Happy Pi Day (1 Viewer)

gurmies · Mar 14, 2012

Pi is exactly 3.

Carrotsticks · Mar 14, 2012

Oh dear...

gurmies · Mar 14, 2012

I think I just created a wormhole...somewhere.

mirakon · Mar 14, 2012

Carrotsticks said:
Pi shows its face in even the most unexpected of problems seemingly totally related to pi.

For example, suppose I have a stick of some unit length.

In front of me, I have horizontal intervals (imagine floorboards) of the same unit length.

When I throw the stick, it can either cross one of the intervals or lie in between two intervals.

The probability of it CROSSING a line is.... you guessed it!

$\frac{2}{\pi}$

So just by throwing sticks, I can approximate pi!

This is more commonly known as Buffon's Needle Problem.

If by guess you mean "tested to the limits of my mathematical power till I gave up and just looked at what you typed" then yes lol

AAEldar · Mar 14, 2012

Carrotsticks said:
Enough to entertain myself for a good few minutes when I'm bored =)

Seems to be just a few more than me... I can only entertain myself for 10 seconds or so hahaha.

Heard many interesting facts relating to pi today though!

Carrotsticks · Mar 14, 2012

gurmies said:
Pi is exactly 3.

AAEldar · Mar 14, 2012

http://www.math.utah.edu/~palais/pi.pdf

Read this ages ago now, pi day reminded me of it!

Nooblet94 · Mar 14, 2012

IamBread said:
3.14159265?

That's all I can do as well. One of my friends could recite 50 digits at one point.

Also, mum made me pie for dinner to celebrate :')

Carrotsticks · Mar 14, 2012

I can also approximate pi if I have a whole lot of coins. Consider the following:

I have two 25c coins (also known as 'quarters' in the US). I flip the two coins simultaneously, then give them both to you.

I continue this process until I either have no coins left, or the number of heads is equal to the number of tails (so as you can imagine, there would be several times when I stop after 1 toss). The moment the number of heads is equal to the number of tails, I record how much money you 'earned' up to that point.

I then take that money and put it back in the bag and repeat this process many times.

Suppose I do this 1000 times. You can actually *very accurately* approximate how many coins I have in the bag without me having to tell you anything.

Let your average 'earnings' be denoted by E.

The number of coins I have in the bag is approximately.....

$\pi \times e^2$

How amazing is that!

mirakon · Mar 14, 2012

Carrotsticks said:
I can also approximate pi if I have a whole lot of coins. Consider the following:

I have two 25c coins (also known as 'quarters' in the US). I flip the two coins simultaneously, then give them both to you.

I continue this process until I either have no coins left, or the number of heads is equal to the number of tails (so as you can imagine, there would be several times when I stop after 1 toss). The moment the number of heads is equal to the number of tails, I record how much money you 'earned' up to that point.

I then take that money and put it back in the bag and repeat this process many times.

Suppose I do this 1000 times. You can actually *very accurately* approximate how many coins I have in the bag without me having to tell you anything.

Let your average 'earnings' be denoted by E.

The number of coins I have in the bag is approximately.....

$\pi \times e^2$

How amazing is that!

Can u link me to the theory behind it please? I like maths even though im not particularly adept at it

Carrotsticks · Mar 14, 2012

More about coins and pi!

Perhaps you are aware of the 'Wallis Product' expression for pi, which is as follows:

$\prod_{n=1}^{\infty} \frac{(2n)^2}{(2n-1)(2n+1)} = \frac{\pi}{2}$

Did you know that it is heavily related to the simple act of flipping coins and observing the most likely outcome? This was actually proved in the 1995 HSC Question 7 (a) and (b).

Carrotsticks · Mar 14, 2012

mirakon said:
Can u link me to the theory behind it please? I like maths even though im not particularly adept at it

Here is a link outlining the same problem. It also provides a proof, which utilises the Wallis Product!

http://edp.org/coinflip.htm

It even has a random-number generator to model the random outcomes of flipping two coins.

You will see that as the number of iterations approaches infinity, the more accurate the approximation becomes.

mirakon · Mar 14, 2012

Thanks!

Carrotsticks · Mar 14, 2012

So far I have been talking about pi popping up in real life, but I left out poor Statistics, arguably THE most important branch of modern Mathematics in society.

There is a famous integral used often in statistics called the Gaussian Integral.

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

After normalisation, is the distributive function for the normal distribution, which is the most prominent distribution in Statistics. Many people also know the normal distribution as the famous 'bell curve' of probability.

Carrotsticks · Mar 14, 2012

There is so much I can say about pi, but the identity which most of us know (and in my opinion, the most beautiful of them all) is Euler's Identity (Note: Euler is pronounced 'Oiler' not 'Youw-ler'. Yes Math Man, I'm talking to you.)

$e^{i \pi} + 1 =0$

mirakon · Mar 14, 2012

Whats the difference between having limits of infinity and just having an indefinite integral?

Carrotsticks said:
So far I have been talking about pi popping up in real life, but I left out poor Statistics, arguably THE most important branch of modern Mathematics in society.

There is a famous integral used often in statistics called the Gaussian Integral.

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

After normalisation, is the distributive function for the normal distribution, which is the most prominent distribution in Statistics. Many people also know the normal distribution as the famous 'bell curve' of probability.

Carrotsticks · Mar 14, 2012

mirakon said:
Whats the difference between having limits of infinity and just having an indefinite integral?

As the term suggests, 'indefinite' integrals do not specify a bound for the area under the curve.

Note that not specifying the bound isn't the same thing as having upper and lower bounds of positive infinity and negative infinity respectively.

mirakon · Mar 14, 2012

Carrotsticks said:
As the term suggests, 'indefinite' integrals do not specify a bound for the area under the curve.

Note that not specifying the bound isn't the same thing as having upper and lower bounds of positive infinity and negative infinity respectively.

Oh ok, but I dont get why they aren't the same thing because isn't infinity conceptually an indefinite value anyway?

mnmaa · Mar 14, 2012

SpiralFlex said:
SPAM spam spam! Still partying on pi day. At 4 AM!

4:00 Am, seriously?

Carrotsticks · Mar 14, 2012

mirakon said:
Oh ok, but I dont get why they aren't the same thing because isn't infinity conceptually an indefinite value anyway?

Oh I understand your confusion.

They are not called indefinite integrals in the sense that the limits themselves are indefinite (taking the definition of 'indefinite' to be 'going on forever')

There is a lot of very *deep* mathematics that goes into the relationship between indefinite and the definite integral, but the essential thing is that indefinite does not imply infinite in this case.

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