Wikipedia:Reference desk/Archives/Mathematics/2006 August 2
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Word Problems
[edit]Hello. Could someone please remind me how to do problems in which it is The train full of school children is heading east on a track X miles long at Y miles per hour. The train full of nuns is heading west towards their inevitable doom on the same tracks at Z miles per hour. At what point will the two trains collide?
Thank you. --Demonesque 05:17, 2 August 2006 (UTC)
- There are several ways you can do this. One way is to figure out how quickly they are approaching each other (in miles per hour); from this you can tell how long it will take until they collide. Another way is to write one equation that gives the position of the first train as a function of time and a second equation that gives the position of the other train as a function of time, and solve the system, eliminating time. —Bkell (talk) 05:44, 2 August 2006 (UTC)
Yeah, I know that I have the information to solve it, I just don't know how to do any of those things you suggested. :( --Demonesque 05:49, 2 August 2006 (UTC)
- Suppose that the first train is traveling at 35 miles per hour, and the second is traveling at 45 miles per hour. How fast are the trains approaching each other? If the track is 160 miles long, then the trains start 160 miles apart. If they are approaching each other at such and such a speed, how long will it take them to collide? If the first train travels at 35 miles per hour for that length of time, where will it be when it smashes into the other train? —Bkell (talk) 05:53, 2 August 2006 (UTC)
Are they approaching each other at 80 miles per hour? Do I add the two values? I'm not sure how to get the speed at which the trains are approaching each other.
- Yes. Velocity is relative. If I'm sitting in the first train, then from my point of view the ground is traveling backwards at 35 miles per hour, and the other train is going 45 miles per hour faster than that, so it's approaching me at 80 miles an hour. —Bkell (talk) 06:01, 2 August 2006 (UTC)
Thank you for spending so much time on helping me. ^^ They'd hit at the 70 mile mark, I believe.
- Yes—70 miles from where the first train started, or 90 miles from where the second train started. —Bkell (talk) 06:05, 2 August 2006 (UTC)
- Another variation to this problem is to have a speedy fly that flies along the track at 320mph between the two trains, reversing back when it touches the front of one of them, and keeping going until the trains collide. What is the total distance the fly will travel?
I derived a formula to help you in the future:
If: "The train full of school children is heading east on a track X miles long at Y miles per hour. The train full of nuns is heading west towards their inevitable doom on the same tracks at Z miles per hour" ...
The distance from where the first train started and where it collided is: (xy/(y+z)) miles
The distance from where the 2nd train started and where it collided is: (xz/(y+z)) miles
Is System F Turing Complete?
[edit]I can't seem to find a simple answer anywhere on this. Is System F or Second-Order Lambda Calculus turing complete? -Exomnium
- System F has the normalization property and can therefore (as noted in that article) not be Turing complete. --LambiamTalk 06:40, 2 August 2006 (UTC)
cube
[edit]A cube is painted blue on all sides.It is then cut into 125 small equal cubes.How many cubes will be there in which no face is coloured?
- You might try looking at this: [1]. I don't think it will tell you the answer (do your own homework!), but it might help. Madmath789 12:50, 2 August 2006 (UTC)
- Here's a hint: 5*5*5 = 125. --198.125.178.207 14:42, 2 August 2006 (UTC)
AnnuaL Interest rate
[edit]What is the annualized rate of return in a discount payment term of 1% 10 days net 30? Please explain the calculation. Thanks
Answer: The usual term you see on bills of lading would be 1/10, n/30, which means by paying 20 days before the amount is due, you save 1% of the invoice price. So to estimate the annualized rate, you multiply 1% by (360/20), which gives you 18%.
Vector normalization
[edit]Though I found an article with the appropiate title in wikipedia, it did not actually go about explaining HOW to normalize a vector, but rather defined it, and described its uses. What techniques are used for normalizing vectors? Relatedly, when using you have a recursively defined vector (in something such as page rank), how does iterating the process bring one closer to the "true" vector. (Is it that the process converges toward some limit for each vector? ) In advance- thanks for the help
--17:24, 2 August 2006 (UTC) Tim
- Can you be more specific? Which kind of vector are you interested in, and which Wikipedia article did you check? Generally, you just calculate the norm of the vector and divide by it. If it's a vector in (the n-dimensional Euclidean space), v = (a1, ..., an), the norm is:
- -- Meni Rosenfeld (talk) 18:00, 2 August 2006 (UTC)
- For the pure mathematicians, we note that we need a normed vector space, not merely a vector space, for normalization to be defined. (Commonly we have more, an inner product space, though not all norms can be expressed as inner products.) Now suppose we are given a vector v, and that we compute its norm to be ℓ = ‖v‖. An essential property of every norm is that it commutes with scaling, so that ‖sv‖ = s‖v‖. Therefore, assuming ℓ ≠ 0, the scaled vector (1/ℓ)v will have unit norm. The only remaining question is how to compute the norm, but that depends on the specific norm being used. In the case of the common Euclidean norm in a finite-dimensional space, ‖(x1,…,xn)‖ = (x12,…,xn2)1/2, the computation is mathematically trivial (though practical computation with floating point numbers raises questions of speed, accuracy, and overflow). --KSmrqT 18:31, 2 August 2006 (UTC)
- Wouldn't that be ‖(x1,…,xn)‖ = (x12 + … + xn2)1/2? -- Meni Rosenfeld (talk) 18:40, 2 August 2006 (UTC)
- No, I'm using the convention that on the right hand side of an equation when I'm distracted a comma indicates summation. ;-)
- Were I not distracted I'd write ‖(x1,…,xn)‖ = (x12+⋯+xn2)1/2, with plus signs and centered dots. :-)
- And I might even generalize to ‖(x1,…,xn)‖p = (|x1|p+⋯+|xn|p)1/p, where p ≥ 1 chooses a norm, with p = ∞ denoting the Chebyshev norm. --KSmrqT 03:32, 3 August 2006 (UTC)
- Wouldn't that be ‖(x1,…,xn)‖ = (x12 + … + xn2)1/2? -- Meni Rosenfeld (talk) 18:40, 2 August 2006 (UTC)
Thanks a lot guys, you've been really helpful! --67.184.219.154 20:20, 2 August 2006 (UTC) Tim
In Dimensional Analysis, what is the dimension of an interest rate?
[edit]In Dimensional analysis, what would the dimension of an interest rate be please?
- 1/T, where T is time. A more difficult question is what the dimension of money is. But using arbitrarily the symbol D, an interest rate of 12 % per year means, for example, ($12/$100)/year, which has dimension (D/D)/T = 1/T. --LambiamTalk 00:45, 3 August 2006 (UTC)
- Why not just think of it as .12/year, which obviously has dimension 1/T because .12 is dimensionless? Tesseran 23:25, 3 August 2006 (UTC)
- Because the other way is more fun :) and also shows to who did not yet realize this that percentages are dimensionless. Interestingly (or not), we find the same dimension as for angular velocity. --23:53, 3 August 2006 (UTC)
- Why not just think of it as .12/year, which obviously has dimension 1/T because .12 is dimensionless? Tesseran 23:25, 3 August 2006 (UTC)
Neural nets v. Multiple regression - which is better?
[edit]I would like to try doing some forecasting of the property (real-estate in American english) market using several different economic time series as the independant variables.
If I used multiple regression, then the forecast would just be the sum of the weights of the variables. Neural nets on the other hand should be smarter, but statistical textbooks never mention them.
When I write neural net, I mean a multi-layered perceptron.
A feature of economic series is that it could be lagged variables that have the greatest predictive power. I wonder how neural nets, or multiple regression for that matter, could be made to cope with this?
I remember reading somewhere that it had been mathematically proved that you only need a maximum of three layers in a neural net. Is this true in practice? And are the three layers just the input, middle, and output layers?
Is multicollinearity a problem with MLP as it is with multiple regression? Although I've recently read that if you just want to do forecasts rather than create explaintory models, multicollinearity dosnt matter.
If anyone knows of any software that allows using a MLP with a spreadsheet or CSV type interface, I'd be very grateful if you could let me know. --62.253.48.74 23:38, 2 August 2006 (UTC)
- I don't know much about multiple regression, but here are some facts about neural nets that may be helpful. Multilayer perceptron networks can classify data with a combination of linear seperators. If your data can't be easily classified in this way, then a perceptron-based network may not be for you.
- I haven't seen the result you speak of about neural network layers, but I do recall that recurrent multi-level tower networks (which have a feedback loop), are equivalent to single-level tower networks.
- The thing about neural networks is that they are relatively simple to program. If you're skilled, you could do this yourself. Dysprosia 08:54, 3 August 2006 (UTC)
Could I make $$$ by playing online with a poker-computer?
[edit]Chess computers can play a very good game. Even a basic chess program can beat me easily.
So, I've been wondering if it would be possible in theory to use a poker-playing software program to play my hand in an online poker game, and make a profit from doing so by being a better player than other people?
Thanks --62.253.48.74 23:48, 2 August 2006 (UTC)
- In theory yes, if the poker program is indeed better than the other people playing, and the other people are dumb enough not to use such a poker computer even though it plays a better game than they do. What I don't know is, how do you know the operators of the online game don't cheat? --LambiamTalk 01:15, 3 August 2006 (UTC)
- I understand that the operators of online games get their profit by taking a "rake" off every pot. So they don't care how much you win - the bigger the win, the bigger the pot, the bigger the rake. But poker as a video game would be different.
- Maybe. But you'll have to design a robust algorithm that can't be predicted easily. If somebody suspected a program, they'll try to baffle it. --ColourBurst 01:40, 3 August 2006 (UTC)
- AFAIK, most casual players will get beaten by a well-written bot, like PokiBot [2]. At this site you can download a client to play against several bots. How to use it in practice is another question, since they won't give you its source code. Most bots you'll find on Internet either suck or are scams [3], so I suggest you to practice your own skill instead. Grue 21:09, 3 August 2006 (UTC)
The huge difference between chess and poker is one is a perfect information system and one is not. See game theory. Evalulating it from this point of view, on-line poker does not have the depth that chess does. In this theory there is still a random element, one that no algorithm can ever predict. So in theory, I can't see how even a well designed software could win consistantly. You could, however, use software as an aid in calculating odds. Real life poker is a different story.--Tmchk 01:20, 4 August 2006 (UTC)
- That is a bit of a non sequitur. All you need is that the program wins more often than the other players. Backgammon involves an element of chance, and yet the best programs outperform the best human opponents. --LambiamTalk 01:40, 4 August 2006 (UTC)
Right, I meant one would not have the same degree of success in cheating with a poker program as with cheating with a chess program. Tmchk 23:07, 5 August 2006 (UTC)