Re: Church-Turing compared to Zuse-Fredkin thesis (two new papers)

From: Lester Zick (lesterDELzick_at_worldnet.att.net)
Date: 02/17/04 

Date: Tue, 17 Feb 2004 15:54:38 GMT

On Tue, 17 Feb 2004 00:33:02 GMT, "Stephen Harris"
<stephen.p.harris@sbcglobal.net> in comp.ai.philosophy wrote:

>
>"Lester Zick" <lesterDELzick@worldnet.att.net> wrote in message
>news:4031179b.35683384@netnews.att.net...
>> On Mon, 16 Feb 2004 17:58:41 GMT, "Stephen Harris"
>> <stephen.p.harris@sbcglobal.net> in comp.ai.philosophy wrote:
>>
>> >
>> >"Lester Zick" <lesterDELzick@worldnet.att.net> wrote in message
>> >news:4030e88c.32841454@netnews.att.net...
>> >> On Sun, 15 Feb 2004 17:53:15 -0800, "Russell Easterly"
>> >> <logiclab@comcast.net> in comp.ai.philosophy wrote:
>> >>
>> >> >
>> >> >"Stephen Harris" <stephen.p.harris@sbcglobal.net> wrote in message
>> >> >news:HqCXb.25527$Co6.14111@newssvr25.news.prodigy.com...
>> >> >>
>> >>
>> >> [. . .]
>> >>
>> >> >Any definable real number is computable to an arbitrary number of
>> >> >finite positions. I am pointing out that if PI and e are "computable"
>> >> >then any irrational number is computable.
>> >> >
>> >> This may or may not be germane and if not I apologize. However I
>> >> wonder if it is generally realized that there are at least two classes
>> >> of irrationals, one with a graphable representation like the square
>> >> root of 2 and others with no graphable representation, transcendentals
>> >> like pi and e. What I'm wondering is whether a graphable solution to
>> >> non transcendentals counts as computable or whether computable only
>> >> refers to turing solutions?
>> >>
>> >> Regards - Lester
>> >>
>> >
>> >Did you know the square root of any prime is irrational?
>> >Here is a website which explains how to do square roots by hand.
>>
>> I didn't know that but it seems plausible. Do you happen to know of
>> any method to do cube roots by hand?
>>
>
>Yes, I've seen it, I once collected a book with the method included.
>Maybe higher roots can be broken down into square and cube. It was
>an older book. They quit teaching square root by hand in the 1960s in the
>US. I found it: http://mathforum.org/library/drmath/view/52605.html
>
>> >http://www.dattalo.com/technical/theory/sqrt.html
>> >
>> >Turing's computability covers anything that a human can
>> >do (in principle) with pencil, paper, and eraser, using a rote
>> >procedure, that does not require ingenuity. There are by
>> >hand algorithms for figuring out the value of Pi, also.
>> >
>> I know there are things like Archimedes' method of inscription and
>> superscription for approximating pi but this is only an approximation
>> and is not an exact graphical representation analogous to the
>> geometric equivalence of a square root.
>>
>
>Ok, I'm not sure what you mean. Doing square root is an approximation
>constrained by the number of decimals you carry out the operation.

Well the distinction I had in mind is that with pi, straight line
approximations never become circular arcs however great the degree of
approximation whereas square roots are always straight lines. So there
is no solution to pi in the form of any straight line but there is for
square roots. It strikes me that this means there are at least two
categorically different forms of irrational: straight line irrationals
and transcendental irrationals.

>An irrational number can be defined as a converging sequence of
>rational numbers. The irrational number serves as the limit.
>Computable numbers are computable because they have structure,
>so that a finite input can compute a potentially infinite output.
>So there is no algorithm to generate a truly random number and
>psuedo-random numbers can be generated by an algorithm.

OK. I think I see the difference. Computability doesn't imply an
actual steady state solution. Yet I would think a RAC solution is a
steady state solution. However I'm getting somewhat beyond my
competence.

>Well, the Archimedes method is visual, at any rate. You can
>start with a unit circle contained by an octagon and measure
>one side with a ruler and multipy it by eight. This will begin
>to approximate Pi. By the time you get a closest fitting polygon
>of 180,000 sides containing the unit circle, each side is very
>short and hard to measure with a ruler, the angle is no longer
>visible and looks just like a circle. The difference between
>the arc length of the circle and the straight edge of the side
>of the polygon are virtually indistinguishable. When the
>construction has the polygon inside the circle you see the
>areas made by the arc diverging from the straight line decrease.
>(The loop of the circle has the polygon side serve as a chord
>and two triangular areas can be constructed within.)
>
>When a circle has radius 1, then the area, Pi * r ^2, equals Pi.
>As the circumscribed n-sided polygon gets larger, its area grows to Pi.
>You have probably seen the graphical proof (using squares) that
>a^2 + b^2 = c^2 which is the basis for the law of sines.
>
>I am gonna hunt the n-sided perimeter/area polygon formula
>and compute it with 180,000 sides and compare it to Pi.
>I feel confident because I've already turned up this post by:
>
>Mark Stark
>"Start with a circle of radius 1, origin at o.
>Inscribe a square.
>Label the ends of one side a and b.
>Triangle aob is a right isosceles triangle including
> 90 degrees of arc.
>Chord ab is the square root of 2 (Pythagoras).
>Divide ab in half and label the midpoint c. Angle oca is 90 deg.
>Draw line oc and extend it to contact the radius at point d.
>Angle dca is 90 deg.
>
>Let chord ab be s1.
>Let chord ad be s2.
>Let line oc = x.
>Let line cd = z.
>Let line ac = y.
>Let x^2 mean x squared.
>Let sqr(x) mean square root of x.
>
>Then:
>
>1 - y^2 = x^2
>1 - x = z
>y = s1 / 2
>s2 = sqr(z^2 + y^2)
>
>A little algebra yields:
>
>1) s2 = sqr(2 - (2 * sqr(1 - (s1^2 / 4))))
>
>Each itteration of formula 1 above, renaming s2 and using it
>as s1, yields a new isosceles triangle and doubles number of
>sides of the inscribed polygon.
>
>At each stage the perimeter is s2 * 2^(n+1) where n = the
>number of itterations (considering the original square as n=1)
>
>After only four itterations the perimeter is already approaching pi."
>
>"The area of a unit circle is PI, whereas the area of a regular n-
>sided polygon inscribed in the circle is (n/2) sin(2PI/n), which
>of course approaches PI as n goes to infinity."
>
>Dr. Math pictures: http://mathforum.org/library/drmath/view/58292.html
>
>http://ouray.cudenver.edu/~rmulmer/archim.html
>SH: Follow this procedure for any polygon, and you find the general formula
>for the perimeter: P = 2 × n × r × sin (180÷n)
>
>You see why I chose 180,000 sides. P = 2 x 180,000 x 1 x .00001745
>= 6.282 since the radius is 1 then the Diameter is 2 so C/2 = P = 3.141
>which is converging toward Pi = 3.1415
>
>This website has an applet with the Archimedes method and
>it converges quite rapidly with only 240 sides. Also a nice diagram.
>
>> I understand what you say with respect to pencil, paper, and eraser
>> but this suggests to me some numerical equivalence rather than a
>> graphical correspondence.
>>
>> Regards - Lester
>>
>
>SH: Computable numbers are defined in terms of a decimal expansion.
>But the decimal equivalences of relatioships in right triangles has a
>geometric basis. The difference between the area of the circle (Pi)
>and the area of the n-sided polygon (approaching Pi) is seen in two
>right triangles which emerge at each side of the n-sides of the polygon.
>I think Archimedes Pi method relies heavily on finding areas of triangles
>and the law of sines (ratios) involving the Pythagorean theorem which
>can be proven by using areas in what seems a graphic manner to me.
>
>I found a website that shows a^2 + b^2 = c^2 using visual areas.
>http://staff.washington.edu/aganse/public.projects/pythproof/pythproof.html
>But it is not as good as the website that I first saw this. Of course areas
>of squares can be divided into areas of two isoceles triangles which is
>the possible connection I see to the graphical views of the geometry
>which is creating the decimal measuements.
>
>Another graphical applet of the above pyth. proof
>http://www.cut-the-knot.org/pythagoras/Proof37.shtml
>and another http://www.math.ubc.ca/~morey/java/pyth/index.html
>
>I found a good picture at http://ouray.cudenver.edu/~rmulmer/archim.html
>as well as how perimeters look at only n=240 with values inside/outside
>
>And Archimedes method interactive applet:
>http://www.math.utah.edu/~alfeld/Archimedes/Archimedes.html
>
>http://sourceforge.net/project/shownotes.php?release_id=132197
>Notes: This program calculates the value of pi. A regular polygon having 'n'
>number of sides is inscribed within a circle of known radius and another
>regular polygon with the same number of sides is circumscribed around
>thecircle. As the value of 'n' is increased, the average value of the
>perimeters of the two regular polygons approach the circumference of the
>circle. The average when divided by the diameter of the circle gives the
>approximate value of Pi.
>
>----------------------------------------------------------------------------
>----
>Changes: For greater accuracy, 'n' has been taken as power of 2. If we join
>the centre of any one of the two polygons to two adjacent vertices of the
>same polygon, the angle subtended between the two straight lines must be
>known if we are to calculate the perimeter. For a 'n' sided polygon (where
>n=2^r), this angle is (360/2^r)degrees=(45/2^(r-3))degrees. Since the value
>of cos 45 degree can be determined without using pi, the cosine of this
>angle can also be deduced by repeatedly using the formula cos
>(x/2)=sqrt(((cos x) +1)/2).
>
>SH: Though I do not agree with every thing you say, it usually seems
>somewhat reasonable and conscientiously considered.
>
Thanks Stephen. I certainly appreciate the information and I think
that what I take away is the idea that turing computability doesn't
produce steady state solutions for the computation of irrationals.

Regards - Lester

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