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Math Help - Euclid and Recursion

  1. #1
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    Euclid and Recursion

    We need to develop a recursive algorithm based on Euclid's proof of XII-2:

    Euclid's Elements, Book XII, Proposition 2

    It needs to be used to approximate the area of the circle of radius. I understand that we are successively inscribing n-gons, but I don't know how to represent this by recursion. Any ideas? Thanks.
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    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by twittytwitter View Post
    We need to develop a recursive algorithm based on Euclid's proof of XII-2:

    Euclid's Elements, Book XII, Proposition 2

    It needs to be used to approximate the area of the circle of radius. I understand that we are successively inscribing n-gons, but I don't know how to represent this by recursion. Any ideas? Thanks.
    Posting links frightens people. Just so you know.
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  3. #3
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    So you're saying this is the better thing to do:

    Circles are to one another as the squares on their diameters.
    Let ABCD and EFGH be circles, and let BD and FH be their diameters.
    I say that the circle ABCD is to the circle EFGH as the square on BD is to the square on FH.


    For, if the square on BD is not to the square on FH as the circle ABCD is to the circle EFGH, then as the square on BD is to the square on FH, the circle ABCD is either to some less area than the circle EFGH, or to a greater area.
    First, let it be in that ratio to a less area S.

    Inscribe the square EFGH in the circle EFGH. Then the inscribed square is greater than the half of the circle EFGH, for if through the points E, F, G, and H we draw tangents to the circle, then the square EFGH is half the square circumscribed about the circle, and the circle is less than the circumscribed square, hence the inscribed square EFGH is greater than the half of the circle EFGH. IV.6
    III.17
    Bisect the circumferences EF, FG, GH, and HE at the points K, L, M, and N. Join EK, KF, FL, LG, GM, MH, HN, and NE.
    Therefore each of the triangles EKF, FLG, GMH, and HNE is also greater than the half of the segment of the circle about it, for if through the points K, L, M, and N we draw tangents to the circle and complete the parallelograms on the straight lines EF, FG, GH, and HE, then each of the triangles EKF, FLG, GMH, and HNE is half of the parallelogram about it, while the segment about it is less than the parallelogram, hence each of the triangles EKF, FLG, GMH, and HNE is greater than the half of the segment of the circle about it. III.17
    Thus, by bisecting the remaining circumferences and joining straight lines, and by doing this repeatedly, we shall leave some segments of the circle which will be less than the excess by which the circle EFGH exceeds the area S.
    For it was proved in the first theorem of the tenth book that if two unequal magnitudes are set out, and if from the greater there is subtracted a magnitude greater than the half, and from that which is left a greater than the half, and if this is done repeatedly, then there will be left some magnitude which is less than the lesser magnitude set out. X.1
    Let segments be left such as described, and let the segments of the circle EFGH on EK, KF, FL, LG, GM, MH, HN, and NE be less than the excess by which the circle EFGH exceeds the area S.
    Therefore the remainder, the polygon EKFLGMHN, is greater than the area S.
    Now inscribe in the circle ABCD the polygon AOBPCQDR similar to the polygon EKFLGMHN.
    Therefore the square on BD is to the square on FH as the polygon AOBPCQDR is to the polygon EKFLGMHN. XII.1
    But the square on BD is to the square on FH as the circle ABCD to the area S, therefore the circle ABCD is to the area S as the polygon AOBPCQDR is to the polygon EKFLGMHN. Therefore, alternately the circle ABCD is to the polygon inscribed in it as the area S is to the polygon EKFLGMHN. V.11
    V.16
    But the circle ABCD is greater than the polygon inscribed in it, therefore the area S is also greater than the polygon EKFLGMHN. But it is also less, which is impossible.
    Therefore the square on BD is to the square on FH not as the circle ABCD is to any area less than the circle EFGH.

    Similarly we can prove that the circle EFGH is to any area less than the circle ABCD not as the square on FH is to the square on BD.

    I say next that neither is the circle ABCD to any area greater than the circle EFGH as the square on BD is to the square on FH.

    For, if possible, let it be in that ratio to a greater area S. Therefore, inversely the square on FH is to the square on DB as the area S is to the circle ABCD.
    But the area S is to the circle ABCD as the circle EFGH is to some area less than the circle ABCD, therefore the square on FH is to the square on BD as the circle EFGH is to some area less than the circle ABCD, which was proved impossible. Therefore the square on BD is to the square on FH not as the circle ABCD to any area greater than the circle EFGH. Lemma
    V.11
    And it was proved that neither is it in that ratio to any area less than the circle EFGH, therefore the square on BD is to the square on FH as the circle ABCD is to the circle EFGH.
    Q.E.D.
    Lemma

    I say that, the area S being greater than the circle EFGH the area S is to the circle ABCD as the circle EFGH is to some area less than the circle ABCD.
    For let it be contrived that the area S is to the circle ABCD as the circle EFGH to the area T.
    I say that the area T is less than the circle ABCD.

    Since the area S is to the circle ABCD as the circle EFGH is to the area T, therefore, alternately the area S is to the circle EFGH as the circle ABCD is to the area T. V.16
    But the area S is greater than the circle EFGH, therefore the circle ABCD is also greater than the area T.
    Hence the area S is to the circle ABCD as the circle EFGH is to some area less than the circle ABCD.
    Therefore, circles are to one another as the squares on their diameters.
    Q.E.D.


    In the last proposition it was shown that similar polygons inscribed in circles are proportional to the squares on the diameters of the circles. By approximating circles closely by similar polygons, the proportion is carried over to the circles.
    The form of the proof is a double proof by contradiction. There are three cases when comparing the ratio of the squares BD:FH to the ratio of the circles ABCD:EFGH. One case is that the ratio of the squares BD:FH equals ABCD:S where S is some area less than circle EFGH. Most of the proof is spent refuting this case. The second case is that the ratio of the squares BD:FH equals ABCD:S where this time S is some area greater than circle EFGH. This is inverted to a statement that the ratio of the squares FH:BD equals EFGH to some area less than circle ABCD, which is the first case already already shown not to occur. That leaves only the third case that the ratio of the squares BD:FH equals the ratio of the circles ABCD:EFGH. (Actually, there is a gap in the proof at this last step; it was never shown that these are the only three cases. It may be true that there are three cases, that the ratio of the squares is greater, less, or equal to the ratio of the circles, but the three cases of in the proof are one step removed from these three cases.)

    Approximation by polygons

    The first case is disposed of by approximating the circles by very close polygons. To begin with a square EFGH is inscribed in the circle EFGH, and it is shown that the remainder is less than half the circle. Next the circumferences are bisected to construct an octagon EKFLGMHN, and the remainder of the circle is shown to be less than half the old remainder. Continuing, polygons of 16, 32, 64, etc., sides are constructed and each one leaves a remainder less than half the previous remainder.
    Now the circle EFGH exceeds the area S by some finite amount, and by the principle of proposition X.1, at some stage mentioned above, the remainder will be less than the excess of circle EFGH over S. For the rest of the proof, that stage is taken as that of the polygon EKFLGMHN, that is, circle

    EFGH polygon EKFLGMHN < circle EFGH area S,
    and so area S < polygon EKFLGMHN.
    The similar polygon AOBPCQDR is inscribed in the other circle ABCD. Then

    circle ABCD : area S = BC2 : FH2
    = polygon AOBPCQDR : polygon EKFLGMHN.
    Alternately, circle ABCD : polygon AOBPCQDR = area S : polygon EKFLGMHN. But circle ABCD > polygon AOBPCQDR, so area S > polygon EKFLGMHN, contradicting the statement above that the area S is less than the polygon.

    Principle of exhaustion

    The approximation of a figure by a sequence of figures inside it is sometimes called the "principle of exhaustion." The important point of this principle is that the sequence of approximations can be made so that the difference between the original figure and the inscribed figure decreases by at least half at each step of the sequence.
    This principle is used in several later propositions in this book. Proposition XII.5 uses it to show that pyramids of the same height with triangular bases are proportional to their bases. The pyramids are approximated by a union of similar triangular prisms. Proposition XII.10 uses the principle of exhaustion to show that a cone inscribed in a cylinder is one-third of the cylinder. The cone is approximated by inscribed pyramids while the cylinder is approximated by inscribed prisms. Pyramids inscribed in cones are similarly used in XII.11 and XII.12. Finally, the principle of exhaustion is used in proposition XII.18 to show spheres are to one another in triplicate ratio of their diameters. There the spheres are exhausted by inscribed polyhedra.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by twittytwitter View Post
    So you're saying this is the better thing to do:
    .
    No.
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