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

From: Stephen Harris (stephen.p.harris_at_sbcglobal.net)
Date: 02/17/04 

Date: Tue, 17 Feb 2004 00:33:02 GMT

"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.

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.

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.
Regards,
Stephen

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