Average Distances From The Origin












0












$begingroup$



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















0












$begingroup$



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$



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







share|cite|improve this question









$endgroup$





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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















-1












$begingroup$

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, $f(m,n) = f(n,m)$...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









-1












$begingroup$

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, $f(m,n) = f(n,m)$...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















-1












$begingroup$

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, $f(m,n) = f(n,m)$...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














-1












-1








-1





$begingroup$

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, $f(m,n) = f(n,m)$...






share|cite|improve this answer









$endgroup$



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, $f(m,n) = f(n,m)$...







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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