## Search found 61 matches

- Sun Jan 15, 2012 6:42 pm UTC
- Forum: Mathematics
- Topic: Compositional square root
- Replies:
**18** - Views:
**3396**

### Re: Compositional square root

The function defined by f(x) = (x-1) 2 is continuous from R to itself, but it has exactly one 2-cycle. Thus, it has no compositional square root. Interesting! Seems possibly related to f having two fixed points This certainly shows my "vague feeling" wrong, and moreso, it shows that conti...

- Sun Jan 15, 2012 3:19 pm UTC
- Forum: Mathematics
- Topic: Compositional square root
- Replies:
**18** - Views:
**3396**

### Re: Compositional square root

How do you combine arbitrary branch structures? That's what I tried to address here: "(this is considering the various preimage depths for type 5, also the preimage level of x, as separate types)". With that I mean how many successive preimages we can take before they're all empty. So, yo...

- Sun Jan 15, 2012 12:27 am UTC
- Forum: Mathematics
- Topic: Compositional square root
- Replies:
**18** - Views:
**3396**

### Re: Compositional square root

I think Deedlit's classification works for general (i.e. not necessarily injective) functions as well. There's just a few more cases to consider. The basic idea is to consider the equivalence relation x~y :<=> There is n, m in N (including 0) such that f^n(x) = f^m(y) Let S(x) be the equivalence cla...

- Fri Jan 13, 2012 4:21 pm UTC
- Forum: Mathematics
- Topic: Help a philosophy student with some math about possibilities
- Replies:
**16** - Views:
**2297**

### Re: Help a philosophy student with some math about possibili

The above is not correct, there are a lot more possibilities added in each step. k n+1 : k n as above for the cases where there's no bracket behind the last symbol. Now we add square brackets around the last symbol. This gives us k n +1 (because the "no brackets at all"-case before the las...

- Tue Jan 10, 2012 5:34 am UTC
- Forum: Mathematics
- Topic: Compositional square root
- Replies:
**18** - Views:
**3396**

### Compositional square root

So, in this post , the construction of a function e 1/2 with e 1/2 (e 1/2 (x))=exp(x) is discussed. I don't think the post refered to there is a solid valid proof of existance, so I tried to formalize the argument, but failed. That irks me especially because it reminded me of the more general proble...

- Mon Jan 09, 2012 8:07 pm UTC
- Forum: Mathematics
- Topic: dimensions: I think I lost a few.
- Replies:
**37** - Views:
**5247**

### Re: dimensions: I think I lost a few.

Well, I at least showed that the set I chose is spanning. Its a whole other level of tedium to show that its linearly independent (you'd need a fair bit more theory about homogeneity to make it work). I think the same argument I gave against your proposed basis spanning k(X) shows this as well: Ass...

- Mon Jan 09, 2012 5:35 pm UTC
- Forum: Mathematics
- Topic: dimensions: I think I lost a few.
- Replies:
**37** - Views:
**5247**

### Re: dimensions: I think I lost a few.

You're absolutely right. It's kinda amazing that I missed that, yet was looking at all the steps in detail to understand what happens. Am I missing something else, or does the proof go through for k(X) over K just as well? I don't see where the specific choice of V is used anywhere... all that seems...

- Mon Jan 09, 2012 2:14 pm UTC
- Forum: Mathematics
- Topic: dimensions: I think I lost a few.
- Replies:
**37** - Views:
**5247**

### Re: dimensions: I think I lost a few.

If you assume choice, finding the B(x) is trivial. In particular, if you have a choice function, f, then {f(i) |i in I} is a basis, and the (not actually multiple) choice that you get back is your original choice function (as a_b(y) = y/f(i)). That's not a basis. These would only generate rational ...

- Mon Jan 09, 2012 12:09 am UTC
- Forum: Mathematics
- Topic: dimensions: I think I lost a few.
- Replies:
**37** - Views:
**5247**

### Re: dimensions: I think I lost a few.

... actually, that proof is also amusing. Having read over it only twice, I don't fully get it. What I should do is try it in a few explicit (decreasingly simple) cases and see what the beta_bi sets look like. (For small enough x_i, the beta_bi look like they should be constructable.) You'd actuall...

- Sun Jan 08, 2012 4:56 am UTC
- Forum: News & Articles
- Topic: Banach Tarski!
- Replies:
**8** - Views:
**2260**

### Re: Banach Tarski!

It hasn't been posted here because it's not news. It's also a rather pointless video with a crappy soundtrack. Well, this isn't for "news". The definition of this forum is "Seen something interesting in the news or on the intertubes? Discuss it here." You're welcome to argue tha...

- Sun Jan 08, 2012 4:43 am UTC
- Forum: News & Articles
- Topic: Banach Tarski!
- Replies:
**8** - Views:
**2260**

### Banach Tarski!

How the heck has this not been posted here yet?

This is the most amazing combination of art and science possible!

http://www.youtube.com/watch?v=uFvokQUHh08

Souce = http://boingboing.net/2012/01/04/banach ... -some.html BoingBoing

This is the most amazing combination of art and science possible!

http://www.youtube.com/watch?v=uFvokQUHh08

Souce = http://boingboing.net/2012/01/04/banach ... -some.html BoingBoing

- Sun Jan 08, 2012 1:26 am UTC
- Forum: Mathematics
- Topic: f(f(x))=exp(x) and extensions to noninteger numbers of f.
- Replies:
**9** - Views:
**3613**

### Re: f(f(x))=exp(x) and extensions to noninteger numbers of f

The problem is not so much whether such functions exist, they do, but whether they can be made unique by extra assumptions. To show that they exist, consider the following: Any number of solutions to f(f(...f(x)))=exp(x) can be generated using the axiom of choice. Suppose that there are n occurrenc...

- Sat Jan 07, 2012 9:04 pm UTC
- Forum: Mathematics
- Topic: Quick, probably simple question from a math contest
- Replies:
**12** - Views:
**1897**

- Sat Jan 07, 2012 5:35 am UTC
- Forum: Individual XKCD Comic Threads
- Topic: 1000: "1000 Comics"
- Replies:
**147** - Views:
**49883**

### Re: 1000: "1000 Comics"

Thank you and congratulations!

- Sat Jan 07, 2012 12:03 am UTC
- Forum: Mathematics
- Topic: dimensions: I think I lost a few.
- Replies:
**37** - Views:
**5247**

### Re: dimensions: I think I lost a few.

Since I believe in AoC, that problem does not arise.

More to the point, though - my explanation was aimed at the OP, whose level of math background suggests that the question doesn't even arise. Obviously, the entire description I gave is no rigorous definition, anyway.

More to the point, though - my explanation was aimed at the OP, whose level of math background suggests that the question doesn't even arise. Obviously, the entire description I gave is no rigorous definition, anyway.

- Fri Jan 06, 2012 4:35 pm UTC
- Forum: Mathematics
- Topic: dimensions: I think I lost a few.
- Replies:
**37** - Views:
**5247**

### Re: dimensions: I think I lost a few.

tomtom2357 wrote:What about infinite dimensional space?

Same thing - that means that a basis has infinitely many elements. The linear combinations just use finitely many of those.

- Fri Jan 06, 2012 4:47 am UTC
- Forum: Mathematics
- Topic: dimensions: I think I lost a few.
- Replies:
**37** - Views:
**5247**

### Re: dimensions: I think I lost a few.

There's a mix-up here. First, we're not really talking mathematics here. Math knows tons of different things called dimensions, and they're not always the same. What we are talking about is "degrees of freedom" in a "common language" definition of dimension. And these only have l...

- Fri Jan 06, 2012 12:02 am UTC
- Forum: Logic Puzzles
- Topic: Plane colorings
- Replies:
**53** - Views:
**25645**

### Re: Plane colorings

"Measurable" is generally taken to mean Lebesgue-measureable; a short sketchy definition for the plane is the infimum of the area of all square coverings (with the area of a square defined as expected) of the set A to be measured; note that being measurable hinges on the additional require...

- Thu Jan 05, 2012 12:29 am UTC
- Forum: Logic Puzzles
- Topic: Plane colorings
- Replies:
**53** - Views:
**25645**

### Re: Plane colorings

Yes I've seen that page, but it doesn't give any more results than my findings. I am thinking about using circles of diameter 1, but having the edges be in a different color. This one won't give a (much) better result; it's been shown that using either polygons or jordan curves (i.e. any sort of &q...

- Wed Jan 04, 2012 6:30 am UTC
- Forum: Logic Puzzles
- Topic: Plane colorings
- Replies:
**53** - Views:
**25645**

### Re: Plane colorings

Spoilers: http://en.wikipedia.org/wiki/Hadwiger-Nelson_problem Thanks for that link; I was not aware of the number of results known and the complexities involved. Apologies for the somewhat extensive "bonus question"! That said - it certainly seems interesting that AoC might play a role i...

- Tue Jan 03, 2012 1:01 am UTC
- Forum: Logic Puzzles
- Topic: Plane colorings
- Replies:
**53** - Views:
**25645**

### Plane colorings

So, I've checked and was surprised that apparently this classic hasn't been posted here yet: Is it possible to color the plane with three colors such that no two points of equal color have distance 1? And since I'm certain that for the regulars, this will be easy to solve, bonus question: What is th...