Wikipedia:Reference desk/Archives/Mathematics/2008 February 22

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February 22

Transfinite number

What's the name of a transfinite number that can have any finite number subtracted from or added to it, with the result being different than the original number? For example, ω wouldn't be it because there's no ω-1. ℵ₀ wouldn't be it because ℵ₀+1=ℵ₀. — Daniel 02:08, 22 February 2008 (UTC)

The term "transfinite number" refers specifically to cardinals and ordinals, neither of which have that property. You may be thinking of the nonstandard integers. Black Carrot (talk) 02:21, 22 February 2008 (UTC)

Indeed, cardinals and ordinals are extensions of the natural numbers, not the integers, so subtraction isn't particularly meaningful (except for undoing addition). --Tango (talk) 11:49, 22 February 2008 (UTC)

Any nonstandard model of arithmetic has numbers with this property. The OP might also be interested in the surreal numbers. Algebraist 12:41, 22 February 2008 (UTC)

A non-diagram solution?

Is there a way to solve the following question without drawing a diagram, ie using combinatorials, permutations, etc. only? "How many ways are there to seat two students in a row of four desks so that there's at least one empty desk between them?" Imagine Reason (talk) 20:02, 22 February 2008 (UTC)

Well, I just solved it without drawing a diagram. There are two ways with exactly one space between them (the other empty desk can be on the left or the right) plus one way with exactly two. Algebraist 20:08, 22 February 2008 (UTC)

I drew a diagram in my head. Does that count? (I got either three or six, depending on whether you count switching the students.) Maybe a bigger example would help. Let's say we have 10 people and 100 desks. How many ways are there to arrange the people in the desks so that any two people have at least one empty desk between them? This is indeed a standard combinatorics problem, and can be solved without drawing diagrams. It does require imagining the situation in your head, but no more. Black Carrot (talk) 20:23, 22 February 2008 (UTC)

Let's say we're only worried about which desks have a person in them and which don't. There are 10 people, so there must be at least 9 desks empty - one between each pair of people. In particular, each person but the one on the right end must have an empty desk to their right. So, the question is really, how do we arrange 10 people (each attached at their right hip to an empty desk) in 91 desks? Black Carrot (talk) 20:40, 22 February 2008 (UTC)

Obviously order matters in this case, so the answer to the original question is six. I'm not sure it's a simple combinatorial question, however, because the one person who doesn't have an empty desk attached to the hip cannot be placed directly to the left of another student, but I don't see a simple way of preventing that from happening. Imagine Reason (talk) 01:55, 23 February 2008 (UTC)

All have a hip attachment, without exception. Add an extra empty desk to compensate, so we have for example 10 students with attachments + 91 loose desks = 10 people + 101 desks total. Since the rightmost desk is always empty, there is a bijection with the lawful arrangements in the original 100-desk formulation.  --Lambiam 23:57, 23 February 2008 (UTC)

You can decide which seats will contain students first (for example, by ignoring order), and then afterward decide which order to assign the students to the chosen seats. —Bkell (talk) 06:27, 23 February 2008 (UTC)

Yes, but there's no way of doing that without drawing diagrams in your head, right? Imagine Reason (talk) 18:17, 23 February 2008 (UTC)

Depends on how you think, I guess. What's wrong with diagrams? The diagram is a way to help you translate the problem into mathematics which you can then solve. The fact that you used a diagram to get there doesn't make the maths any less mathematical. --Tango (talk) 21:54, 23 February 2008 (UTC)

Here's the thing about rearrangement. I start by putting the students in some order, maybe alphabetical. Then I let them sit wherever they want, keeping to that order. 9 of the students have empty desks attached to them during that. Once I've counted the options without rearrangement, if there are 10 students, I can multiply by 10 factorial to get the number with rearrangement. I start by worrying only about whether a desk is empty or full, then later worry about who it is that's filling it. Black Carrot (talk) 19:28, 24 February 2008 (UTC)

Changing coins

How many ways can you change a five cent coin (nickel) into one cent coins and/or two cent coins?.

The answer is of course 3 ways. (1+1+1+1+1,1+1+1+2,1+2+2)

But I cannot figure out how many ways there are of changing $100 dollar note (10000 cents) into one cent coins and/or two cent coins. Exhaustive counting is way too exhaustive for me. The only thing I could think of is writing a computer program but I dont think that is how mathematicians in the 19th century would solve this problem. 122.107.226.136 (talk) 22:43, 22 February 2008 (UTC)

Hint: Once you choose how many 2-cent coins to use, the number of 1-cent coins is determined. -- Meni Rosenfeld (talk) 22:53, 22 February 2008 (UTC)

I figure out how to do that. If you have n of n+1 cents ($n\in$ even numbers), then you have ${\frac {n}{2}}+1$ ways to do it. In the question, you have 10000 cents, so you have 5001 ways. Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) 02:49, 23 February 2008 (UTC)

Thanks. Now how many ways are there to change $100 dollar note (10000 cents) into one cent and/or two cent and/or five cent coins. This is even more difficult than just one cent and two cent coins. 122.107.226.136 (talk) 03:43, 23 February 2008 (UTC)

Hint: Once you choose how many 5-cent coins to use, the number of ways of changing the remainder into 1-cent and 2-cent coins is determined. Michael Slone (talk) 03:53, 23 February 2008 (UTC)

More ODE madness

Last question of my assignment and I can't do it at all!

The differential equation is ${\frac {\mathrm {d} y}{\mathrm {d} t}}=y(y-1)$ with initial condition $y(0)=2$.

I've been taught solving by separating variables and also the integrating factor method, but I can't see how to apply either here! One attempt got me 1 = -2 which was most disconcerting! I'm really looking for hints just to get me started.

Thanks all Psymun (talk) 23:40, 22 February 2008 (UTC)

What goes wrong with separating the variables and integrating? At worst, it's integration by parts, and there is probably a better method I'm not seeing (in my defence, it is nearly midnight here). --Tango (talk) 23:54, 22 February 2008 (UTC)

It looks like partial fractions to me. --Tardis (talk) 23:57, 22 February 2008 (UTC)

Oh, yeah... Never did like partial fractions! --Tango (talk) 00:14, 23 February 2008 (UTC)

Incidentally, I just checked the method I used (not int. by parts, that was a terrible idea!) by using partial fractions, and it turns out I actually it right - I used a change of variables y->1/y->1-1/y. --Tango (talk) 00:17, 23 February 2008 (UTC)

Right! That's odd, I haven't been taught it, but it won't do any harm for me to look it up. Why did you change the varibles, Tango? Psymun (talk) 10:47, 23 February 2008 (UTC)

Changing variables is a standard method for integrating and is useful in a wide range of cases. It's not really the best way for this problem, partial fractions is better, but I use change of variables more often so that's what I'm more comfortable with. As for how I chose what to change them to - lucky (educated) guess! --Tango (talk) 14:04, 23 February 2008 (UTC)

OP again. I'm sorry chaps, but I'm still a bit confused as to how I get $y(t)$ here - I can't see how integrating as a partial fraction would work here; I know how to integrate rational functions but I haven't been taught it as a method of solving ODEs! I wanted to try separating but I have no independent variable to separate, nor can I get it in the form $dy/dt+P(t)y=Q(t)$. This question is horrible - but thank you for everyone's suggestions so far! What a wonderful resource - I need to look for some questions I can help with! :S Psymun (talk) 10:47, 24 February 2008 (UTC)

To separate the variables, just divide each side by y(y-1) and multiply both sides by dt. --Tango (talk) 12:19, 24 February 2008 (UTC)

Good lord, you may have just saved my bacon! I'm working through it... Psymun (talk) 14:07, 24 February 2008 (UTC)

Ok!! How does $y=\left({1-{\frac {1}{2}}e^{t}}\right)^{-1}$ sound, noting the initial condition? Psymun (talk) 14:38, 24 February 2008 (UTC)

That's what I got. --Tango (talk) 14:55, 24 February 2008 (UTC)

Hurrah!!!! xDDDDD Psymun (talk) 15:18, 24 February 2008 (UTC)