Estimating the value of pi by a geometric method
The circumference of the circle
is smaller than the perimeter
of the outer hexagon,
and bigger than the perimeter
of the i...
The circumference of the circle
is smaller than the perimeter
of the outer hexagon,
and bigger than the perimeter
of the i...
The circumference of the circle
is smaller than the perimeter
of the outer hexagon,
and bigger than the perimeter
of the i...
The length of each side is 2R Tan 30O
The length of each side is 2R Tan 30O
2R Tan 30O

So, the perimeter of the outer hexagon is
6 x 2R Tan 30O = 12R Tan 30O

...
The length of each side is 2R Tan 30O
2R Tan 30O

So, the perimeter of the outer hexagon is
6 x 2R Tan 30O = 12R Tan 30O

...
The length of each side is 2R Tan 30O
2R Tan 30O

So, the perimeter of the outer hexagon is
6 x 2R Tan 30O = 12R Tan 30O

...
The length of each side is 2R Tan 30O
2R Tan 30O

So, the perimeter of the outer hexagon is
6 x 2R Tan 30O = 12R Tan 30O

...
I can improve on the estimate by doubling the number of sides.
R

2R Tan 15O

2R Sin15O

I can improve on the estimate by doubling the number of sides.

Inside wedge

Outside wedge
R

12 Sin 15O < p < 12 Tan 15O
3.106 < p < 3.21

2R Tan 15O

2R Sin15O

I can improve on the estimate by doubling the numb...
In general, for an n-sided polygon inside the
circle and another one outside of the circle,

n x Sin (180/n) < p < n x Tan...
In general, for an n-sided polygon inside the
circle and another one outside of the circle,

n x Sin (180/n) < p < n x Tan...
In general, for an n-sided polygon inside the
circle and another one outside of the circle,

n x Sin (180/n) < p < n x Tan...
This may seem like the end of the story but
there is a problem with this process:

sides
6
12
24
48
96
192
384
768
1,536
3...
This may seem like the end of the story but
there is a problem with this process:

When a calculator or spreadsheet genera...
This may seem like the end of the story but
there is a problem with this process:

When a calculator or spreadsheet genera...
You have to have a way of generating SIN and TAN values
that doesn’t rely on pi. Here’s a way:
You can use geometry to get...
...and then you can use the half
angle formulas to generate all the
SINs and TANs you need after that.
So starting from 60 degrees, you can calculate the Sin, Cos and Tan
of increasingly small angles
(as long as you have a go...
Presentations on related subjects by the same author:
A formula for pi
http://www.slideshare.net/yaherglanite/a-formula-fo...
[END]

David C, Canterbury NZ, 2009
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Proof that pi = 3.141....

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A geometric method for localising the value of Pi

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Proof that pi = 3.141....

  1. 1. Estimating the value of pi by a geometric method
  2. 2. The circumference of the circle is smaller than the perimeter of the outer hexagon, and bigger than the perimeter of the inner hexagon.
  3. 3. The circumference of the circle is smaller than the perimeter of the outer hexagon, and bigger than the perimeter of the inner hexagon. R R R The perimeter of the inner hexagon is exactly 6 radii. R R R
  4. 4. The circumference of the circle is smaller than the perimeter of the outer hexagon, and bigger than the perimeter of the inner hexagon. R R R The perimeter of the inner hexagon is exactly 6 radii. R R R The perimeter of the outer hexagon is harder to work out.
  5. 5. The length of each side is 2R Tan 30O
  6. 6. The length of each side is 2R Tan 30O 2R Tan 30O So, the perimeter of the outer hexagon is 6 x 2R Tan 30O = 12R Tan 30O 2R Tan 30O
  7. 7. The length of each side is 2R Tan 30O 2R Tan 30O So, the perimeter of the outer hexagon is 6 x 2R Tan 30O = 12R Tan 30O R R R The circle is sandwiched between the inner hexagon and the outer hexagon. R R R 2R Tan 30O
  8. 8. The length of each side is 2R Tan 30O 2R Tan 30O So, the perimeter of the outer hexagon is 6 x 2R Tan 30O = 12R Tan 30O R R R The circle is sandwiched between the inner hexagon and the outer hexagon. R R R 2R Tan 30O 6R < Circumference < 12RTan 30O
  9. 9. The length of each side is 2R Tan 30O 2R Tan 30O So, the perimeter of the outer hexagon is 6 x 2R Tan 30O = 12R Tan 30O R R R The circle is sandwiched between the inner hexagon and the outer hexagon. R R R 2R Tan 30O 6R < Circumference < 12RTan 30O C = 2pR 3 < p < 3.4
  10. 10. I can improve on the estimate by doubling the number of sides.
  11. 11. R 2R Tan 15O 2R Sin15O I can improve on the estimate by doubling the number of sides. Inside wedge Outside wedge
  12. 12. R 12 Sin 15O < p < 12 Tan 15O 3.106 < p < 3.21 2R Tan 15O 2R Sin15O I can improve on the estimate by doubling the number of sides. Inside wedge Outside wedge
  13. 13. In general, for an n-sided polygon inside the circle and another one outside of the circle, n x Sin (180/n) < p < n x Tan (180/n)
  14. 14. In general, for an n-sided polygon inside the circle and another one outside of the circle, n x Sin (180/n) < p < n x Tan (180/n) As the number of sides increase, the value of p is squeezed more tightly by the upper and lower limits.
  15. 15. In general, for an n-sided polygon inside the circle and another one outside of the circle, n x Sin (180/n) < p < n x Tan (180/n) sides 6 12 24 48 96 192 384 768 1,536 3,072 6,144 p min 3.000000 3.105829 3.132629 3.139350 3.141032 3.141452 3.141558 3.141584 3.141590 3.141592 3.141593 p max 3.464102 3.215390 3.159660 3.146086 3.142715 3.141873 3.141663 3.141610 3.141597 3.141594 3.141593
  16. 16. This may seem like the end of the story but there is a problem with this process: sides 6 12 24 48 96 192 384 768 1,536 3,072 6,144 p min 3.000000 3.105829 3.132629 3.139350 3.141032 3.141452 3.141558 3.141584 3.141590 3.141592 3.141593 p max 3.464102 3.215390 3.159660 3.146086 3.142715 3.141873 3.141663 3.141610 3.141597 3.141594 3.141593
  17. 17. This may seem like the end of the story but there is a problem with this process: When a calculator or spreadsheet generates the SIN or TAN of an angle, it is using a formula that depends on the value of pi.
  18. 18. This may seem like the end of the story but there is a problem with this process: When a calculator or spreadsheet generates the SIN or TAN of an angle, it is using a formula that depends on the value of pi. You can’t use a formula that relies on pi to estimate pi! That’s circular reasoning.
  19. 19. You have to have a way of generating SIN and TAN values that doesn’t rely on pi. Here’s a way: You can use geometry to get the COS of 60 degrees.
  20. 20. ...and then you can use the half angle formulas to generate all the SINs and TANs you need after that.
  21. 21. So starting from 60 degrees, you can calculate the Sin, Cos and Tan of increasingly small angles (as long as you have a good way of calculating square roots).
  22. 22. Presentations on related subjects by the same author: A formula for pi http://www.slideshare.net/yaherglanite/a-formula-for-pi Deriving the circle area formula http://www.slideshare.net/yaherglanite/area-of-circle-proof-1789707 Calculating the area of a circle without using Pi (and without using a calculator) http://www.slideshare.net/yaherglanite/area-of-a-circle-simplification
  23. 23. [END] David C, Canterbury NZ, 2009
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