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JASONQUANTUM1
Hello all,

My name is Jason D. Padgett, I'm a math/physics student in Washington state and I would like to float this idea. I believe that I have found where Pi physically ends. Since the Planck constant is the smallest interval of space that can be measured and still be relative then Pi would physically end at:

Two times the interval from -1 to 1 of the square root of 1-x^2 dx with
delta t=b-a/h.

I apologize for typing this out but the computer I'm on doesn't have equation writing capabilities. In the above equation h=Planck length.

To calculate pi to any decimal we can use a trig formula:

f(x)=xsin(pi/x) As x approaches infinity f(x) will approach Pi. In this equation x stand for the number of sides to a circle. But again, this will physically end when each side to the circle equals one Planck length.

For those who argue that the Planck length may not be the smallest measurement I would answer that this equation will work as long as there is ANY limit to how small you can measure space.

What do you think?

Sincerely,

Jason D. Padgett
JASONQUANTUM1@YAHOO.COM
Sapo
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TheDoc
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JASONQUANTUM1
One last thing...if you have a graphing calculator graph f(x)=xsin(Pi/x) and you can test this equation for yourself so that you will know it is correct.

Again, this is where Pi as a shape, expanding closer and closer to a circle would physically end (when each side to the circle is one Planck length). Or in integral form when delta t=b-a/h

Jason
TheDoc
QUOTE (JASONQUANTUM1+Mar 12 2008, 01:27 AM)
One last thing...if you have a graphing calculator graph f(x)=xsin(Pi/x) and you can test this equation for yourself so that you will know it is correct.

Again, this is where Pi as a shape, expanding closer and closer to a circle would physically end (when each side to the circle is one Planck length). Or in integral form when delta t=b-a/h

Jason

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AlphaNumeric
I explained this to you before Jason.

f(x) = x.sin(pi/x)

Take a Taylor expansion :

sin(pi/x) = pi/x + O(1/x²)

Therefore

x.sin(pi/x) = pi + O(1/x)

Since O(1/x) -> 0 as x->infinity, x.sin(pi/x) ->pi

This isn't novel, it's something any 17 year old school pupil should know how to do in their sleep.

The problem is, it doesn't tell you how to calculate pi. The formula is nothing but a limiting process which generalises to :

lim x->infinity of x.sin(k/x) = k

If you want to compute the expansion for pi then you can work out the Taylor series for arctan(x), then work out 4arctan(1) = pi.

Or use one of these.

Physics doesn't come into mathematical proofs. Pi is mathematically proven to be irrational and transcendental. Not only does it not have a terminating decimal expansion, we know the expansion doesn't repeat and we know pi isn't the zero of a polynomial with integer coefficents.
JASONQUANTUM1
What I'm trying to show is a connection between mathematics and physics. While the sequence of number in Pi will 'mathematically' go to infinity and the sequence of numbers will never repeat because as you inscribe more and more polygons within a circle you are always creating a new and unique are and circumference, I would still think that since there is a physical limit to the measurement of space then 'in the universe of physics' Pi would end when the sub intervals=Planck length.

I agree with you that Pi mathematically will go to infinity but what I'm talking about is limiting the mathematics to the physical limitations of our universe.

Sincerely,

Jason
TheDoc
QUOTE (JASONQUANTUM1+Mar 12 2008, 01:49 AM)
<incoherent dribble snipped>

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AlphaNumeric
QUOTE (JASONQUANTUM1+Mar 12 2008, 02:49 AM)
While the sequence of number in Pi will 'mathematically' go to infinity and the sequence of numbers will never repeat because as you inscribe more and more polygons within a circle you are always creating a new and unique are and circumference,

There's no reason to think that just because you construct pi via an interative process then it's irrational. I can construct any rational number you like through an interative processes, which gives me a new number every step but whose limit is a particular number.

The proof of Pi's irrationality is a little more involved than that. See here.
QUOTE (JASONQUANTUM1+Mar 12 2008, 02:49 AM)
I would still think that since there is a physical limit to the measurement of space then 'in the universe of physics' Pi would end when the sub intervals=Planck length.
Then you haven't found 'an end to pi', because that's a mathematical claim. You have found the very obvious fact that our ability to measure reality is not perfect. Even ignoring quantum mechanics and Planck lengths, our instruments are only so good. We can only time down to certain intervals, we can only measure down to certain lengths. Already we knowpi so well that we'd know the circumference of the universe to within a Planck length, if it were a circle and we knew it's radius. That's only about 100 orders of magnitude. All you need is the first 100 digits of pi. We know the first trillion.

So we know all we need to know about pi in terms of physics, but there's so much more to it than that.
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