# Quick's quick question

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• Jul 2nd 2006, 05:22 PM
Quick
Quick's quick question
is there a fraction form of pi? if there isn't, then how do you find the number? (besides measuring to an infinitely small measurement of course :D )
• Jul 2nd 2006, 07:33 PM
CaptainBlack
Quote:

Originally Posted by Quick
is there a fraction form of pi? if there isn't, then how do you find the number? (besides measuring to an infinitely small measurement of course :D )

$\pi$ is irrational first proven by Lambert in 1760 (in fact it is
transcendental - that is not even algebraic).

Rational approximations to $\pi$ of arbitary precission can be
found by a number of methods, the earliest probably being that of
Archimedes obtained by trapping the area of the unit circle between the
areas of inscribed and circumscribed regular polygons.

RonL
• Jul 2nd 2006, 08:44 PM
ThePerfectHacker
CaptainBlack you just need to make a simple explanation complicated. Listen to me Quick.
--
Okay, first there are not $a,b$ such as,
$\pi =\frac{a}{b}$- this means that $\pi$ is irrational. This was proven in 1776 (correction to CaptainBlack) by Lambert.

The other fact that $\pi$ is transcendental was proven (I think in 1882) by Louiville. Meaning there is no polynomial equation with integers such has it as its solution. For example, $\sqrt{2}$ is irrational but, $x^2-2=0$ is a polynomial which has it as its root.

---
(CaptainBlack is having a bad day cuz, poor England lost.
I am having a great day cuz, my team, Germany, is in the semifinals and are probably winning.)
• Jul 2nd 2006, 08:47 PM
ThePerfectHacker
To approximate $\pi$ there are several ways. My favorite is the use of "countinued fractions" it produces the best possible approximate with a certain denominator. Another way, is through "infinite series", for example, Ramanajuan (let him rest in peace) produced the best series ever. It gives approximately 8 decimals places each time used.
• Jul 2nd 2006, 09:45 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
(CaptainBlack is having a bad day cuz, poor England lost.

1. It was Saturday, all the posts concerned are dated today on my machine
which is Monday here. Football is not such a big factor in my life that I am
going to sulk about a result for more than about 10 minutes (after all our
local team is Portsmouth, so we are used to losing down here)

2. I thought they played well after Rooney was sent off - showing that
England is lucky to have gotten rid of Ericson at last.

3. They have nothing to be ashamed of, they played well, and the result
is a lottery - but what ya gonna do.

RonL
• Jul 2nd 2006, 09:53 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
This was proven in 1776 (correction to CaptainBlack) by Lambert.

My searches give 1761, 1768, 1765, ..

(can't find the original reference I had which gave 1760 :mad: )

RonL
• Jul 3rd 2006, 06:44 AM
ThePerfectHacker
Accoriding to Wikipedia it was 1761, that is what I would accept. I can promise that I seen this number change many times, which is perhaps why you choose 1760. I was right about 1882, that I seen everywhere to be the same. Interesting to why, many authors change this date.
• Jul 3rd 2006, 07:37 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Accoriding to Wikipedia it was 1761, that is what I would accept. I can promise that I seen this number change many times, which is perhaps why you choose 1760. I was right about 1882, that I seen everywhere to be the same. Interesting to why, many authors change this date.

I thought at one point this might have been due to Britain and British North
America not being on the Gregorian calendar, but that change took place in
1752, so its not that :mad:

RonL
• Jul 3rd 2006, 12:45 PM
Soroban
Like TPHacker, I prefer the continued fractions approach.

You can create your own rational approximation to $\pi.$

We have: . $\pi\;=\;3.141592654...\;=\;3 + 0.141592654...$

. . The reciprocal of $0.141592654$ is: . $\frac{1}{0.141592654}\;=\;7.062573306$

And we have: . $\pi \;= \;3 + \frac{1}{7 + 0.062573306}$

. . If we disregard the decimal, we have: . $\pi \;\approx\;3 + \frac{1}{7} \;= \;\frac{22}{7}$ . . . an old friend.

We have: . $\pi \:\approx\:\frac{22}{7} \:=$ 3.14 $285714...$

The reciprocal of $0.62583306$ is: . $\frac{1}{0.062573306} \:=\:15.99659441$

And we have: . $\pi \;=\;3 + \frac{1}{7 + \frac{1}{15 + 0.99659441}}$

. . If we disrgard the decimal, we have: . $\pi \:\approx\:3 + \frac{1}{7 + \frac{1}{15}}\:=\:\frac{333}{106}$

We have: . $\pi \:\approx\:\frac{333}{106}\:=$ 3.1415 $09434...$

The reciprocal of $0.99659441$ is: . $\frac{1}{0.99659441}\:=\:1.003417228$

And we have: . $\pi \;= \;3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1+ 0.003417228}}}}$

. . If we disregard the decimal, we have: . $\pi \:\approx\:3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1}}}} \:=\:\frac{355}{113}$

We have: . $\pi \:\approx\:\frac{355}{113}\:=$ 3.141592 $92$

This is the next popular approximation, noting that we have: . $1\text{-}1\text{-}3\text{-}3\text{-}5\text{-}5$
. . and we have accuracy to the nearest millionth.

You can take this further if you like . . . I'll wait in the car.
• Jul 3rd 2006, 02:08 PM
CaptainBlack
Quote:

Originally Posted by Soroban
Like TPHacker, I prefer the continued fractions approach.

You can create your own rational approximation to $\pi.$

We have: . $\pi\;=\;3.141592654...\;=\;3 + 0.141592654...$

. . The reciprocal of $0.141592654$ is: . $\frac{1}{0.141592654}\;=\;7.062573306$

And we have: . $\pi \;= \;3 + \frac{1}{7 + 0.062573306}$

. . If we disregard the decimal, we have: . $\pi \;\approx\;3 + \frac{1}{7} \;= \;\frac{22}{7}$ . . . an old friend.

We have: . $\pi \:\approx\:\frac{22}{7} \:=$ 3.14 $285714...$

The reciprocal of $0.62583306$ is: . $\frac{1}{0.062573306} \:=\:15.99659441$

And we have: . $\pi \;=\;3 + \frac{1}{7 + \frac{1}{15 + 0.99659441}}$

. . If we disrgard the decimal, we have: . $\pi \:\approx\:3 + \frac{1}{7 + \frac{1}{15}}\:=\:\frac{333}{106}$

We have: . $\pi \:\approx\:\frac{333}{106}\:=$ 3.1415 $09434...$

The reciprocal of $0.99659441$ is: . $\frac{1}{0.99659441}\:=\:1.003417228$

And we have: . $\pi \;= \;3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1+ 0.003417228}}}}$

. . If we disregard the decimal, we have: . $\pi \:\approx\:3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1}}}} \:=\:\frac{355}{113}$

We have: . $\pi \:\approx\:\frac{355}{113}\:=$ 3.141592 $92$

This is the next popular approximation, noting that we have: . $1\text{-}1\text{-}3\text{-}3\text{-}5\text{-}5$
. . and we have accuracy to the nearest millionth.

You can take this further if you like . . . I'll wait in the car.

Now I expect you will think I'm being picky, but I find something unsatisfying
in the construction the continued fraction expansion from the decimal expansion.

The rational approximations should be (at least if it is going to satisfy my
sense of aesthetics) produced directly from the definition and/or some
closed form expressible property of $\pi$.

RonL
• Jul 3rd 2006, 02:11 PM
Quick
Quick's Approximation
While I was reading your posts, I came up with my own approximation..which is $\pi\approx3.14159265358979$

I got it by using this formula, which I created for this purpose.
$\frac{n}{d}\left(d\sin\frac{360}{2n}\right)$
this formula is for any regular polygon, d over 2 is the distance from the center of the polygon to any corner and n is the number of sides.

note: I used $n=9\cdot10^{99}$ which is the closest to a circle I could get with excel.

How accurate is my approximation?
• Jul 3rd 2006, 03:55 PM
ThePerfectHacker
Pi does have a number of interesing appearances.

$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...=\frac{\pi}{4}$

$\zeta(2)=1+\frac{1}{2^2}+\frac{1}{3^2}+...=\frac{ \pi }{6}$

$\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}$

$\Gamma(1/2)=\sqrt{\pi}$
• Jul 3rd 2006, 04:20 PM
Jameson
Aren't the last two kind of repetitive PH?

$\Gamma(z)=2\int_{0}^{\infty}e^{-t^2}t^{2z-1}dz$
• Jul 3rd 2006, 04:46 PM
Soroban
Another representation . . .

$\frac{4}{\pi} \;= \;1 + \frac{1^2}{2 + \frac{3^2}{2 + \frac{5^2}{2 + \frac{7^2}{2 + \cdots} }}}}$

I found some history on various approximations.

$3$ . . . . . value implied by I Kings vii, 23

$\frac{22}{7}$ . . . . upper bound found by Archimedes, 3rd century B.C.

$\frac{333}{106}$ . . . lower found found by Adriaan Anthoniszoon, ca. 1583

$\frac{355}{113}$ . . . found by Valentinus Otho, 1573
. . . . . . also Anthoniszoon, Metius and Viete (all 16th century)

Further approximations were found by Johann Heinrich Lambert (1728 - 1777)

$\frac{103,993}{33,102} \quad\frac{104,348}{33,215} \quad\frac{208,341}{66,317} \quad\frac{312,689}{99,532} \quad\frac{833,719}{265,381} \quad etc.
$

• Jul 3rd 2006, 07:28 PM
ThePerfectHacker
Quote:

Originally Posted by Soroban
$3$ . . . . . value implied by I Kings vii, 23

I believe it is
Kings 1:22:7

And he made a molten sea, ten cubits from the one brim to the other; it was round all about, and its height was five cubits; a line of thirty cubits did compass it around

It is possible that we disagree perhaps, you follow the King James Version, I was following the original Bible Text.

Which makes it interesing cuz, if you divide the verse by the chapter you get, 22/7.
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