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Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
Proof that pi = 3.141....
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Proof that pi = 3.141....

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A geometric method for localising the value of Pi

A geometric method for localising the value of Pi

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  • 1. Estimating the value of pi by a geometric method
  • 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. 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. 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. The length of each side is 2R Tan 30O
  • 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. 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. 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. 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. I can improve on the estimate by doubling the number of sides.
  • 11. R 2R Tan 15O 2R Sin15O I can improve on the estimate by doubling the number of sides. Inside wedge Outside wedge
  • 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. 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. 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. 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. 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. 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. 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. 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. ...and then you can use the half angle formulas to generate all the SINs and TANs you need after that.
  • 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. 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. [END] David C, Canterbury NZ, 2009

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