# this thread is for Rpenner

rpenner: I don't know why you waste your time with stuff like this.
To claim my attempts to educate are a waste of time would make me think you think Jason.Marshall is incapable of learning or that learning is valueless or that my time is too valuable to be spend answering Jason.Marshall's questions. Is this your position? Even if it is a waste of my time, what business is it of yours who have never found value in my posts?
One can check that a disk is circular with say a compass, then measure its diameter and circumference.
How are physical measurements, with necessarily finite precision, going to establish that the ratio of circumference to diameter is exactly π? What is needed are procedures that generalize to any level of precision. Archimedes gave us such a method, but needed regular 96-gons to demonstrate 3 + 10/71 < π < 22/7. But lots of numbers satisfy 3 + 10/71 < x < 22/7, for example: 355/113 and all numbers of the form (22 n+223)/(7 n+71) where n > 0. Today's modern methods give higher precision answers with much less labor than those of Archimedes but in doing so leave the student without connection to the geometrical problem. So I use antique methods in modern notation. Post #6 gives a result from analysis that π ≠ 3.125, post #14 gives a demonstration of how to proceed from Pythagoras via Newton to a ratio of circumference to diameter of a circle of any radius and relates that to an infinite sum of positive quantities, which prove 3.125 < π. Posts #15 and #18 demonstrate that in the limit of arbitrarily many sides, the perimeter and area of polygons approach a common behavior which is that of the circle, but also that they all share a relation of 2 A = r P. Archimedes developed the first topic:
http://itech.fgcu.edu/faculty/clindsey/mhf4404/archimedes/archimedes.html
With facile threads like this, I wouldn't be surprised if the person asking the question was pimping his own wares.
The book is puerile and wrong, demonstrating its wrongness is not necessarily trivial depending on the education level of the reader and one would have to have only a superficial understanding of the topic to think the value of π could be settled with a single physical measurement. You have not been elected to judge Jason.Marshall for the rest of us, but instead we are all free to judge his motivation and understanding by his interactions on this forum.

With facile threads like this, I wouldn't be surprised if the person asking the question was pimping his own wares.
Am I alone in seeing a certain irony in this?

Nope!

Hey I took a quick look at your explanations it looks thorough as usual what I would expect from you that is why I asked you specifically to answer my question. Right now am currently working on another project so I wont be able accurately respond with full comprehension of everything you are expressing mathematically as accurately as I would like to at the moment but once I do I will let you know what I have concluded and thanks for taking to time to explain yourself thoroughly and objectively and actually taking the question seriously.

One of my faves

Based on this, you need to add perfectly about 1160 4625 terms before you can prove that $$\frac{25}{8} \lt \pi$$

Whew, tedious, and exactly why we have mathematical expressions that converge to pi faster.

That's why I gave the sine method first because it requires many less terms to check.

N.B.
$$\begin{pmatrix} n - \frac{1}{2} \\ - \frac{1}{2} \end{pmatrix} = \frac{\Gamma \left(n + \frac{1}{2}\right)}{\Gamma \left( \frac{1}{2}\right) \, \Gamma \left(n + 1\right) } =\frac{(2n-1)!!}{(2n)!!} = \prod_{k=1}^{n} \frac{2k-1}{2k}$$

I coded up a python version 2.6 script that does this calculation with exact arithmetic with ratios of integers almost 6800 digits long. python is installed by default on current Macintoshs and is available for free for Windows and Linux. This program prints in approximate decimal for visual convenience.

Code:
import fractions
sum = fractions.Fraction()
binomial = fractions.Fraction(1,1)
term = binomial * 2
sum += term

print str(0) + " -> " + str(float(sum))

for n in range(1,4625) :
binomial = binomial * ( 2 * n - 1 ) / ( 2 * n )
term = binomial * 2 / ( 2 * n + 1 )
sum += term
print str(n) + " -> " + str(float(sum))

print len(str(sum.numerator))
print len(str(sum.denominator))
print sum > fractions.Fraction(25,8)
print sum.limit_denominator(10**6)
print float(sum - fractions.Fraction(25,8))
print float(sum - sum.limit_denominator(10**6))
with (partial) output of :
Code:
0 -> 2.0
1 -> 2.33333333333
2 -> 2.48333333333
3 -> 2.57261904762
4 -> 2.63338293651
5 -> 2.67812725469
6 -> 2.71283278354
7 -> 2.74076247104
8 -> 2.76386607283
9 -> 2.78338929189
...
4616 -> 3.12498611799
4617 -> 3.12498791614
4618 -> 3.12498971371
4619 -> 3.1249915107
4620 -> 3.1249933071
4621 -> 3.12499510292
4622 -> 3.12499689815
4623 -> 3.12499869281
4624 -> 3.12500048688
6795
6795
True
1604619/513478
4.86876896068e-07
1.11950146867e-12
where 1604619/513478 is a better approximation to the sum of 4625 terms than 25/8 is.

By contrast, a nearly identical perl computation, which uses IEEE approximate math, gives: "4624 -> 3.12500048687689" which is in good agreement, but not all the digits may be trusted. //Edit: and as it turns out, the correct rounded version ends 690 not 689.
Code:
my $sum = 0; my$b = 1;
$sum += (2 *$b) ;
print 0, " -> ", $sum, "\n"; for my$n ( 1..4624 ) {
$b *= ( 2 *$n - 1) / ( 2 * $n );$sum += (2 * $b) / ( 2 *$n + 1);
print $n, " -> ",$sum, "\n";
}

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Just one more quick question Rpenner the human calculator lol is there a reacurring pattern in the square root of "10368"

is there a reacurring pattern in the square root of "10368"
There are recurring patterns in every irrational quadratic surd if you know where to look.
For example $$\frac{1}{x/72 - 1} = x/72 + 1$$ has as it's solutions $$x = \pm \sqrt{10368}$$.
This follows from $$a(0) = \frac{1}{2}, a(n+1) = \frac{1}{2 + a(n)}, 1 + \lim_{n\to\infty} a(n) = \sqrt{2}$$.
Another recurrence is $$a(0) = 0, a(n+1) = \frac{1}{ \frac{202}{167} + \frac{1}{202 + a(n) } }, 101 + \lim_{n\to\infty} a(n) = \sqrt{10368}$$.

There of course is no recurring pattern in the decimal digits of 10368.
Rpenner the human calculator lol
As with Farsight, theorist-constant12345 and others, what you find remarkable, physicists and to a greater degree mathematicians will find ordinary and professionally required.
Just one more quick question
Did you actually have a question?

Nice exactly, I was thinking of it in a simpler form the side c hypotenuese/ it's side a or b .When side a and b or equal lengths is the symbolic representation for the exact ratio of the square of two, just wanted to see how it would be expressed correctly but I knew this one would be easy for you that's why I said it was a quick question lol thanks bro.

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Take your pick: $$72^2 + 72^2 = \left(\sqrt{10368}\right)^2$$ or $$72^2 + \left(\sqrt{10368}\right)^2 = \left(\sqrt{15552}\right)^2$$
But these are not primitive triangles, but scaled up versions of $$1^2 + 1^2 = 2$$ and $$1^2 + 2 = 3$$.
Then $$6^2 + \left(\sqrt{10368}\right)^2 = 102^2$$ is a scaled up version of $$1^2 + \left(\sqrt{288}\right)^2 = 17^2$$.

More interesting for the lack of common factors between terms are:
$$49^2 + \left(\sqrt{10368}\right)^2 = 113^2$$ and
$$2591^2 + \left(\sqrt{10368}\right)^2 = 2593^2$$

But I would not call that a recurrence.

To claim my attempts to educate are a waste of time would make me think you think Jason.Marshall is incapable of learning or that learning is valueless or that my time is too valuable to be spend answering Jason.Marshall's questions. Is this your position? Even if it is a waste of my time, what business is it of yours who have never found value in my posts?

How are physical measurements, with necessarily finite precision, going to establish that the ratio of circumference to diameter is exactly π? What is needed are procedures that generalize to any level of precision. Archimedes gave us such a method, but needed regular 96-gons to demonstrate 3 + 10/71 < π < 22/7. But lots of numbers satisfy 3 + 10/71 < x < 22/7, for example: 355/113 and all numbers of the form (22 n+223)/(7 n+71) where n > 0. Today's modern methods give higher precision answers with much less labor than those of Archimedes but in doing so leave the student without connection to the geometrical problem. So I use antique methods in modern notation. Post #6 gives a result from analysis that π ≠ 3.125, post #14 gives a demonstration of how to proceed from Pythagoras via Newton to a ratio of circumference to diameter of a circle of any radius and relates that to an infinite sum of positive quantities, which prove 3.125 < π. Posts #15 and #18 demonstrate that in the limit of arbitrarily many sides, the perimeter and area of polygons approach a common behavior which is that of the circle, but also that they all share a relation of 2 A = r P. Archimedes developed the first topic:
http://itech.fgcu.edu/faculty/clindsey/mhf4404/archimedes/archimedes.html

The book is puerile and wrong, demonstrating its wrongness is not necessarily trivial depending on the education level of the reader and one would have to have only a superficial understanding of the topic to think the value of π could be settled with a single physical measurement. You have not been elected to judge Jason.Marshall for the rest of us, but instead we are all free to judge his motivation and understanding by his interactions on this forum.

Well said.

Also, I venture to suggest, one attribute of mathematicians who love their subject is that they find it hard to resist the challenge of proving or disproving a mathematical contention, purely for the fun of the exercise. Mathematicians, in my limited experience, are often playful. Which I think is exactly as it should be.

I have learnt a bit from this thread, even if others have not.

Here is something I was thinking about but cant seem to understand why its most likely another easy question for you Rpenner so here we go I noticed that if you multiply the radian by 180 degrees let this be quantity #1 you get the quantity of radian square times pi let this be quantity#2, both quantity #1 and#2 or exactly equal??

... if you multiply the radian by 180 degrees let this be quantity #1 you get the quantity of radian square times pi let this be quantity#2, both quantity #1 and#2 or exactly equal??
Could you write that in equation form? The actual numbers?

radian equals 180/pi lets call it r so...r * 180 = X then r ^2 *pi = Y then X=Y

radian equals 180/pi lets call it r so...r * 180 = X then r ^2 *pi = Y then X=Y
"Hala-kin"

radian equals 180/pi lets call it r so...r * 180 = X then r ^2 *pi = Y then X=Y

That looks pretty meaningless. Multiplying and dividing the same numbers...

No arc length and you seem to just use circle decoy.

radian equals 180/pi lets call it r so...r * 180 = X then r ^2 *pi = Y then X=Y

post deleted

radian equals 180/pi lets call it r so...r * 180 = X then r ^2 *pi = Y then X=Y
Um....in an equation, you use numbers, letters and an equals sign. Can you not write that in math?

Anyway, putting it together, it looks like you mean this:

r * 180 = pi * r^2
solving for r:
r= 180/pi

So what?

To claim my attempts to educate are a waste of time would make me think you think Jason Marshall is incapable of learning or that learning is valueless or that my time is too valuable to be spend answering Jason Marshall's questions. Is this your position?
Actually no, my position is that Jason Marshall is some kind of plant, and this thread is the rpenner smoke and mirrors show. Because the simple answer is to tell "Jason Marshall" to go measure the diameter and circumference of a disk.

Actually no, my position is that Jason Marshall is some kind of plant, and this thread is the rpenner smoke and mirrors show. Because the simple answer is to tell "Jason Marshall" to go measure the diameter and circumference of a disk.

I guess my post two was too much of a dry hint.