Does this limit $lim_{ntoinfty}sum_{i=0}^n 1/n sqrt{1 - i^2/n^2}$ converge to $pi/4$?
$begingroup$
While trying to find an approximate area of a quarter of a circle by splicing it into small rectangles and summing their areas I've reached a point where I have this formula:
$$sum_{i=0}^n 1/n sqrt{1 - i^2/n^2}$$
Writing quick program and calculating the sum with n = 100, 100, 1000, 10k, 100k items suggest this sum converges to $pi/4$, however I have no idea why. I've tried to search for known series converging to $pi/4$ but nothing seems to resemble above formula.
Please note that in this question I'm not interested in what I was initially for, i.e. the area of a quarter of a circle. This was merely an exercise to show my nephew how we can approximate certain things.
sequences-and-series limits pi
asked Nov 29 '18 at 9:11
JohnnyJohnny
82
$endgroup$
add a comment |
$begingroup$
While trying to find an approximate area of a quarter of a circle by splicing it into small rectangles and summing their areas I've reached a point where I have this formula:
$$sum_{i=0}^n 1/n sqrt{1 - i^2/n^2}$$
Writing quick program and calculating the sum with n = 100, 100, 1000, 10k, 100k items suggest this sum converges to $pi/4$, however I have no idea why. I've tried to search for known series converging to $pi/4$ but nothing seems to resemble above formula.
Please note that in this question I'm not interested in what I was initially for, i.e. the area of a quarter of a circle. This was merely an exercise to show my nephew how we can approximate certain things.
sequences-and-series limits pi
asked Nov 29 '18 at 9:11
JohnnyJohnny
82
$endgroup$
2
$begingroup$
This is a Riemann sum.
$endgroup$
– Surb
Nov 29 '18 at 9:15
add a comment |
$begingroup$
While trying to find an approximate area of a quarter of a circle by splicing it into small rectangles and summing their areas I've reached a point where I have this formula:
$$sum_{i=0}^n 1/n sqrt{1 - i^2/n^2}$$
Writing quick program and calculating the sum with n = 100, 100, 1000, 10k, 100k items suggest this sum converges to $pi/4$, however I have no idea why. I've tried to search for known series converging to $pi/4$ but nothing seems to resemble above formula.
Please note that in this question I'm not interested in what I was initially for, i.e. the area of a quarter of a circle. This was merely an exercise to show my nephew how we can approximate certain things.
sequences-and-series limits pi
asked Nov 29 '18 at 9:11
JohnnyJohnny
82
$endgroup$
While trying to find an approximate area of a quarter of a circle by splicing it into small rectangles and summing their areas I've reached a point where I have this formula:
$$sum_{i=0}^n 1/n sqrt{1 - i^2/n^2}$$
Writing quick program and calculating the sum with n = 100, 100, 1000, 10k, 100k items suggest this sum converges to $pi/4$, however I have no idea why. I've tried to search for known series converging to $pi/4$ but nothing seems to resemble above formula.
Please note that in this question I'm not interested in what I was initially for, i.e. the area of a quarter of a circle. This was merely an exercise to show my nephew how we can approximate certain things.
sequences-and-series limits pi
sequences-and-series limits pi
asked Nov 29 '18 at 9:11
JohnnyJohnny
82
asked Nov 29 '18 at 9:11
JohnnyJohnny
82
asked Nov 29 '18 at 9:11
JohnnyJohnny
82
asked Nov 29 '18 at 9:11
JohnnyJohnny
82
asked Nov 29 '18 at 9:11
JohnnyJohnny
82
82
2
$begingroup$
This is a Riemann sum.
$endgroup$
– Surb
Nov 29 '18 at 9:15
add a comment |
2
$begingroup$
This is a Riemann sum.
$endgroup$
– Surb
Nov 29 '18 at 9:15
2
2
$begingroup$
This is a Riemann sum.
$endgroup$
– Surb
Nov 29 '18 at 9:15
$begingroup$
This is a Riemann sum.
$endgroup$
– Surb
Nov 29 '18 at 9:15
add a comment |
2 Answers
2
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$begingroup$
The sum is nothing but a Riemann sum for $int_0^{1}sqrt{1-t^{2}}, dt$. You can evaluate this by making the substitution $t=sin, theta$ and using the formula $2 cos ^{2}, theta =1+cos, (2theta)$ and you will get $pi /4$.
answered Nov 29 '18 at 9:17
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
$endgroup$
add a comment |
$begingroup$
This directly leads
from fundamental theorem of calculus
(https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus)
and
the definition of Reimannian sum
(https://en.wikipedia.org/wiki/Riemann_sum)
answered Nov 29 '18 at 9:24
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The sum is nothing but a Riemann sum for $int_0^{1}sqrt{1-t^{2}}, dt$. You can evaluate this by making the substitution $t=sin, theta$ and using the formula $2 cos ^{2}, theta =1+cos, (2theta)$ and you will get $pi /4$.
answered Nov 29 '18 at 9:17
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
$endgroup$
add a comment |
$begingroup$
The sum is nothing but a Riemann sum for $int_0^{1}sqrt{1-t^{2}}, dt$. You can evaluate this by making the substitution $t=sin, theta$ and using the formula $2 cos ^{2}, theta =1+cos, (2theta)$ and you will get $pi /4$.
answered Nov 29 '18 at 9:17
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
$endgroup$
add a comment |
$begingroup$
The sum is nothing but a Riemann sum for $int_0^{1}sqrt{1-t^{2}}, dt$. You can evaluate this by making the substitution $t=sin, theta$ and using the formula $2 cos ^{2}, theta =1+cos, (2theta)$ and you will get $pi /4$.
answered Nov 29 '18 at 9:17
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
$endgroup$
The sum is nothing but a Riemann sum for $int_0^{1}sqrt{1-t^{2}}, dt$. You can evaluate this by making the substitution $t=sin, theta$ and using the formula $2 cos ^{2}, theta =1+cos, (2theta)$ and you will get $pi /4$.
answered Nov 29 '18 at 9:17
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
answered Nov 29 '18 at 9:17
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
answered Nov 29 '18 at 9:17
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
answered Nov 29 '18 at 9:17
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
53.2k32055
add a comment |
add a comment |
$begingroup$
This directly leads
from fundamental theorem of calculus
(https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus)
and
the definition of Reimannian sum
(https://en.wikipedia.org/wiki/Riemann_sum)
answered Nov 29 '18 at 9:24
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
This directly leads
from fundamental theorem of calculus
(https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus)
and
the definition of Reimannian sum
(https://en.wikipedia.org/wiki/Riemann_sum)
answered Nov 29 '18 at 9:24
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
This directly leads
from fundamental theorem of calculus
(https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus)
and
the definition of Reimannian sum
(https://en.wikipedia.org/wiki/Riemann_sum)
answered Nov 29 '18 at 9:24
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
This directly leads
from fundamental theorem of calculus
(https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus)
and
the definition of Reimannian sum
(https://en.wikipedia.org/wiki/Riemann_sum)
answered Nov 29 '18 at 9:24
Mostafa AyazMostafa Ayaz
15.3k3939
answered Nov 29 '18 at 9:24
Mostafa AyazMostafa Ayaz
15.3k3939
answered Nov 29 '18 at 9:24
Mostafa AyazMostafa Ayaz
15.3k3939
answered Nov 29 '18 at 9:24
Mostafa AyazMostafa Ayaz
15.3k3939
15.3k3939
add a comment |
add a comment |
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2
$begingroup$
This is a Riemann sum.
$endgroup$
– Surb
Nov 29 '18 at 9:15