Mixing
How does the value of the following integral change when we scale the shape by a factor $k$?
$begingroup$
Let $U$ be a bounded open subset of an $n$-dimensional euclidean space endowed with the usual topology and the usual metric $d$. Now, let
$$sigma(U)=iint_U d(x,y)dxdy$$
Suppose we scale the space by a factor of $k$ from the origin. Let $k U$ be the new shape.
My question is how does $sigma(k U)$ change with respect to $sigma(U)$. I have doubt with two results. Is $sigma(kU)=kcdotsigma(U)$ or does it depend on its dimension, i.e. $sigma(kU)=k^ncdotsigma(U)$, or it is something else?
I have doubts because every distance in the new shape is scaled by $k$ with respect to the old shape, but the area that gets scaled by a factor of $k^n$. Any help would be appreciated. Thanks.
calculus metric-spaces
asked Jan 14 at 13:48
GarmekainGarmekain
1,432720
$endgroup$
add a comment |
$begingroup$
Let $U$ be a bounded open subset of an $n$-dimensional euclidean space endowed with the usual topology and the usual metric $d$. Now, let
$$sigma(U)=iint_U d(x,y)dxdy$$
Suppose we scale the space by a factor of $k$ from the origin. Let $k U$ be the new shape.
My question is how does $sigma(k U)$ change with respect to $sigma(U)$. I have doubt with two results. Is $sigma(kU)=kcdotsigma(U)$ or does it depend on its dimension, i.e. $sigma(kU)=k^ncdotsigma(U)$, or it is something else?
I have doubts because every distance in the new shape is scaled by $k$ with respect to the old shape, but the area that gets scaled by a factor of $k^n$. Any help would be appreciated. Thanks.
calculus metric-spaces
asked Jan 14 at 13:48
GarmekainGarmekain
1,432720
$endgroup$
$begingroup$
I think there is no "obvious" relation between $sigma(U)$ and $sigma(kU)$ in general.
$endgroup$
– nicomezi
Jan 14 at 13:57
add a comment |
$begingroup$
Let $U$ be a bounded open subset of an $n$-dimensional euclidean space endowed with the usual topology and the usual metric $d$. Now, let
$$sigma(U)=iint_U d(x,y)dxdy$$
Suppose we scale the space by a factor of $k$ from the origin. Let $k U$ be the new shape.
My question is how does $sigma(k U)$ change with respect to $sigma(U)$. I have doubt with two results. Is $sigma(kU)=kcdotsigma(U)$ or does it depend on its dimension, i.e. $sigma(kU)=k^ncdotsigma(U)$, or it is something else?
I have doubts because every distance in the new shape is scaled by $k$ with respect to the old shape, but the area that gets scaled by a factor of $k^n$. Any help would be appreciated. Thanks.
calculus metric-spaces
asked Jan 14 at 13:48
GarmekainGarmekain
1,432720
$endgroup$
Let $U$ be a bounded open subset of an $n$-dimensional euclidean space endowed with the usual topology and the usual metric $d$. Now, let
$$sigma(U)=iint_U d(x,y)dxdy$$
Suppose we scale the space by a factor of $k$ from the origin. Let $k U$ be the new shape.
My question is how does $sigma(k U)$ change with respect to $sigma(U)$. I have doubt with two results. Is $sigma(kU)=kcdotsigma(U)$ or does it depend on its dimension, i.e. $sigma(kU)=k^ncdotsigma(U)$, or it is something else?
I have doubts because every distance in the new shape is scaled by $k$ with respect to the old shape, but the area that gets scaled by a factor of $k^n$. Any help would be appreciated. Thanks.
calculus metric-spaces
calculus metric-spaces
asked Jan 14 at 13:48
GarmekainGarmekain
1,432720
asked Jan 14 at 13:48
GarmekainGarmekain
1,432720
asked Jan 14 at 13:48
GarmekainGarmekain
1,432720
asked Jan 14 at 13:48
GarmekainGarmekain
1,432720
asked Jan 14 at 13:48
GarmekainGarmekain
1,432720
1,432720
$begingroup$
I think there is no "obvious" relation between $sigma(U)$ and $sigma(kU)$ in general.
$endgroup$
– nicomezi
Jan 14 at 13:57
add a comment |
$begingroup$
I think there is no "obvious" relation between $sigma(U)$ and $sigma(kU)$ in general.
$endgroup$
– nicomezi
Jan 14 at 13:57
$begingroup$
I think there is no "obvious" relation between $sigma(U)$ and $sigma(kU)$ in general.
$endgroup$
– nicomezi
Jan 14 at 13:57
$begingroup$
I think there is no "obvious" relation between $sigma(U)$ and $sigma(kU)$ in general.
$endgroup$
– nicomezi
Jan 14 at 13:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose $U$ is our shape of dimension $n$ and each of its points will have coordinates $x=(x_1,ldots,x_n)$. Then the new shape $kU$ will have points with coordinates $x'=(x_1',ldots,x_n')$ with the relations
$$x_i'=kx_i$$
$$dx_i'=kdx_i$$
So $dx'=prod_{i=1}^ndx_i'=prod_{i=1}^nk dx_i=k^ndx$, and because $x_i=x_i'/k$, the shape $kU$ gets transformed to ${xmid exists x'in kU:x=x'/k}=U$, so we have that:
begin{align}
sigma(kU) &= iint_{kU}d(x',y')dx' dy' \
&= iint_U d(kx,ky)k^ndx k^ndy \
&= iint_U kd(x,y)k^{2n}dx dy \
&= k^{2n+1}iint_Ud(x,y)dx dy \
sigma(kU)&= k^{2n+1}sigma(U)
end{align}
answered Jan 14 at 21:00
GarmekainGarmekain
1,432720
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
Suppose $U$ is our shape of dimension $n$ and each of its points will have coordinates $x=(x_1,ldots,x_n)$. Then the new shape $kU$ will have points with coordinates $x'=(x_1',ldots,x_n')$ with the relations
$$x_i'=kx_i$$
$$dx_i'=kdx_i$$
So $dx'=prod_{i=1}^ndx_i'=prod_{i=1}^nk dx_i=k^ndx$, and because $x_i=x_i'/k$, the shape $kU$ gets transformed to ${xmid exists x'in kU:x=x'/k}=U$, so we have that:
begin{align}
sigma(kU) &= iint_{kU}d(x',y')dx' dy' \
&= iint_U d(kx,ky)k^ndx k^ndy \
&= iint_U kd(x,y)k^{2n}dx dy \
&= k^{2n+1}iint_Ud(x,y)dx dy \
sigma(kU)&= k^{2n+1}sigma(U)
end{align}
answered Jan 14 at 21:00
GarmekainGarmekain
1,432720
$endgroup$
add a comment |
$begingroup$
Suppose $U$ is our shape of dimension $n$ and each of its points will have coordinates $x=(x_1,ldots,x_n)$. Then the new shape $kU$ will have points with coordinates $x'=(x_1',ldots,x_n')$ with the relations
$$x_i'=kx_i$$
$$dx_i'=kdx_i$$
So $dx'=prod_{i=1}^ndx_i'=prod_{i=1}^nk dx_i=k^ndx$, and because $x_i=x_i'/k$, the shape $kU$ gets transformed to ${xmid exists x'in kU:x=x'/k}=U$, so we have that:
begin{align}
sigma(kU) &= iint_{kU}d(x',y')dx' dy' \
&= iint_U d(kx,ky)k^ndx k^ndy \
&= iint_U kd(x,y)k^{2n}dx dy \
&= k^{2n+1}iint_Ud(x,y)dx dy \
sigma(kU)&= k^{2n+1}sigma(U)
end{align}
answered Jan 14 at 21:00
GarmekainGarmekain
1,432720
$endgroup$
add a comment |
$begingroup$
Suppose $U$ is our shape of dimension $n$ and each of its points will have coordinates $x=(x_1,ldots,x_n)$. Then the new shape $kU$ will have points with coordinates $x'=(x_1',ldots,x_n')$ with the relations
$$x_i'=kx_i$$
$$dx_i'=kdx_i$$
So $dx'=prod_{i=1}^ndx_i'=prod_{i=1}^nk dx_i=k^ndx$, and because $x_i=x_i'/k$, the shape $kU$ gets transformed to ${xmid exists x'in kU:x=x'/k}=U$, so we have that:
begin{align}
sigma(kU) &= iint_{kU}d(x',y')dx' dy' \
&= iint_U d(kx,ky)k^ndx k^ndy \
&= iint_U kd(x,y)k^{2n}dx dy \
&= k^{2n+1}iint_Ud(x,y)dx dy \
sigma(kU)&= k^{2n+1}sigma(U)
end{align}
answered Jan 14 at 21:00
GarmekainGarmekain
1,432720
$endgroup$
Suppose $U$ is our shape of dimension $n$ and each of its points will have coordinates $x=(x_1,ldots,x_n)$. Then the new shape $kU$ will have points with coordinates $x'=(x_1',ldots,x_n')$ with the relations
$$x_i'=kx_i$$
$$dx_i'=kdx_i$$
So $dx'=prod_{i=1}^ndx_i'=prod_{i=1}^nk dx_i=k^ndx$, and because $x_i=x_i'/k$, the shape $kU$ gets transformed to ${xmid exists x'in kU:x=x'/k}=U$, so we have that:
begin{align}
sigma(kU) &= iint_{kU}d(x',y')dx' dy' \
&= iint_U d(kx,ky)k^ndx k^ndy \
&= iint_U kd(x,y)k^{2n}dx dy \
&= k^{2n+1}iint_Ud(x,y)dx dy \
sigma(kU)&= k^{2n+1}sigma(U)
end{align}
answered Jan 14 at 21:00
GarmekainGarmekain
1,432720
edited Feb 4 at 14:24
answered Jan 14 at 21:00
GarmekainGarmekain
1,432720
answered Jan 14 at 21:00
GarmekainGarmekain
1,432720
answered Jan 14 at 21:00
GarmekainGarmekain
1,432720
1,432720
add a comment |
add a comment |
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$begingroup$
I think there is no "obvious" relation between $sigma(U)$ and $sigma(kU)$ in general.
$endgroup$
– nicomezi
Jan 14 at 13:57