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Is this series for Pi correct? And who has done it before?

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6

$begingroup$

The idea was to use an infinite series of triangles. The red then green then the... to get the area of this sector then the area of the circle is 16 times this. If it is a unit circle than area should equal Pi.

Here is the series I got using Pythagorean’s theorem , is it correct?

$$beginalign
A&=3r^2 + 12sum_ n=0^infty2^n-1x_nleft(1-sqrtr^2-fracx_n^24right),\ x_0&=rsqrt2-sqrt3
,\x_n+1&=sqrt2r^2-2rsqrtr^2-fracx_n^24
endalign$$

So-for-a-unit-circle

$$beginalign
pi&=3 + 12sum_ n=0^infty2^n-1x_nleft(1-sqrt1-fracx_n^24right),\ x_0&=sqrt2-sqrt3
,\x_n+1&=sqrt2-2sqrt1-fracx_n^24
endalign$$

sequences-and-series geometry proof-verification

share|cite|improve this question

edited Jun 16 at 17:14

Bodi Osman

asked Jun 16 at 14:23

Bodi OsmanBodi Osman

3844 bronze badges

New contributor

Bodi Osman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Looks like Viete's formula, except you are starting with a triangle instead of a square.
  $endgroup$
  – Wojowu
  Jun 16 at 14:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. Please type your posts using MathJax to format the math.
  $endgroup$
  – saulspatz
  Jun 16 at 14:48
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Archimedes did it first!
  $endgroup$
  – herb steinberg
  Jun 16 at 16:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  "There was more imagination in Arquimedes' head than in Homer's" (Voltaire)
  $endgroup$
  – Piquito
  Jun 16 at 17:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is called the Method of Exhaustion, and as noted, it is thousands of years old.
  $endgroup$
  – Eric Lippert
  Jun 16 at 22:35
  

  
  
  
  

 | 
show 3 more comments

6

$begingroup$

The idea was to use an infinite series of triangles. The red then green then the... to get the area of this sector then the area of the circle is 16 times this. If it is a unit circle than area should equal Pi.

Here is the series I got using Pythagorean’s theorem , is it correct?

$$beginalign
A&=3r^2 + 12sum_ n=0^infty2^n-1x_nleft(1-sqrtr^2-fracx_n^24right),\ x_0&=rsqrt2-sqrt3
,\x_n+1&=sqrt2r^2-2rsqrtr^2-fracx_n^24
endalign$$

So-for-a-unit-circle

$$beginalign
pi&=3 + 12sum_ n=0^infty2^n-1x_nleft(1-sqrt1-fracx_n^24right),\ x_0&=sqrt2-sqrt3
,\x_n+1&=sqrt2-2sqrt1-fracx_n^24
endalign$$

sequences-and-series geometry proof-verification

share|cite|improve this question

edited Jun 16 at 17:14

Bodi Osman

asked Jun 16 at 14:23

Bodi OsmanBodi Osman

3844 bronze badges

New contributor

Bodi Osman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Looks like Viete's formula, except you are starting with a triangle instead of a square.
  $endgroup$
  – Wojowu
  Jun 16 at 14:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. Please type your posts using MathJax to format the math.
  $endgroup$
  – saulspatz
  Jun 16 at 14:48
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Archimedes did it first!
  $endgroup$
  – herb steinberg
  Jun 16 at 16:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  "There was more imagination in Arquimedes' head than in Homer's" (Voltaire)
  $endgroup$
  – Piquito
  Jun 16 at 17:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is called the Method of Exhaustion, and as noted, it is thousands of years old.
  $endgroup$
  – Eric Lippert
  Jun 16 at 22:35
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

The idea was to use an infinite series of triangles. The red then green then the... to get the area of this sector then the area of the circle is 16 times this. If it is a unit circle than area should equal Pi.

Here is the series I got using Pythagorean’s theorem , is it correct?

$$beginalign
A&=3r^2 + 12sum_ n=0^infty2^n-1x_nleft(1-sqrtr^2-fracx_n^24right),\ x_0&=rsqrt2-sqrt3
,\x_n+1&=sqrt2r^2-2rsqrtr^2-fracx_n^24
endalign$$

So-for-a-unit-circle

$$beginalign
pi&=3 + 12sum_ n=0^infty2^n-1x_nleft(1-sqrt1-fracx_n^24right),\ x_0&=sqrt2-sqrt3
,\x_n+1&=sqrt2-2sqrt1-fracx_n^24
endalign$$

sequences-and-series geometry proof-verification

share|cite|improve this question

edited Jun 16 at 17:14

Bodi Osman

asked Jun 16 at 14:23

Bodi OsmanBodi Osman

3844 bronze badges

New contributor

Bodi Osman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Jun 16 at 17:14

Bodi Osman

asked Jun 16 at 14:23

Bodi OsmanBodi Osman

3844 bronze badges

New contributor

Bodi Osman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Jun 16 at 17:14

Bodi Osman

asked Jun 16 at 14:23

Bodi OsmanBodi Osman

3844 bronze badges

New contributor

Bodi Osman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Jun 16 at 14:23

Bodi OsmanBodi Osman

3844 bronze badges

New contributor

Bodi Osman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Jun 16 at 14:23

Bodi OsmanBodi Osman

3844 bronze badges

2 Answers
2

active

oldest

votes

3

$begingroup$

Numerical calculation suggests that you have made an error (or that I have, of course.)

from math import sqrt, pi

xs = [sqrt(2-sqrt(3))]

def f(x):
 return sqrt(2-2*sqrt(1-x*x/4))

def a(x):
 return 1 - sqrt(1-x*x/4)

for _ in range(50):
 xs.append(f(xs[-1]))

answer = 3+sum(2**(n-1)*xs[n]*a(xs[n]) for n in range(50))

print("answer=", answer)
print((pi-3)/(answer-3))

produces the output

answer= 3.0117993877991496
11.999999999999849

Have you dropped a factor of $12$ before the sum, perhaps?

share|cite|improve this answer

answered Jun 16 at 17:04

saulspatzsaulspatz

20.4k44 gold badges1616 silver badges3636 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes it seems that is the case
  $endgroup$
  – Bodi Osman
  Jun 16 at 17:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Fixed it thank you so much
  $endgroup$
  – Bodi Osman
  Jun 16 at 17:14
  

  
  
  
  

add a comment | 

1

$begingroup$

Note that, in
$x_n+1
=sqrt2-2sqrt1-fracx_n^24
$
if
$x_n = 2sin(t)
$
then

$beginarray\
x_n+1
&=sqrt2-2sqrt1-fracx_n^24\
&=sqrt2-2sqrt1-sin^2(t)\
&=sqrt2-2cos(t)\
&=2sqrtdfrac1-cos(t)2\
&=2sin(dfract2)\
endarray
$

share|cite|improve this answer

answered Jun 16 at 19:52

marty cohenmarty cohen

77.6k55 gold badges4949 silver badges133133 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Excuse my ignorance, but what do you mean I should notethis?
  $endgroup$
  – Bodi Osman
  Jun 16 at 23:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just that your iteration divides the angle by 2, so it goes from an n-gon to a 2n-gon.
  $endgroup$
  – marty cohen
  Jun 17 at 18:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah yes thank you
  $endgroup$
  – Bodi Osman
  Jun 17 at 19:34
  

  
  
  
  

add a comment | 

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3

$begingroup$

Numerical calculation suggests that you have made an error (or that I have, of course.)

from math import sqrt, pi

xs = [sqrt(2-sqrt(3))]

def f(x):
 return sqrt(2-2*sqrt(1-x*x/4))

def a(x):
 return 1 - sqrt(1-x*x/4)

for _ in range(50):
 xs.append(f(xs[-1]))

answer = 3+sum(2**(n-1)*xs[n]*a(xs[n]) for n in range(50))

print("answer=", answer)
print((pi-3)/(answer-3))

produces the output

answer= 3.0117993877991496
11.999999999999849

Have you dropped a factor of $12$ before the sum, perhaps?

share|cite|improve this answer

answered Jun 16 at 17:04

saulspatzsaulspatz

20.4k44 gold badges1616 silver badges3636 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes it seems that is the case
  $endgroup$
  – Bodi Osman
  Jun 16 at 17:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Fixed it thank you so much
  $endgroup$
  – Bodi Osman
  Jun 16 at 17:14
  

  
  
  
  

add a comment | 

3

$begingroup$

Numerical calculation suggests that you have made an error (or that I have, of course.)

from math import sqrt, pi

xs = [sqrt(2-sqrt(3))]

def f(x):
 return sqrt(2-2*sqrt(1-x*x/4))

def a(x):
 return 1 - sqrt(1-x*x/4)

for _ in range(50):
 xs.append(f(xs[-1]))

answer = 3+sum(2**(n-1)*xs[n]*a(xs[n]) for n in range(50))

print("answer=", answer)
print((pi-3)/(answer-3))

produces the output

answer= 3.0117993877991496
11.999999999999849

Have you dropped a factor of $12$ before the sum, perhaps?

share|cite|improve this answer

answered Jun 16 at 17:04

saulspatzsaulspatz

20.4k44 gold badges1616 silver badges3636 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes it seems that is the case
  $endgroup$
  – Bodi Osman
  Jun 16 at 17:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Fixed it thank you so much
  $endgroup$
  – Bodi Osman
  Jun 16 at 17:14
  

  
  
  
  

add a comment | 

$begingroup$

Numerical calculation suggests that you have made an error (or that I have, of course.)

from math import sqrt, pi

xs = [sqrt(2-sqrt(3))]

def f(x):
 return sqrt(2-2*sqrt(1-x*x/4))

def a(x):
 return 1 - sqrt(1-x*x/4)

for _ in range(50):
 xs.append(f(xs[-1]))

answer = 3+sum(2**(n-1)*xs[n]*a(xs[n]) for n in range(50))

print("answer=", answer)
print((pi-3)/(answer-3))

produces the output

answer= 3.0117993877991496
11.999999999999849

Have you dropped a factor of $12$ before the sum, perhaps?

share|cite|improve this answer

answered Jun 16 at 17:04

saulspatzsaulspatz

20.4k44 gold badges1616 silver badges3636 bronze badges

$endgroup$

from math import sqrt, pi

xs = [sqrt(2-sqrt(3))]

def f(x):
 return sqrt(2-2*sqrt(1-x*x/4))

def a(x):
 return 1 - sqrt(1-x*x/4)

for _ in range(50):
 xs.append(f(xs[-1]))

answer = 3+sum(2**(n-1)*xs[n]*a(xs[n]) for n in range(50))

print("answer=", answer)
print((pi-3)/(answer-3))

answer= 3.0117993877991496
11.999999999999849

share|cite|improve this answer

answered Jun 16 at 17:04

saulspatzsaulspatz

20.4k44 gold badges1616 silver badges3636 bronze badges

answered Jun 16 at 17:04

saulspatzsaulspatz

20.4k44 gold badges1616 silver badges3636 bronze badges

answered Jun 16 at 17:04

saulspatzsaulspatz

20.4k44 gold badges1616 silver badges3636 bronze badges

1

$begingroup$

Note that, in
$x_n+1
=sqrt2-2sqrt1-fracx_n^24
$
if
$x_n = 2sin(t)
$
then

$beginarray\
x_n+1
&=sqrt2-2sqrt1-fracx_n^24\
&=sqrt2-2sqrt1-sin^2(t)\
&=sqrt2-2cos(t)\
&=2sqrtdfrac1-cos(t)2\
&=2sin(dfract2)\
endarray
$

share|cite|improve this answer

answered Jun 16 at 19:52

marty cohenmarty cohen

77.6k55 gold badges4949 silver badges133133 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Excuse my ignorance, but what do you mean I should notethis?
  $endgroup$
  – Bodi Osman
  Jun 16 at 23:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just that your iteration divides the angle by 2, so it goes from an n-gon to a 2n-gon.
  $endgroup$
  – marty cohen
  Jun 17 at 18:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah yes thank you
  $endgroup$
  – Bodi Osman
  Jun 17 at 19:34
  

  
  
  
  

add a comment | 

1

$begingroup$

Note that, in
$x_n+1
=sqrt2-2sqrt1-fracx_n^24
$
if
$x_n = 2sin(t)
$
then

$beginarray\
x_n+1
&=sqrt2-2sqrt1-fracx_n^24\
&=sqrt2-2sqrt1-sin^2(t)\
&=sqrt2-2cos(t)\
&=2sqrtdfrac1-cos(t)2\
&=2sin(dfract2)\
endarray
$

share|cite|improve this answer

answered Jun 16 at 19:52

marty cohenmarty cohen

77.6k55 gold badges4949 silver badges133133 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Excuse my ignorance, but what do you mean I should notethis?
  $endgroup$
  – Bodi Osman
  Jun 16 at 23:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just that your iteration divides the angle by 2, so it goes from an n-gon to a 2n-gon.
  $endgroup$
  – marty cohen
  Jun 17 at 18:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah yes thank you
  $endgroup$
  – Bodi Osman
  Jun 17 at 19:34
  

  
  
  
  

add a comment | 

$begingroup$

Note that, in
$x_n+1
=sqrt2-2sqrt1-fracx_n^24
$
if
$x_n = 2sin(t)
$
then

$beginarray\
x_n+1
&=sqrt2-2sqrt1-fracx_n^24\
&=sqrt2-2sqrt1-sin^2(t)\
&=sqrt2-2cos(t)\
&=2sqrtdfrac1-cos(t)2\
&=2sin(dfract2)\
endarray
$

share|cite|improve this answer

answered Jun 16 at 19:52

marty cohenmarty cohen

77.6k55 gold badges4949 silver badges133133 bronze badges

$endgroup$

share|cite|improve this answer

answered Jun 16 at 19:52

marty cohenmarty cohen

77.6k55 gold badges4949 silver badges133133 bronze badges

answered Jun 16 at 19:52

marty cohenmarty cohen

77.6k55 gold badges4949 silver badges133133 bronze badges

answered Jun 16 at 19:52

marty cohenmarty cohen

77.6k55 gold badges4949 silver badges133133 bronze badges