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Pi concept

  1. Nov 10, 2005 #1
    How was the irrational number pi invented or how did man reach upon it? I have studied elementary calculus to understand the basic kinematics. Infact I know only the formulae to find for integration and derivative of some standard functions. The limits and such things weren't taught to me. It will be taught a year later. From this knowledge I myself proved the area of a circle and surface area of sphere, cylinder and cone and also their volume. But in finding the circumference of the circle, the formula that circumference of a circle is 2pi*r was uses. However I thought of a lot of ways but still cannot reach or use calculus to verify the formula for the circumference of a circle. I also felt the number pi like something heavenly and couldn't find any property in a curve which could have led to its invention. Is this because my present information on the subject is not sufficient or can I reach pi just from this information?
     
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  3. Nov 10, 2005 #2
    Pi was first theoretically studied by Archimedes's method of exhaustion (a precursor to calculus), in this case by studying the areas and perimeters of regular polygons that inscribed and circumscribed the circle. As mathematics evolved, more algebraic series for pi became common.
     
  4. Nov 10, 2005 #3

    Integral

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    [itex] \pi [/itex] is now and always has been simply the ratio of the circumfrence of a circle to its diameter.

    By definition

    [tex] \pi = \frac C D = \frac C {2R} [/tex]

    Thus

    [tex] C = 2 \pi R [/tex]
     
  5. Nov 12, 2005 #4

    HallsofIvy

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    It was recognized very early that the ratio of the circumference to its diameter (easier to measure the radius) was a constant. There is a reference in the bible to a kettle having circumference three time its diameter- not a bad estimate for those days. As pointed out above it was Archimedes who got the first really good approximations to pi. Euclid earlier showed that the ratio of circumference to radius is the same for all circles (essentially that all circles are "similar") by dividing two circles into n triangles (select n equally spaced points on the circumference, draw the radii and chords), showing that corresponding triangles in the two circles were similar and therefore the ratio of total of all of the bases to the radii must be the same. Then arguing that, since, as n got larger, the total bases come closer and closer to the circumference the same must be true of the ratio of the circumference to the radius. That's similar to the process of "exhaustion" Archimedes used and a primitive limit process.

    If you've taken enough calculus to be able to find the volume of sphere and cone, you should also know that, if y= f(x) then the length of the curve is given by
    [tex]\int_{x_0}^{x_1}\sqrt{1+ \left(\frac{df}{dx}\right)^2}dx[/tex]
    For the upper half of a circle, of radius R, [itex]f(x)= \sqrt{R^2- x^2}[/itex] so [itex]\frac{df}{dx}= \frac{x}{\sqrt{R^2- x^2}}[/itex].

    The integral for the arc length (of the hemisphere) is
    [tex]R\int_{-R}^R\frac{dx}{\sqrt{R^2- x^2}}[/itex]

    You will need a trig substitution to do that and make use of the fact that [itex]cos(\pi)[/itex]= 1. Since often sine and cosine are defined in terms of a circle, that is, in a sense "circular reasoning", but it is possible to define sine and cosine independently of a circle (for example as solutions to the differential equation y"= -y with specific initial values) and show that the ratio of circumference to radius is a constant for all circles, that constant being half the period of sine and cosine.
     
    Last edited: Nov 12, 2005
  6. Nov 12, 2005 #5

    dx

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    Im not sure if what Integral said was what you were looking for. Did you want a proof that the ratio of the circumference to the diameter of a circle is always the same constant? You can verify this by drawing two arbitrary concentric polygons and imagining that the number of sides approaches infinity. Now you can think of them as circles and since you can show that the ratio of the circumference and diameter is the same for both the polygons (using the properties of similar triangles), it follows that the ratio of the circumference to the diameter of the two circles must be the same. And since the two polygons we chose were arbitrary, it follows that the ratio of the circumference to the diameter of ANY two circles is the same. This means that ALL circles have the same circumference/diameter ratio. We call this ratio pi.
     
  7. Nov 12, 2005 #6
    dx-not only that they are in proportion, bot also that ratio is 2pi
     
  8. Nov 12, 2005 #7

    Tide

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    Vaishakh,

    The ratio of circumference to diameter is defined to be [itex]\pi[/itex].
     
  9. Nov 13, 2005 #8

    HallsofIvy

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    No, the ratio of circumference to diameter, which is what dx said, is [itex]\pi[/itex]. The ratio of circumference to radius is [itex]2\pi[/itex].
     
  10. Nov 13, 2005 #9
    don't teach me such ratios. i am not that fool even if i don't know these facts of pi. my idea behind that post was to make dx understand what all i meant from the initial post. Hall, i don't expect you to do such a thing after that fantastic explanation in the first post. infact you destroyed your praise.
    lets not discuss on what others write. only continue the discussion if soeone has something new or more clearer. this is thereason i didn't react to integral
     
  11. Nov 13, 2005 #10

    Tide

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    vaishakh,

    I think Halls' reply was completely suitable in light of your post #6. Don't be so quick on the trigger and consider that you may not have expressed yourself as clearly as you intended in #6.
     
  12. Nov 13, 2005 #11

    dx

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    vaishakh-"How was the irrational number pi invented or how did man reach upon it?"

    First, man realized that the ratio of the circumference to the diameter of any circle is the same. So they named it [tex] \pi [/tex]. Since

    [tex] \frac{C}{D} = \frac{C}{2R} = \pi [/tex]

    [tex] C = 2\pi{R} [/tex]
     
    Last edited: Nov 13, 2005
  13. Nov 14, 2005 #12

    HallsofIvy

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    Vaishakh- dx said "This means that ALL circles have the same circumference/diameter ratio. We call this ratio pi."

    Your response was "dx- not only that they are in proportion, bot also that ratio is 2pi"

    I was pointing out that dx was correct- the ratio of circumference to diameter is pi, not 2pi. I thought perhaps you were thinking of the ratio of circumference to radius.
     
  14. Nov 15, 2005 #13

    arildno

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    The number pi gained vastly in importance when Archimedes managed to prove that the proportionality constant between the circumference of the circle and its diameter was, in fact, the same proportionality constant existing between a circle's area and the square of its radius.

    Pi would never have gained its status unless that elegant result held.
     
  15. Nov 15, 2005 #14

    mathwonk

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    you are probably right arildno, but the same exhaustion argument given above also proves that the area of a circle is half the product of the circumference and the radius, so the equality of the constant ratios is a consequence.

    the result about the areas is also sort of obvious if you think of a circle as a triangle with base equal to its circumference and height equal to its radius. Of course that is archimedes' proof of both results cited above.

    I find it rather more surprizing that the same constant arises in the formula relating the surface area and volume of a sphere to its radius,

    not to mention in the formula for the sum of the series
    1/1^2 + 1/2^2 + 1/3^2 + ....+1/n^2+........,

    or in that of other such "even" values of the zeta function.
     
    Last edited: Nov 15, 2005
  16. Nov 16, 2005 #15

    dx

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    Can we prove that we can do this?
     
  17. Nov 16, 2005 #16

    arildno

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    I fully agree with you; Archimedes' clever proof makes the result obvious; before that, however, I would assume the equality of these constants (i.e, both being the number pi) was unobvious. Thus, by welding together these constants as one in his elegant argument, Archimedes presumably made [itex]\pi[/itex] seem a lot more important than before.
    Again agreed, with this [itex]\pi[/itex] is soaring above all the other dumb numbers..
    and have now shown itself to be practically divine..


    Archimedes' proof was the first step (of many thereafter) in revealing the glory of [itex]\pi[/itex]
     
  18. Nov 16, 2005 #17
    It should be pointed out that, although Aristotle mentioned the incommensurability of pi (i.e. that is irrational), it was not proven (as far as anyone knows) until Lambert in 1766, the proof of which was incomplete until Legendre completed (with a lemma that shows that certain infinite continued fractions are irrational) in 1806.

    As for the so-called divinity of the origins of pi, I'm rather unconvinced, particularly with regard to higher dimensional volume vs. surface area, since the lower dimensional cases can always be found by taking slices of higher dimensional cases and using some calculus. I.e. the ubiquity of pi is an intrinsic element of Euclidean geometry in any dimension.

    Of course, if you tried the same thing in other geometries, you're going to get a different constant -- if you get a constant at all (you need similarities in the geometry).
     
  19. Nov 16, 2005 #18
    Actually, comtemporous to this estimate, the Egyptians were quite a bit more accurate: circa 1900BC, they estimated pi at (16/9)^2, which is about 3.16.

    The Israelites didn't seem to go for fractions all that much.
     
  20. Nov 16, 2005 #19
    why? once you assume the circumference of a circle to be 2pi*r we get all of the above results. infact i have two methods through which we can find out the area of circle from its circumference as per first post.
     
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