Select 6 mirrors on 6 faces of a room (a cube),when you go to the center, how many selves you can see from...
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Select $6$ mirrors on $6$ faces of a room (a cube),when you go to the center, how many selves can you see in one mirror ?
geometry physics
edited Nov 18 at 4:00
AryanSonwatikar
819
asked Nov 18 at 3:42
Alexander Lau
557
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Select $6$ mirrors on $6$ faces of a room (a cube),when you go to the center, how many selves can you see in one mirror ?
geometry physics
edited Nov 18 at 4:00
AryanSonwatikar
819
asked Nov 18 at 3:42
Alexander Lau
557
4
Why do you say select? A cube has $6$ faces, so you are choosing them all. If the mirrors cover the face, you will see an infinite number of images because you are between two parallel mirrors.
– Ross Millikan
Nov 18 at 3:55
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Select $6$ mirrors on $6$ faces of a room (a cube),when you go to the center, how many selves can you see in one mirror ?
geometry physics
edited Nov 18 at 4:00
AryanSonwatikar
819
asked Nov 18 at 3:42
Alexander Lau
557
Select $6$ mirrors on $6$ faces of a room (a cube),when you go to the center, how many selves can you see in one mirror ?
geometry physics
geometry physics
edited Nov 18 at 4:00
AryanSonwatikar
819
asked Nov 18 at 3:42
Alexander Lau
557
edited Nov 18 at 4:00
AryanSonwatikar
819
asked Nov 18 at 3:42
Alexander Lau
557
edited Nov 18 at 4:00
AryanSonwatikar
819
edited Nov 18 at 4:00
AryanSonwatikar
819
edited Nov 18 at 4:00
AryanSonwatikar
819
819
asked Nov 18 at 3:42
Alexander Lau
557
asked Nov 18 at 3:42
Alexander Lau
557
asked Nov 18 at 3:42
Alexander Lau
557
557
4
Why do you say select? A cube has $6$ faces, so you are choosing them all. If the mirrors cover the face, you will see an infinite number of images because you are between two parallel mirrors.
– Ross Millikan
Nov 18 at 3:55
add a comment |
4
Why do you say select? A cube has $6$ faces, so you are choosing them all. If the mirrors cover the face, you will see an infinite number of images because you are between two parallel mirrors.
– Ross Millikan
Nov 18 at 3:55
4
4
Why do you say select? A cube has $6$ faces, so you are choosing them all. If the mirrors cover the face, you will see an infinite number of images because you are between two parallel mirrors.
– Ross Millikan
Nov 18 at 3:55
Why do you say select? A cube has $6$ faces, so you are choosing them all. If the mirrors cover the face, you will see an infinite number of images because you are between two parallel mirrors.
– Ross Millikan
Nov 18 at 3:55
add a comment |
1 Answer
1
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oldest
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0
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The short answer here is that you'll get the same result as if you'd be searching for the visible points in a primitive cubical lattice, when looking from the origin (which clearly has to be a lattice point itself).
$ $
You have to observe that a point is visible exactly when its coordinates (for comfort all being integers) would be coprime, i.e. when having $P=(x,y,z)$ and $gcd(x,y,z)=1$.
Next consider probabilities. Let be $p$ some prime number. For having $p$ dividing some integer number $x$, i.e. $p|x$, the probability is $1/p$. Thus for having all three, $p|x$, $p|y$, and $p|z$, the probability then is $1/p^3$. Thus the searched for probability happens to be
$$prod_p left(1-frac1{p^3}right)$$
If you want to calculate that thing, you'd first have to have a look onto Euler's product formula and Riemann's zeta function.
$$zeta(s)=sum_{ngeq 1}frac1{n^s}=prod_pleft(sum_{kgeq 0}frac1{p^{ks}}right)=prod_pfrac1{1-p^s}$$
The one but last equality is the fundamental theorem of arithmetic that every integer has a unique prime factorization. Thus multiplying out the product would produce any $1/n^s$ exactly once. - The last one then is just the geometric series formula.
Thus your searched for probability happens to be
$$prod_p left(1-frac1{p^3}right)=frac1{zeta(3)}approx 0.83$$
--- rk
answered Nov 18 at 9:21
Dr. Richard Klitzing
1,1866
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The short answer here is that you'll get the same result as if you'd be searching for the visible points in a primitive cubical lattice, when looking from the origin (which clearly has to be a lattice point itself).
$ $
You have to observe that a point is visible exactly when its coordinates (for comfort all being integers) would be coprime, i.e. when having $P=(x,y,z)$ and $gcd(x,y,z)=1$.
Next consider probabilities. Let be $p$ some prime number. For having $p$ dividing some integer number $x$, i.e. $p|x$, the probability is $1/p$. Thus for having all three, $p|x$, $p|y$, and $p|z$, the probability then is $1/p^3$. Thus the searched for probability happens to be
$$prod_p left(1-frac1{p^3}right)$$
If you want to calculate that thing, you'd first have to have a look onto Euler's product formula and Riemann's zeta function.
$$zeta(s)=sum_{ngeq 1}frac1{n^s}=prod_pleft(sum_{kgeq 0}frac1{p^{ks}}right)=prod_pfrac1{1-p^s}$$
The one but last equality is the fundamental theorem of arithmetic that every integer has a unique prime factorization. Thus multiplying out the product would produce any $1/n^s$ exactly once. - The last one then is just the geometric series formula.
Thus your searched for probability happens to be
$$prod_p left(1-frac1{p^3}right)=frac1{zeta(3)}approx 0.83$$
--- rk
answered Nov 18 at 9:21
Dr. Richard Klitzing
1,1866
add a comment |
up vote
0
down vote
The short answer here is that you'll get the same result as if you'd be searching for the visible points in a primitive cubical lattice, when looking from the origin (which clearly has to be a lattice point itself).
$ $
You have to observe that a point is visible exactly when its coordinates (for comfort all being integers) would be coprime, i.e. when having $P=(x,y,z)$ and $gcd(x,y,z)=1$.
Next consider probabilities. Let be $p$ some prime number. For having $p$ dividing some integer number $x$, i.e. $p|x$, the probability is $1/p$. Thus for having all three, $p|x$, $p|y$, and $p|z$, the probability then is $1/p^3$. Thus the searched for probability happens to be
$$prod_p left(1-frac1{p^3}right)$$
If you want to calculate that thing, you'd first have to have a look onto Euler's product formula and Riemann's zeta function.
$$zeta(s)=sum_{ngeq 1}frac1{n^s}=prod_pleft(sum_{kgeq 0}frac1{p^{ks}}right)=prod_pfrac1{1-p^s}$$
The one but last equality is the fundamental theorem of arithmetic that every integer has a unique prime factorization. Thus multiplying out the product would produce any $1/n^s$ exactly once. - The last one then is just the geometric series formula.
Thus your searched for probability happens to be
$$prod_p left(1-frac1{p^3}right)=frac1{zeta(3)}approx 0.83$$
--- rk
answered Nov 18 at 9:21
Dr. Richard Klitzing
1,1866
add a comment |
up vote
0
down vote
up vote
0
down vote
The short answer here is that you'll get the same result as if you'd be searching for the visible points in a primitive cubical lattice, when looking from the origin (which clearly has to be a lattice point itself).
$ $
You have to observe that a point is visible exactly when its coordinates (for comfort all being integers) would be coprime, i.e. when having $P=(x,y,z)$ and $gcd(x,y,z)=1$.
Next consider probabilities. Let be $p$ some prime number. For having $p$ dividing some integer number $x$, i.e. $p|x$, the probability is $1/p$. Thus for having all three, $p|x$, $p|y$, and $p|z$, the probability then is $1/p^3$. Thus the searched for probability happens to be
$$prod_p left(1-frac1{p^3}right)$$
If you want to calculate that thing, you'd first have to have a look onto Euler's product formula and Riemann's zeta function.
$$zeta(s)=sum_{ngeq 1}frac1{n^s}=prod_pleft(sum_{kgeq 0}frac1{p^{ks}}right)=prod_pfrac1{1-p^s}$$
The one but last equality is the fundamental theorem of arithmetic that every integer has a unique prime factorization. Thus multiplying out the product would produce any $1/n^s$ exactly once. - The last one then is just the geometric series formula.
Thus your searched for probability happens to be
$$prod_p left(1-frac1{p^3}right)=frac1{zeta(3)}approx 0.83$$
--- rk
answered Nov 18 at 9:21
Dr. Richard Klitzing
1,1866
The short answer here is that you'll get the same result as if you'd be searching for the visible points in a primitive cubical lattice, when looking from the origin (which clearly has to be a lattice point itself).
$ $
You have to observe that a point is visible exactly when its coordinates (for comfort all being integers) would be coprime, i.e. when having $P=(x,y,z)$ and $gcd(x,y,z)=1$.
Next consider probabilities. Let be $p$ some prime number. For having $p$ dividing some integer number $x$, i.e. $p|x$, the probability is $1/p$. Thus for having all three, $p|x$, $p|y$, and $p|z$, the probability then is $1/p^3$. Thus the searched for probability happens to be
$$prod_p left(1-frac1{p^3}right)$$
If you want to calculate that thing, you'd first have to have a look onto Euler's product formula and Riemann's zeta function.
$$zeta(s)=sum_{ngeq 1}frac1{n^s}=prod_pleft(sum_{kgeq 0}frac1{p^{ks}}right)=prod_pfrac1{1-p^s}$$
The one but last equality is the fundamental theorem of arithmetic that every integer has a unique prime factorization. Thus multiplying out the product would produce any $1/n^s$ exactly once. - The last one then is just the geometric series formula.
Thus your searched for probability happens to be
$$prod_p left(1-frac1{p^3}right)=frac1{zeta(3)}approx 0.83$$
--- rk
answered Nov 18 at 9:21
Dr. Richard Klitzing
1,1866
answered Nov 18 at 9:21
Dr. Richard Klitzing
1,1866
answered Nov 18 at 9:21
Dr. Richard Klitzing
1,1866
answered Nov 18 at 9:21
Dr. Richard Klitzing
1,1866
1,1866
add a comment |
add a comment |
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4
Why do you say select? A cube has $6$ faces, so you are choosing them all. If the mirrors cover the face, you will see an infinite number of images because you are between two parallel mirrors.
– Ross Millikan
Nov 18 at 3:55