INFINITE POINT SOLUTION ANALOGY



                                                                           Contains Copyrighted Material

                                                                 Copyright 1995
                                                                      Protection laws apply

Iterative Euclidean Bisection which leads to Trisection is presented in this treastise.

In this following analysis, "iterative" is taken to mean a recurring math process, or repeated rounds of geometric operation.

Bisection is considered to be one of the most rudimentary Euclidean geometric practices, conducted simply using a straight edge and compass.  This operation divides any given angle into two equal parts.   Performing bisection on either of these resultant angles again divides it into two more equal parts; or one-fourth of the original known angle.   There shouldn't be any argument as to the fact that such iterative analysis is entirely Euclidean.

So then, the question arises as to how many iterations can be performed before an analysis may be considered non-Euclidean?   Obviously, an infinite number may be performed simply because each operation employs only a straight edge and compass.

It seems logical to consider that Euclid's constraints were relegated to identifying new locii from known coordinates by using only a straight edge and compass during such determinations.   Iteration, as such, has no adverse impact on such determinations when it is conducted with the sole use of straight edge and compass.   Hence, Euclid's constraints are not violated by the priocess of iteration.

The abstraction which appears in this following analysis is whether or not an infinite number of bisector iterations may be qualify as non-Euclidean. Naturally, this possibly raises a whole new line of questioning.   Considering a known angle, 3 θ of, say 60 degrees, the cosine of such angle is the very well known fraction of 1/2.   Being that the cosine of 2 θ is irrational, it appears that going through the bisection process an infinite number of times may open up a new portal which then may be used for determining such illusive quantities.

Moreover, considering this dilemma as it might have surfaced during Euclid's day, one might then have proposed that a finite number of iterations would have to produce adequate, quantifyable trisection results.

How many of these iterations is necessary to produce such accuracy.   Well, for starters, this analysis demonstrates that just twenty-five iterations results in an accuracy of one in ten-million -- one considered to be far superior to using compass and straight edge instruments in the first place!

Hence, performing such iterative analysis with a straight edge and compass would merely result in retracing the same line over and over again, perhaps after as few as just five iterations -- thereby yielding a very sound, practical, and accurate result.

                                                      THE ANALYSIS

A geometric progression is a series of terms connected by a constant multiplier.   For purposes of examination, the following geometric progression consisting of an "n" number of terms is presented:

3 θ - 3 θ / 2 + 3 θ / 4 - 3 θ / 8 .. + 3 θ / (2n-2) - 3 θ / (2n-1) [1]

Dividing the second term by the first term of the equation yields the contant multiplier "m":

m = (- 3 θ / 2) / 3 θ

     = - 1/2

Hence, each factor is turn is multiplied by this common ratio.   As such, the third term is computed by multiplying - 3 θ / 2 by a minus 1/2 to yield + 3 θ / 4.

From examination, it becomes obvious that the denominator of each term is equal to a power of two that is one less than the numerical sequence of the term.  Hence, the fourth term denominator is equal to:

2[4-1] = 2[3]

         = 8

Therefore, the last term, or nth term denominator is equal to:

       2[n-1]

And, counting backwards, the second to last and third to last terms are 2[n-2] and 2[n-3], respectively.

Now, for the sum of a geometric progression, the common knowledge formula is as follows:

s = f{m[n] - 1} / (m - 1)

where

f = the first term

Using this formula, the sum of the first four terms of Equation 1 is:

s  = 3 θ {- 1/2[4] - 1} / (- 1/2 - 1)

    = 3 θ (+ 1/16 - 1) / (- 3/2)

    = 3 θ (- 15/16) / (- 3/2)

    = 3 θ (5/8)

    = 15 θ / 8

To verify, the first four terms of Equation 1 are added to produce:

(3 θ - 3 θ / 2) + (3 θ / 4 - 3 θ / 8) = 3 θ / 2 + 3 θ / 8

                                                     = 12 θ/ 8 + 3 θ / 8

                                                     = 15 θ / 8

When m is a fraction, as is - 1/2 in Equation 1, and the number of terms "n" approaches infinity, the term m[n] approaches 0.   As such, for an infinite number of terms, the above equation for the sum of a geometric progression reduces to:

s   = f (0 - 1) / (m - 1)

    = f / (1 - m)

Applying this to Equation 1 results in the following:

s  = 3 θ / [1 - (-1/2)]

    = 3 θ / (1 + 1/2)

    = 3 θ / (3/2)

    = 2 θ

Hence, the "2 θ" term is one of the trisectors of the first term, 3 θ since:

3 θ - 2 θ = θ.

Figure 1 depicts Equation 1 as a sequence of bisector operations constructed in opposite directions to one another.  The convention used is that positive quantities are plotted in the counterclockwise direction and negative quanitites are affixed in the clockwise direction.

First, the angle 3 θ is plotted in the counterclockwise direction.  The - 3 θ / 2 term of Equation 1 is applied in the clockwise direction.  Being one-half the quantity of the first term, 3 θ, it represents a bisector.   Its negative sign indicates that the end-line location is a subtraction of one-half of the original angle.

Hence, the end-line location of the second term represents the absolute sum of the first two terms as follows:

3 θ - 3 θ / 2 = 3 θ / 2

The positive 3 θ / 4 third term represents the counterclockwise bisector of the difference between last two end-line locations as follows:

1/2[3 θ - 3 θ / 2)] = + 3 θ / 4

The counterclockwise rotation signifies that the 3 θ / 4 term is to be added in order to produce a new end-line location of:

+ 3 θ / 2 + 3 θ / 4 = 9 θ / 4

Since, the first term above, 3 θ / 2, is the sum of the first two terms of Equation 1, the 9 θ / 4 represents the absolute sum of the first three terms of the geometric series.

In Figure 1 these absolute quantities appear in bold letters at the end of the line segments to which they refer.

Likewise, the negative 3 θ / 8 fourth term represents the clockwise bisector of the difference between the last two constructed end-line locations, as follows:

1/2[+ 9 θ / 4 - (+ 3 θ / 2)] = 1/2(3 θ / 4)

                                          = 3 θ/8

The clockwise rotation signifies that the 3 θ / 8 term is to be subtracted in order to produce a new end-line location of:

9 θ / 4 - 3 θ / 8 = 15 θ / 8

Since, the first term above, 9 θ / 4, is the sum of the first three terms of Equation 1, the 15 θ / 8 represents the sum of the first four terms of the geometric series.

Therefore, the geometric series of Equation 1 is analogous to a series of consecutive bisections in opposite directions, where the end-line locators, or bisectors, represent absolute sums when they are represented as last terms of the series.

As such, completing this operation an infinite number of times will produce an end-line locator, or bisector which aligns at 2 θ, or the trisector of the first term.

The accuracy of this method as a trisecting medium is shown below.   Table I shows that after twenty bisections, the initial angle is trisected to five decimal places.   The effectiveness of this method is clear since just eight trisections yields trisection down to two decimal places.  Twenty-five trisections yields trisection down to seven decimal places for an overall accuracy of one in ten-million.

 

                                                      TABLE I, TRISECTION ACCURACY PLOT


n      sum

1      3 θ = 3.000000 θ

2      3 θ - 3 θ / 2 = 3 θ / 2 = 1.500000 θ

3      3 θ / 2 + 3 θ / 4 = 9 θ / 4 = 2.250000 θ

4      9 θ / 4 - 3 θ / 8 = 15 θ / 8 = 1.875000 θ

5      15 θ / 8 + 3 θ / 16 = 33 θ / 16 = 2.062500 θ

6      33 θ / 16 - 3 θ / 32 = 63 θ / 32 = 1.968750 θ

7      63 θ / 32 + 3 θ / 64 = 129 θ / 64 = 2.015625 θ

8      129 θ / 64 - 3 θ / 128 = 255 θ / 128 = 1.9921875 θ

9      255 θ / 128 + 3 θ / 256 = 513 θ / 256 = 2.00390625  θ

10      513 θ / 256 - 3 θ / 512 = 1023 θ / 512 = 1.998046875 θ

11      1023 θ / 512 + 3 θ / 1024 = 2049 θ / 1024 = 2.000976563 θ

12      2049 θ / 1024 - 3 θ / 2048 = 4095 θ / 2048 = 1.999511719 θ

13      4095 θ/ / 2048 + 3 θ / 4096 = 8193 θ / 4096 = 2.000244141 θ

14      8193 θ / 4096 - 3 θ / 8192 = 16383 θ / 8192 = 1.99987793 θ

15      16383 θ / 8192 + 3 θ / 16384 = 32769 θ / 16384 = 2.000061035 θ

16      32769 θ / 16384 - 3 θ / 32768 = 65535 θ / 32768 = 1.999969482 θ

17      65535 θ / 32768 + 3 θ / 65536 = 131073 θ / 65536 = 2.000015259 θ

18      131073 θ / 65536 - 3 θ / 131072 = 262143 θ / 131072 = 1.999992371 θ

19      262143 θ / 131072 + 3 θ / 262144 = 524289 θ / 262144 = 2.000003815  θ

20      524289 θ  /  262144 - 3 θ  /  524288 = 1048575 / 524288 = 1.999998093 θ

21      1048575 θ  /  524288 - 3 θ  /  1048576 = 2097153 / 1048576 = 2.000000954 θ

22      2097153 θ  /  1048576 - 3 θ  /  2097152 = 4194303 / 2097152 = 1.999999523 θ

23      4194303 θ  /  2097152 - 3 θ  /  4194304 = 8388609 / 4194304 = 2.000000238 θ

24      8388609 θ  /  4194304 - 3 θ  /  8388608 = 16777215 / 8388608 = 1.999999881 θ

25      16777215 θ  /  8388608 - 3 θ  /  16777216 = 33554433 / 16777216 = 2.00000006 θ










Free Digits Counter
Sites Counter