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does this formula about hyper-pyramids work? and if so why
How come the Euclidian distance for n-space involves only squares?Area of twisted torusFractals - when the number of seed shapes that can fit into the scaled-up copy is non-integer.What is this myth/legend and origin of related ideas?Proof check: proving a neighborhood is an open set?Gluing diagrams: is it possible to glue a surface with itself in the same point? how is the diagram drawn?Trying to follow proof in Royden that every metric space is normalClosed subset of compact metric space contained in an open setUnderstanding Linear Algebra Geometrically - Reference RequestDifferentiating areas of polygons with respect to its inscribed circle
$begingroup$
you start with a 0d shape, a point. how much "point" is there? well it's just 1, the point has value 1. then you crate another point, this point can be everywhere in space, except for the exact same spot as the last point. if you connect the new point to the old point you got a 1d shape, a linesegment. how long is the linesegment? first you find the shortest distance between the new point and the old shape, which also was a point. we'll call the distance A. then you multiply it by the value the last shape had, which was 1. then you multiply it by $1/dimension$. the dimension is 1 so the problem is $A*1*(1/1)=lenght$.
now let's add a new point. this point can be anywhere in space, except for on the line create by the last points, in other word, you have to move in a new dimension. if you connect the new point to the old points you got a 2d shap, a triangle. what is the area of the triangle? first you find the shortest distance between the new point and the old shape, the line. we'll call the distance B. then you multiply it by the value the last shape had, which i called lenght. then you then you multiply it by "1/dimension". the dimension is 2 so the problem is $B*lenght*(1/2)=area$.
as you can see the process i identical, if you repeat it again you get $C*area*1/3=volume$. you could also write it as $A*B*C*1/3!$ if you repeat it again you would get $D*volume*(1/4)=4dvolume$
the formula is shortest distance times last shape times 1/dimension. if you want to be more general you can open up the "last shape"s and say the formuale is all the shortest distances multiplied with each other times 1/dimension!
so my question is, is this true for the 4th dimension and any other dimension? and why is it true? why is it even true for the first 3?
my third question is the one im wondering most about, because i feel like i need to understand why it works in 2d and 3d before i can understand why it works in any d. i've seen a proof for the formula of a pyramid, but it didnt seem to have any connections to the formula of a triangle.
as you maybe could tell, i havent actually red about anything of this, i just noticed the similarity between the triangle and pyramid formula and tried to generalise it. so if what i am describing has a name, or if there are texts describing it i would like to see it. thank you
general-topology reference-request
$endgroup$
add a comment |
$begingroup$
you start with a 0d shape, a point. how much "point" is there? well it's just 1, the point has value 1. then you crate another point, this point can be everywhere in space, except for the exact same spot as the last point. if you connect the new point to the old point you got a 1d shape, a linesegment. how long is the linesegment? first you find the shortest distance between the new point and the old shape, which also was a point. we'll call the distance A. then you multiply it by the value the last shape had, which was 1. then you multiply it by $1/dimension$. the dimension is 1 so the problem is $A*1*(1/1)=lenght$.
now let's add a new point. this point can be anywhere in space, except for on the line create by the last points, in other word, you have to move in a new dimension. if you connect the new point to the old points you got a 2d shap, a triangle. what is the area of the triangle? first you find the shortest distance between the new point and the old shape, the line. we'll call the distance B. then you multiply it by the value the last shape had, which i called lenght. then you then you multiply it by "1/dimension". the dimension is 2 so the problem is $B*lenght*(1/2)=area$.
as you can see the process i identical, if you repeat it again you get $C*area*1/3=volume$. you could also write it as $A*B*C*1/3!$ if you repeat it again you would get $D*volume*(1/4)=4dvolume$
the formula is shortest distance times last shape times 1/dimension. if you want to be more general you can open up the "last shape"s and say the formuale is all the shortest distances multiplied with each other times 1/dimension!
so my question is, is this true for the 4th dimension and any other dimension? and why is it true? why is it even true for the first 3?
my third question is the one im wondering most about, because i feel like i need to understand why it works in 2d and 3d before i can understand why it works in any d. i've seen a proof for the formula of a pyramid, but it didnt seem to have any connections to the formula of a triangle.
as you maybe could tell, i havent actually red about anything of this, i just noticed the similarity between the triangle and pyramid formula and tried to generalise it. so if what i am describing has a name, or if there are texts describing it i would like to see it. thank you
general-topology reference-request
$endgroup$
add a comment |
$begingroup$
you start with a 0d shape, a point. how much "point" is there? well it's just 1, the point has value 1. then you crate another point, this point can be everywhere in space, except for the exact same spot as the last point. if you connect the new point to the old point you got a 1d shape, a linesegment. how long is the linesegment? first you find the shortest distance between the new point and the old shape, which also was a point. we'll call the distance A. then you multiply it by the value the last shape had, which was 1. then you multiply it by $1/dimension$. the dimension is 1 so the problem is $A*1*(1/1)=lenght$.
now let's add a new point. this point can be anywhere in space, except for on the line create by the last points, in other word, you have to move in a new dimension. if you connect the new point to the old points you got a 2d shap, a triangle. what is the area of the triangle? first you find the shortest distance between the new point and the old shape, the line. we'll call the distance B. then you multiply it by the value the last shape had, which i called lenght. then you then you multiply it by "1/dimension". the dimension is 2 so the problem is $B*lenght*(1/2)=area$.
as you can see the process i identical, if you repeat it again you get $C*area*1/3=volume$. you could also write it as $A*B*C*1/3!$ if you repeat it again you would get $D*volume*(1/4)=4dvolume$
the formula is shortest distance times last shape times 1/dimension. if you want to be more general you can open up the "last shape"s and say the formuale is all the shortest distances multiplied with each other times 1/dimension!
so my question is, is this true for the 4th dimension and any other dimension? and why is it true? why is it even true for the first 3?
my third question is the one im wondering most about, because i feel like i need to understand why it works in 2d and 3d before i can understand why it works in any d. i've seen a proof for the formula of a pyramid, but it didnt seem to have any connections to the formula of a triangle.
as you maybe could tell, i havent actually red about anything of this, i just noticed the similarity between the triangle and pyramid formula and tried to generalise it. so if what i am describing has a name, or if there are texts describing it i would like to see it. thank you
general-topology reference-request
$endgroup$
you start with a 0d shape, a point. how much "point" is there? well it's just 1, the point has value 1. then you crate another point, this point can be everywhere in space, except for the exact same spot as the last point. if you connect the new point to the old point you got a 1d shape, a linesegment. how long is the linesegment? first you find the shortest distance between the new point and the old shape, which also was a point. we'll call the distance A. then you multiply it by the value the last shape had, which was 1. then you multiply it by $1/dimension$. the dimension is 1 so the problem is $A*1*(1/1)=lenght$.
now let's add a new point. this point can be anywhere in space, except for on the line create by the last points, in other word, you have to move in a new dimension. if you connect the new point to the old points you got a 2d shap, a triangle. what is the area of the triangle? first you find the shortest distance between the new point and the old shape, the line. we'll call the distance B. then you multiply it by the value the last shape had, which i called lenght. then you then you multiply it by "1/dimension". the dimension is 2 so the problem is $B*lenght*(1/2)=area$.
as you can see the process i identical, if you repeat it again you get $C*area*1/3=volume$. you could also write it as $A*B*C*1/3!$ if you repeat it again you would get $D*volume*(1/4)=4dvolume$
the formula is shortest distance times last shape times 1/dimension. if you want to be more general you can open up the "last shape"s and say the formuale is all the shortest distances multiplied with each other times 1/dimension!
so my question is, is this true for the 4th dimension and any other dimension? and why is it true? why is it even true for the first 3?
my third question is the one im wondering most about, because i feel like i need to understand why it works in 2d and 3d before i can understand why it works in any d. i've seen a proof for the formula of a pyramid, but it didnt seem to have any connections to the formula of a triangle.
as you maybe could tell, i havent actually red about anything of this, i just noticed the similarity between the triangle and pyramid formula and tried to generalise it. so if what i am describing has a name, or if there are texts describing it i would like to see it. thank you
general-topology reference-request
general-topology reference-request
asked Mar 29 at 3:35
MettekMettek
537
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Yes, it works. I don't know how to justify it without the concept of integration. When you make a triangle, the length perpendicular to the altitude grows linearly with distance, from zero at the vertex you draw the altitude from to the length of the other side. If the altitude is $h$ and the other side is $s$ you can think of the area as the integral $int_0^hfracxshdx=sh(frac 12x^2)|_0^1=frac 12sh$. In three dimensions the shape perpendicular to the altitude is a triangle, a two dimensional shape. Its area grows as $x^2$, which integrates to $frac 13x^3$ and delivers the $frac 13$ factor you seek. In $n$ dimensions, the area perpendicular to the altitude grows as $x^n-1$ and delivers the $frac 1n$ factor. The result is that the volume of an $n$ dimensional simplex is $frac 1n!$ of the volume of the enclosing $n$ dimensional box.
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$begingroup$
Yes, it works. I don't know how to justify it without the concept of integration. When you make a triangle, the length perpendicular to the altitude grows linearly with distance, from zero at the vertex you draw the altitude from to the length of the other side. If the altitude is $h$ and the other side is $s$ you can think of the area as the integral $int_0^hfracxshdx=sh(frac 12x^2)|_0^1=frac 12sh$. In three dimensions the shape perpendicular to the altitude is a triangle, a two dimensional shape. Its area grows as $x^2$, which integrates to $frac 13x^3$ and delivers the $frac 13$ factor you seek. In $n$ dimensions, the area perpendicular to the altitude grows as $x^n-1$ and delivers the $frac 1n$ factor. The result is that the volume of an $n$ dimensional simplex is $frac 1n!$ of the volume of the enclosing $n$ dimensional box.
$endgroup$
add a comment |
$begingroup$
Yes, it works. I don't know how to justify it without the concept of integration. When you make a triangle, the length perpendicular to the altitude grows linearly with distance, from zero at the vertex you draw the altitude from to the length of the other side. If the altitude is $h$ and the other side is $s$ you can think of the area as the integral $int_0^hfracxshdx=sh(frac 12x^2)|_0^1=frac 12sh$. In three dimensions the shape perpendicular to the altitude is a triangle, a two dimensional shape. Its area grows as $x^2$, which integrates to $frac 13x^3$ and delivers the $frac 13$ factor you seek. In $n$ dimensions, the area perpendicular to the altitude grows as $x^n-1$ and delivers the $frac 1n$ factor. The result is that the volume of an $n$ dimensional simplex is $frac 1n!$ of the volume of the enclosing $n$ dimensional box.
$endgroup$
add a comment |
$begingroup$
Yes, it works. I don't know how to justify it without the concept of integration. When you make a triangle, the length perpendicular to the altitude grows linearly with distance, from zero at the vertex you draw the altitude from to the length of the other side. If the altitude is $h$ and the other side is $s$ you can think of the area as the integral $int_0^hfracxshdx=sh(frac 12x^2)|_0^1=frac 12sh$. In three dimensions the shape perpendicular to the altitude is a triangle, a two dimensional shape. Its area grows as $x^2$, which integrates to $frac 13x^3$ and delivers the $frac 13$ factor you seek. In $n$ dimensions, the area perpendicular to the altitude grows as $x^n-1$ and delivers the $frac 1n$ factor. The result is that the volume of an $n$ dimensional simplex is $frac 1n!$ of the volume of the enclosing $n$ dimensional box.
$endgroup$
Yes, it works. I don't know how to justify it without the concept of integration. When you make a triangle, the length perpendicular to the altitude grows linearly with distance, from zero at the vertex you draw the altitude from to the length of the other side. If the altitude is $h$ and the other side is $s$ you can think of the area as the integral $int_0^hfracxshdx=sh(frac 12x^2)|_0^1=frac 12sh$. In three dimensions the shape perpendicular to the altitude is a triangle, a two dimensional shape. Its area grows as $x^2$, which integrates to $frac 13x^3$ and delivers the $frac 13$ factor you seek. In $n$ dimensions, the area perpendicular to the altitude grows as $x^n-1$ and delivers the $frac 1n$ factor. The result is that the volume of an $n$ dimensional simplex is $frac 1n!$ of the volume of the enclosing $n$ dimensional box.
answered Mar 29 at 4:24
Ross MillikanRoss Millikan
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