then all of your geometric intuition about signed areas and things being "under" curves flies out the window.
So how could you get around this? Well I'm not going to say there is not some extremely clever and brilliant definition lurking out there, but why not just pull apart your codomain until each piece looks like stuff you are used to? Why reinvent the wheel when it comes to interpretations of what the sums of strange hyeprvolumes are when we can simply iterate what we already know on multiple dimensions? And going back to my earlier comment, what possible other definition can we come up with that is compatible with our intution for the case of $m=1$?
answered Dec 20 '18 at 21:26
DeficientMathDudeDeficientMathDude
1212
$endgroup$
1
$begingroup$
Why reinvent the wheel? Because when you integrate functions with values in infinite-dimensional spaces (e.g., values in a Banach space that is infinite-dimensional) there are no natural coordinates at all. Conceptually, the definition of integration of functions with values in a normed vector space should not involve coordinates in the definition, but instead prove that as a theorem. In introductory courses, using coordinatewise integration as a definition of integration of functions with values in $mathbf R^n$ is concrete. But for later generalizations, this approach is not good.
$endgroup$
– KCd
Dec 22 '18 at 4:14
add a comment |
$begingroup$
Perhaps I'm missing the point of your quest for intuition, but we wish to define integrals component-wise because each component "should be independent" from one another. Having $mathbb{R}$ as a codomain is really special; when you try to move towards $mathbb{R}^m$ you lose the ability to totally order things in a natural way, and then all of your geometric intuition about signed areas and things being "under" curves flies out the window.
So how could you get around this? Well I'm not going to say there is not some extremely clever and brilliant definition lurking out there, but why not just pull apart your codomain until each piece looks like stuff you are used to? Why reinvent the wheel when it comes to interpretations of what the sums of strange hyeprvolumes are when we can simply iterate what we already know on multiple dimensions? And going back to my earlier comment, what possible other definition can we come up with that is compatible with our intution for the case of $m=1$?
answered Dec 20 '18 at 21:26
DeficientMathDudeDeficientMathDude
1212
$endgroup$
1
$begingroup$
Why reinvent the wheel? Because when you integrate functions with values in infinite-dimensional spaces (e.g., values in a Banach space that is infinite-dimensional) there are no natural coordinates at all. Conceptually, the definition of integration of functions with values in a normed vector space should not involve coordinates in the definition, but instead prove that as a theorem. In introductory courses, using coordinatewise integration as a definition of integration of functions with values in $mathbf R^n$ is concrete. But for later generalizations, this approach is not good.
$endgroup$
– KCd
Dec 22 '18 at 4:14
add a comment |
$begingroup$
Perhaps I'm missing the point of your quest for intuition, but we wish to define integrals component-wise because each component "should be independent" from one another. Having $mathbb{R}$ as a codomain is really special; when you try to move towards $mathbb{R}^m$ you lose the ability to totally order things in a natural way, and then all of your geometric intuition about signed areas and things being "under" curves flies out the window.
So how could you get around this? Well I'm not going to say there is not some extremely clever and brilliant definition lurking out there, but why not just pull apart your codomain until each piece looks like stuff you are used to? Why reinvent the wheel when it comes to interpretations of what the sums of strange hyeprvolumes are when we can simply iterate what we already know on multiple dimensions? And going back to my earlier comment, what possible other definition can we come up with that is compatible with our intution for the case of $m=1$?
answered Dec 20 '18 at 21:26
DeficientMathDudeDeficientMathDude
1212
$endgroup$
Perhaps I'm missing the point of your quest for intuition, but we wish to define integrals component-wise because each component "should be independent" from one another. Having $mathbb{R}$ as a codomain is really special; when you try to move towards $mathbb{R}^m$ you lose the ability to totally order things in a natural way, and then all of your geometric intuition about signed areas and things being "under" curves flies out the window.
So how could you get around this? Well I'm not going to say there is not some extremely clever and brilliant definition lurking out there, but why not just pull apart your codomain until each piece looks like stuff you are used to? Why reinvent the wheel when it comes to interpretations of what the sums of strange hyeprvolumes are when we can simply iterate what we already know on multiple dimensions? And going back to my earlier comment, what possible other definition can we come up with that is compatible with our intution for the case of $m=1$?
answered Dec 20 '18 at 21:26
DeficientMathDudeDeficientMathDude
1212
answered Dec 20 '18 at 21:26
DeficientMathDudeDeficientMathDude
1212
answered Dec 20 '18 at 21:26
DeficientMathDudeDeficientMathDude
1212
answered Dec 20 '18 at 21:26
DeficientMathDudeDeficientMathDude
1212
1212
1
$begingroup$
Why reinvent the wheel? Because when you integrate functions with values in infinite-dimensional spaces (e.g., values in a Banach space that is infinite-dimensional) there are no natural coordinates at all. Conceptually, the definition of integration of functions with values in a normed vector space should not involve coordinates in the definition, but instead prove that as a theorem. In introductory courses, using coordinatewise integration as a definition of integration of functions with values in $mathbf R^n$ is concrete. But for later generalizations, this approach is not good.
$endgroup$
– KCd
Dec 22 '18 at 4:14
add a comment |
1
$begingroup$
Why reinvent the wheel? Because when you integrate functions with values in infinite-dimensional spaces (e.g., values in a Banach space that is infinite-dimensional) there are no natural coordinates at all. Conceptually, the definition of integration of functions with values in a normed vector space should not involve coordinates in the definition, but instead prove that as a theorem. In introductory courses, using coordinatewise integration as a definition of integration of functions with values in $mathbf R^n$ is concrete. But for later generalizations, this approach is not good.
$endgroup$
– KCd
Dec 22 '18 at 4:14
1
1
$begingroup$
Why reinvent the wheel? Because when you integrate functions with values in infinite-dimensional spaces (e.g., values in a Banach space that is infinite-dimensional) there are no natural coordinates at all. Conceptually, the definition of integration of functions with values in a normed vector space should not involve coordinates in the definition, but instead prove that as a theorem. In introductory courses, using coordinatewise integration as a definition of integration of functions with values in $mathbf R^n$ is concrete. But for later generalizations, this approach is not good.
$endgroup$
– KCd
Dec 22 '18 at 4:14
$begingroup$
Why reinvent the wheel? Because when you integrate functions with values in infinite-dimensional spaces (e.g., values in a Banach space that is infinite-dimensional) there are no natural coordinates at all. Conceptually, the definition of integration of functions with values in a normed vector space should not involve coordinates in the definition, but instead prove that as a theorem. In introductory courses, using coordinatewise integration as a definition of integration of functions with values in $mathbf R^n$ is concrete. But for later generalizations, this approach is not good.
$endgroup$
– KCd
Dec 22 '18 at 4:14
add a comment |
$begingroup$
The idea of component-wise integrals has a concrete geometrical meaning: to decompose a function curve by projecting it onto planes, then to take the (signed) areas of these projections in their respective planes.
For example, let's take a look at a function v(t)=(x(t),y(t)). In the picture below, the two components of the integral of v are the area of the red curve in its horizontal plane and the area of the yellow curve in its vertical plane.
Another way to define the integral of v would be to take the integral of the norms ||g(t)||, but I guess component-wise integrals have more respect for structure and thus have nicer algebraic properties.
We can also look at a physical interpretation, to make this even more concrete. In the above, interpret t as time, and v(t) as the velocity of an object O moving within some plane, so that x(t) is its horizontal velocity and y(t) is its vertical velocity, at any instant t. Then the integral of x is the horizontal distance traveled by O, and the integral of y is the vertical distance traveled by O, so that the component-wise integral of v is the 2D distance traveled by O. So with component-wise integrals, the integral of velocity remains the distance, even in multidimensional situations.
(Illustration produced thanks to Octave and this script.)
answered Dec 22 '18 at 20:52
FornostFornost
335
$endgroup$
add a comment |
$begingroup$
The idea of component-wise integrals has a concrete geometrical meaning: to decompose a function curve by projecting it onto planes, then to take the (signed) areas of these projections in their respective planes.
For example, let's take a look at a function v(t)=(x(t),y(t)). In the picture below, the two components of the integral of v are the area of the red curve in its horizontal plane and the area of the yellow curve in its vertical plane.
Another way to define the integral of v would be to take the integral of the norms ||g(t)||, but I guess component-wise integrals have more respect for structure and thus have nicer algebraic properties.
We can also look at a physical interpretation, to make this even more concrete. In the above, interpret t as time, and v(t) as the velocity of an object O moving within some plane, so that x(t) is its horizontal velocity and y(t) is its vertical velocity, at any instant t. Then the integral of x is the horizontal distance traveled by O, and the integral of y is the vertical distance traveled by O, so that the component-wise integral of v is the 2D distance traveled by O. So with component-wise integrals, the integral of velocity remains the distance, even in multidimensional situations.
(Illustration produced thanks to Octave and this script.)
answered Dec 22 '18 at 20:52
FornostFornost
335
$endgroup$
add a comment |
$begingroup$
The idea of component-wise integrals has a concrete geometrical meaning: to decompose a function curve by projecting it onto planes, then to take the (signed) areas of these projections in their respective planes.
For example, let's take a look at a function v(t)=(x(t),y(t)). In the picture below, the two components of the integral of v are the area of the red curve in its horizontal plane and the area of the yellow curve in its vertical plane.
Another way to define the integral of v would be to take the integral of the norms ||g(t)||, but I guess component-wise integrals have more respect for structure and thus have nicer algebraic properties.
We can also look at a physical interpretation, to make this even more concrete. In the above, interpret t as time, and v(t) as the velocity of an object O moving within some plane, so that x(t) is its horizontal velocity and y(t) is its vertical velocity, at any instant t. Then the integral of x is the horizontal distance traveled by O, and the integral of y is the vertical distance traveled by O, so that the component-wise integral of v is the 2D distance traveled by O. So with component-wise integrals, the integral of velocity remains the distance, even in multidimensional situations.
(Illustration produced thanks to Octave and this script.)
answered Dec 22 '18 at 20:52
FornostFornost
335
$endgroup$
The idea of component-wise integrals has a concrete geometrical meaning: to decompose a function curve by projecting it onto planes, then to take the (signed) areas of these projections in their respective planes.
For example, let's take a look at a function v(t)=(x(t),y(t)). In the picture below, the two components of the integral of v are the area of the red curve in its horizontal plane and the area of the yellow curve in its vertical plane.
Another way to define the integral of v would be to take the integral of the norms ||g(t)||, but I guess component-wise integrals have more respect for structure and thus have nicer algebraic properties.
We can also look at a physical interpretation, to make this even more concrete. In the above, interpret t as time, and v(t) as the velocity of an object O moving within some plane, so that x(t) is its horizontal velocity and y(t) is its vertical velocity, at any instant t. Then the integral of x is the horizontal distance traveled by O, and the integral of y is the vertical distance traveled by O, so that the component-wise integral of v is the 2D distance traveled by O. So with component-wise integrals, the integral of velocity remains the distance, even in multidimensional situations.
(Illustration produced thanks to Octave and this script.)
answered Dec 22 '18 at 20:52
FornostFornost
335
edited Dec 22 '18 at 22:19
answered Dec 22 '18 at 20:52
FornostFornost
335
answered Dec 22 '18 at 20:52
FornostFornost
335
answered Dec 22 '18 at 20:52
FornostFornost
335
335
add a comment |
add a comment |
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