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October 9

Copper price

BENCHMARK copper on the London Metal Exchange closed at $US5,135 a tonne. Please help me, I want to find out how much a kilogram of copper costs 175.45.116.59 (talk) 04:54, 9 October 2015 (UTC)

The London Metal Exchange uses the metric tonne, which makes this nice and simple. A tonne is 1,000 kilograms, so to work out the price per kilogram, just divide the price per tonne by 1,000 - in other words, replace the comma with a decimal point. 1 kg of copper costs $5.135. Smurrayinchester 08:10, 9 October 2015 (UTC)

Just to clarify, "metric tonne" is a redundancy. There is only one unit spelled tonne, which as Smurray says is a metric one equal to 1,000 kilograms. There are several units called ton and, since "tonne" and "ton" are commonly pronounced the same, "metric ton" is sometimes used as another name for "tonne" in a context where people may speak also of other kinds of tons. --174.88.134.156 (talk) 02:50, 10 October 2015 (UTC)

Laplace transform

What is the most complete laplace transform table on web?--95.250.182.104 (talk) 14:43, 9 October 2015 (UTC)

I'm not sure about "on the web", but Erdelyi's "Tables of integral transforms" contains about 100 pages of tables of Laplace transforms. Sławomir
Biały 14:59, 9 October 2015 (UTC)

Together, these [1] [2] [3] [4] are fairly comprehensive. While not a table per se, Wolfram alpha can give many Laplace transforms. SemanticMantis (talk) 15:47, 9 October 2015 (UTC)

Those are missing Bessel functions, Gamma functions, elliptic functions, parabolic cylinder functions, hyperbolic functions, hypergeometric functions, and Legendre functions. Erdelyi also includes Laplace transforms of orthogonal polynomials. Sławomir
Biały 14:41, 10 October 2015 (UTC)

Where should I have gotten gas?

I figured out eight years ago when I bought my current car that it got 22 miles to the gallon. I haven't determined whether this is still the case, but let's assume it is. Let us also assume consistent gas mileage regardless of whether I am in the city or the country, or whether I am running my air condtioning or not.

I am approximately 176 miles from the beach. After 70 miles, with my usual route, I cross the border from North Carolina into South Carolina. Regular unleaded gas which may contain ethanol (referred to as "gas" from now on) in the last town before the border was around $2.09.9 (".9" is assumed from now on). At the border, gas was $1.94. South Carolina gas taxes are lower. In the first town, about 15 miles from the border, one place is $1.89, though it is cash only (I try to avoid using cash but it wouldn't have been impossible) and there is a higher price if I use my credit card. One place is $1.87 but there's no place to wash my hands after I finish. Another place is $1.89 but I wasn't close enough to see if that was a credit price. 20 miles farther down the road is the place where I ate lunch, and gas was $1.89. But I figured I could do better. About 20 miles from the beach, one place was $1.79 but I recalled that it was cash only. Another place was $1.79 but it was on the wrong side of the road and the clerk gave me attitude about filling up, explaining how easy it was to "pay at the pump". Well, maybe for her. The one place that had the lowest price some years ago was $1.83, about 15 miles from the beach. I decided it was probably going to go up from there and I filled up.

I was so wrong. In fact, the price kept going down and several places at the beach were $1.79. This has never happened. Still, I was gambling because I had never seen prices continue to go down. One place was $1.78 as I was going home, but I couldn't benefit. The two places that were $1.79 were the same, even though some places were a penny higher and some were a penny lower. I still couldn't benefit. But I returned to that last town and had a decision to make. Now I could have gone to that one other place that was $1.89. By this time I was going to benefit at least a little. Instead, I stopped at the place where I would pay cash and just got milk to eat with my lunch bought 20 miles away. I just decided to wait because gas would be that low soon.

Gas went back up, then down. This week with the "low fuel" light on, I passed a place that was $1.98, though most places were much higher. I got $20 worth.

So are there any mathematical formulas that might tell me what the best move would have been?— Vchimpanzee • talk • contributions • 18:28, 9 October 2015 (UTC)

This sounds like an optimal stopping problem. It's not the secretary problem, since the prices aren't randomly ordered, but that article may be worth a read anyway. -- BenRG (talk) 21:49, 9 October 2015 (UTC)

I've given this problem lots of thought previously. The main issue is that you always have incomplete information. Even if you knew what every gasoline price was in driving range yesterday, they are subject to change on a daily basis, and they don't always change proportionately either. Another consideration is that if gas prices are likely to go down over the next few days, you don't want to fill up until it does, whereas if prices are going up, you want to fill up right now.

Also note that the fuel you burn to get to a further gas station with lower prices isn't the only cost. There's also wear and tear on the car/depreciation, the value of your time, the chance you'll get a ticket, etc. I've pretty much concluded that I should try to gas up when on the way to somewhere else, and I've learned there's a small set of gas stations which typically have the lowest prices, so I try to gas up at one of those when I am near.

I've also thought it would be great to have an app that could tell you the best gas station to use, but there the problem is getting every gas station, even those with high prices, to send it an update of their prices every day. About the only way I can see to get past that would be a camera aimed at every gas station sign, that does character recognition to determine the current prices. StuRat (talk) 01:11, 10 October 2015 (UTC)

• By the way, you're allowed to write "$2.099". It's a decimal fraction like any other. —Tamfang (talk) 05:55, 10 October 2015 (UTC)

Thanks, optimal stopping was what I was looking for. Anyway, since there's all that extra information and only a finite number of stations, plus the fact that they COULD have changed stations, there's probably not much point.— Vchimpanzee • talk • contributions • 15:46, 10 October 2015 (UTC)

October 11

In a city, how do you see more people, walking around or standing?

In a city, how do you see more people, walking around or standing on a concrete place? Suppose the same number of people come and go from any side. And all move at the same speed as you.--Jubilujj 2015 (talk) 19:01, 11 October 2015 (UTC)

I would say more walking. Under the conditions you give, assume the city is also uniformly dense of people and infinite, or at least large enough to make no difference. Then you will have the same density of people at any time. But the rate at which you meet them will depend on their speed. The faster they are going, relative to you, on average, the more you will meet.

If you are stationary they all have the same speed relative to you. If you are walking then some appear faster some slower but it is not symmetric. Those travelling in the same direction as you seem slower, even stationary. Those travelling the opposite direction seem faster, up to twice as fast. But those moving at right angles seem faster, about √2 as fast. More seem faster than slower so you meet more of them, in theory.--JohnBlackburne^words_deeds 19:59, 11 October 2015 (UTC)

In general you should have people seen when walking / people seen when stationary = ${\frac {\int _{||v||=1}||v-{\hat {x}}||\,dv}{\int _{||v||=1}\,dv}}$ = ${\frac {\int _{0}^{\pi }(\sin \theta )^{d-2}{\sqrt {2+2\cos \theta }}\,d\theta }{\int _{0}^{\pi }(\sin \theta )^{d-2}\,d\theta }}$. In 1, 2, 3, 4 dimensions I get 1, 4/π, 4/3, 64/15π ≈ 1, 1.27, 1.33, 1.36. It looks like it converges to √2 in the d → ∞ limit. -- BenRG (talk) 22:12, 13 October 2015 (UTC)

Your 2-dimensional result only applies to the squares in the city, with the added assumption that people walk randomly over any point and with any direction. Most of the walking in a city is done along 1-dimensional routes, such as sidewalks. Therefore, the actual ratio should be close to 1. In other words, it's almost a wash. — Sebastian 23:48, 13 October 2015 (UTC)

It looks like you've assumed that all the people are converging to a point, assuming I understand what the "v" means. That doesn't seem like a very realistic matter model governing "people". Presumably "people" are not infinitely compressible. I think a better matter model would be that people are totally incompressible. In which case, a local solution is fairly obvious: just go against the flow of traffic. Or have I missed something? Sławomir
Biały 00:57, 14 October 2015 (UTC)

v is the velocity of other people relative to the ground. Yours is either ${\hat {x}}$ or 0. I assumed that the people are noninteracting (pass through each other and you, or are so sparse that they may as well be treated that way), their velocities are uniformly distributed over the (d−1)-sphere, and people with a given velocity are more or less uniformly distributed through space, so the rate at which you see new people with that velocity is proportional to their speed relative to you. Obviously this is unrealistic; I thought the problem was intended that way. I think JohnBlackburne made the same assumptions. -- BenRG (talk) 03:06, 14 October 2015 (UTC)

Oh, and we're of course interpreting the condition "And all move at the same speed as you" as not applying to the v=0 case, or else you would not meet anybody new. — Sebastian 23:53, 13 October 2015 (UTC)

October 12

Worst possible champion record in a multiple round-robin tournament

(Trying to figure out how good a chance some teams I'm following can win their tournaments...) Consider a multiple round-robin tournament with n players/teams and r rounds, so that each player/team plays each other player/team exactly r times, for a total of r(n - 1) matches per player/team. Assuming no tied games/matches are possible, at the conclusion of the tournament, what is the least possible number of wins sufficient to be an untied champion? —SeekingAnswers (reply) 21:25, 12 October 2015 (UTC)

Not 100% clear on what you mean. For a round robin tournament with 5 teams there are 10 games and you can have win records of 3-2-2-2-1 in which the first team wins the tournament with 3 wins. But 3 wins does not guarantee winning the tournament since you can also have 3-4-2-1-0. So are you asking what is the the smallest number of wins you can have while winning the tournament or what is the number of wins needed to guarantee winning the the tournament? In the first case, if r(n-1) is even then it's fairly easy to come up with an n-way tie of r(n-1)/2 wins each, and by reversing one game you could have a tournament winner with r(n-1)/2+1 wins. If all teams have less than that then it must an n-way tie so the answer is r(n-1)/2+1. If r(n-1) is odd and every team has at most (r(n-1)+1)/2 wins then you must have an n/2-way tie and again it's easy to come up with a way that this is possible. So you can win the tournament with (r(n-1)+3)/2 but not less assuming n>2. If n=2 then the tournament is just a best of r series and the answer is obvious. The second interpretation of the question seems a bit more tricky. If there is only one round then it's pretty clear a team would need to win all n-1 games to guarantee a win, otherwise another team could win all of its games and still win the tournament. --RDBury (talk) 22:22, 12 October 2015 (UTC)

Thanks. I was originally asking about the first/former interpretation, but now that you've mentioned it, I've given some thought to the second/latter interpretation, along lines suggested by your explanation above, and here follows what I think is the answer to the second/latter interpretation. Call our potential champion Team A, and label a potential rival as Team B. In the worst case for Team A, Team B wins all games against the other teams besides Team A. This means Team A must also win all games against all opponents other than Team B, so that means Team A must win at least (r - 1)(n - 1) games. Added to that, to be an untied champion, Team A must win over half of its games against Team B, which is n / 2 games if n is even and (n + 1) / 2 games if n is odd. So I think the answer to the second/latter interpretation is (r - 1)(n - 1) + n / 2 if n is even and (r - 1)(n - 1) + (n + 1) / 2 if n is odd. —SeekingAnswers (reply) 00:06, 14 October 2015 (UTC)

October 13

Percent of a percent

Hi newbie here. I'm playing this RPG that offers many different shields, and I was wondering if one shield blocks an 15% of an attack 22% of the time, but the other blocks 22% of an attack 15% of the time, which one is better? Thanks --2connect4 (talk) 11:30, 13 October 2015 (UTC)

Both shields block the same amount of damage on average (3.3%), so to a first order approximation, they are equivalent.

To a second order approximation, you want to look at the variance. The first shield has a lower variance - standard deviation of 6.21%, whereas the 2nd shield is 7.85%. In general, this means that against stronger opponents, you want the 2nd shield - it will increase the chance of a lucky victory. But against weaker opponents, you prefer the 1st shield - it will decrease the chance of an accidental defeat.

There are other considerations, such as rounding issues caused by short, insufficiently random battles. Suppose you expect to face opponents that deal 100 damage per attack. If you have 335 HP - then with the first shield, even if it successfully blocks all attacks, you will still be defeated in 4 hits, just as if you had no shield at all (85*4 = 340). But with the 2nd shield, you have a chance (1.4%) to block some attacks and last 5 hits. If you had 356 HP, the 1st shield would have the advantage - with either shield, you need to block at least 3 attacks to last 5 hits, and the 1st shield has a better chance for that (4.5% vs. 1.4%). Either shield can be better depending on your HP.

It's impossible to tell whether any of these considerations are important enough to bother with, without knowing more about the game. But my guess is that not - you're best off simply comparing average damage reduction, calculated by multiplying the two figures. -- Meni Rosenfeld (talk) 12:04, 13 October 2015 (UTC)

Yeah, that's what happens in a game with a lot of customization. Thank you so very much for all your help, you opened my eyes to so much lol. I'm not really sure what you mean by approximation, is this the right wikipedia article to read about? Orders of approximation --2connect4 (talk) 01:29, 14 October 2015 (UTC)

Anyone old enough to answer this?

I am trying to remember the name of those wonderful swirly patterns the edges of which are recursive repeats of themselves as you zoom in on them. I used to have some fascinating programs on old 3.5" disks that drew them for you but I had to scrap them when PCs had evolved so far that they wouldn't run them any more.

Can anyone remember what the patterns were called? The name is (annoyingly) on the tip of my tongue.

Thank you Gurumaister (talk) 13:11, 13 October 2015 (UTC)

Mandelbrot set is probably what you're looking for. I'm not that old! :-) --LarryMac | Talk 13:18, 13 October 2015 (UTC)

You are absolutely right!! And if you aren't that old then you are probably more maths aware than I am. Thank you!! Gurumaister (talk) 13:25, 13 October 2015 (UTC)

Mandelbrot set and many, many other fractals. --CiaPan (talk) 14:45, 13 October 2015 (UTC)

I am that old, and the program I was running off 3.5" disks back in the day was Fractint. Their project page is at http://www.fractint.org/ and under http://www.fractint.org/ftp/current/ you will find directories with the latest versions of the program to run under DOS, Windows, and Linux. I don't believe that Fractint has an Android app. I run the free and open source app Fractoid on my phone and tablet, but there are a lot of options out there and I've not done much comparison. As CiaPan alludes, you would be better served by searching on "fractal" instead of "Mandelbrot set", as most fractal generators can display many types of fractals, though the Mandelbrot set is the best known. -- ToE 01:11, 14 October 2015 (UTC)

IMHO, Fractint is still king. It builds without a lot of dependencies, the source code is easily hackable. Sławomir
Biały 01:36, 14 October 2015 (UTC)

October 14

Graphing a linear equation

I am looking for a website that does the following. (1) I enter a linear equation (e.g., y = 2x + 5) and the website graphs the correct line. Or (2) I enter the data points (ordered pairs such as (1,1), (2,2), (3,3), etc.) and the website graphs the correct line. Also, I'd like to be able to print the results. Finally, is there a way in Excel (or Word) to do something similar?

The gist of my problem is this. I have to graph several equations. And I'd want to submit the work so that it is neat (i.e., computer-generated or word-processed) and not hand-written. Any suggestions? I want it to look like the graphs of linear equations that you might see in a textbook, and not one that is written by hand. Like, for example, the two graphs shown at the very bottom of this page: [5]. Or this page: [6]. There must be something out there that does this. That makes this task easier. Any suggestions? Thank you. — Preceding unsigned comment added by 2602:252:D13:6D70:19F4:A168:19D0:22A1 (talk) 04:26, 14 October 2015 (UTC)