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May 3

Draupner Wave

I have read your article on this topic. At the end it is noted: "Engineer Paul Taylor estimated the Draupner wave was a one in 200,000 wave."

My question is: how do people come to such a conclusion? What is the math involved in estimating the chances of X happening being 1 in Z (e.g. 1 in a million)? 99.250.117.26 (talk) 02:18, 3 May 2011 (UTC)[reply]

Following a link at Draupner wave gives the text of a paper by Paul Taylor, which reads as something written in advance of an actual presentation. It contains (to me) advanced hydrodynamics, so unless you are an expert in this field I suspect that there is no simple answer to your question. There is an email address given for Paul at the University of Oxford, so maybe you'd be better directing your question directly to him.Semiable (talk) 20:41, 3 May 2011 (UTC)[reply]

Taking loose change to Coinstar vs. rolling my own

OK, so I have a coin jar that I have been putting loose change in to for about a year. I'm ready to cashout to fund an annual camping trip, and am debating on whether to use a coin counting service (like Coinstar), or roll my own change and cash it out at the bank at no fee. So here are the factors:

I have roughly an equal count (500) of the all usual coins. Quarters ($125 total value / $10 value per roll / 40 coin count per roll), dimes ($50 total value / $5 value per roll / 50 coin count per roll), nickels ($25 total value / $2 value per roll / 40 coin count per roll), and pennies ($5 total value / $0.50 total value per roll / 50 coin count per roll)
Coinstar keeps 9.8% of the total value.
It takes me about 3 minutes to complete a single roll of any denomination.
I value my time (based on our income) at about $0.80 per minute.
Exclude the time (gas, mileage, etc) it takes to go to Coinstar or the bank, as I do both of these things anyway in the course of a regular day.

My wife suggests I should roll the quarters myself, and take everything else to Coinstar. I think I would be better off doing it all myself. Is there any combination of taking some to Coinstar/some to the bank that makes the most sense from an overall value perspective? (I hope I provided enough info. If not, let me know, and I'll provide more info. Quinn ^{✩ STARRY NIGHT} 03:03, 3 May 2011 (UTC))[reply]

Does your bank not accept deposits of coins? Surely your bank has a coin-counting machine. —Bkell (talk) 03:30, 3 May 2011 (UTC)[reply]

If I understand your question correctly: No. My bank will not accept bulk loose change. You have to roll it and write your account number on the rolls in case there is any discrepancy. Now, I could take $10 of quarters and cash it out with no problem, but not $300 of loose change and say "Will you count this for me please?" Quinn ^{✩ STARRY NIGHT} 03:58, 3 May 2011 (UTC)[reply]

From those figures, it seems it's worth to take all the coins to Coinstar. Hmm, wait, it depends on how much you value proving your wife wrong. Beware, though, because she'll suggest that you could have rolled all the quarters in the time you spend by asking this question and reading all the replies. – b_jonas 08:12, 3 May 2011 (UTC)[reply]

Right. Even for quarters, making a roll of quarters yourself will cost you $2.4, while Coinstar will take only $1. -- Meni Rosenfeld (talk) 08:27, 3 May 2011 (UTC)[reply]

Oh, a related thought experiment. As you pay for something in a shop, you accidentally drop a $100 bill to the floor. How much would you have to earn if leaning down to pick up the bill wasn't worth your time to do it? – b_jonas 12:41, 3 May 2011 (UTC)[reply]

Assuming it takes 2 seconds to pick it up...and figuring for an 8 hour work day...I figure you'd need to make a little over $500,000,000 a year for it not to be worth it. Wish I had that problem :) Quinn ^{✩ STARRY NIGHT} 15:22, 3 May 2011 (UTC)[reply]

Do you have any kids? Ask them to make the quarter rolls. – b_jonas 12:43, 3 May 2011 (UTC) [reply]

Research assistants are also good for this sort of thing... Sławomir Biały (talk) 12:46, 3 May 2011 (UTC)[reply]

Indeed, but a small child might even enjoy it. – b_jonas 12:51, 3 May 2011 (UTC) [reply]

Another perspective: Valuing your time based on your salary is problematic. Will coin rolling be done instead of working for wages? If not, then I think your scheme vastly overrates the cost of your time. See opportunity cost. If it were me, I'd roll while engaging in some hands-free activity, such as watching a movie. This way, there is very low opportunity cost, and it is much easier to beat the coinstar rate. SemanticMantis (talk) 15:46, 3 May 2011 (UTC)[reply]

I would just find a bank that does accept loose coins. That way, I would solve this problem for the times it will arise in the future as well. Googlemeister (talk) 16:06, 3 May 2011 (UTC)[reply]

Do any banks do that for such large amounts? I believe in the UK it is normal to have to sort large amounts of coins yourself before depositing them (although we use special bags, rather than rolls). --Tango (talk) 23:33, 3 May 2011 (UTC)[reply]

I bank with HSBC, and they have a really cool machine. You put your card in and it starts the machine running. You pour your change into a little tray that's fixed on the right side by a hinge. You lift the tray and the change slides through a hole that, by the noise and vibration of the machine, goes along a conveyor belt and through a sorting process. You repeat the process until all your change has gone. (It drops Euros, Cents, damaged English coins, etc, out of the bottom into a reject box.) Once you're done it gives you a print out and deposits the money into your account, commission free, within 24 hours. It's great fun, and it's commission free. It's amazing how much your change can build up to. I got £27.40 (€31.16 or $45.40) the last time I used it. I was very pleased with myself; although it didn't last very long. — Fly by Night (talk) 00:07, 4 May 2011 (UTC)[reply]

I've rolled a lot of coins by hand in my day. With a little practice you can do it a lot faster than 3 minutes per roll. Even 1 minute per roll is pretty slow. I wouldn't pay that outrageous Coinstar fee, which is so high it's offensive. 10:34, 4 May 2011 (UTC) —Preceding unsigned comment added by 69.111.194.167 (talk)

I have taken almost $2000 in coins to my bank (Wells Fargo) at one time, and they have this device that sorts and counts them, just like coinstar, without charging you for it. Took about 30 minutes. They are somewhat common in the US, but not all banks have them. Googlemeister (talk) 14:34, 4 May 2011 (UTC)[reply]

Not an answer to the immediate problem, but I try to eliminate change before it builds up. At first I did this by trying to fish exact change out of my pocket whenever paying a cashier, but now I have a second option. My grocery store has automatic lanes, where they accept however much change you care to give them. After I dump all the loose change from my pocket into it, I can pay the remainder with a credit card. StuRat (talk) 21:58, 6 May 2011 (UTC)[reply]

Are NURBS curves the most general class of curves that form a closed set under projective & perspective transforms?

I guess the set of curves full-stop is closed under such transforms, but I mean the most general subset of these. I'm interested in the geometry of $\mathbb {RP} ^{3}$ .--Leon (talk) 09:47, 3 May 2011 (UTC)[reply]

Multiplying by eight

If you add the digits together of the results of multiplying by eight until you get a single number (eg. 8, 16, 24, 32, 40, 48), the answers form the pattern 9,8,7,6,5,4,3,2,1, (the first 'run' obviously starts at 8). Why does this happen? 80.176.84.184 (talk) 11:02, 3 May 2011 (UTC)[reply]

Adding the digits together in this fashion is the same as reducing mod 9 (except using 9 in place of 0). 8 is -1 mod 9. So every time you add 8, you subtract one from the total.--Antendren (talk) 11:43, 3 May 2011 (UTC)[reply]

First you need to prove that the sum of a number's digits, and the number itself, have the same remainder when divided by nine. Let n be the number whose k^th digit from the right is n_k. You can write

n=n_{1}+10n_{2}+100n_{3}+\cdots +10^{\ell -1}n_{\ell }\,.

If you use modular arithmetic, and reduce modulo nine, i.e. work out the remainder after diving by nine you see that

(n_{1}+10n_{2}+\cdots +10^{\ell -1}n_{\ell })\!\!\!\mod 9=(n_{1}+n_{2}+\cdots +n_{\ell })\!\!\!\mod 9\,.

That proves that the sum of a number's digits, and the number itself, have the same remainder when divided by nine. The final step is to apply Antendren's argument above, i.e. adding eight or subtract one does the same things to a number's remainder after division by nine. — Fly by Night (talk) 13:41, 3 May 2011 (UTC)[reply]

On r- Chromatic Graphs

Hi. For k,l positive integers let h(k,l) be the least integer with the property that in graph on h(k,l) vertices either there is a closed circuit of k or fewer lines or that the graph contains l independent points. It is given that for sufficiently large l, h(k,l)>l^1+1/2k. Now how do we conclude from here that for all r, there is an r-chromatic graph with no k-gon in it. Clearly a graph on floor(r^1+1/2k) vertices will not have a k-gon in it, but how can we construct it in such a way that it is also r-chromatic?-Shahab (talk) 11:45, 3 May 2011 (UTC)[reply]

In effect I am trying to conclude that for every two integers m, n ≥ 2, there exists an n-chromatic graph of girth m.-Shahab (talk) 05:01, 4 May 2011 (UTC)[reply]

number theory

Suppose I have two numbers xy^3 and (xy)^2, whose largest apparent common factor is xy^2. Suppose that I am also given that their greatest common factor is n. Does it follow that xy^2 is a factor of n, and why? This was recently tested on an ACT practice test, but I didn't get the problem. Thanks. 72.128.95.0 (talk) 22:51, 3 May 2011 (UTC)[reply]

Maybe a look at a more general problem will help: is it true that every common divisor of two natural numbers is a divisor of their Greatest common divisor? --Martynas Patasius (talk) 00:05, 4 May 2011 (UTC)[reply]

First of all, and it's only a small point, we often use letters like k, m and n for integers, while x and y are usually used for real numbers. Of course it doesn't matter; but it looks more normal. What you want to use here is prime power decomposition. Every positive whole number has a prime power decomposition; i.e. we can write it as the product of prime numbers to certain powers, e.g. 1176 = 2³·3¹·7². In general:

m=p_{1}^{\alpha _{1}}\ldots p_{k}^{\alpha _{k}}\,,

where each p_i is a prime number and each α_j is a positive integer. If you want to compare two whole numbers you decompose them as a product of primes, but here's the key: as a product of the same primes while some of the α_j's equal to zero. For example

1176=2^{3}\cdot 3^{1}\cdot 5^{0}\cdot 7^{2}\,,

1350=2^{1}\cdot 3^{3}\cdot 5^{2}\cdot 7^{0}\,.

Doing this allows you to calculate the greatest common divisor and the least common multiple. For example 1176 is divisible by 2³ while 1350 is divisible by 2¹; it follows that both 1176 and 1350 are divisible by 2¹. Next, 1176 is divisible by 3¹ while 1350 is divisible by 3³; it follows that both 1176 and 1350 are divisible by 3¹. Next, 1176 is divisible by 5⁰ while 1350 is divisible by 5²; it follows that both 1176 and 1350 are divisible by 5⁰. Finally, 1176 is divisible by 7² while 1350 is divisible by 7⁰; it follows that both 1176 and 1350 are divisible by 7⁰. That means that

\gcd(1176,1350)=2^{1}\cdot 3^{1}\cdot 5^{0}\cdot 7^{0}=6\,.

The general method would be the same. So we break our two numbers up as follows:

m=p_{1}^{\alpha _{1}}\ldots p_{k}^{\alpha _{k}}\,,

n=p_{1}^{\beta _{1}}\ldots p_{k}^{\beta _{k}}\,\,

where the p's are prime numbers and the α's and β's are whole numbers greater than equal to zero. It follows that

\gcd(m,n)=p_{1}^{\min(\alpha _{1},\beta _{1})}\ldots p_{k}^{\min(\alpha _{k},\beta _{k})}\,,

I'm sure that you'll be able to answer your question using this method. Good luck. — Fly by Night (talk) 02:31, 4 May 2011 (UTC)[reply]