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May 18

How can Pi exist.

I was thinking a while back about this. If Pi is 3.14 and so on but NEVER ends then it doesnt make sense. Look at it like this(yes im resorting to childish terms) remember how in school you learned fractions, with a pie. 1 was a whole pi, 2 half and so on. Take Pi over 1. that would still be pi but it how would that look? 3 whole pies, not the complcated part, then u have the .14159................... the pie would never stop going around, filling up the tray. It would be very very small increments but it would still be moving. Therefore it would eventually fill up,seeings that Pi is infinitum. This has been meesing with my mind for a while now. Everyone i explain this to understands where im coming from, but cant give an answer

76.111.150.73 (talk) 13:20, 18 May 2008 (UTC)Rob[reply]

Well, I'm afraid your perfect record has been broken, because now there's someone you've explained this to who doesn't understand where you're coming from, namely me.

Are you asking how it is that the decimal representation 3.141592654... can go on forever and never quite equal exactly π, or are you making a little joke about π and pie? —Steve Summit (talk) 13:44, 18 May 2008 (UTC)[reply]

You know that the decimal expansion of 1/3 is 0.333..., and that of 1/7 is 0.14285714..., right? Do these give you the same problems in comprehension? Oliphaunt (talk) 13:48, 18 May 2008 (UTC)[reply]

I think I see what you're saying. Pi is the sum of an infinite number of numbers (3+1/10+4/100+1/1000+...), and you're wondering how come that doesn't add up to infinity? It's a good question. Put simply, the numbers are getting smaller and smaller fast enough that you never get past a certain amount (pi). Consider this sum: 1+1/2+1/4+1/8+... That, it turns out, adds up to 2. You can see how it works like this: Consider a box 2 metres long. You put something 1 metre long in the box, that leaves 1 metre left. You then put something 1/2 a metre long in the box, that leaves 1/2 a metre left. Then you add 1/4, and that leaves 1/4 left, and so on. Because the amounts you're adding are getting smaller and smaller, there is always room to add the next one. That amount left gets smaller and smaller, and after an infinite number of steps it's 0, so the total length of all the things is the length of the box: 2. Pi is a little trickier, because it doesn't follow a nice pattern (it's irrational), but the concept is the same. Does that make sense? --Tango (talk) 13:58, 18 May 2008 (UTC)[reply]

Ohh! One could make a very similar example for pi using the Method of exhaustion. Unfortunately the article doesn’t have the standard illustrations, but using the particular summation as the area added in each n-gon you get the same illustration. GromXXVII (talk) 16:44, 18 May 2008 (UTC)[reply]

Take a look at the articles on 0.999... and Zeno's paradoxes too, because I think they are related to your question. • Anakin ^(talk) 15:13, 18 May 2008 (UTC)[reply]

No I wasnt making a joke on the words, i understand that even after many many decimal places its still not even to be considered, but still it gos on FOREVER so there fore why wouldnt all the decimals eventually add up to a whole? —Preceding unsigned comment added by 76.111.150.73 (talk) 23:48, 18 May 2008 (UTC)[reply]

76.111.150.73 (talk) 23:49, 18 May 2008 (UTC)Rob[reply]

Why do you expect the decimals to add up to 1 as opposed to, say, 1/2? If an infinite number of decimals can add up to a finite number (which they can, as I demonstrated above), then can add up to any finite number (less than or equal to 1). In this case, they add up to the fractional part of pi. --Tango (talk) 23:56, 18 May 2008 (UTC)[reply]

I dont expect it to add to anything, If it were infinite then why would it not continiously go and go and go. like after years and years wouldnt it be 4 or even as you said 3.5 or anything like it??

Im not trying to argue im trying to understand 76.111.150.73 (talk) 00:07, 19 May 2008 (UTC)Rob[reply]

Sorry, I misunderstood you there - I thought you were expecting it to stop at a whole, you mean eventually reach a whole and keep going? Did you understand my explanation above about putting things in a box? --Tango (talk) 00:11, 19 May 2008 (UTC)[reply]

Actually the box thing is just about exactly what i mean. Only problem is a box has a closing point. Pi and numbers do not so wouldnt it overflow? Is this even a rational question or am i just making myself look stupid. 76.111.150.73 (talk) 00:14, 19 May 2008 (UTC)Rob[reply]

No, it's not a dumb question. I, myself, have also pondered this same situation. Just think of it like this: There are some infinte series (

\lim _{n\to \infty }\sum _{k=0}^{n}{\frac {1}{2^{k}}}=2

as Tango said) that converge to a real number from an infinite series of terms. On the other hand, you have series that diverge. Such as

\lim _{n\to \infty }\sum _{k=1}^{n}k=\infty

. Pi is kind of like a convergent series in that there is an infinite quantity of numbers that does not amount to infinity. Does that help at all? Zrs 12 (talk) 01:27, 19 May 2008 (UTC)[reply]

A slight misunderstanding. Tango referred to the series

\sum _{k=0}^{\infty }{\frac {1}{2^{k}}}=2.

The series

\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}

is the Basel problem. It also converges, but not towards 2, but rather towards

{\frac {\pi ^{2}}{6}}\approx 1.64493.

Bo Jacoby (talk) 10:16, 19 May 2008 (UTC).[reply]

Oops, sorry about that. Typo. I fixed it now. -- Zrs 12 (talk) 19:37, 19 May 2008 (UTC)[reply]

Sorry I misunderstood your question at first. You asked, "after years and years wouldn't it be 4 or even as you said 3.5?" Well, no.

My explanation wasn't very good... The box has a closing point simply because I already knew the answer so picked a box the right size (I was just demonstrating that my answer was right, rather than calculating the answer). There was no end to the items I was adding to the box, but they still fit in the box because they got smaller. If I had a box pi metres long and added items 3 metres long, then a tenth of a metre long, then 4/100 m long, the 1/1000 m long, and so on, there would still always be room for the next item and it would never overflow (I used the example of powers of two because it's easier to see, but it works the same way). --Tango (talk) 14:45, 19 May 2008 (UTC)[reply]

We start with 3. *If* we added a bunch of 9's we could get to 4, but we don't: the first thing we add is a 1, or 1/10. So now we've got 3.1.

If we added a bunch more 9's we could get to 3.2, but we don't: the next thing we add is a 4, or 4/100. So now we've got 3.14.

If we added a bunch more 9's we could get to 3.15, but we don't: the next thing we add is a 1, or 1/1000. So now we've got 3.141.

If we added a bunch more 9's we could get to 3.142, but we don't: the next thing we add is a 5, or 5/10000. So now we've got 3.1415.

If we added a bunch more 9's we could get to 3.1416. We do add one 9, but after that we add a 2, so we're stuck with 3.141592. And so on.

Not only will we never get to 3.5 or 4, it looks like we'll never get to 3.2, or even 3.1416.

Remember that "I'm thinking of a number between 1 and 100" game? You guess 50, and I say it's higher. You guess 75, and I say it's lower. So now you know it's between 50 and 75. Each guess you make narrows it down -- and your guesses will never get outside of the interval 50..75.

Suppose after a few more guesses you get it narrowed down to the interval 61..63. You triumphantly guess "62", but I say "too high". I cheated: the number I'm thinking of isn't an integer, and evidently it's some fraction between 61 and 62.

So you guess "61.5" and I say "too high", and you guess "61.3" and I say "too low", and you guess "61.4" and I say "too high", and you guess "61.35" and I say "too low", and you guess "61.37" and I say "too high", and you guess "61.36" and I say "too low", and you guess "61.365" and I say "too low", and so on. We can go on like that forever, but unless I cheat again, the number will never get outside the range 61..62, and in fact it will never get very far from 61.365. —Steve Summit (talk) 04:36, 19 May 2008 (UTC)[reply]

Your question is a valid one, but it has a good answer. Go look at Zeno's paradox (the "dichotomy paradox" section), which is essentially equivalent to the question you've raised. kfgauss (talk) 05:38, 19 May 2008 (UTC)[reply]

It might be worthwhile looking at the ratio test, for example, for some sense of general rules when adding little pieces does or does not result in infinity. --Fangz (talk) 10:22, 19 May 2008 (UTC)[reply]

Ok I guess i understand now, the increase is only fractions of the last fraction. I think i get it. Thanks. BringBackZero (talk) 17:28, 19 May 2008 (UTC)Rob[reply]

By the way if this question seems irrational to some of you, its coming from a 15 year old, so dont flip out BringBackZero (talk) 17:34, 19 May 2008 (UTC)Rob[reply]

Not at all - it's a difficult thing to get your head round. I didn't learn the underlying maths properly until the first year of University. --Tango (talk) 19:02, 19 May 2008 (UTC)[reply]

Ha! If it seems "irrational." LOL. Irrational numbers are strange, but e^{i π} = -1, which may be even stranger (though I'm not sure we should jump the gun on what it may or may not prove). --Prestidigitator (talk) 22:04, 19 May 2008 (UTC)[reply]

Parity of powers

In parity (mathematics) it states that a fractional number, by definition is not odd or even. But I was considering the fact that square numbers have the same parity as their square root. I.e., square an odd number, you get an odd number, and square an even number, you get an even number. The same seems to apply to other powers. E.g.,

powers of 2: 2, 4, 8, 16, 32, 64... are all even, and

powers of 3: 3, 9, 27, 81, 243, 729... are all odd.

The only case this definitely doesn't work for is $x^{0}$ , because you still get 1 when x is even. But I was wondering, why shouldn't this work for fractional or negative powers. E.g., √2 is irrational, but if the pattern holds it should be even. Similarly for $2^{-1}=0.5$ . I haven't been able to find any conflicts where a power of an even number is equal to a power of an odd number, so is there any other reason why this might not work? • Anakin ^(talk) 13:59, 18 May 2008 (UTC)[reply]

If

x=\log _{2}3

, then

2^{x}=3

, and you have a power of 2 which is clearly odd. I don't see how "a is even if there is an odd integer b and rational numbers c and d where

c\neq 0

such that

x=2^{c}b^{d}

" is a plausible extension of "a is even if there is an integer k such that

a=2k

". -- Meni Rosenfeld (talk) 14:24, 18 May 2008 (UTC)[reply]

(ec). The rule you desecribed above works if the powers are integers and greater than zero. For example, it is possible to solve for x if 2^0.5 = 3^x but 2 is even and 3 is odd. Also, even and odd have quite specific definitions, you should have a look at them. Zain Ebrahim (talk) 14:27, 18 May 2008 (UTC)[reply]

Thank you for these answers. The fact that

x=\log _{2}3

gives a power of 2 which is odd does clearly invalidate my rule. That kind of means that x can't be even or odd (and it is irrational), which makes it less of a surprise that √2 can't be even or odd. The fact that an even number must be divisible by 2 is also a blow, because √2 (1.41421...) clearly isn't. This gives me a few things to think about though, thanks. • Anakin ^(talk) 15:07, 18 May 2008 (UTC)[reply]

There are similar concepts that sometimes come up. For instance in

\mathbb {Z} \left[{\frac {1+{\sqrt {-1}}}{2}}\right]=\left\{a+b{\frac {1+{\sqrt {-1}}}{2}}|a,b\in \mathbb {Z} \right\}

there is another form of parity: in whether or not the coefficients are integral, or have a denominator of 2. GromXXVII (talk) 16:52, 18 May 2008 (UTC)[reply]

Coin toss

If you have 7 coins and you toss them together, what is the probability of getting 3 heads? What is the probability of getting all heads?

What is the method of solving this? Can you do this: 7C3/2^7 for the first question? —Preceding unsigned comment added by 68.99.185.240 (talk) 17:32, 18 May 2008 (UTC)[reply]

Hello. Yes, you are correct that that would work. For the second, just replace 7C3 with 7C7 = 1. -mattbuck (Talk) 17:52, 18 May 2008 (UTC)[reply]

Note that, assuming ideal coins so Pr(heads) = 1/2 for one toss,

{\tbinom {7}{3}}/2^{7}

is the probability of getting exactly three heads, not at least three heads. --Prestidigitator (talk) 17:12, 19 May 2008 (UTC)[reply]

probability

If you draw a 5-card hand, what is the probability of getting exactly three 7's?

I was wondering if you could simply find the number of combinations for which you have three 7's, and divide it by the total number of 5-card combinations? So... (4C3*48C2)/(52C5)??? Does this mean that we are considering each 5-card hand to have equal probability of coming up? Is the problem more complex than that? —Preceding unsigned comment added by 68.99.185.240 (talk) 19:23, 18 May 2008 (UTC)[reply]

Yes, that should work: indeed the probability of each hand is the same. GromXXVII (talk) 20:30, 18 May 2008 (UTC)[reply]

Test

My friend has a vocabulary test. His study sheet has 75 words that might be on the test. 15 of the words will be randomly selected to actually be on the test. And among the 15, he is required to only define 6. I know that in order to have a 100% chance of getting all six, you only need to know 66 words. However, it seems there is a 99.99999% chance of getting all 6 with only 65 and so on. How do you calculate the probability as a function of target (number of words you want to get right) and work (number of words you need to study)? --MagneticFlux (talk) 20:08, 18 May 2008 (UTC)[reply]

Looks like a hypergeometric distribution to me. 75 words with the number you memorize “defective”, and a sample of 15. GromXXVII (talk) 20:34, 18 May 2008 (UTC)[reply]

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