March 14th (3.14)

jinydu · **Joined:** Sun, 7 Nov 2004 08:21:37 UTC **Posts:** 2987

kippah wrote:

bugzpodder wrote:

its the circumference of a unit circle, thats how it is defined.

pi is half the circumference of the unit circle .. ?

Yeah.

cos 0 = 1
cos pi = -1 not = 1

heardie · **Joined:** Thu, 31 Jul 2003 12:36:00 UTC **Posts:** 1737

unit circle has radious 1...if you consider the 'unit' circle to be diamater 1 then ...

jinydu · **Joined:** Sun, 7 Nov 2004 08:21:37 UTC **Posts:** 2987

heardie wrote:

unit circle has radious 1...if you consider the 'unit' circle to be diamater 1 then ...

This goes back to the original question then. Why do we use the diameter instead of the radius to define pi?

Surely, there's got to be a reason, right? Does it make a certain, important, class of equations easier? Because, the three cube roots of unity are:

1
e^(i*2pi/3)
e^(i*4pi/3)

Aesthetically, there's an extra factor of two hanging around...

kippah · **Posted:** Tue, 15 Mar 2005 08:41:07 UTC

Factors of two are generally easier to work with than factors of 1/2, take one of my favourite results as an example:

$$\int_{-\infty}^\infty e^{-t^2}\text{d}t=\sqrt{\pi}$

Take out a 1/2 and you get $$\sqrt{\frac{\pi}{2}}$ which just looks ugly in my opinion.

It's a moot point really, I just don't think there's anything wrong with the current definition. Leave me be with my 2pi circumference

jinydu · **Joined:** Sun, 7 Nov 2004 08:21:37 UTC **Posts:** 2987

kippah wrote:

Factors of two are generally easier to work with than factors of 1/2, take one of my favourite results as an example:

$$\int_{-\infty}^\infty e^{\frac{-t^2}{2}}\text{d}t=\sqrt{\pi}$

Actually, Mathematica 4 says that

$$\int_{-\infty}^\infty e^{\frac{-t^2}{2}}\text{d}t=\sqrt{2\pi}$
$$\int_{-\infty}^\infty e^{{-t^2}}\text{d}t=\sqrt{\pi}$

But I guess you're right in that an extra factor of 1/2 is less pleasing than an extra factor of 2.

Arguably though, since 2 and 1/2 are such simple numbers, it doesn't matter too much. I was just curious...

kippah · **Posted:** Tue, 15 Mar 2005 08:50:26 UTC

True, I'll edit my post. Fuzzy memory :roll:

Also $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}\text{d}x\text{d}y = \pi$

CHAVS · **Posted:** Tue, 15 Mar 2005 19:01:08 UTC

kippah wrote:

True, I'll edit my post. Fuzzy memory :roll:

Also $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}\text{d}x\text{d}y = \pi$

Also,

$$\int\limits_{-\infty}^{\infty}\cdots\int\limits_{-\infty}^{\infty}e^{-x_1^2-x_2^2-\cdots-x_n^2}{dx_1dx_2\cdots{dx_n}}=\pi^{\frac{n}{2}}$

Prove by induction . . .

robert@fm · **Posted:** Fri, 22 Apr 2005 13:51:21 UTC

robert@fm wrote:

Such a pity that it's only Pi Day in North America.

(Elsewhere in the world, where they use European or International date format, it's 14/3, not 3/14.)

I've just remembered that in countries which use European date format, Pi Day is the 22nd July (22/7).

Quote:

However, one can celebrate Pi Day by downloading the ZX-Spectrum* classic Pimania from World of Spectrum (along with an emulator if needed). (Automata Software did a number of other Pi games, all of which are also there.)

*TS2068 in the USA.

The "date" portion of the solution to Pimania was 22 July, for the reason stated above.

Shadow · **Posted:** Fri, 22 Apr 2005 21:41:50 UTC

Kwyjibo wrote:

Hey Matthew!

Did you know that in 1899 the Indiana Legislature almost passed a bill
requiring that all Indiana schools teach pi to be exactly 4???

Wonder what the world would be like today if it had passed....

why 4!? It's closer to 3. :roll:

Some people in government are just total morons.

eblaska · **Joined:** Fri, 22 Jul 2005 05:23:58 UTC **Posts:** 257

bump...

it's coming.

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9642

It's Pi Day's Eve. Fun facts:

(1) Albert Einstein's birthday falls on Pi Day
(2) Pi Day is the same day as Steak and BJ Day, the males' answer to Valentine's Day
(3) Those of you not in America may instead celebrate the alternative: Pi Approximation Day

Benny · **Posted:** Tue, 14 Mar 2006 14:07:16 UTC

$e^{ - \left( {x_1 ^2 + ... + x_n ^2 } \right)} = \prod\limits_{i = 1}^n {e^{ - x_i ^2 } }$

So can't you just 'factor' the n-fold integral and calculate it as $I = \left( {\int\limits_{ - \infty }^\infty {e^{ - x^2 } dx} } \right)^n$ or is that a fudge method?

eblaska · **Joined:** Fri, 22 Jul 2005 05:23:58 UTC **Posts:** 257

HAPPY PI DAY!

bugzpodder · **Joined:** Sun, 4 May 2003 16:04:19 UTC **Posts:** 2906

Benny wrote:

$e^{ - \left( {x_1 ^2 + ... + x_n ^2 } \right)} = \prod\limits_{i = 1}^n {e^{ - x_i ^2 } }$

So can't you just 'factor' the n-fold integral and calculate it as $I = \left( {\int\limits_{ - \infty }^\infty {e^{ - x^2 } dx} } \right)^n$ or is that a fudge method?

A similar identity:
http://www.sosmath.com/CBB/viewtopic.ph ... ar+algebra

skeeter · **Posted:** Tue, 14 Mar 2006 23:49:32 UTC

Matthew wrote:

It's Pi Day's Eve. Fun facts:

(2) Pi Day is the same day as Steak and BJ Day, the males' answer to Valentine's Day

you should have seen the look on mi esposa's face when I mentioned Steaks and a BJ ... :lol:

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March 14th (3.14)

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