User:Harishamur/Pi-Sequence
This is not a Wikipedia article: It is an individual user's work-in-progress page, and may be incomplete and/or unreliable. For guidance on developing this draft, see Wikipedia:So you made a userspace draft. Find sources: Google (books · news · scholar · free images · WP refs) · FENS · JSTOR · TWL |
Pi-Sequence
Geometry of regular polygons provides a method to identify a sequence of real numbers starting with 2√2 and converging to π. For simplicity we call this sequence π-sequence.
Consider a regular polygon with even number of sides(RPES) inscribed in a circle. Each such polygon has a diagonal which divides the area of the polygon into two equal parts. This is analogous to diameter dividing the area of a circle into two equal parts.We set:
R2n = Perimeter of RPES of sides 2n/ Length of the diagonal of the RPES We claim that the formula for R is given by
As n tends to infinity , the number of sides of RPES increases, but the size of each side goes on decreasing and since RPES is inscribed in a circle it ultimately coincides with the circle.
We give illustrations which set the pattern for arriving at the general formula.
For n=2, RPES is a square . Consider half the part of a square with side a. We need to find the length of the diagonal. Let O be the centre of the diagonal, we denote its length as 2x. Draw OM perpendicular to side AB. Clearly AM=a/2 and AMO is a right angled triangle. Hence x= a/2sin45. Thus
Formula (1) gives the same result for n=2,
For n=3, RPES is a hexagon. The part of the hexagon under the semicircle is divided into 3 equilateral triangles each of side a . The angle subtended at the centre by each triangle is 600. We need to find the length 2x of the diagonal AD. Draw OM perpendicular to the side AB.Then
Formula (1) gives this for n=3.
The pattern of the geometrical proof is the same for octagon(n=4), decagon(n=5) and so on.
Using the formula(1) we give a few more computations to show that the terms in the sequence gradually go on increasing and reach towards the limit π=3.142. . .
R8= 8 * sin(22.5) = 3.061 R10 = 10 * sin(18) = 3.09 R12 = 12 * sin(15) = 3.105 R14 = 14 * sin(12.8) = 3.110 R16 = 16 * sin(11.25) = 3.121 . . . . . . . . . . . . R90 = 90 * sin(2) = 3.1401 R180 = 180 * sin(1) = 3.1410 So the π-sequence is: 2√2, 3, 3.061, 3.09, 3.105,. . . 3.1401,. . . 3.1410. . . π
References
[edit]External links
[edit]