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Definition of double integral in $epsilon - delta$ terms.
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When we define double integral for function $f$, we usually start with it's definition on a rectangle $R=[a,b]times[c,d]$ which can be split into bunch of smaller rectangles $[a,b]=[x_0,x_1]cup[x_1,x_2]cup...cup[x_n-1,x_n]$ and $[c,d]=[y_0,y_1]cup[y_1,y_2]cup...cup[y_n-1,y_n]$ then forming a sum
where $Delta x_i=x_i-x_i-1$ and $Delta y_j=y_j-y_j-1$ and $(epsilon_i, gamma_j)$ is point inside rectangle $R_ij$
Now, if we choose the rectangle with greatest diameter (let that diameter be $d_ij$ ) and if we what to find limit of the given sum when $d_ij rightarrow 0 $, then if that sum has a finite limit regardless of how we split the intervals $[a,b]$ and $[c,d]$ and which points inside $R_ij$ rectangles we choose, then, that limit of the sum is called DOUBLE INTEGRAL of $f$ on $R$.
However, i am supposed to understand the definition of double integral in $epsilon - delta$ terms.
Defining double integral this way makes sense since this is obviously a limit and we know that limits can be definied in $epsilon - delta$ terms.
There it goes:
Number $I$ is called double integral of fucntion $f$ on rectangle $R$ if for every $epsilon > 0$ there is $delta >0$ such that if we split intervals $[a,b]$ and $[c,d]$ and we get $d_ij<delta$, then, regardles of points chosen following inequality is true:
$$| sum_i=1^m sum_j=1^n f(epsilon_i, gamma_j)Delta x_i Delta y_j - I | < epsilon$$
Now, this should't be difficult to understand, but i struggle with it very much, since i cannot understand why is it required that we have to split given intervals in such way that $d_ij<delta$, why this is not working if we split the intervals in a such way that $d_ij>delta$ (AGAIN: $d_ij$ is a diameter of largest rectangle we got when we split the two intervals given). Any help is appreciated!
calculus multivariable-calculus
asked Aug 6 at 14:38
cdummie
694412
add a comment |Â
up vote
1
down vote
favorite
When we define double integral for function $f$, we usually start with it's definition on a rectangle $R=[a,b]times[c,d]$ which can be split into bunch of smaller rectangles $[a,b]=[x_0,x_1]cup[x_1,x_2]cup...cup[x_n-1,x_n]$ and $[c,d]=[y_0,y_1]cup[y_1,y_2]cup...cup[y_n-1,y_n]$ then forming a sum
where $Delta x_i=x_i-x_i-1$ and $Delta y_j=y_j-y_j-1$ and $(epsilon_i, gamma_j)$ is point inside rectangle $R_ij$
Now, if we choose the rectangle with greatest diameter (let that diameter be $d_ij$ ) and if we what to find limit of the given sum when $d_ij rightarrow 0 $, then if that sum has a finite limit regardless of how we split the intervals $[a,b]$ and $[c,d]$ and which points inside $R_ij$ rectangles we choose, then, that limit of the sum is called DOUBLE INTEGRAL of $f$ on $R$.
However, i am supposed to understand the definition of double integral in $epsilon - delta$ terms.
Defining double integral this way makes sense since this is obviously a limit and we know that limits can be definied in $epsilon - delta$ terms.
There it goes:
Number $I$ is called double integral of fucntion $f$ on rectangle $R$ if for every $epsilon > 0$ there is $delta >0$ such that if we split intervals $[a,b]$ and $[c,d]$ and we get $d_ij<delta$, then, regardles of points chosen following inequality is true:
$$| sum_i=1^m sum_j=1^n f(epsilon_i, gamma_j)Delta x_i Delta y_j - I | < epsilon$$
Now, this should't be difficult to understand, but i struggle with it very much, since i cannot understand why is it required that we have to split given intervals in such way that $d_ij<delta$, why this is not working if we split the intervals in a such way that $d_ij>delta$ (AGAIN: $d_ij$ is a diameter of largest rectangle we got when we split the two intervals given). Any help is appreciated!
calculus multivariable-calculus
asked Aug 6 at 14:38
cdummie
694412
2
You might want to remove the ambiguity of notation: $d$ is used for the right hand endpoint of the interval $[c,d]$; and it is also used for the maximum diameter of the rectangles $R_ij$.
– Lee Mosher
Aug 6 at 14:41
1
@Arthur: It sounds like you are referring to Fubini's Theorem, which equates certain multiple integrals to iterated integrals. I would say that it functions well as a mathematical theorem, but quite poorly as a philosophy: defining a multiple integral as an iterated integral is a bad idea.
– Lee Mosher
Aug 6 at 14:52
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
When we define double integral for function $f$, we usually start with it's definition on a rectangle $R=[a,b]times[c,d]$ which can be split into bunch of smaller rectangles $[a,b]=[x_0,x_1]cup[x_1,x_2]cup...cup[x_n-1,x_n]$ and $[c,d]=[y_0,y_1]cup[y_1,y_2]cup...cup[y_n-1,y_n]$ then forming a sum
where $Delta x_i=x_i-x_i-1$ and $Delta y_j=y_j-y_j-1$ and $(epsilon_i, gamma_j)$ is point inside rectangle $R_ij$
Now, if we choose the rectangle with greatest diameter (let that diameter be $d_ij$ ) and if we what to find limit of the given sum when $d_ij rightarrow 0 $, then if that sum has a finite limit regardless of how we split the intervals $[a,b]$ and $[c,d]$ and which points inside $R_ij$ rectangles we choose, then, that limit of the sum is called DOUBLE INTEGRAL of $f$ on $R$.
However, i am supposed to understand the definition of double integral in $epsilon - delta$ terms.
Defining double integral this way makes sense since this is obviously a limit and we know that limits can be definied in $epsilon - delta$ terms.
There it goes:
Number $I$ is called double integral of fucntion $f$ on rectangle $R$ if for every $epsilon > 0$ there is $delta >0$ such that if we split intervals $[a,b]$ and $[c,d]$ and we get $d_ij<delta$, then, regardles of points chosen following inequality is true:
$$| sum_i=1^m sum_j=1^n f(epsilon_i, gamma_j)Delta x_i Delta y_j - I | < epsilon$$
Now, this should't be difficult to understand, but i struggle with it very much, since i cannot understand why is it required that we have to split given intervals in such way that $d_ij<delta$, why this is not working if we split the intervals in a such way that $d_ij>delta$ (AGAIN: $d_ij$ is a diameter of largest rectangle we got when we split the two intervals given). Any help is appreciated!
calculus multivariable-calculus
asked Aug 6 at 14:38
cdummie
694412
When we define double integral for function $f$, we usually start with it's definition on a rectangle $R=[a,b]times[c,d]$ which can be split into bunch of smaller rectangles $[a,b]=[x_0,x_1]cup[x_1,x_2]cup...cup[x_n-1,x_n]$ and $[c,d]=[y_0,y_1]cup[y_1,y_2]cup...cup[y_n-1,y_n]$ then forming a sum
where $Delta x_i=x_i-x_i-1$ and $Delta y_j=y_j-y_j-1$ and $(epsilon_i, gamma_j)$ is point inside rectangle $R_ij$
Now, if we choose the rectangle with greatest diameter (let that diameter be $d_ij$ ) and if we what to find limit of the given sum when $d_ij rightarrow 0 $, then if that sum has a finite limit regardless of how we split the intervals $[a,b]$ and $[c,d]$ and which points inside $R_ij$ rectangles we choose, then, that limit of the sum is called DOUBLE INTEGRAL of $f$ on $R$.
However, i am supposed to understand the definition of double integral in $epsilon - delta$ terms.
Defining double integral this way makes sense since this is obviously a limit and we know that limits can be definied in $epsilon - delta$ terms.
There it goes:
Number $I$ is called double integral of fucntion $f$ on rectangle $R$ if for every $epsilon > 0$ there is $delta >0$ such that if we split intervals $[a,b]$ and $[c,d]$ and we get $d_ij<delta$, then, regardles of points chosen following inequality is true:
$$| sum_i=1^m sum_j=1^n f(epsilon_i, gamma_j)Delta x_i Delta y_j - I | < epsilon$$
Now, this should't be difficult to understand, but i struggle with it very much, since i cannot understand why is it required that we have to split given intervals in such way that $d_ij<delta$, why this is not working if we split the intervals in a such way that $d_ij>delta$ (AGAIN: $d_ij$ is a diameter of largest rectangle we got when we split the two intervals given). Any help is appreciated!
calculus multivariable-calculus
asked Aug 6 at 14:38
cdummie
694412
edited Aug 7 at 6:32
asked Aug 6 at 14:38
cdummie
694412
asked Aug 6 at 14:38
cdummie
694412
asked Aug 6 at 14:38
cdummie
694412
694412
2
You might want to remove the ambiguity of notation: $d$ is used for the right hand endpoint of the interval $[c,d]$; and it is also used for the maximum diameter of the rectangles $R_ij$.
– Lee Mosher
Aug 6 at 14:41
1
@Arthur: It sounds like you are referring to Fubini's Theorem, which equates certain multiple integrals to iterated integrals. I would say that it functions well as a mathematical theorem, but quite poorly as a philosophy: defining a multiple integral as an iterated integral is a bad idea.
– Lee Mosher
Aug 6 at 14:52
add a comment |Â
2
You might want to remove the ambiguity of notation: $d$ is used for the right hand endpoint of the interval $[c,d]$; and it is also used for the maximum diameter of the rectangles $R_ij$.
– Lee Mosher
Aug 6 at 14:41
1
@Arthur: It sounds like you are referring to Fubini's Theorem, which equates certain multiple integrals to iterated integrals. I would say that it functions well as a mathematical theorem, but quite poorly as a philosophy: defining a multiple integral as an iterated integral is a bad idea.
– Lee Mosher
Aug 6 at 14:52
2
2
You might want to remove the ambiguity of notation: $d$ is used for the right hand endpoint of the interval $[c,d]$; and it is also used for the maximum diameter of the rectangles $R_ij$.
– Lee Mosher
Aug 6 at 14:41
You might want to remove the ambiguity of notation: $d$ is used for the right hand endpoint of the interval $[c,d]$; and it is also used for the maximum diameter of the rectangles $R_ij$.
– Lee Mosher
Aug 6 at 14:41
1
1
@Arthur: It sounds like you are referring to Fubini's Theorem, which equates certain multiple integrals to iterated integrals. I would say that it functions well as a mathematical theorem, but quite poorly as a philosophy: defining a multiple integral as an iterated integral is a bad idea.
– Lee Mosher
Aug 6 at 14:52
@Arthur: It sounds like you are referring to Fubini's Theorem, which equates certain multiple integrals to iterated integrals. I would say that it functions well as a mathematical theorem, but quite poorly as a philosophy: defining a multiple integral as an iterated integral is a bad idea.
– Lee Mosher
Aug 6 at 14:52
add a comment |Â
1 Answer
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why this is not working if we split the intervals in a such way that $d > delta$
It's not clear what you mean by “not working†here. Do you mean, why is it necessary that $d<delta$? Otherwise, the condition is not equivalent to the definition.
I mean, it's possible that for some $epsilon$, there exists a partition with $d > delta$ and the difference between the Riemann sum and $I$ is smaller than $epsilon$. But if you are looking for functions which satisfy that condition for all partitions with $d>delta$, you're going to eliminate a lot of integrable functions.
Compare it to the $epsilon$-$delta$ definition of limit. We say $lim_xto af(x) = L$ if for all $epsilon > 0$ there exists a $delta > 0$ such that $0 < | x- a | < delta implies |f(x) - L| < epsilon$. Why do we rule out points $x$ with $|x - a| > delta$? Without it, the statement that $lim_xto a x = a$ wouldn't be true.
answered Aug 6 at 15:05
Matthew Leingang
15k12143
i can imagine in my head what happens when we are talking about limits defined that way, it is just that when we reach the point from the different sides we will end up with the same value despite of the way we approach it, but what is going on here, in this specific case, what if $d_ij>delta$ (ive changed symbol from $d$ to $d_ij$ , see Lee Moshers comment) i need to understand what geometrically happens there.
– cdummie
Aug 6 at 15:20
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
why this is not working if we split the intervals in a such way that $d > delta$
It's not clear what you mean by “not working†here. Do you mean, why is it necessary that $d<delta$? Otherwise, the condition is not equivalent to the definition.
I mean, it's possible that for some $epsilon$, there exists a partition with $d > delta$ and the difference between the Riemann sum and $I$ is smaller than $epsilon$. But if you are looking for functions which satisfy that condition for all partitions with $d>delta$, you're going to eliminate a lot of integrable functions.
Compare it to the $epsilon$-$delta$ definition of limit. We say $lim_xto af(x) = L$ if for all $epsilon > 0$ there exists a $delta > 0$ such that $0 < | x- a | < delta implies |f(x) - L| < epsilon$. Why do we rule out points $x$ with $|x - a| > delta$? Without it, the statement that $lim_xto a x = a$ wouldn't be true.
answered Aug 6 at 15:05
Matthew Leingang
15k12143
i can imagine in my head what happens when we are talking about limits defined that way, it is just that when we reach the point from the different sides we will end up with the same value despite of the way we approach it, but what is going on here, in this specific case, what if $d_ij>delta$ (ive changed symbol from $d$ to $d_ij$ , see Lee Moshers comment) i need to understand what geometrically happens there.
– cdummie
Aug 6 at 15:20
add a comment |Â
up vote
0
down vote
why this is not working if we split the intervals in a such way that $d > delta$
It's not clear what you mean by “not working†here. Do you mean, why is it necessary that $d<delta$? Otherwise, the condition is not equivalent to the definition.
I mean, it's possible that for some $epsilon$, there exists a partition with $d > delta$ and the difference between the Riemann sum and $I$ is smaller than $epsilon$. But if you are looking for functions which satisfy that condition for all partitions with $d>delta$, you're going to eliminate a lot of integrable functions.
Compare it to the $epsilon$-$delta$ definition of limit. We say $lim_xto af(x) = L$ if for all $epsilon > 0$ there exists a $delta > 0$ such that $0 < | x- a | < delta implies |f(x) - L| < epsilon$. Why do we rule out points $x$ with $|x - a| > delta$? Without it, the statement that $lim_xto a x = a$ wouldn't be true.
answered Aug 6 at 15:05
Matthew Leingang
15k12143
i can imagine in my head what happens when we are talking about limits defined that way, it is just that when we reach the point from the different sides we will end up with the same value despite of the way we approach it, but what is going on here, in this specific case, what if $d_ij>delta$ (ive changed symbol from $d$ to $d_ij$ , see Lee Moshers comment) i need to understand what geometrically happens there.
– cdummie
Aug 6 at 15:20
add a comment |Â
up vote
0
down vote
up vote
0
down vote
why this is not working if we split the intervals in a such way that $d > delta$
It's not clear what you mean by “not working†here. Do you mean, why is it necessary that $d<delta$? Otherwise, the condition is not equivalent to the definition.
I mean, it's possible that for some $epsilon$, there exists a partition with $d > delta$ and the difference between the Riemann sum and $I$ is smaller than $epsilon$. But if you are looking for functions which satisfy that condition for all partitions with $d>delta$, you're going to eliminate a lot of integrable functions.
Compare it to the $epsilon$-$delta$ definition of limit. We say $lim_xto af(x) = L$ if for all $epsilon > 0$ there exists a $delta > 0$ such that $0 < | x- a | < delta implies |f(x) - L| < epsilon$. Why do we rule out points $x$ with $|x - a| > delta$? Without it, the statement that $lim_xto a x = a$ wouldn't be true.
answered Aug 6 at 15:05
Matthew Leingang
15k12143
why this is not working if we split the intervals in a such way that $d > delta$
It's not clear what you mean by “not working†here. Do you mean, why is it necessary that $d<delta$? Otherwise, the condition is not equivalent to the definition.
I mean, it's possible that for some $epsilon$, there exists a partition with $d > delta$ and the difference between the Riemann sum and $I$ is smaller than $epsilon$. But if you are looking for functions which satisfy that condition for all partitions with $d>delta$, you're going to eliminate a lot of integrable functions.
Compare it to the $epsilon$-$delta$ definition of limit. We say $lim_xto af(x) = L$ if for all $epsilon > 0$ there exists a $delta > 0$ such that $0 < | x- a | < delta implies |f(x) - L| < epsilon$. Why do we rule out points $x$ with $|x - a| > delta$? Without it, the statement that $lim_xto a x = a$ wouldn't be true.
answered Aug 6 at 15:05
Matthew Leingang
15k12143
answered Aug 6 at 15:05
Matthew Leingang
15k12143
answered Aug 6 at 15:05
Matthew Leingang
15k12143
answered Aug 6 at 15:05
Matthew Leingang
15k12143
15k12143
i can imagine in my head what happens when we are talking about limits defined that way, it is just that when we reach the point from the different sides we will end up with the same value despite of the way we approach it, but what is going on here, in this specific case, what if $d_ij>delta$ (ive changed symbol from $d$ to $d_ij$ , see Lee Moshers comment) i need to understand what geometrically happens there.
– cdummie
Aug 6 at 15:20
add a comment |Â
i can imagine in my head what happens when we are talking about limits defined that way, it is just that when we reach the point from the different sides we will end up with the same value despite of the way we approach it, but what is going on here, in this specific case, what if $d_ij>delta$ (ive changed symbol from $d$ to $d_ij$ , see Lee Moshers comment) i need to understand what geometrically happens there.
– cdummie
Aug 6 at 15:20
i can imagine in my head what happens when we are talking about limits defined that way, it is just that when we reach the point from the different sides we will end up with the same value despite of the way we approach it, but what is going on here, in this specific case, what if $d_ij>delta$ (ive changed symbol from $d$ to $d_ij$ , see Lee Moshers comment) i need to understand what geometrically happens there.
– cdummie
Aug 6 at 15:20
i can imagine in my head what happens when we are talking about limits defined that way, it is just that when we reach the point from the different sides we will end up with the same value despite of the way we approach it, but what is going on here, in this specific case, what if $d_ij>delta$ (ive changed symbol from $d$ to $d_ij$ , see Lee Moshers comment) i need to understand what geometrically happens there.
– cdummie
Aug 6 at 15:20
add a comment |Â
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2
You might want to remove the ambiguity of notation: $d$ is used for the right hand endpoint of the interval $[c,d]$; and it is also used for the maximum diameter of the rectangles $R_ij$.
– Lee Mosher
Aug 6 at 14:41
1
@Arthur: It sounds like you are referring to Fubini's Theorem, which equates certain multiple integrals to iterated integrals. I would say that it functions well as a mathematical theorem, but quite poorly as a philosophy: defining a multiple integral as an iterated integral is a bad idea.
– Lee Mosher
Aug 6 at 14:52