/sci/ - Science & Math

>>4325556
Works for me. Having the non-computationally-challenging proof would be nice, though.

I'm pretty sure I remember this being true for diffeomorphisms (or what was true was that <span class="math">||x||_X\to \infty\Rightarrow ||f(x)||_Y\to\infty[/spoiler] for <span class="math">f:X\to Y[/spoiler] a diffeormophism between Banach spaces <span class="math">X[/spoiler] and <span class="math">Y[/spoiler], which looks equivalent to what you're trying to show).

I don't remember if that used the differentiality or just the homeomophism properties. And I don't remember the proof either...

>>4325539
Bah...
At least OP realized he didn't read the formula properly. I'm too lazy to care about what reasoning mistake I'm doing here.

>>4325526
Oh so you think the "<span class="math">2n![/spoiler]" should be "<span class="math">(2n)![/spoiler]" and it's a mistake in the formula? (Or am I missing something?)

>>4325507
Well that isn't correct, since it should be <span class="math">(2n)![/spoiler] then... My bad.

Then I would get this 1/2 out of the series, make <span class="math">\pi^{2n}[/spoiler] a <span class="math">(\pi^2)^n[/spoiler], then get the <span class="math">(-1)^n[/spoiler] in it to have <span class="math">(-\pi^2)^n[/spoiler], and... that looks nice but the series for exp(x) only works if <span class="math">|x|<1[/spoiler] and here we have <span class="math">x=-2\pi[/spoiler].

Damn. It's not about series (or I'm just very bad at those problems). I'm disappointed.

>>4325505
Yeah I would also think that. But now we have to prove it. I agree that if it's not that, then the gaps will most likely not all be "aligned" properly and some values will fall between a <span class="math">\frac{m+1}{1994}[/spoiler] and then next <span class="math">\frac{m+1}{1993}[/spoiler].

>>4325493
Clues:
<span class="math">2n![/spoiler] means that you're looking at something that takes half the terms of an exponential integer series.
<span class="math">(-1)^nx^{2n}[/spoiler] confirms that.
Things that are half the terms of exponential (skipping each other term) are Re and Im of an imaginary exponential: cosines and sines.
Figure out which one it is, plug pi in it, win.

>>4325487
Ah, you beat me to it.

Bounding from the other side:
<span class="math">n\leq 1993+1994[/spoiler]
If <span class="math">n\geq 1993+1994[/spoiler], then we take <span class="math">k[/spoiler] = <span class="math">2m+1[/spoiler] and <div class="math">\frac{m}{1993}=\frac{2m}{1993+1993}< \frac{2m+1}{1993+1994}<\frac{2m+2}{1994+1994}=\frac{m+1}{1994}</div>

>>4323181
>I'm trying but it's too disorganized for me to follow ><
I can give you that.

>>4323176
We didn't use that one post of yours in the end, so it doesn't matter, does it? Or is the inequality from the previous posts wrong?

>>4323150
Hey there ponymath. Frustrated that you're a bit late today maybe? Wanna peer review the rest of the thread so that we know if we've done other mistakes?

>>4323128
Okay, cool!
So we have <span class="math">n < m(2+\sqrt{3}) < n + 1 \Rightarrow \lfloor(n+1)(2+\sqrt{3})\rfloor - \lfloor n(2 + \sqrt{3})\rfloor = 3[/spoiler]. Let's prove that it's enough for the reverse implication. Actually that's easier than what I had in mind (assuming I'm correct). The distance between "2"s is either 3 or 4 (easy to prove by induction). We already have a lot of "2"s indices, which are the <span class="math">n[/spoiler] so that <span class="math">\lfloor(n+1)(2+\sqrt{3})\rfloor - \lfloor n(2 + \sqrt{3})\rfloor = 3[/spoiler], and those are all the <span class="math">\lfloor m(2+\sqrt{3})\rfloor[/spoiler], by your previous post. We don't have enough margin between two of these to put another <span class="math">2[/spoiler].

(Right?)

Np. That's the kind of mistake that is easy to make: understanding how it works but going to fast on the simple parts because they are boring.

Anyway, I don't have any simple realistic approach here. We need to prove the equivalence between <span class="math">n < m(2+\sqrt{3}) < n + 1[/spoiler] and <span class="math">\lfloor(n+1)(2+\sqrt{3})\rfloor - \lfloor n(2 + \sqrt{3})\rfloor = 3[/spoiler]. The little gain I can think of is that proving either the direct or the reverse implication should be enough because we know how many <span class="math">n[/spoiler]'s we have that satisfy either of these equations, so I think any direction we prove is enough, if you see what I mean.

>>4323020
Oh but the distance should be 3, not 2.

Okay, okay.

So we're supposed to show that if <span class="math">n < m(2+\sqrt{3}) < n + 1[/spoiler], then <span class="math">\lfloor(n+1)(2+\sqrt{3})\rfloor - \lfloor n(2 + \sqrt{3})\rfloor = 2[/spoiler]. Actually that can't be right.
<span class="math">\lfloor(n+1)(2+\sqrt{3})\rfloor\geq (n+1)(2+\sqrt{3})-1[/spoiler]
<span class="math">\lfloor n(2 + \sqrt{3})\rfloor \leq n(2 + \sqrt{3})[/spoiler]
So the difference is at least <span class="math">(n+1)(2+\sqrt{3})-1 - n(2 + \sqrt{3})=1+\sqrt{3}>2[/spoiler].

>>4322821
Yeah but we're down to something algebraic that is equivalent to the statement we're trying to prove. That's basically the point at which I stop caring ^^'

>>4322793
Kinda cool. I didn't think it would be this quick. Nice job!

>>4322743
Oh okay. So now we have a real proof of what <span class="math">r[/spoiler] is, instead of a strong intuition. How do we go on from here?

>>4322720
Hi!
Hmm, <span class="math">r=2+\sqrt{3}[/spoiler] for sure. I don't have any idea about the perfect ratio though if that's what you're talking about.

>>4322431
Google knows 3.73205 as an approximation of the "sacred ratio" <span class="math">2+\sqrt{3}[/spoiler] (what the hell is the sacred ratio?). Now hoping that <span class="math">r=2+\sqrt{3}[/spoiler], even though that breaks my proof by periodicity, but because that sounds cute.

>>4322431
Oh and, the fact that it looked rational in the previous post was good because it would have meant the sequence is periodic, and proving that the property is true on a period would have been not only easy (by an algo) but enough to prove the whole result (by periodicity). So I'm back to square <span class="math">\approx 0[/spoiler].

>>4322330
Okay, so that post was stupid. First, I'm not talking about <span class="math">r[/spoiler] but about its inverse. And second, I thought my algo had converged but it hadn't, and the precision was bad. Better precision on <span class="math">r[/spoiler] is <span class="math">3.732051[/spoiler], which doesn't look like anything special to me.

>>4322388
The formulation is a bit weird. Basically, you take the first sequence. It starts with a 2. That means the 2nd sequence starts with 2332 (the space between the 2s is "2"). Then there's a 3. That means that after 2332, there are 3 3s before the next 2. So the 2nd sequence is 23323332. Then again a 3, so the sequence is 233233323332. And the sequence defines itself like that. You read it and while reading it, you add stuff at the end.

>("Too-lazy-to-think" approach, starting now)
Fuck, now this doesn't work anymore. Coding is done, thinking is required. Help me, /sci/!