1. ## My very, very basic questions about pi

To calculate pi from scratch, I imagine that you have to either:

a. Physically draw a circle using a compass or the old pen tied to a pin method... and manually measure the Circumference (with a piece of string? ) and Diameter and then apply the Pi = C/d formula OR

b. (The only part of the Wikipedia article on pi that I understand, but only vaguely at that) Draw a regular polygon with as many sides as possible. Calculate Pi using the polygon's C and d. Or maybe use an area formula for the polygon--would that be more accurate?--then use Pi = A/r squared.

c. Use a computer to draw the polygon. The more sides to the polygon, the more accurate the resulting Pi.

Well, how do people actually go about getting a better and better approximation of pi?

And are the decimal places in 3.141592654 (from my calculator) accepted as the first few decimal places of the very best approximation, or can these first few numbers still change / be refined?

Originally Posted by Lilalena
Well, how do people actually go about getting a better and better approximation of pi?
Mostly they use series expressions for pi. Lots of examples of series are on the wikipedia page. Here's one:

$\pi = 4\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\dots\right)$

This gets more and more accurate as you take more and more terms in the series. For example, the first term gives:

$\pi = 4$

The first two terms give:

$\pi = 4\left(1-\frac{1}{3}\right) = 4\left(\frac{2}{3}\right) = \frac{8}{3} = 2.666666666$

The first three terms give:

$\pi=\frac{8}{3} + 4\times \frac{1}{5} = 3.46.....$

This particular series requires a LOT of terms to get more and more decimal digits of pi correct, but some of the other series on the wikipedia page get more digits with many fewer terms (in technical terms, the series converge more rapidly than the example I have used here).

And are the decimal places in 3.141592654 (from my calculator) accepted as the first few decimal places of the very best approximation, or can these first few numbers still change / be refined?
If we go just a few more places we have:

$\pi = 3.141592653 589792...$

$\pi = 3.141592654$

As you can see, your calculator's value rounds up at the last displayed digit. However, this approximation of pi is accurate except at the very last digit.

I'm not entirely sure what you mean about changing or refining the first few numbers. Pi is what it is. You can round it off at any number of decimal digits you want, but up to the point where you choose to stop taking the digits, the digits must agree with everybody else's value of pi.

3. Originally Posted by Lilalena
Well, how do people actually go about getting a better and better approximation of pi?
Pi can be calculated to an arbitrary degree of accuracy, given any random procedure which can be matched to the positive integers, because the probability of any two randomly selected positive integers being co-prime, is 6/pi^2.

4. ughaibu:

Perhaps an example would be good. Also, it is possible that not everybody knows what "co-prime" means.

5. Originally Posted by James R
ughaibu:

Perhaps an example would be good. Also, it is possible that not everybody knows what "co-prime" means.
Here is a site which provides suitable numbers: http://www.random.org/integers/ Two integers are "co-prime" if they have no common factor, and here's a short proof of the relevant result: http://www.physics.harvard.edu/acade...week/sol44.pdf

6. Originally Posted by James R

Mostly they use series expressions for pi. Lots of examples of series are on the wikipedia page. Here's one:

$\pi = 4\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\dots\right)$

This gets more and more accurate as you take more and more terms in the series. For example, the first term gives:

$\pi = 4$

The first two terms give:

$\pi = 4\left(1-\frac{1}{3}\right) = 4\left(\frac{2}{3}\right) = \frac{8}{3} = 2.666666666$

The first three terms give:

$\pi=\frac{8}{3} + 4\times \frac{1}{5} = 3.46.....$

This particular series requires a LOT of terms to get more and more decimal digits of pi correct, but some of the other series on the wikipedia page get more digits with many fewer terms (in technical terms, the series converge more rapidly than the example I have used here).

Questions:

1. Why should these numbers end up giving you pi? Has it to do with polygons i.e. is your example related to this Wikipedia entry?

Another geometry-based approach, attributed to Archimedes,[33] is to calculate the perimeter, Pn, of a regular polygon with n sides circumscribed around a circle with diameter d. Then compute the limit of a sequence as n increases to infinity:

\pi = \lim_{n \to \infty}\frac{P_{n}}{d}.

This sequence converges because the more sides the polygon has, the smaller its maximum distance from the circle.

2. Or is your example a 'non-geometric' approach. From Wikipedia:
mathematicians often prefer to define π without reference to geometry, instead selecting one of its analytic properties as a definition.

- I was shocked by this line actually. How do you divorce pi from geometry?
What are the uses of pi outside of geometry?

* * *

Originally Posted by James R

I'm not entirely sure what you mean about changing or refining the first few numbers. Pi is what it is. You can round it off at any number of decimal digits you want, but up to the point where you choose to stop taking the digits, the digits must agree with everybody else's value of pi.

I was wondering if the scenario that someone turns up and claims pi is for instance 3.14142631118741 etc.. giving completely different digits from the accepted ones, and says he can prove it is the best approximation -- is such a scenario possible at all? Or are the first few digits of pi pretty much set in stone by now?

I mean how do people prove that their approach to calculating pi is the one that has the best chance of yielding the closest approximation?

7. Originally Posted by ughaibu
Here is a site which provides suitable numbers: http://www.random.org/integers/ Two integers are "co-prime" if they have no common factor, and here's a short proof of the relevant result: http://www.physics.harvard.edu/acade...week/sol44.pdf
Cool idea..! I would presume the number approached is dependent upon the range from which the integers are drawn. It would be interesting to see how the range and ultimate accuracy are related...

8. By using analytic geometry, one can discover a scheme to approximate $2 \pi$ to any degree necessary. This method is attributed to Archimedes.
Take a circle of radius 1. Construct the circumscribing and inscribing equilateral triangles, which have perimeters of $6\sqrt{3}$ and $3 \sqrt{3}$, respectively. Divide the angles in half and construct the regular circumscribing and inscribing hexagons, with perimeters of $4\sqrt{3}$ and 6, respectively. Divide the angles in half and construct the regular circumscribing and inscribing 12-gons, with perimeters of $24(2 -\sqrt{3})$ and $6(\sqrt{6}- \sqrt{2})$, respectively. It turns out that there is a relation between these numbers for regular polygons of n sides and regular polygons of 2n sides.
$\begin{eqnarray} f_{\tiny n} & = & \sqrt{ \frac{2 x_{\tiny n} }{x_{\tiny n} \; + \; y_{\tiny n} }} \\ x_{\tiny 2n} & = & y_{\tiny n} f_{\tiny n}^2 \\ y_{\tiny 2n} & = & y_{\tiny n} f_{\tiny n} \end{eqnarray}$
$\begin{array}{rccr} n & x_n & y_n & f_n \\ 3 & 6 \sqrt{3} & 3\sqrt{3} & \frac{2}{\sqrt{3}} \; \approx \; 1.1547 \\ 6 & 4 \sqrt{3} & 6 & \sqrt{6} - \sqrt{2} \; \approx \; 1.0353 \\ 12 & 48 - 24 \sqrt{3} & 6 \sqrt{6} - 6 \sqrt{2} & \sqrt{16 - 10 \sqrt{2} + 8 sqrt{3} - 6 sqrt{6}} \; \approx \; 1.0086 \\ 24 & 48 ( \sqrt{6} + \sqrt{2} - \sqrt{3} - 2 ) & 24 \sqrt{ 2 - \sqrt{2 + \sqrt{3}}} & \frac{2}{ \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}} } \; \approx \; 1.0021 \\ 48 & 96 \sqrt{ \frac{2-\sqrt{2+\sqrt{2+\sqrt{3}}} }{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}} & 48 \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}} & \frac{2}{ \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}} \; \approx \; 1.0005 \\ 96 & 192 \sqrt{ \frac{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{3}}}}}} & 96 \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}} & \dots \end{array}$

So this method converges on $\pi$ from above and from below, but you need the equivalent of nearly a 100-sided polygon to get just a few digits of accuracy. Archimedes then proved that $\frac{223}{71} \quad \lt \quad 48 \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}} \quad \lt \quad \pi \quad \lt \quad 96 \sqrt{ \frac{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{3}}}}}} \quad \lt \quad \frac{22}{7}$ which is a useful result we use today.

http://mathworld.wolfram.com/ArchimedesAlgorithm.html

$\pi$ is also the unique smallest positive solution to the equation $\sin x = 0$. Using analysis, we can use this definition and write it like this: $\pi = \inf \left{ x \in \mathbb{R}^{+} \, : \; \sin x = 0 \right} = \inf \left{ x \in \mathbb{Q}^{+} \, : \; \sin x < 0 \right}$

$\sin$ here is divorced from geometry by its definition in terms of analysis: $\sin x = \frac{e^{\tiny ix} \; - \; e^{\tiny -ix}}{2i} = \sum_{k=0}^{\infty} \frac{(-1)^{\tiny k} x^{\tiny 2k+1}}{(2k+1)!} = x - \frac{x^{\tiny 3}}{3!} + \frac{x^{\tiny 5}}{5!} - \frac{x^{\tiny 7}}{7!} + \dots$

9. Thanks rpenner. When I need to calculate pi I prefer my method though:

Start with $e^{{\pi}{i}}+1=0$, and simply solve for $\pi$

10. You won't know which solution you will get since all numbers of the form $\pm(2n+1)\pi$ are solutions for $e^{ix} = -1$

11. Originally Posted by rpenner
You won't know which solution you will get since all numbers of the form $\pm(2n+1)\pi$ are solutions for $e^{ix} = -1$
Drat!

12. Originally Posted by Lilalena
Well, how do people actually go about getting a better and better approximation of pi?
by using a method known as an arithmetic series.

And are the decimal places in 3.141592654 (from my calculator) accepted as the first few decimal places of the very best approximation, or can these first few numbers still change / be refined?
the numbers you quoted are exact and will never change except for rounding.

13. Originally Posted by leopold
the numbers you quoted are exact and will never change except for rounding.
If you write a number down as a decimal and round it off it is not $\pi$. It may be a good approximation to $\pi$ but $\pi$ is defined to be the exact number that is given by the ratio of a circles circumference to it's diameter.

14. . . . so . . .is the value of pi the same in 'curled-up' dimensions (ala string theory spaces)?

15. Yes, and the value of 1 is the same also.

16. Originally Posted by rpenner
Yes, and the value of 1 is the same also.
...but Prom said
pi is defined to be the exact number that is given by the ratio of a circles circumference to it's diameter.
...with no mention of "flat" space; is that part of the definition? Because the ratio of circle's circumference to its diameter could take on any value.

17. Originally Posted by RJBeery
...with no mention of "flat" space; is that part of the definition? Because the ratio of circle's circumference to its diameter could take on any value.
yes because a circle is based on, and derived by, plane geometry.

it seems to me if you want to talk "curved space" then you must resort to something other than plane geometry, spherical geometry perhaps.

18. Originally Posted by leopold
yes because a circle is based on, and derived by, plane geometry.

it seems to me if you want to talk "curved space" then you must resort to something other than plane geometry, spherical geometry perhaps.
You can have non-Euclidean plane geometry. You usage of "flat" differs from mine in this context.

19. Originally Posted by wlminex
. . . so . . .is the value of pi the same in 'curled-up' dimensions (ala string theory spaces)?
Rpenner . . . let me restate . . . . is the geometric construction and mathematical solution for pi, the same in string theory space as it is in "flat" geometric space?

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•