Does this limit $lim_{ntoinfty}sum_{i=0}^n 1/n sqrt{1 - i^2/n^2}$ converge to $pi/4$?
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 at 9:11
Johnny
82
add a comment |
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 at 9:11
Johnny
82
2
This is a Riemann sum.
– Surb
Nov 29 at 9:15
add a comment |
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 at 9:11
Johnny
82
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 at 9:11
Johnny
82
asked Nov 29 at 9:11
Johnny
82
asked Nov 29 at 9:11
Johnny
82
asked Nov 29 at 9:11
Johnny
82
asked Nov 29 at 9:11
Johnny
82
82
2
This is a Riemann sum.
– Surb
Nov 29 at 9:15
add a comment |
2
This is a Riemann sum.
– Surb
Nov 29 at 9:15
2
2
This is a Riemann sum.
– Surb
Nov 29 at 9:15
This is a Riemann sum.
– Surb
Nov 29 at 9:15
add a comment |
2 Answers
2
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oldest
votes
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 at 9:17
Kavi Rama Murthy
48.9k31854
add a comment |
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 at 9:24
Mostafa Ayaz
13.7k3836
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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votes
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 at 9:17
Kavi Rama Murthy
48.9k31854
add a comment |
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 at 9:17
Kavi Rama Murthy
48.9k31854
add a comment |
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 at 9:17
Kavi Rama Murthy
48.9k31854
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 at 9:17
Kavi Rama Murthy
48.9k31854
answered Nov 29 at 9:17
Kavi Rama Murthy
48.9k31854
answered Nov 29 at 9:17
Kavi Rama Murthy
48.9k31854
answered Nov 29 at 9:17
Kavi Rama Murthy
48.9k31854
48.9k31854
add a comment |
add a comment |
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 at 9:24
Mostafa Ayaz
13.7k3836
add a comment |
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 at 9:24
Mostafa Ayaz
13.7k3836
add a comment |
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 at 9:24
Mostafa Ayaz
13.7k3836
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 at 9:24
Mostafa Ayaz
13.7k3836
answered Nov 29 at 9:24
Mostafa Ayaz
13.7k3836
answered Nov 29 at 9:24
Mostafa Ayaz
13.7k3836
answered Nov 29 at 9:24
Mostafa Ayaz
13.7k3836
13.7k3836
add a comment |
add a comment |
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2
This is a Riemann sum.
– Surb
Nov 29 at 9:15