Average Distances From The Origin
$begingroup$
Consider all points $(m,n)$ where $m$ and $n$ are natural numbers,
$mleq10$, and $nleq5$. What is the average distances from each point
to the origin $(0,0)$?
I know how to find the distance $d$ between the points $(x_1,y_1)$ and $(x_2,y_2)$;
$d=sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$.
In this case, the first point is $(m,n)$ and the second point is the origin, therefore the distance, $d=sqrt{m^2+n^2}$.
It is annoying to find all distances, summing them, and dividing them by the number of distances, is not it annoying?
What if the given conditions, $mleq10$ and $nleq5$, were changed to $mleq49$ and $nleq36$?!
geometry coordinate-systems
asked Dec 5 '18 at 15:58
Hussain-AlqatariHussain-Alqatari
3217
$endgroup$
add a comment |
$begingroup$
Consider all points $(m,n)$ where $m$ and $n$ are natural numbers,
$mleq10$, and $nleq5$. What is the average distances from each point
to the origin $(0,0)$?
I know how to find the distance $d$ between the points $(x_1,y_1)$ and $(x_2,y_2)$;
$d=sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$.
In this case, the first point is $(m,n)$ and the second point is the origin, therefore the distance, $d=sqrt{m^2+n^2}$.
It is annoying to find all distances, summing them, and dividing them by the number of distances, is not it annoying?
What if the given conditions, $mleq10$ and $nleq5$, were changed to $mleq49$ and $nleq36$?!
geometry coordinate-systems
asked Dec 5 '18 at 15:58
Hussain-AlqatariHussain-Alqatari
3217
$endgroup$
$begingroup$
I don't think there's a quick way, It's easy with a computer program. For very large $m$ and $n$ you could get an approximation using calculus.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 16:04
$begingroup$
Even with small values of $m$ and $n$, it is annoying. Say $mleq5$, and $nleq2$. You will calculate $10$ distances. As you said it is easy by programming, but without that, can not we? :(
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:08
add a comment |
$begingroup$
Consider all points $(m,n)$ where $m$ and $n$ are natural numbers,
$mleq10$, and $nleq5$. What is the average distances from each point
to the origin $(0,0)$?
I know how to find the distance $d$ between the points $(x_1,y_1)$ and $(x_2,y_2)$;
$d=sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$.
In this case, the first point is $(m,n)$ and the second point is the origin, therefore the distance, $d=sqrt{m^2+n^2}$.
It is annoying to find all distances, summing them, and dividing them by the number of distances, is not it annoying?
What if the given conditions, $mleq10$ and $nleq5$, were changed to $mleq49$ and $nleq36$?!
geometry coordinate-systems
asked Dec 5 '18 at 15:58
Hussain-AlqatariHussain-Alqatari
3217
$endgroup$
Consider all points $(m,n)$ where $m$ and $n$ are natural numbers,
$mleq10$, and $nleq5$. What is the average distances from each point
to the origin $(0,0)$?
I know how to find the distance $d$ between the points $(x_1,y_1)$ and $(x_2,y_2)$;
$d=sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$.
In this case, the first point is $(m,n)$ and the second point is the origin, therefore the distance, $d=sqrt{m^2+n^2}$.
It is annoying to find all distances, summing them, and dividing them by the number of distances, is not it annoying?
What if the given conditions, $mleq10$ and $nleq5$, were changed to $mleq49$ and $nleq36$?!
geometry coordinate-systems
geometry coordinate-systems
asked Dec 5 '18 at 15:58
Hussain-AlqatariHussain-Alqatari
3217
asked Dec 5 '18 at 15:58
Hussain-AlqatariHussain-Alqatari
3217
asked Dec 5 '18 at 15:58
Hussain-AlqatariHussain-Alqatari
3217
asked Dec 5 '18 at 15:58
Hussain-AlqatariHussain-Alqatari
3217
asked Dec 5 '18 at 15:58
Hussain-AlqatariHussain-Alqatari
3217
3217
$begingroup$
I don't think there's a quick way, It's easy with a computer program. For very large $m$ and $n$ you could get an approximation using calculus.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 16:04
$begingroup$
Even with small values of $m$ and $n$, it is annoying. Say $mleq5$, and $nleq2$. You will calculate $10$ distances. As you said it is easy by programming, but without that, can not we? :(
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:08
add a comment |
$begingroup$
I don't think there's a quick way, It's easy with a computer program. For very large $m$ and $n$ you could get an approximation using calculus.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 16:04
$begingroup$
Even with small values of $m$ and $n$, it is annoying. Say $mleq5$, and $nleq2$. You will calculate $10$ distances. As you said it is easy by programming, but without that, can not we? :(
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:08
$begingroup$
I don't think there's a quick way, It's easy with a computer program. For very large $m$ and $n$ you could get an approximation using calculus.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 16:04
$begingroup$
I don't think there's a quick way, It's easy with a computer program. For very large $m$ and $n$ you could get an approximation using calculus.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 16:04
$begingroup$
Even with small values of $m$ and $n$, it is annoying. Say $mleq5$, and $nleq2$. You will calculate $10$ distances. As you said it is easy by programming, but without that, can not we? :(
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:08
$begingroup$
Even with small values of $m$ and $n$, it is annoying. Say $mleq5$, and $nleq2$. You will calculate $10$ distances. As you said it is easy by programming, but without that, can not we? :(
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is a quick solution for a related problem, even though it's not quite what you are asking. If you are willing to compute squares of distances of all points, then you get
$$
begin{split}
f(m,n)
&= sum_{i=1}^m sum_{k=1}^n left(i^2 + k^2right) \
&= sum_{i=1}^m sum_{k=1}^n i^2 + sum_{i=1}^m sum_{k=1}^n k^2 \
&= n sum_{i=1}^m i^2 + m sum_{k=1}^n k^2 \
&= n frac{m(m+1)(2m+1)}{6} + m frac{n(n+1)(2n+1)}{6} \
&= frac{mn}{6} left[(m+1)(2m+1) + (n+1)(2n+1)right]
end{split}
$$
As expected, $f(m,n)$ is symmetric in $m,n$, in other words, $f(m,n) = f(n,m)$...
answered Dec 5 '18 at 16:11
gt6989bgt6989b
33.9k22455
$endgroup$
$begingroup$
This is not related. However, it is new for me, I like it. But should not it be $f(m-1,n-1)$ since $m$ and $n$ are natural numbers (non-zero positive integers)?
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:17
$begingroup$
@Hussain-Alqatari do you mean you want to include $0$? It is easy to change the above to accommodate for that if you like
$endgroup$
– gt6989b
Dec 5 '18 at 16:37
$begingroup$
What you have posted (including $0$). My problem (excluding $0$ because they are natural numbers). So, should it be $f(m-1,n-1)$?
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:41
$begingroup$
@Hussain-Alqatari I don't understand. My post does not include zero, since both sums start with min index $1$, so the points $(1,1),(1,2),(2,1)$ are included but $(0,x),(x,0)$ are not included
$endgroup$
– gt6989b
Dec 5 '18 at 16:45
$begingroup$
Yes, you are right. What about my original problem? No way? :(
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:46
|
show 1 more comment
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$begingroup$
There is a quick solution for a related problem, even though it's not quite what you are asking. If you are willing to compute squares of distances of all points, then you get
$$
begin{split}
f(m,n)
&= sum_{i=1}^m sum_{k=1}^n left(i^2 + k^2right) \
&= sum_{i=1}^m sum_{k=1}^n i^2 + sum_{i=1}^m sum_{k=1}^n k^2 \
&= n sum_{i=1}^m i^2 + m sum_{k=1}^n k^2 \
&= n frac{m(m+1)(2m+1)}{6} + m frac{n(n+1)(2n+1)}{6} \
&= frac{mn}{6} left[(m+1)(2m+1) + (n+1)(2n+1)right]
end{split}
$$
As expected, $f(m,n)$ is symmetric in $m,n$, in other words, $f(m,n) = f(n,m)$...
answered Dec 5 '18 at 16:11
gt6989bgt6989b
33.9k22455
$endgroup$
$begingroup$
This is not related. However, it is new for me, I like it. But should not it be $f(m-1,n-1)$ since $m$ and $n$ are natural numbers (non-zero positive integers)?
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:17
$begingroup$
@Hussain-Alqatari do you mean you want to include $0$? It is easy to change the above to accommodate for that if you like
$endgroup$
– gt6989b
Dec 5 '18 at 16:37
$begingroup$
What you have posted (including $0$). My problem (excluding $0$ because they are natural numbers). So, should it be $f(m-1,n-1)$?
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:41
$begingroup$
@Hussain-Alqatari I don't understand. My post does not include zero, since both sums start with min index $1$, so the points $(1,1),(1,2),(2,1)$ are included but $(0,x),(x,0)$ are not included
$endgroup$
– gt6989b
Dec 5 '18 at 16:45
$begingroup$
Yes, you are right. What about my original problem? No way? :(
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:46
|
show 1 more comment
$begingroup$
There is a quick solution for a related problem, even though it's not quite what you are asking. If you are willing to compute squares of distances of all points, then you get
$$
begin{split}
f(m,n)
&= sum_{i=1}^m sum_{k=1}^n left(i^2 + k^2right) \
&= sum_{i=1}^m sum_{k=1}^n i^2 + sum_{i=1}^m sum_{k=1}^n k^2 \
&= n sum_{i=1}^m i^2 + m sum_{k=1}^n k^2 \
&= n frac{m(m+1)(2m+1)}{6} + m frac{n(n+1)(2n+1)}{6} \
&= frac{mn}{6} left[(m+1)(2m+1) + (n+1)(2n+1)right]
end{split}
$$
As expected, $f(m,n)$ is symmetric in $m,n$, in other words, $f(m,n) = f(n,m)$...
answered Dec 5 '18 at 16:11
gt6989bgt6989b
33.9k22455
$endgroup$
$begingroup$
This is not related. However, it is new for me, I like it. But should not it be $f(m-1,n-1)$ since $m$ and $n$ are natural numbers (non-zero positive integers)?
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:17
$begingroup$
@Hussain-Alqatari do you mean you want to include $0$? It is easy to change the above to accommodate for that if you like
$endgroup$
– gt6989b
Dec 5 '18 at 16:37
$begingroup$
What you have posted (including $0$). My problem (excluding $0$ because they are natural numbers). So, should it be $f(m-1,n-1)$?
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:41
$begingroup$
@Hussain-Alqatari I don't understand. My post does not include zero, since both sums start with min index $1$, so the points $(1,1),(1,2),(2,1)$ are included but $(0,x),(x,0)$ are not included
$endgroup$
– gt6989b
Dec 5 '18 at 16:45
$begingroup$
Yes, you are right. What about my original problem? No way? :(
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:46
|
show 1 more comment
$begingroup$
There is a quick solution for a related problem, even though it's not quite what you are asking. If you are willing to compute squares of distances of all points, then you get
$$
begin{split}
f(m,n)
&= sum_{i=1}^m sum_{k=1}^n left(i^2 + k^2right) \
&= sum_{i=1}^m sum_{k=1}^n i^2 + sum_{i=1}^m sum_{k=1}^n k^2 \
&= n sum_{i=1}^m i^2 + m sum_{k=1}^n k^2 \
&= n frac{m(m+1)(2m+1)}{6} + m frac{n(n+1)(2n+1)}{6} \
&= frac{mn}{6} left[(m+1)(2m+1) + (n+1)(2n+1)right]
end{split}
$$
As expected, $f(m,n)$ is symmetric in $m,n$, in other words, $f(m,n) = f(n,m)$...
answered Dec 5 '18 at 16:11
gt6989bgt6989b
33.9k22455
$endgroup$
There is a quick solution for a related problem, even though it's not quite what you are asking. If you are willing to compute squares of distances of all points, then you get
$$
begin{split}
f(m,n)
&= sum_{i=1}^m sum_{k=1}^n left(i^2 + k^2right) \
&= sum_{i=1}^m sum_{k=1}^n i^2 + sum_{i=1}^m sum_{k=1}^n k^2 \
&= n sum_{i=1}^m i^2 + m sum_{k=1}^n k^2 \
&= n frac{m(m+1)(2m+1)}{6} + m frac{n(n+1)(2n+1)}{6} \
&= frac{mn}{6} left[(m+1)(2m+1) + (n+1)(2n+1)right]
end{split}
$$
As expected, $f(m,n)$ is symmetric in $m,n$, in other words, $f(m,n) = f(n,m)$...
answered Dec 5 '18 at 16:11
gt6989bgt6989b
33.9k22455
answered Dec 5 '18 at 16:11
gt6989bgt6989b
33.9k22455
answered Dec 5 '18 at 16:11
gt6989bgt6989b
33.9k22455
answered Dec 5 '18 at 16:11
gt6989bgt6989b
33.9k22455
33.9k22455
$begingroup$
This is not related. However, it is new for me, I like it. But should not it be $f(m-1,n-1)$ since $m$ and $n$ are natural numbers (non-zero positive integers)?
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:17
$begingroup$
@Hussain-Alqatari do you mean you want to include $0$? It is easy to change the above to accommodate for that if you like
$endgroup$
– gt6989b
Dec 5 '18 at 16:37
$begingroup$
What you have posted (including $0$). My problem (excluding $0$ because they are natural numbers). So, should it be $f(m-1,n-1)$?
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:41
$begingroup$
@Hussain-Alqatari I don't understand. My post does not include zero, since both sums start with min index $1$, so the points $(1,1),(1,2),(2,1)$ are included but $(0,x),(x,0)$ are not included
$endgroup$
– gt6989b
Dec 5 '18 at 16:45
$begingroup$
Yes, you are right. What about my original problem? No way? :(
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:46
|
show 1 more comment
$begingroup$
This is not related. However, it is new for me, I like it. But should not it be $f(m-1,n-1)$ since $m$ and $n$ are natural numbers (non-zero positive integers)?
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:17
$begingroup$
@Hussain-Alqatari do you mean you want to include $0$? It is easy to change the above to accommodate for that if you like
$endgroup$
– gt6989b
Dec 5 '18 at 16:37
$begingroup$
What you have posted (including $0$). My problem (excluding $0$ because they are natural numbers). So, should it be $f(m-1,n-1)$?
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:41
$begingroup$
@Hussain-Alqatari I don't understand. My post does not include zero, since both sums start with min index $1$, so the points $(1,1),(1,2),(2,1)$ are included but $(0,x),(x,0)$ are not included
$endgroup$
– gt6989b
Dec 5 '18 at 16:45
$begingroup$
Yes, you are right. What about my original problem? No way? :(
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:46
$begingroup$
This is not related. However, it is new for me, I like it. But should not it be $f(m-1,n-1)$ since $m$ and $n$ are natural numbers (non-zero positive integers)?
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:17
$begingroup$
This is not related. However, it is new for me, I like it. But should not it be $f(m-1,n-1)$ since $m$ and $n$ are natural numbers (non-zero positive integers)?
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:17
$begingroup$
@Hussain-Alqatari do you mean you want to include $0$? It is easy to change the above to accommodate for that if you like
$endgroup$
– gt6989b
Dec 5 '18 at 16:37
$begingroup$
@Hussain-Alqatari do you mean you want to include $0$? It is easy to change the above to accommodate for that if you like
$endgroup$
– gt6989b
Dec 5 '18 at 16:37
$begingroup$
What you have posted (including $0$). My problem (excluding $0$ because they are natural numbers). So, should it be $f(m-1,n-1)$?
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:41
$begingroup$
What you have posted (including $0$). My problem (excluding $0$ because they are natural numbers). So, should it be $f(m-1,n-1)$?
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:41
$begingroup$
@Hussain-Alqatari I don't understand. My post does not include zero, since both sums start with min index $1$, so the points $(1,1),(1,2),(2,1)$ are included but $(0,x),(x,0)$ are not included
$endgroup$
– gt6989b
Dec 5 '18 at 16:45
$begingroup$
@Hussain-Alqatari I don't understand. My post does not include zero, since both sums start with min index $1$, so the points $(1,1),(1,2),(2,1)$ are included but $(0,x),(x,0)$ are not included
$endgroup$
– gt6989b
Dec 5 '18 at 16:45
$begingroup$
Yes, you are right. What about my original problem? No way? :(
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:46
$begingroup$
Yes, you are right. What about my original problem? No way? :(
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:46
|
show 1 more comment
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$begingroup$
I don't think there's a quick way, It's easy with a computer program. For very large $m$ and $n$ you could get an approximation using calculus.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 16:04
$begingroup$
Even with small values of $m$ and $n$, it is annoying. Say $mleq5$, and $nleq2$. You will calculate $10$ distances. As you said it is easy by programming, but without that, can not we? :(
$endgroup$
– Hussain-Alqatari
Dec 5 '18 at 16:08