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Calculating pi

  1. Apr 24, 2007 #1
    i think found a formula to calculate pic (almost). Problem is, it has pi in it. If you are working in radians, it is Ntan(π/N), where N is the number of sides of a polygon that is close to a circle. as n approaches infinity, the function approaches π. Is it cheating if i simply change it to degrees (Ntan(180°/N)), or is this a legitimate way to calculate π?

    and sorry for the lack of proper terms...
  2. jcsd
  3. Apr 24, 2007 #2
    Your talking Archimedes, who considered the inequality: Ksin(pi/K)<pi<Ktan(pi/K), and used K=96 to arrive at the value 223/71<pi<22/7.

    However this came about by geometric considerations. Using L Hospital rule Lim [tex]\frac{tan(pi/N)}{1/N} [/tex] just puts us back where we begain, employing the symbol pi. Pi was defined by Euclid, but that did not produce a value for it.
    Last edited: Apr 24, 2007
  4. Apr 25, 2007 #3

    Gib Z

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    I actually did that myself when I was a bit younger, but more so in a bid to find the area of an n-sided regular polygon. Of course, when I took my formulas limit of sides to infinity and perpendicular distance from the centre to each side as 1, it gave me a limit that approached Pi.

    However, mine was the sin version, not the tan one.
  5. Apr 25, 2007 #4
    2 questions

    two questions:

    1) how do you simplify "Ksin(pi/K)<pi<Ktan(pi/K)"?

    2) is it cheating to convert to degrees? http://en.wikipedia.org/wiki/Radians" [Broken], after all, are based on the length of the arc disected, which is based on pi, so its kind of circular. But is it an actual derivation to use degrees, or am i somehow cheating(using pi in the work, just hidden) again?

    thanks for your help!
    Last edited by a moderator: May 2, 2017
  6. Apr 26, 2007 #5
    mr200backstrok: 1) how do you simplify "Ksin(pi/K)<pi<Ktan(pi/K)"?
    2) is it cheating to convert to degrees? radians, after all,

    1)To get a value for pi, we want an upper and lower limit. In the triangles created by Archimedes the small value is less than the arc and the larger value, the tangent, is extended beyond the arc. The matter is a case of construction, Archimedes using 96 such triangles.

    In the Calculus we always employ radiants. From the standpoint of the unit circle as we construct an angle from the center, say pi/n, the arc cut out of the circle is of that exact length. If we were working with degrees, say u, we would cut an arc in the circle of (2pi)(u/360).
  7. Apr 26, 2007 #6

    Gib Z

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    1) how do you simplify "Ksin(pi/K)<pi<Ktan(pi/K)"?

    See what happens when you take that in the limit as k approaches infinity.
  8. Apr 26, 2007 #7
    ok i know it would work if you tried it on a calculator, but if you had no calculator, and no idea what pi was, how would you solve Ksin(pi/K)<pi<Ktan(pi/K)? would you just guess and check?

    And if i use degrees in those equations ( ksin(180°/K)<pi<Ktan(180°/K) ), can i consider that a true derivation?

    Basically I'm trying to pretend that I'm Archimedes, and find the value of pi without modern knowledge or technology, aside from finding sin and tan.

    BTW, how would I calculate sin or tan without a calculator? is there a way or is it just a pain?
  9. Apr 27, 2007 #8
    In the case above it would not matter if one used radians or degrees.

    mr200backstrok: how would you solve Ksin(pi/K)<pi<Ktan(pi/K)? would you just guess and check?

    Hopefully this can be seen without a drawing. I'll take a real simple case: Divide a circle into 6 parts and form the triangles. In this case, we have a 60 degree angle in each of these equilateral triangles.

    NOw in this case from the center of the circle bisect the angle and form a right angle at the base of the triangle. Now this is easy to work with. Assuming the radius is 1 for simplicity, we are looking at the sign of 30 degrees, which was determined by Euclid to be 1/2. Thus 1 is the base of the triangle we just bisected, and since we have 6 such triangles, the length of these =6.

    Now while the base of the tirangles are all inside the arc of the circle, if we move to the tangent, looking at it the same way, which is outside the arc, we have 2/sqrt3 for the base of the triangle and multiplying by 6 we go all around the circle.

    Thus we have shown: 6<2pi<12/sqrt3=6.93. For the next case, we could look at 12 triangles of 30 degrees and bisect from the center, producing a 15 degree angle. We can find that from the half angle formula.

    2sin15 =[tex]\sqrt{2-\sqrt3}[/tex]. [tex]tan 15 = 2-\sqrt3 [/tex] So the problem now is transformed into dealing with square roots, and continuing in a similar manner.
    Last edited: Apr 27, 2007
  10. May 5, 2007 #9
    Going on with the above, we have sin(180/96) =[tex]\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}/2[/tex]

    The cos(180/96) =[tex]\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}/2[/tex]

    From which, at least using the TI-86, and multiplying by 96, we arrive to four decimals at a sin vaule 3.1410 and a tangent value of 3.1427, thus to two decimals pi = 3.14.
    Last edited: May 5, 2007
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