prooving volume formular for simplices
$begingroup$
let $ 1leq n $ and $a_1,...,a_n mathbb{R}^{+} $
A simplex $ sigma $ ist given by $ sigma := [0,a_1 e^1, .., a_n e^n ] $
I want to proove that
$$ int_ {sigma } 1= frac{1}{n!} prod_{j=1}^n a_j $$
per induction.
first, I wanted to clarify the formula for lower n.
for $ n=1 , [0,e^1]$
you get $int_{[0,e^1] }1= int_0^{e^1} dx= e^1$
and set $e^1 = 1$
for $n=2, [0,e^1,e^2]$
you get
$int_{[0,e^1,e^2]} 1 = int_0^{e^1} int_0^y dx dy = frac{(e^1)^2}{2} $
for $ n=3, [0,e^1,e^2,e^3] $
you get, that
$ int_{[0,e^1,e^2,e^3]} 1= int_0^{e^1} int_0^z int_0^y dxdydz= frac{(e^1)^3}{6} $
my question is also, if this is formally right? fells kinda wrong
so
$int_{[0,e^1,e^n,e^{n+1}]} 1 = int_0^{e1} int_0^{x_{n+1}}....int_0^{x_2} dx_1...dx_n dx_{n+1} $
$= int_0^{e^1} frac{1}{n!} x_{n+1}^n dx_{n+1} = frac{e^{n+1}}{(n+1)!} $
and also set $ e^{n+1}=1$
..again..is this formally correct?
integration multivariable-calculus simplex
asked Jan 15 at 14:34
wondering1123wondering1123
14911
$endgroup$
add a comment |
$begingroup$
let $ 1leq n $ and $a_1,...,a_n mathbb{R}^{+} $
A simplex $ sigma $ ist given by $ sigma := [0,a_1 e^1, .., a_n e^n ] $
I want to proove that
$$ int_ {sigma } 1= frac{1}{n!} prod_{j=1}^n a_j $$
per induction.
first, I wanted to clarify the formula for lower n.
for $ n=1 , [0,e^1]$
you get $int_{[0,e^1] }1= int_0^{e^1} dx= e^1$
and set $e^1 = 1$
for $n=2, [0,e^1,e^2]$
you get
$int_{[0,e^1,e^2]} 1 = int_0^{e^1} int_0^y dx dy = frac{(e^1)^2}{2} $
for $ n=3, [0,e^1,e^2,e^3] $
you get, that
$ int_{[0,e^1,e^2,e^3]} 1= int_0^{e^1} int_0^z int_0^y dxdydz= frac{(e^1)^3}{6} $
my question is also, if this is formally right? fells kinda wrong
so
$int_{[0,e^1,e^n,e^{n+1}]} 1 = int_0^{e1} int_0^{x_{n+1}}....int_0^{x_2} dx_1...dx_n dx_{n+1} $
$= int_0^{e^1} frac{1}{n!} x_{n+1}^n dx_{n+1} = frac{e^{n+1}}{(n+1)!} $
and also set $ e^{n+1}=1$
..again..is this formally correct?
integration multivariable-calculus simplex
asked Jan 15 at 14:34
wondering1123wondering1123
14911
$endgroup$
1
$begingroup$
One way to do this is to construct a linear map that converts the simplex into the standard simplex and use the change of variables formula.
$endgroup$
– mlainz
Jan 16 at 11:03
$begingroup$
The reasonning behind the edit is perfectly fine, but the notations are a bit off. As far as I understand, your $e^i$ are vectors, so they shouldn't be included in your integral bounds. Basically, you should consider that $int_{[0,e^1]} f = int_0^1 f(x) dx$ is a definition. Thus, you do not need to "set e^{n+1} = 1". (which I would consider nonsensical in this context)
$endgroup$
– G. Fougeron
Jan 16 at 15:19
add a comment |
$begingroup$
let $ 1leq n $ and $a_1,...,a_n mathbb{R}^{+} $
A simplex $ sigma $ ist given by $ sigma := [0,a_1 e^1, .., a_n e^n ] $
I want to proove that
$$ int_ {sigma } 1= frac{1}{n!} prod_{j=1}^n a_j $$
per induction.
first, I wanted to clarify the formula for lower n.
for $ n=1 , [0,e^1]$
you get $int_{[0,e^1] }1= int_0^{e^1} dx= e^1$
and set $e^1 = 1$
for $n=2, [0,e^1,e^2]$
you get
$int_{[0,e^1,e^2]} 1 = int_0^{e^1} int_0^y dx dy = frac{(e^1)^2}{2} $
for $ n=3, [0,e^1,e^2,e^3] $
you get, that
$ int_{[0,e^1,e^2,e^3]} 1= int_0^{e^1} int_0^z int_0^y dxdydz= frac{(e^1)^3}{6} $
my question is also, if this is formally right? fells kinda wrong
so
$int_{[0,e^1,e^n,e^{n+1}]} 1 = int_0^{e1} int_0^{x_{n+1}}....int_0^{x_2} dx_1...dx_n dx_{n+1} $
$= int_0^{e^1} frac{1}{n!} x_{n+1}^n dx_{n+1} = frac{e^{n+1}}{(n+1)!} $
and also set $ e^{n+1}=1$
..again..is this formally correct?
integration multivariable-calculus simplex
asked Jan 15 at 14:34
wondering1123wondering1123
14911
$endgroup$
let $ 1leq n $ and $a_1,...,a_n mathbb{R}^{+} $
A simplex $ sigma $ ist given by $ sigma := [0,a_1 e^1, .., a_n e^n ] $
I want to proove that
$$ int_ {sigma } 1= frac{1}{n!} prod_{j=1}^n a_j $$
per induction.
first, I wanted to clarify the formula for lower n.
for $ n=1 , [0,e^1]$
you get $int_{[0,e^1] }1= int_0^{e^1} dx= e^1$
and set $e^1 = 1$
for $n=2, [0,e^1,e^2]$
you get
$int_{[0,e^1,e^2]} 1 = int_0^{e^1} int_0^y dx dy = frac{(e^1)^2}{2} $
for $ n=3, [0,e^1,e^2,e^3] $
you get, that
$ int_{[0,e^1,e^2,e^3]} 1= int_0^{e^1} int_0^z int_0^y dxdydz= frac{(e^1)^3}{6} $
my question is also, if this is formally right? fells kinda wrong
so
$int_{[0,e^1,e^n,e^{n+1}]} 1 = int_0^{e1} int_0^{x_{n+1}}....int_0^{x_2} dx_1...dx_n dx_{n+1} $
$= int_0^{e^1} frac{1}{n!} x_{n+1}^n dx_{n+1} = frac{e^{n+1}}{(n+1)!} $
and also set $ e^{n+1}=1$
..again..is this formally correct?
integration multivariable-calculus simplex
integration multivariable-calculus simplex
asked Jan 15 at 14:34
wondering1123wondering1123
14911
asked Jan 15 at 14:34
wondering1123wondering1123
14911
edited Jan 16 at 12:13
wondering1123
asked Jan 15 at 14:34
wondering1123wondering1123
14911
asked Jan 15 at 14:34
wondering1123wondering1123
14911
asked Jan 15 at 14:34
wondering1123wondering1123
14911
14911
1
$begingroup$
One way to do this is to construct a linear map that converts the simplex into the standard simplex and use the change of variables formula.
$endgroup$
– mlainz
Jan 16 at 11:03
$begingroup$
The reasonning behind the edit is perfectly fine, but the notations are a bit off. As far as I understand, your $e^i$ are vectors, so they shouldn't be included in your integral bounds. Basically, you should consider that $int_{[0,e^1]} f = int_0^1 f(x) dx$ is a definition. Thus, you do not need to "set e^{n+1} = 1". (which I would consider nonsensical in this context)
$endgroup$
– G. Fougeron
Jan 16 at 15:19
add a comment |
1
$begingroup$
One way to do this is to construct a linear map that converts the simplex into the standard simplex and use the change of variables formula.
$endgroup$
– mlainz
Jan 16 at 11:03
$begingroup$
The reasonning behind the edit is perfectly fine, but the notations are a bit off. As far as I understand, your $e^i$ are vectors, so they shouldn't be included in your integral bounds. Basically, you should consider that $int_{[0,e^1]} f = int_0^1 f(x) dx$ is a definition. Thus, you do not need to "set e^{n+1} = 1". (which I would consider nonsensical in this context)
$endgroup$
– G. Fougeron
Jan 16 at 15:19
1
1
$begingroup$
One way to do this is to construct a linear map that converts the simplex into the standard simplex and use the change of variables formula.
$endgroup$
– mlainz
Jan 16 at 11:03
$begingroup$
One way to do this is to construct a linear map that converts the simplex into the standard simplex and use the change of variables formula.
$endgroup$
– mlainz
Jan 16 at 11:03
$begingroup$
The reasonning behind the edit is perfectly fine, but the notations are a bit off. As far as I understand, your $e^i$ are vectors, so they shouldn't be included in your integral bounds. Basically, you should consider that $int_{[0,e^1]} f = int_0^1 f(x) dx$ is a definition. Thus, you do not need to "set e^{n+1} = 1". (which I would consider nonsensical in this context)
$endgroup$
– G. Fougeron
Jan 16 at 15:19
$begingroup$
The reasonning behind the edit is perfectly fine, but the notations are a bit off. As far as I understand, your $e^i$ are vectors, so they shouldn't be included in your integral bounds. Basically, you should consider that $int_{[0,e^1]} f = int_0^1 f(x) dx$ is a definition. Thus, you do not need to "set e^{n+1} = 1". (which I would consider nonsensical in this context)
$endgroup$
– G. Fougeron
Jan 16 at 15:19
add a comment |
1 Answer
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oldest
votes
$begingroup$
My two cents : first do the linear, diagonal change of variables $y_i=a_i * x_i$. The determinant of the jacobian matrix of this transformation is uniform, and its value is $prod_i a_i$. The pre-image of your simplex is then the so-called "standard" simplex $sigma_0 = [0,e^1,dots,e^n]$, so that all you are left to prove is $int_{sigma_0} 1 = frac{1}{n!}$. This is done using recurrence on the dimension $n$.
Edit : This is exactly what was already suggested in the comments, only a tiny bit more verbose.
answered Jan 16 at 11:29
G. FougeronG. Fougeron
409213
$endgroup$
$begingroup$
thank you ! I have edited my question a bit. Can you expain how to do the diagonal change of variables in this case? Dont understand how to get there
$endgroup$
– wondering1123
Jan 16 at 12:15
$begingroup$
@wondering1123 : Do you feel like your question has been solved ? If so, please consider accepting the answer.
$endgroup$
– G. Fougeron
Jan 21 at 12:00
add a comment |
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$begingroup$
My two cents : first do the linear, diagonal change of variables $y_i=a_i * x_i$. The determinant of the jacobian matrix of this transformation is uniform, and its value is $prod_i a_i$. The pre-image of your simplex is then the so-called "standard" simplex $sigma_0 = [0,e^1,dots,e^n]$, so that all you are left to prove is $int_{sigma_0} 1 = frac{1}{n!}$. This is done using recurrence on the dimension $n$.
Edit : This is exactly what was already suggested in the comments, only a tiny bit more verbose.
answered Jan 16 at 11:29
G. FougeronG. Fougeron
409213
$endgroup$
$begingroup$
thank you ! I have edited my question a bit. Can you expain how to do the diagonal change of variables in this case? Dont understand how to get there
$endgroup$
– wondering1123
Jan 16 at 12:15
$begingroup$
@wondering1123 : Do you feel like your question has been solved ? If so, please consider accepting the answer.
$endgroup$
– G. Fougeron
Jan 21 at 12:00
add a comment |
$begingroup$
My two cents : first do the linear, diagonal change of variables $y_i=a_i * x_i$. The determinant of the jacobian matrix of this transformation is uniform, and its value is $prod_i a_i$. The pre-image of your simplex is then the so-called "standard" simplex $sigma_0 = [0,e^1,dots,e^n]$, so that all you are left to prove is $int_{sigma_0} 1 = frac{1}{n!}$. This is done using recurrence on the dimension $n$.
Edit : This is exactly what was already suggested in the comments, only a tiny bit more verbose.
answered Jan 16 at 11:29
G. FougeronG. Fougeron
409213
$endgroup$
$begingroup$
thank you ! I have edited my question a bit. Can you expain how to do the diagonal change of variables in this case? Dont understand how to get there
$endgroup$
– wondering1123
Jan 16 at 12:15
$begingroup$
@wondering1123 : Do you feel like your question has been solved ? If so, please consider accepting the answer.
$endgroup$
– G. Fougeron
Jan 21 at 12:00
add a comment |
$begingroup$
My two cents : first do the linear, diagonal change of variables $y_i=a_i * x_i$. The determinant of the jacobian matrix of this transformation is uniform, and its value is $prod_i a_i$. The pre-image of your simplex is then the so-called "standard" simplex $sigma_0 = [0,e^1,dots,e^n]$, so that all you are left to prove is $int_{sigma_0} 1 = frac{1}{n!}$. This is done using recurrence on the dimension $n$.
Edit : This is exactly what was already suggested in the comments, only a tiny bit more verbose.
answered Jan 16 at 11:29
G. FougeronG. Fougeron
409213
$endgroup$
My two cents : first do the linear, diagonal change of variables $y_i=a_i * x_i$. The determinant of the jacobian matrix of this transformation is uniform, and its value is $prod_i a_i$. The pre-image of your simplex is then the so-called "standard" simplex $sigma_0 = [0,e^1,dots,e^n]$, so that all you are left to prove is $int_{sigma_0} 1 = frac{1}{n!}$. This is done using recurrence on the dimension $n$.
Edit : This is exactly what was already suggested in the comments, only a tiny bit more verbose.
answered Jan 16 at 11:29
G. FougeronG. Fougeron
409213
answered Jan 16 at 11:29
G. FougeronG. Fougeron
409213
answered Jan 16 at 11:29
G. FougeronG. Fougeron
409213
answered Jan 16 at 11:29
G. FougeronG. Fougeron
409213
409213
$begingroup$
thank you ! I have edited my question a bit. Can you expain how to do the diagonal change of variables in this case? Dont understand how to get there
$endgroup$
– wondering1123
Jan 16 at 12:15
$begingroup$
@wondering1123 : Do you feel like your question has been solved ? If so, please consider accepting the answer.
$endgroup$
– G. Fougeron
Jan 21 at 12:00
add a comment |
$begingroup$
thank you ! I have edited my question a bit. Can you expain how to do the diagonal change of variables in this case? Dont understand how to get there
$endgroup$
– wondering1123
Jan 16 at 12:15
$begingroup$
@wondering1123 : Do you feel like your question has been solved ? If so, please consider accepting the answer.
$endgroup$
– G. Fougeron
Jan 21 at 12:00
$begingroup$
thank you ! I have edited my question a bit. Can you expain how to do the diagonal change of variables in this case? Dont understand how to get there
$endgroup$
– wondering1123
Jan 16 at 12:15
$begingroup$
thank you ! I have edited my question a bit. Can you expain how to do the diagonal change of variables in this case? Dont understand how to get there
$endgroup$
– wondering1123
Jan 16 at 12:15
$begingroup$
@wondering1123 : Do you feel like your question has been solved ? If so, please consider accepting the answer.
$endgroup$
– G. Fougeron
Jan 21 at 12:00
$begingroup$
@wondering1123 : Do you feel like your question has been solved ? If so, please consider accepting the answer.
$endgroup$
– G. Fougeron
Jan 21 at 12:00
add a comment |
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$begingroup$
One way to do this is to construct a linear map that converts the simplex into the standard simplex and use the change of variables formula.
$endgroup$
– mlainz
Jan 16 at 11:03
$begingroup$
The reasonning behind the edit is perfectly fine, but the notations are a bit off. As far as I understand, your $e^i$ are vectors, so they shouldn't be included in your integral bounds. Basically, you should consider that $int_{[0,e^1]} f = int_0^1 f(x) dx$ is a definition. Thus, you do not need to "set e^{n+1} = 1". (which I would consider nonsensical in this context)
$endgroup$
– G. Fougeron
Jan 16 at 15:19