• Welcome to the Internet Infidels Discussion Board.

Happy Pi Day

 
 
 
lpetrich

lpetrich

Contributor
Joined
Jul 27, 2000
Messages
26,850
Location
Eugene, OR
Gender
Male
Basic Beliefs
Atheist
Happy Pi Day: 3-14-2018

Called that because of what its digits are.

Pi is the ratio of a circle's circumference to its diameter, its distance around it to its distance across it.

Some other names for it are Archimedes's number and the circle constant.

I've seen an argument that we ought to use 2*pi instead, the ratio of a circle's circumference to its radius (half its diameter) instead. I've even seen a name for it: tau. But it may be too late now.

Archimedes was notable for finding a way to find the value of pi. He did it by making a circle and then constructing a series of regular polygons, one just inside the circle and the other the same shape, but just outside the circle. Each polygon is constructed from the previous one by dividing its sides into two equal halves and then moving out the division points to make the new polygon regular. Translating into algebra, his method uses some trigonometric identities:

tan(pi/3) = sqrt(3) -- triangle
tan(pi/4) = 1 -- square
tan(pi/5) = sqrt(5 - 2*Sqrt[5]) -- pentagon
tan(pi/6) = 1/sqrt(3) -- hexagon

tan(a/2) = sin(a)/(1 + cos(a))
sin(a) < a < tan(a)

Archimedes found that (223/71) < pi < (22/7).

Over the last five centuries or so, mathematicians have found another way to calculate pi: infinite series. That is the sum of an infinite number of numbers, and sometimes the product of them. Inverse trigonometric functions are a rather obvious source of ways to calculate pi, and a simple one is
pi = 4 * ( 1 - 1/3 + 1/5 - 1/7 + ... )
from arctan(x) = x - x3/3 + x5/5 - x7/7 + ...

That does not converge very fast, but this formula converges much faster:
pi = 4 * ( 4*arctan(1/5) - arctan(1/239) )

and some mathematicians have discovered even faster-converging ones.

Mathematicians have also devised iterative formulas in addition to Archimedes's ones, like the Gauss-Legendre arithmetic-geometric-mean one:

Initial: a = 1, b = 1/sqrt(2), t = 1/4, p = 1
Next: a' = (a + b)/2, b' = sqrt(a*b), t' = t - p*(a - a')2, p' = 2p
Estimate of pi: (a + b)2/(4t)

So far, they have gotten up to 10 trillion (1013) digits.


But is there an end? Mathematicians have proved that pi is irrational and transcendental, meaning that pi's digits never become a repeating sequence as one continues with them. The proof that pi is irrational is somewhat complicated, but it should be easy to follow for someone with experience with calculus, and it works much like the proof that the square root of 2 is irrational. It shows that if pi was rational, that a contradiction would result.

The proof that pi is transcendental is more difficult, but that result has an implication for one of the unsolved mathematical problems of antiquity: squaring the circle with only a ruler and compass. This problem is finding the size of a square with the same area as some circle, and it is equivalent to finding the value of sqrt(pi). Ruler-and-compass solutions are equivalent to doing a finite number of arithmetic operations and square roots. Archimedes introduced a marked ruler or neusis, and it could do cube roots. That enabled solution of duplicating the cube and trisecting the angle, though not of squaring the circle. Ruler-and-compass numbers and marked-ruler-and-compass numbers are all algebraic. They can all be expressed as solutions of integer-coefficient polynomial equations. But pi is transcendental, and that means that it is not algebraic. Thus, squaring the circle cannot be done with (marked-)ruler-and-compass numbers.
 
We need to set a new start date for year numbering, perhaps using the date of the accession of Valentinian III as Emperor of Rome as our datum, to replace the purported date of the birth of Jesus, so that today can be 3.14.1593
 
Pie-I-have-Eaten.jpg
 
I was thinking about Pi and significant digits... as in, how many digits of Pi do you NEED?

So, I made a thought experiment...

What problem requires the maximum number of digits to solve, and then how many digits would be sufficient to solve it?

Here is the problem I devised:


How many digits of Pi would be needed to have a sufficiently accurate plot of the circumference of a circle with a radius of 27.4 billion light years, within 1 plank distance? 27.4 bly represents "across the universe", and 1 plank distance represents the highest level of granularity.
 
Malintent said:
I was thinking about Pi and significant digits... as in, how many digits of Pi do you NEED?

So, I made a thought experiment...

What problem requires the maximum number of digits to solve, and then how many digits would be sufficient to solve it?

Here is the problem I devised:


How many digits of Pi would be needed to have a sufficiently accurate plot of the circumference of a circle with a radius of 27.4 billion light years, within 1 plank distance? 27.4 bly represents "across the universe", and 1 plank distance represents the highest level of granularity.
Click to expand...

Problems arise when you need to chain computations together. Your approximation for pi might be fine for this particular calculation, but then if you try to use THAT number in another computation, and another, and another, the error can grow and grow until you are off by a lot.
 
I've always had trouble conceptualizing significant digits... at least, insofar as understanding how "more is always better" is incorrect.

I'm trying t understand what the "maximum error" can be (for the purpose of finding a maximum number of useful significant digits.

When looking at vectors (distances and directions - calculated from an arc), the higher the magnitude (distance) the greater the magnification of error.

So, when looking at the max distance (disallowing infinity) as being "the diameter of the universe", and the minimum distance as "the plank length", what is the minimum number of digits of Pi needed for the error to be eliminated (lower than 1 plank length)?

My understanding from another poster is that it has already been calculated to some trillions of digits. Is that sufficient? Even close? Intuitively, I feel like trillions of digits is insufficient... that the error would still be thousands of light years off. I would like to quantify for the purposes of setting an "ultimate goal" for calculation of Pi.

Has it been mathematically proven that Pi repeats forever? It seems it should be. I am curious what happens to the number as we approach (and pass?) that maximum number of significant digits in this universe.
 
Malintent said:
I've always had trouble conceptualizing significant digits... at least, insofar as understanding how "more is always better" is incorrect.

I'm trying t understand what the "maximum error" can be (for the purpose of finding a maximum number of useful significant digits.

When looking at vectors (distances and directions - calculated from an arc), the higher the magnitude (distance) the greater the magnification of error.

So, when looking at the max distance (disallowing infinity) as being "the diameter of the universe", and the minimum distance as "the plank length", what is the minimum number of digits of Pi needed for the error to be eliminated (lower than 1 plank length)?

My understanding from another poster is that it has already been calculated to some trillions of digits. Is that sufficient? Even close? Intuitively, I feel like trillions of digits is insufficient... that the error would still be thousands of light years off. I would like to quantify for the purposes of setting an "ultimate goal" for calculation of Pi.
Click to expand...

You wouldn't even need 100 digits to get that precision for one calculation. More than one calculation will be subject to cumulative round-off errors that can grow exponentially, or at least much faster than you'd expect. Numerical analysis of round-off error and stability is more important than a single 'ultimate precision goal'.

Has it been mathematically proven that Pi repeats forever? It seems it should be. I am curious what happens to the number as we approach (and pass?) that maximum number of significant digits in this universe.
Click to expand...

The decimal expansion of pi never repeats and never ends, it is an irrational number.
 
Perhaps my thought experiment is flawed... but here is what I am thinking...

an arc has a length that is calculated based on the radius of the arc and Pi.

If the radius of that arc is 27.4 bly, how many digits of Pi is needed for the accuracy of the calculation of the length of that arc to be within 1 plank distance?


You have an infinitely powerful laser that emits a 1 particle wide beam of light that is unaffected by gravity or matter, and arrives at the end of the universe instantaneously. A magic laser.

You stand on one side of the universe and shine the laser in a direction. The laser instantly hits the other side of the universe.
At what degree of precision must you be able to rotate the laser pointer such that the endpoint of the laser moves only 1 plank length from its original point? Disregard uncertainty and relativity... this is an Euclidean geometry problem at unreasonable scale.

It is starting to sound paradoxical.. How can you move less than 1 plank distance (1 plank radian - is that a thing?) such that the other end so far away only moves less than 1 plank distance? This is sort of what I mean by an upper limit of the useful number of digits of Pi... calculating fractional units of Plank is meaningless.
 
Malintent said:
Perhaps my thought experiment is flawed... but here is what I am thinking...

an arc has a length that is calculated based on the radius of the arc and Pi.

If the radius of that arc is 27.4 bly, how many digits of Pi is needed for the accuracy of the calculation of the length of that arc to be within 1 plank distance?
Click to expand...

I answered that already - fewer than 100 digits. I get around 60 or so, but you aren't asking the right questions, so the answer isn't super relevant.

You have an infinitely powerful laser that emits a 1 particle wide beam of light that is unaffected by gravity or matter, and arrives at the end of the universe instantaneously. A magic laser.

You stand on one side of the universe and shine the laser in a direction. The laser instantly hits the other side of the universe.

At what degree of precision must you be able to rotate the laser pointer such that the endpoint of the laser moves only 1 plank length from its original point? Disregard uncertainty and relativity... this is an Euclidean geometry problem at unreasonable scale.
Click to expand...

Same answer.

It is starting to sound paradoxical.. How can you move less than 1 plank distance (1 plank radian - is that a thing?) such that the other end so far away only moves less than 1 plank distance? This is sort of what I mean by an upper limit of the useful number of digits of Pi... calculating fractional units of Plank is meaningless.
Click to expand...

That isn't what the Planck length means. And you're still missing the point that single-calculation round-off errors aren't the issue here.
 
Malintent said:
Perhaps my thought experiment is flawed... but here is what I am thinking...

an arc has a length that is calculated based on the radius of the arc and Pi.

If the radius of that arc is 27.4 bly, how many digits of Pi is needed for the accuracy of the calculation of the length of that arc to be within 1 plank distance?


You have an infinitely powerful laser that emits a 1 particle wide beam of light that is unaffected by gravity or matter, and arrives at the end of the universe instantaneously. A magic laser.

You stand on one side of the universe and shine the laser in a direction. The laser instantly hits the other side of the universe.
At what degree of precision must you be able to rotate the laser pointer such that the endpoint of the laser moves only 1 plank length from its original point? Disregard uncertainty and relativity... this is an Euclidean geometry problem at unreasonable scale.

It is starting to sound paradoxical.. How can you move less than 1 plank distance (1 plank radian - is that a thing?) such that the other end so far away only moves less than 1 plank distance? This is sort of what I mean by an upper limit of the useful number of digits of Pi... calculating fractional units of Plank is meaningless.
Click to expand...

Reality isn't granular.

As your thought experiment implies, the Planck length is not the smallest possible subdivision of space; you can in principle produce a measurable change by moving a distance which despite being literally immeasurably small, is calculable and far smaller than your proposed minimum distance. Angles (and therefore arcs) can be arbitrarily small, and an infinite number of possible angles exist between any two positions separated by a finite angle, just as an infinite number of reals exist between any chosen pair of reals. There's no number so small that it cannot be divided by two.
 
Consider a number in some base with form

FSD1.FSD2, infinite number of repeats of FSD3

where FSD1, FSD2, and FSD3 are finite sequences of digits in some number base. It is easy to prove that that number is a rational number.

So (repeating decimals) -> (rational number)

Proving the opposite direction is more difficult: (rational number) -> (repeating decimals). One can use Euler's totient theorem for that. That theorem is

a^(phi(n)) = 1 mod n

for relatively prime (coprime) positive integers a and n. Here, phi(n) is the Euler totient function, the number of positive integers relatively prime to n.

For n having prime factors p1, p2, ..., pk, with exponents m1, m2, ..., mk, n = p1^m1 * p2^m2 * ... * pk^mk
phi(n) = phi(p1^m1) * phi(p2^m2) * ... * phi(pk^mk)
phi(p^m) = p^(m-1) * (p - 1)


So an irrational number will always have a nonrepeating decimal representation and a nonrepeating counterpart in every other number base.
 
Proving Pi is Irrational: a step-by-step guide to a “simple proof” requiring only high school calculus – Mind Your Decisions

He starts with a proof that the square root of 2 is irrational, introducing it to introduce the idea of proof by contradiction.

Let sqrt(2) = a/b where a and b are relatively prime positively integers. This makes their fraction be in lowest terms.

Then, 2 = a^2/b^2 and a^2 = 2*b^2. This makes a divisible by 2, or a = 2*c for positive integer c. Then 4*c^2 = 2*b^2 or 2*c^2 = b^2, making b divisible by 2, and contradicting the initial premise. Thus, sqrt(2) is irrational. This proof also works on every positive integer that is not the square of some positive integer.


The proof that pi is irrational. Suppose that pi = a/b, where a and b are relatively prime positive integers. Define a function
f(x) = x^n * (a - b*x)^n / n!

and find the integral pint = integral of f(x)*sin(x) over x from 0 to pi

Define function g(x) = f(x) - f(2)(x) - f(4)(x) + ... + (-1)^n*f(2n)(x) (any more will be zero)
where f(k)(x) = derivative k of f(x)

We find
integral of f(x)*sin(x) = (first derivative of g(x))*sin(x) - g(x)*cos(x)

g(x) was selected so that g(x) + (second derivative of g(x)) = f(x)

Plugging in the bounds of integration gives the result that pi is some integer.

However, for increasing n, f(x) is bounded from above over the domain of integration by a^n/n!. This means that 0 < pint < 1 and that pint is not an integer. This contradiction means that pi is irrational.
 
lpetrich said:
Consider a number in some base with form

FSD1.FSD2, infinite number of repeats of FSD3

where FSD1, FSD2, and FSD3 are finite sequences of digits in some number base. It is easy to prove that that number is a rational number.

So (repeating decimals) -> (rational number)

Proving the opposite direction is more difficult: (rational number) -> (repeating decimals). One can use Euler's totient theorem for that. That theorem is

a^(phi(n)) = 1 mod n

for relatively prime (coprime) positive integers a and n. Here, phi(n) is the Euler totient function, the number of positive integers relatively prime to n.

For n having prime factors p1, p2, ..., pk, with exponents m1, m2, ..., mk, n = p1^m1 * p2^m2 * ... * pk^mk
phi(n) = phi(p1^m1) * phi(p2^m2) * ... * phi(pk^mk)
phi(p^m) = p^(m-1) * (p - 1)


So an irrational number will always have a nonrepeating decimal representation and a nonrepeating counterpart in every other number base.
Click to expand...

You can avoid the totient function and prime factorization, and prove it simply using the pigeonhole principle.

Take a/b and look at the the sequence of fractions a/b, 10a/b, 100a/b, ..., etc. Since these are all fractions with denominator b, their fractional parts must be one of 0, 1/b, 2/b, ..., (b-1)/b. There are only b possible fractional parts, so at some point we have to get two values with the same remainder, meaning that the decimal expansions past some point are just shifted copies of each other, and must be repeating. This also gives (for free) that, for a rational number a/b, the repeat must start before the (b-1)st digit past the decimal point, and can be at most b-1 digits long.
 
The  Lindemann–Weierstrass theorem implies that pi is transcendental, meaning non-algebraic. Every algebraic number satisfies some polynomial equation with integer coefficients.

This theorem has two forms.

If a1, a2, ..., an are algebraic numbers that are linearly independent over the rational numbers, then e^a1, e^a2, ..., e^an are algebraically independent over the rational numbers.

If a1, a2, ..., an are distinct algebraic numbers, then e^a1, e^a2, ..., e^an are linearly independent over the algebraic numbers.


The numbers a1, a2, ..., an are linearly independent over some domain X if there are no coefficients c1, c2, ..., cn in X such that a1*c1 + a2*c2 + ... + an*cn = 0

The numbers a1, a2, ..., an are algebraically independent over some domain X if there is no polynomial in them with coefficients in X that will equal zero.


For e, take {e, 1} = {e^1, e^0}. Since 0 and 1 are both algebraic, there is thus no nontrivial polynomial in e with rational coefficients that equals zero. Thus, e is transcendental.

For pi, take this famous identity: e^(pi*i) + 1 = 0. It is equivalent to e^(pi*i) + e^0 = 0. We know that i is algebraic, from i^2 = -1, and if pi is algebraic, then that equation would not be possible. Thus, pi is transcendental.
 
Can't recall where I just heard it very recently, but in a video the claim was put forth that for most science all that was required was 30 places. I seem to recall hearing for that for most "normal" applications less than 10 will get you quite good results. I guess it depends on what you're doing, and as someone else pointed out, how one measurement/calculation might get re-used.
 
You must log in or register to reply here.
Back
Top Bottom