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August 14

Entangled slinky — mathematical explanation?

Can someone explain to me how slinkies get entangled? What I find puzzling is not the fact that they get entangled, but that once they do, they can't seem to be untangled by simple twisting. The only times that I was able to untangle tangled-up slinkies and coil phone cords, I did it by slowly and painstakingly shift the "knot", turn by turn, to the end. I got a feeling that that's not the right way to do it, and that the slinky/cord in question didn't get tangled up in that manner in the first place. (That is, I don't remember creating a "knot" somehow and shifting it turn by turn until the "knot" ended up somewhere in the middle of the slinky/cord.)

To get a better feel for it, get hold of a beaded necklace (like the ones they fling around at Mardi Gras). Lay it out flat and straight. Twist the two strands together at one point, then let go. They'll stay together. If you try to pull them apart, likely nothing will happen; you have to untwist them in exactly the same way, at the same point. Now, do the same thing with a slinky. The reason that undoing it again is harder, is that where they intertwine wanders up and down the slinky as the spirals shift against each other. However, if you pull apart what you can, you'll find that there are a few specific points (or, in the case you're describing, one specific point) where they're actually locked together. Treat each end of the slinky, from the last knot onwards, like a giant bead. Twist them around each other in the right direction, and they'll come right apart. Black Carrot 05:14, 14 August 2007 (UTC)[reply]

19

which numbers can 19 be divided into? —The preceding unsigned comment was added by 76.64.52.63 (talk • contribs) 01:57, August 14, 2007 (UTC) – Please sign your posts!

What kinds of questions are these? Anything that is divisible by 38, 19, or 23, depending on the problem. 1 * 23, 2 *23, 3 * 23, etc. Gscshoyru 01:59, 14 August 2007 (UTC)[reply]

If you mean which numbers they are divisible by (their divisors) then see Table of divisors. PrimeHunter 02:08, 14 August 2007 (UTC)[reply]

interesing point there; "divided into" meaning "is an even divisor of" as in 20 can be divided into 40, 60, etc.; or "divided into" meaning is the product of, as in 20 can be divided into 2*10. Gzuckier 17:01, 14 August 2007 (UTC)[reply]

20 can also be divided into 10+5+3+2. Tesseran 19:18, 15 August 2007 (UTC)[reply]

Surely this is asking for the answers to homework problems! Dougmerritt 18:41, 14 August 2007 (UTC)[reply]

Where can you learn math online?

I would like to know of any sites out there that are free that you can learn math and that it is easily explained to you. Sort of like that language program, Rosetta Stone where it is easy to learn a language on that program. So anyone know of any? Thank you. Bond Extreme 02:46, 14 August 2007 (UTC)[reply]

It would help to know what kind of math you want to learn. Arithmetic? Algebra? Geometry? Calculus? The easiest advice is, type in whatever you want and the word "tutorial" into google. Black Carrot 05:17, 14 August 2007 (UTC)[reply]

Basically High School Algebra and Geometry and College Algebra.Bond Extreme 05:52, 14 August 2007 (UTC)[reply]

A search for algebra geometry gives [1] as the first result, which looks pretty good. Searching for high school algebra and following some links gets me to [2] and [3]. Searching for college algebra tutorial gives [4] and [5]. That should get you started. Others may have better suggestions when they wake up. Black Carrot 07:37, 14 August 2007 (UTC)[reply]

I'd recommend MIT's OpenCourseWare program, and if you get bored of that out sister projects Wikibooks and Wikiversity. Although you should keep in mind, these are generally going to be college level courses, so they won't be very 'easy', they should be complete and useful. Your local Library should also have material like this, and depending on where you live a Community College might have cheap courses you could take. --Lu c id 07:41, 14 August 2007 (UTC)[reply]

I don't think the MIT Open Couse Ware covers High School math. Donald Hosek 17:26, 14 August 2007 (UTC)[reply]

Which is why I said these are generally going to be college level courses *wink* --Lu c id

I'd like to throw in a caveat emptor on the Wiki side of things. Huge chunks of those tutorials haven't even been written yet. Black Carrot 18:18, 14 August 2007 (UTC)[reply]

Sadly, this is true. Sadly, the other Wikiprojects don't have the publicity, because they don't have the content, because they don't have the writers, because they don't have the publicity... not to mention it's a lot more work to make a consistent book than it is an encyclopedia article.--Lu c id 00:52, 15 August 2007 (UTC)[reply]

Happily this is NOT true if Wiki has the management skill. Is there any social club (URL please) where we can find potential co-authors to help proofreading, checking, etc.? We have the contents. We have the writers. A lot of work has been done but not up to the level of being books. We need a meeting place (URL please). twma 07:33, 15 August 2007 (UTC)[reply]

I've been working on the Wikiversity/Wikibooks coverage of basic math as we speak. I'm currently focused on the Elementary School Math, [6], which is just now getting into reasonable shape, but will then go on to working on the High School Math [7] subjects you just mentioned. Take a look at what's already there and let us know if any of that meets your needs. StuRat 07:43, 18 August 2007 (UTC)[reply]

Pell's equation

Is there a recursive function to find solutions to the generalized Pell equation, $ax^{2}-by^{2}=c$ , if a solution to the function has already been found? Indeed123 17:58, 14 August 2007 (UTC)[reply]

Yes. See Pell's equation#Solution technique. nadav (talk) 01:41, 15 August 2007 (UTC)[reply]

My answer was incomplete. To get solutions to the "generalized" equation, you have to of course transform the usual Pell equation solutions. The procedure appears on mathworld. nadav (talk) 05:49, 16 August 2007 (UTC)[reply]

Yes, if you have a solution (x,y) to

ax^{2}-by^{2}=c

and you also have a solution (l,m) to

l^{2}-abm^{2}=1

then you can generate an infinite family of (x,y) solutions by recursively using the transformations

x\rightarrow lx+bmy

y\rightarrow amx+ly

Gandalf61 12:39, 16 August 2007 (UTC)[reply]