/sci/ - Science & Math

>>4720956
Meh. I only had one topology class, and it was the senior-level class, and I hadn't really had an introduction to the ideas, so I didn't really get as much out of it as I could have and I ended up with a high B or something since I was so excited about doing computer science.

I was kind of a screw-off, and now I'm doing automata theory in grad school, which has lots of algebra and rarely requires a lot of topology. Although, if I were more topologically minded, I would probably think about the topology of memory access or something stupid like that.

Did you know the number of strings at each length of an unambiguous context-free grammar is necessarily an algebraic formal power series? I didn't, until a few months ago.

Okay, so, let's play with this shit.

First of all, M and N have the same number of connected components. Don't believe me? Count them, motherfucker. Deform g(f(M)) into id_m continuously! Then for each connected component, pick one. f(that point) has to be in a distinct component from f(any point from another component). OR ELSE YOU GET RAPED. There's NO way to deform two points in the same component to two points in a different component. Therefore, f and g are bijections between the equivalence classes of points under path-connectedness. (I guess this might require axiom of choice - you're going to let me have axiom of choice, aren't you? Fuck you, I'm using axiom of choice, and you can go to hell.) So now we can think of f as a union of disjoint maps from a set of disjoint connected closed manifolds whose union is M to a set of disjoint connected closed manifolds in N. I do this because I want connected closed manifolds.

Oh, nigger, what did we just do? We just reduced it to a problem about CONNECTED closed aspherical manifolds. And what's special about connected closed aspherical manifolds? They're FUCKING CONNECTED. FUCK YEAH!

You're welcome, OP.

>>4720788
This is not true at all. I was once a math freshman, and I couldn't do this then.

>>4687396
The first post I was discussing a case which only applies when i+1 is odd. The second post, that one, applies when i+1 is even. Together, they make a valid induction argument that applies for all n.

>>4687380
This is only a clarification of the other person's post, not a full proof. We can already disregard cases where i+1 is even, since we proved it for the odd component of i+1, and we know that mod i+1, the sequence will just be constant mod that odd component, and constant mod a power of 2 (eventually 0 mod the power of 2) so it'll be constant mod i+1 by the Chinese Remainder Theorem.

>>4684827
This induction argument was already a proof, you guys.

It's constant mod 1, since it's always 0.
It's eventually constant mod i+1, by strong induction. Suppose the sequence is eventually constant mod each integer between 1 and i. Since <span class="math"> 1 \leq \phi (i+1) \leq i, a_{k+1} = 2^{a_k} \equiv 2^{a_k\ mod\ \phi (i+1)} (mod\ i+1) [/spoiler] where by inductive hypothesis <span class="math"> a_k\ mod\ \phi(i+1)[/spoiler] is eventually constant, making the sequence eventually constant.

Sorry about all the edits to clean it up.

>>4665428
Scientist, let's ignore this guy and have ourselves a real conversation.

Basically, it comes down to knowing this list: http://en.wikipedia.org/wiki/Essential_nutrient

That's something I said a bit earlier, but no one seemed to make comment. If you have these things, so it would seem, you have the nutrition needed to stay alive.

>>4628760
Why would the collision necessarily fling us into space? More likely the Sun keeps all its pieces and we just get a chance to spread into another galaxy.

>>4628755
Your body is at a higher temperature than the air around you; you're constantly shedding heat.

>>4626562
The reason the number of multiples of p between 1 and n matters is that the multiplicity of p as a factor of n! is one for every multiple of p, and one more for every multiple of p^2, and one more for every multiple of p^3, and so on. The multiplicity of p in n! is actually <span class="math">\sum_{i=1} \lfloor \frac {n}{p^i} \rfloor[/spoiler]

>>4626498
I mean other way around.

Let p be a prime, and let p^k be the largest power of p in n, and let p^j be the largest power of p in m.
If p^1 divides m and n, then the number of multiples of p in between 1 and n is equal to the number of multiples of p in between 1 and m, plus the number of multiples of p in between 1 and n - m, because the multiples of p between m+1 and n line up with the multiples of p between 1 and n-m. This applies for every power of p except for ones that divide n but not m; in those cases, there's one more between 1 and n than between 1 and m and 1 and n-m combined. In these cases, each power contributes one more factor of p to n! than to m!*(n-m)!, which compensates for dividing by all the factors of p in n that aren't in m and therefore aren't in GCD(m,n)

Show that <span class="math">\frac{gcd(m,n)}{n} nCm[/spoiler] is an integer for all <span class="math"> n \geq m \geq 1[/spoiler].

Basically the idea is that for any maximal prime power factor of n, the multiplicity of it as a factor of (n-1)! minus as a factor of m! (n-m)! is bounded below by the multiplicity in GCD(m,n).

>>4605540
Thank you for restating the question. That's not a proof.

>>4602673
"It's too damn simple" is a good thing. This is beautiful.

I should start reading these threads for comedy. The first four posts in this thread are the funniest thing I've read all week. First, a blatant logical failure, then someone ignores that troll and points out the pointlessness of saging a sticky, and then someone complaining about tripfags, and then EK.

>>4602034
Now this is a troll.

>>4601588
It could easily be a mathy report. I wrote my area paper just now in LaTeX.

>>4601591
Of course he's that much of a cunt.

>>4583915
Oh, dammit, yeah, I tried to evaluate the sum without thinking or doing anything else. My confidence was well-placed, though; you did it!

Now I sleep. Oh god, I need to sleep.

The sum in question is a sum over all n of the sum from m = ceil(n/2) to 2n of x^m y^n. This can be rewritten as (x^(2n)-x^(ceil(n/2))/(x-1) y^n, which looks horrible and stupid, but we factor it out again and it looks something like (x^2y)^n - (x^1/2 y)^n /(x-1) which then becomes something miserable and I don't even care any more. I'm sorry, guys, I just did taxes and I didn't get enough sleep and I have confidence in youuuuuuuuuuuuuu!

The limit is only well-defined if it's well-defined for all possible x = x(t), y=y(t), where as t approaches 1, x(t), y(t) approach 1. Basically, it's only well-defined if all possible ratios of x to y as they approach (1,1). First, we have to evaluate the sum. Let's do it like this:

>>4550620
Nope, not brute force. That would be distinctively inelegant. You just had to bait me into posting a full solution.

1+a+b+c and 3+a+3b+c are either both even or both odd (since the difference 2+2b is even); in the case that they're both even, they'd both be 0 mod 4, and hence 2+2b is also congruent to 0 mod 4. If they're both odd, they'd both be 1 mod 4 and 2+2b would still be congruent to 0 mod 4.

<span class="math">2+2b\equiv0[/spoiler]
<span class="math">2b\equiv2[/spoiler]

If <span class="math">c\equiv0[/spoiler], then <span class="math">2b+c\equiv2[/spoiler]. If <span class="math">c\equiv1[/spoiler], then <span class="math">2b+c\equiv3[/spoiler]. Either way, <span class="math">2b+c\equiv8+4a+2b+c[/spoiler] cannot be a square, and so a contradiction is obtained.

This would be pretty easy to prove by contradiction. Suppose there exist integers a, b, c such that for all positive integer n, <span class="math">\sqrt{n^3+an^2+bn+c}[/spoiler] is integer.

Let n take on the values from 1 to 4; then 1+a+b+c, 8+4a+2b+c, 27+9a+3b+c and 64+16a+4b+c are all squares. One property of squares is that they're all congruent to 0 or 1, modulo 4. Reducing these mod 4 gives 1+a+b+c, 2b+c, 3+a+3b+c and c.

And it's not hard from here to prove that it's impossible for all four of these to simultaneously be congruent to either 0 or 1, though I'm too lazy to work it out now.

>>4533267

Fuck, stupid lack of compiler errors! Moot, we want compiler errors!

A determinant path, he means, is a permutation of the rows relative to the columns. The determinant is the sum over all permutations <span class="math">\epsilon[/spoiler] of the product over all i of <span class="math">\sigma(\epsilon)a_{i \epsilon(i)}[/spoiler]. One of these products, for a particular permutation epsilon, is a "path" because of how high schoolers sometimes learn to evaluate determinants, and the sum is odd iff the number of nonzero "paths" is odd.

Ha, the determinant is also the permanent. Wonder if that's even useful.