Mixing
Munkres-Analysis on Manifolds: Theorem 20.1
Clash Royale CLAN TAG#URR8PPP
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I am studying Analysis on Manifolds by Munkres. I have a problem with a proof in section 20:
It states that:
Let $A$ be an $n$ by $n$ matrix. Let $h:mathbbR^nto mathbbR^n$ be the linear transformation $h(x)=A x$. Let $S$ be a rectifiable set (the boundary of $S=BdS$ has measure $0$) in $mathbbR^n$. Then $v(h(S))=|det A|v(S)$ ($v=$volume).
The author starts his proof by considering tha case of $A$ being a non-singular matrix (invertible).
I think I understand his steps in that case (I basically had to prove that $h(int S)=int$ $h(S)$ and $h(S)$ is rectifiable, if anybody knows a way this statements are proven autumatically please tell me).
He proceeds by considering the case where $A$ is singular, so $det A=0$. He tries to show now that $v(T)=0$. He states that since $S$ is bounded so is $h(S)$ (I think thats true because $|h(x)-h(a)|leq n|A||x-a|$ for each $x$ in $S$ and fixed a in $S$, if there is again a better explanation please tell me).
Then he says that $h(mathbbR^n)=V$ with $dim V=p<n$ and that $V$ has measure $0$ (for each $ε>0$ it can be covered by countably many open rectangles of total volume less than $ε$), a statemant that I have no clue how to prove. Then he says that the closure of $h(S)=cl(h(S))$ is closed and bounded and has neasure $0$ (of course $cl((h(S))$ is closed but why is it bounded with measure $0$?). Then makes an addition step (which I understand) and proves the theorem for that case too.
Cound someone help me clarify the points of the proof that I don't understand? Thank you in advance!
linear-algebra multivariable-calculus linear-transformations determinant lebesgue-measure
asked Aug 2 at 14:40
ΒαÃίλη ΒοÃÂλγαÃÂηÂ
86
add a comment |Â
up vote
1
down vote
favorite
I am studying Analysis on Manifolds by Munkres. I have a problem with a proof in section 20:
It states that:
Let $A$ be an $n$ by $n$ matrix. Let $h:mathbbR^nto mathbbR^n$ be the linear transformation $h(x)=A x$. Let $S$ be a rectifiable set (the boundary of $S=BdS$ has measure $0$) in $mathbbR^n$. Then $v(h(S))=|det A|v(S)$ ($v=$volume).
The author starts his proof by considering tha case of $A$ being a non-singular matrix (invertible).
I think I understand his steps in that case (I basically had to prove that $h(int S)=int$ $h(S)$ and $h(S)$ is rectifiable, if anybody knows a way this statements are proven autumatically please tell me).
He proceeds by considering the case where $A$ is singular, so $det A=0$. He tries to show now that $v(T)=0$. He states that since $S$ is bounded so is $h(S)$ (I think thats true because $|h(x)-h(a)|leq n|A||x-a|$ for each $x$ in $S$ and fixed a in $S$, if there is again a better explanation please tell me).
Then he says that $h(mathbbR^n)=V$ with $dim V=p<n$ and that $V$ has measure $0$ (for each $ε>0$ it can be covered by countably many open rectangles of total volume less than $ε$), a statemant that I have no clue how to prove. Then he says that the closure of $h(S)=cl(h(S))$ is closed and bounded and has neasure $0$ (of course $cl((h(S))$ is closed but why is it bounded with measure $0$?). Then makes an addition step (which I understand) and proves the theorem for that case too.
Cound someone help me clarify the points of the proof that I don't understand? Thank you in advance!
linear-algebra multivariable-calculus linear-transformations determinant lebesgue-measure
asked Aug 2 at 14:40
ΒαÃίλη ΒοÃÂλγαÃÂηÂ
86
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 2 at 14:43
Thank you for the information. I will start practising learning it.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 14:53
You know that $h(S)$ is bounded, so of course it's closure is as well. Also you know that $h(S)$ must be contained in $V$, thus its closure is contained in the closure of V which is V which has measure 0. Hence the closure of $h(S)$ also has measure $0$.
– Shalop
Aug 2 at 15:15
I guess the set of the accumulation points of a bounded set is bounded is intuitively obvious but I didn't know that the closure of a vector space is the vector space itself? Because I don;t have the mathematical training to prove the latter claim (I am a physicist) do you know where I can find a proof?
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 18:59
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am studying Analysis on Manifolds by Munkres. I have a problem with a proof in section 20:
It states that:
Let $A$ be an $n$ by $n$ matrix. Let $h:mathbbR^nto mathbbR^n$ be the linear transformation $h(x)=A x$. Let $S$ be a rectifiable set (the boundary of $S=BdS$ has measure $0$) in $mathbbR^n$. Then $v(h(S))=|det A|v(S)$ ($v=$volume).
The author starts his proof by considering tha case of $A$ being a non-singular matrix (invertible).
I think I understand his steps in that case (I basically had to prove that $h(int S)=int$ $h(S)$ and $h(S)$ is rectifiable, if anybody knows a way this statements are proven autumatically please tell me).
He proceeds by considering the case where $A$ is singular, so $det A=0$. He tries to show now that $v(T)=0$. He states that since $S$ is bounded so is $h(S)$ (I think thats true because $|h(x)-h(a)|leq n|A||x-a|$ for each $x$ in $S$ and fixed a in $S$, if there is again a better explanation please tell me).
Then he says that $h(mathbbR^n)=V$ with $dim V=p<n$ and that $V$ has measure $0$ (for each $ε>0$ it can be covered by countably many open rectangles of total volume less than $ε$), a statemant that I have no clue how to prove. Then he says that the closure of $h(S)=cl(h(S))$ is closed and bounded and has neasure $0$ (of course $cl((h(S))$ is closed but why is it bounded with measure $0$?). Then makes an addition step (which I understand) and proves the theorem for that case too.
Cound someone help me clarify the points of the proof that I don't understand? Thank you in advance!
linear-algebra multivariable-calculus linear-transformations determinant lebesgue-measure
asked Aug 2 at 14:40
ΒαÃίλη ΒοÃÂλγαÃÂηÂ
86
I am studying Analysis on Manifolds by Munkres. I have a problem with a proof in section 20:
It states that:
Let $A$ be an $n$ by $n$ matrix. Let $h:mathbbR^nto mathbbR^n$ be the linear transformation $h(x)=A x$. Let $S$ be a rectifiable set (the boundary of $S=BdS$ has measure $0$) in $mathbbR^n$. Then $v(h(S))=|det A|v(S)$ ($v=$volume).
The author starts his proof by considering tha case of $A$ being a non-singular matrix (invertible).
I think I understand his steps in that case (I basically had to prove that $h(int S)=int$ $h(S)$ and $h(S)$ is rectifiable, if anybody knows a way this statements are proven autumatically please tell me).
He proceeds by considering the case where $A$ is singular, so $det A=0$. He tries to show now that $v(T)=0$. He states that since $S$ is bounded so is $h(S)$ (I think thats true because $|h(x)-h(a)|leq n|A||x-a|$ for each $x$ in $S$ and fixed a in $S$, if there is again a better explanation please tell me).
Then he says that $h(mathbbR^n)=V$ with $dim V=p<n$ and that $V$ has measure $0$ (for each $ε>0$ it can be covered by countably many open rectangles of total volume less than $ε$), a statemant that I have no clue how to prove. Then he says that the closure of $h(S)=cl(h(S))$ is closed and bounded and has neasure $0$ (of course $cl((h(S))$ is closed but why is it bounded with measure $0$?). Then makes an addition step (which I understand) and proves the theorem for that case too.
Cound someone help me clarify the points of the proof that I don't understand? Thank you in advance!
linear-algebra multivariable-calculus linear-transformations determinant lebesgue-measure
asked Aug 2 at 14:40
ΒαÃίλη ΒοÃÂλγαÃÂηÂ
86
edited Aug 2 at 15:02
asked Aug 2 at 14:40
ΒαÃίλη ΒοÃÂλγαÃÂηÂ
86
asked Aug 2 at 14:40
ΒαÃίλη ΒοÃÂλγαÃÂηÂ
86
asked Aug 2 at 14:40
ΒαÃίλη ΒοÃÂλγαÃÂηÂ
86
86
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 2 at 14:43
Thank you for the information. I will start practising learning it.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 14:53
You know that $h(S)$ is bounded, so of course it's closure is as well. Also you know that $h(S)$ must be contained in $V$, thus its closure is contained in the closure of V which is V which has measure 0. Hence the closure of $h(S)$ also has measure $0$.
– Shalop
Aug 2 at 15:15
I guess the set of the accumulation points of a bounded set is bounded is intuitively obvious but I didn't know that the closure of a vector space is the vector space itself? Because I don;t have the mathematical training to prove the latter claim (I am a physicist) do you know where I can find a proof?
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 18:59
add a comment |Â
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 2 at 14:43
Thank you for the information. I will start practising learning it.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 14:53
You know that $h(S)$ is bounded, so of course it's closure is as well. Also you know that $h(S)$ must be contained in $V$, thus its closure is contained in the closure of V which is V which has measure 0. Hence the closure of $h(S)$ also has measure $0$.
– Shalop
Aug 2 at 15:15
I guess the set of the accumulation points of a bounded set is bounded is intuitively obvious but I didn't know that the closure of a vector space is the vector space itself? Because I don;t have the mathematical training to prove the latter claim (I am a physicist) do you know where I can find a proof?
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 18:59
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 2 at 14:43
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 2 at 14:43
Thank you for the information. I will start practising learning it.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 14:53
Thank you for the information. I will start practising learning it.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 14:53
You know that $h(S)$ is bounded, so of course it's closure is as well. Also you know that $h(S)$ must be contained in $V$, thus its closure is contained in the closure of V which is V which has measure 0. Hence the closure of $h(S)$ also has measure $0$.
– Shalop
Aug 2 at 15:15
You know that $h(S)$ is bounded, so of course it's closure is as well. Also you know that $h(S)$ must be contained in $V$, thus its closure is contained in the closure of V which is V which has measure 0. Hence the closure of $h(S)$ also has measure $0$.
– Shalop
Aug 2 at 15:15
I guess the set of the accumulation points of a bounded set is bounded is intuitively obvious but I didn't know that the closure of a vector space is the vector space itself? Because I don;t have the mathematical training to prove the latter claim (I am a physicist) do you know where I can find a proof?
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 18:59
I guess the set of the accumulation points of a bounded set is bounded is intuitively obvious but I didn't know that the closure of a vector space is the vector space itself? Because I don;t have the mathematical training to prove the latter claim (I am a physicist) do you know where I can find a proof?
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 18:59
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
As Munkers explains, since $det A=0$, $h(S)(=T)$ is contained in a vactor space of dimension smaller than $n$. But any subset of such a vector space has measure $0$.
answered Aug 2 at 14:47
José Carlos Santos
112k1696172
So if V is a subspace of R^n and dimV<n then V has measure 0? Do you know where I can find a proof? Also how dows the fact that T has measure 0 imply that cl T has measure 0 (and thus I guess it is also bounded)?
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 14:57
@ΒαÃίληÂΒοÃÂλγαÃÂη The fact that if $V$ is a vector subspace of $mathbbR^n$, then its Lebesgue measure is $0$ is a standard measure theory fact. I don't have a textbook at hand right now to tell you where to find it. And, since $V$ is closed, $overline Tsubsetoverline V=V$. Finally, $T$ is bounded because $S$ is bounded and linear maps map bounded sets onto bounded sets.
– José Carlos Santos
Aug 2 at 15:01
I have now experiance with Measure Theory (I am a physicist). Should you find anywhere proves for the above, I would appreciate it. Thank you very much for your answers.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 15:07
@ΒαÃίληÂΒοÃÂλγαÃÂη If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Aug 2 at 15:09
I clicked on the up-pointer but it can not accept it. It says something about reputation.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 18:47
 |Â
show 7 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
As Munkers explains, since $det A=0$, $h(S)(=T)$ is contained in a vactor space of dimension smaller than $n$. But any subset of such a vector space has measure $0$.
answered Aug 2 at 14:47
José Carlos Santos
112k1696172
So if V is a subspace of R^n and dimV<n then V has measure 0? Do you know where I can find a proof? Also how dows the fact that T has measure 0 imply that cl T has measure 0 (and thus I guess it is also bounded)?
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 14:57
@ΒαÃίληÂΒοÃÂλγαÃÂη The fact that if $V$ is a vector subspace of $mathbbR^n$, then its Lebesgue measure is $0$ is a standard measure theory fact. I don't have a textbook at hand right now to tell you where to find it. And, since $V$ is closed, $overline Tsubsetoverline V=V$. Finally, $T$ is bounded because $S$ is bounded and linear maps map bounded sets onto bounded sets.
– José Carlos Santos
Aug 2 at 15:01
I have now experiance with Measure Theory (I am a physicist). Should you find anywhere proves for the above, I would appreciate it. Thank you very much for your answers.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 15:07
@ΒαÃίληÂΒοÃÂλγαÃÂη If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Aug 2 at 15:09
I clicked on the up-pointer but it can not accept it. It says something about reputation.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 18:47
 |Â
show 7 more comments
up vote
0
down vote
accepted
As Munkers explains, since $det A=0$, $h(S)(=T)$ is contained in a vactor space of dimension smaller than $n$. But any subset of such a vector space has measure $0$.
answered Aug 2 at 14:47
José Carlos Santos
112k1696172
So if V is a subspace of R^n and dimV<n then V has measure 0? Do you know where I can find a proof? Also how dows the fact that T has measure 0 imply that cl T has measure 0 (and thus I guess it is also bounded)?
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 14:57
@ΒαÃίληÂΒοÃÂλγαÃÂη The fact that if $V$ is a vector subspace of $mathbbR^n$, then its Lebesgue measure is $0$ is a standard measure theory fact. I don't have a textbook at hand right now to tell you where to find it. And, since $V$ is closed, $overline Tsubsetoverline V=V$. Finally, $T$ is bounded because $S$ is bounded and linear maps map bounded sets onto bounded sets.
– José Carlos Santos
Aug 2 at 15:01
I have now experiance with Measure Theory (I am a physicist). Should you find anywhere proves for the above, I would appreciate it. Thank you very much for your answers.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 15:07
@ΒαÃίληÂΒοÃÂλγαÃÂη If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Aug 2 at 15:09
I clicked on the up-pointer but it can not accept it. It says something about reputation.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 18:47
 |Â
show 7 more comments
up vote
0
down vote
accepted
up vote
0
down vote
accepted
As Munkers explains, since $det A=0$, $h(S)(=T)$ is contained in a vactor space of dimension smaller than $n$. But any subset of such a vector space has measure $0$.
answered Aug 2 at 14:47
José Carlos Santos
112k1696172
As Munkers explains, since $det A=0$, $h(S)(=T)$ is contained in a vactor space of dimension smaller than $n$. But any subset of such a vector space has measure $0$.
answered Aug 2 at 14:47
José Carlos Santos
112k1696172
answered Aug 2 at 14:47
José Carlos Santos
112k1696172
answered Aug 2 at 14:47
José Carlos Santos
112k1696172
answered Aug 2 at 14:47
José Carlos Santos
112k1696172
112k1696172
So if V is a subspace of R^n and dimV<n then V has measure 0? Do you know where I can find a proof? Also how dows the fact that T has measure 0 imply that cl T has measure 0 (and thus I guess it is also bounded)?
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 14:57
@ΒαÃίληÂΒοÃÂλγαÃÂη The fact that if $V$ is a vector subspace of $mathbbR^n$, then its Lebesgue measure is $0$ is a standard measure theory fact. I don't have a textbook at hand right now to tell you where to find it. And, since $V$ is closed, $overline Tsubsetoverline V=V$. Finally, $T$ is bounded because $S$ is bounded and linear maps map bounded sets onto bounded sets.
– José Carlos Santos
Aug 2 at 15:01
I have now experiance with Measure Theory (I am a physicist). Should you find anywhere proves for the above, I would appreciate it. Thank you very much for your answers.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 15:07
@ΒαÃίληÂΒοÃÂλγαÃÂη If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Aug 2 at 15:09
I clicked on the up-pointer but it can not accept it. It says something about reputation.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 18:47
 |Â
show 7 more comments
So if V is a subspace of R^n and dimV<n then V has measure 0? Do you know where I can find a proof? Also how dows the fact that T has measure 0 imply that cl T has measure 0 (and thus I guess it is also bounded)?
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 14:57
@ΒαÃίληÂΒοÃÂλγαÃÂη The fact that if $V$ is a vector subspace of $mathbbR^n$, then its Lebesgue measure is $0$ is a standard measure theory fact. I don't have a textbook at hand right now to tell you where to find it. And, since $V$ is closed, $overline Tsubsetoverline V=V$. Finally, $T$ is bounded because $S$ is bounded and linear maps map bounded sets onto bounded sets.
– José Carlos Santos
Aug 2 at 15:01
I have now experiance with Measure Theory (I am a physicist). Should you find anywhere proves for the above, I would appreciate it. Thank you very much for your answers.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 15:07
@ΒαÃίληÂΒοÃÂλγαÃÂη If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Aug 2 at 15:09
I clicked on the up-pointer but it can not accept it. It says something about reputation.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 18:47
So if V is a subspace of R^n and dimV<n then V has measure 0? Do you know where I can find a proof? Also how dows the fact that T has measure 0 imply that cl T has measure 0 (and thus I guess it is also bounded)?
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 14:57
So if V is a subspace of R^n and dimV<n then V has measure 0? Do you know where I can find a proof? Also how dows the fact that T has measure 0 imply that cl T has measure 0 (and thus I guess it is also bounded)?
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 14:57
@ΒαÃίληÂΒοÃÂλγαÃÂη The fact that if $V$ is a vector subspace of $mathbbR^n$, then its Lebesgue measure is $0$ is a standard measure theory fact. I don't have a textbook at hand right now to tell you where to find it. And, since $V$ is closed, $overline Tsubsetoverline V=V$. Finally, $T$ is bounded because $S$ is bounded and linear maps map bounded sets onto bounded sets.
– José Carlos Santos
Aug 2 at 15:01
@ΒαÃίληÂΒοÃÂλγαÃÂη The fact that if $V$ is a vector subspace of $mathbbR^n$, then its Lebesgue measure is $0$ is a standard measure theory fact. I don't have a textbook at hand right now to tell you where to find it. And, since $V$ is closed, $overline Tsubsetoverline V=V$. Finally, $T$ is bounded because $S$ is bounded and linear maps map bounded sets onto bounded sets.
– José Carlos Santos
Aug 2 at 15:01
I have now experiance with Measure Theory (I am a physicist). Should you find anywhere proves for the above, I would appreciate it. Thank you very much for your answers.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 15:07
I have now experiance with Measure Theory (I am a physicist). Should you find anywhere proves for the above, I would appreciate it. Thank you very much for your answers.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 15:07
@ΒαÃίληÂΒοÃÂλγαÃÂη If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Aug 2 at 15:09
@ΒαÃίληÂΒοÃÂλγαÃÂη If my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Aug 2 at 15:09
I clicked on the up-pointer but it can not accept it. It says something about reputation.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 18:47
I clicked on the up-pointer but it can not accept it. It says something about reputation.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 18:47
 |Â
show 7 more comments
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 2 at 14:43
Thank you for the information. I will start practising learning it.
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 14:53
You know that $h(S)$ is bounded, so of course it's closure is as well. Also you know that $h(S)$ must be contained in $V$, thus its closure is contained in the closure of V which is V which has measure 0. Hence the closure of $h(S)$ also has measure $0$.
– Shalop
Aug 2 at 15:15
I guess the set of the accumulation points of a bounded set is bounded is intuitively obvious but I didn't know that the closure of a vector space is the vector space itself? Because I don;t have the mathematical training to prove the latter claim (I am a physicist) do you know where I can find a proof?
– Î’αÃίλη ΒοÃÂλγαÃÂηÂ
Aug 2 at 18:59