/sci/ - Science & Math

>>4339262
No I'm talking about the 4 triangles around the carpet. I'm too lazy to draw that, be <span class="math">c[/spoiler] is the side of the carpet.

>>4339260
Actually I'm missing something, am I not?

I'll go with the bet that this isn't homework (could be some extra credit thing, though, I guess).

It's probably not the best way to solve this, but I think you can do something by noticing that the 4 right triangles are similar triangles. If you write <span class="math">a[/spoiler] the horizontal side of the top left triangle, <span class="math">b[/spoiler] its vertical side and <span class="math">c[/spoiler] its hypotenuse, and <span class="math">a'[/spoiler], <span class="math">b'[/spoiler] and <span class="math">c'[/spoiler] the matching sides on one of the small triangles, then you have:
<span class="math">a+b'=5[/spoiler]
<span class="math">b'+a=13[/spoiler]
<span class="math">c'=1[/spoiler]
<span class="math">\frac{a'}{a}=\frac{b'}{b}=\frac{c'}{c}[/spoiler]

You're looking for <span class="math">c[/spoiler], and the remainder is boring algebra.

>>4339219
Now you're going to have to figure out a way to convert this guy to deep sea exploration.

Several friends of mine have worked in thing that maybe you call computational biology, one in his master thesis and another in his PhD thesis. The master thesis was about DNA sequencing: you probably know the domain better than I, so I won't try to explain anything, but it's quite cool.
The PhD thesis was about modeling, simulating and verifying biological regulation networks (just roughly copying the title here), for instance, using stochastic petri nets and temporal automata asa models for protein interactions in cells and validating with statical, dynamical, and quantitative dynamical analysis.
I'm not sure if the biological part is big in these examples though, the two guys are both from a pure math/CS background.

And about "MadScientist", http://archive.installgentoo.net/sci/?task=search2&ghost=&search_text=&search_username=m
adscientist&search_tripcode=!!Q11PG81nz2n&search_media_hash=&search_del=dontcare&sea
rch_int=dontcare&search_ord=new

(sorry about the long link)

>>4338877
Hey, looks like you got yourself a tripcode :)
Also good job with the proof, which is clearly beyond my limited abilities in analysis.

>>4332945
Anyway, when <span class="math">z>1[/spoiler], the upper bound of the integral can be reduced to <span class="math">1/z[/spoiler] (otherwise <span class="math">p_Z(z)=0[/spoiler]):
<div class="math">p_Z(z)=\int_0^{1/z} y \mathrm{d}y=\frac{1}{2z^2}</div>

Actually finding the PDF was not too hard.

>>4332944
Thanks! I feel a bit dumb...

>>4332926
>non-measurable
I meant "zero measure".

First of all, let's put some bounds on the integral. We want <span class="math">p_Y(y)\neq 0[/spoiler] so we need <span class="math">y\in(0,1)[/spoiler].

So for <span class="math">0\leq z\leq 1[/spoiler], <span class="math">p_Z(z)=\int_0^1 yP_X(yz) \mathrm{d}y[/spoiler]
And because <span class="math">z\in(0,1)[/spoiler], the fact that <span class="math">y\in(0,1)[/spoiler] also gives <span class="math">yz\in(0,1)[/spoiler], so <span class="math">p_Z(z)=\int_0^1 y \mathrm{d}y = 1/2[/spoiler].

Also:
>>4332936
I guess I could try to go more to the point.

>>4332929
>>4332933
>>4332939
Cool, I have a fan club. Do you read the Putnam thread because you want to participate (in which case, by all means, do it, and constructively tell me if I'm annoying, I listen to intelligible criticism), or just to spit on me (in which case, well do it anyway, it's not like a I care).

So let's try to show that the PDF is what I wrote in the first answer.

We are writing <span class="math">z=x/y[/spoiler]. Now we want the PDF <span class="math">p_Z(z)[/spoiler] at a given point <span class="math">z[/spoiler]. To do that, we write <span class="math">x[/spoiler] in terms of <span class="math">y[/spoiler] knowing <span class="math">z[/spoiler] (I guess we could write <span class="math">y[/spoiler] in terms of <span class="math">x[/spoiler] too but there would be a quotient, let's work with a product instead). <span class="math">x=yz[/spoiler]. So now, we use the PDFs of <span class="math">X[/spoiler] and <span class="math">Y[/spoiler]:
<span class="math">p_Z(z)=\int yP_X(yz)P_Y(y) \mathrm{d}y[/spoiler]

For <span class="math">z<0[/spoiler], either <span class="math">y>0[/spoiler] and then <span class="math">yz<0[/spoiler], <span class="math">P_X(yz)=0[/spoiler], or <span class="math">y<0[/spoiler] and then <span class="math">P(y)=0[/spoiler], so anyway the product of the two is <span class="math">0[/spoiler], and so is <span class="math">p_Z(z)[/spoiler]. Also, f.u., non-measurable cases.

From there yeah, the fact that I spoiled two cases will probably help a bit so do the rest, my bad... (but it shouldn't have been hard anyway).

Also it looks like I've screwed something up about the rest of the work, because the result I get in the end is actually greater than one...

>>4332883
Okay. Sorry about spoiling the PDF then. We still don't have the proof about that the PDF is correct though, so even if I gave the answer, we can still work on the proof.

So anyway, I've "cheated" again now since I used wolfram to compute the integral because I'm too lazy to do the simplifications by myself:
<div class="math">\int_{\frac{2k +1}{2}}^{\frac{2k +3}{2}}\frac{ \mathrm{d}z}{2z^2}= \frac{2}{4k^2+8k+3}</div>

So what we want now is
<div class="math">\frac{1}{4} + \sum_{k=1}^{\infty} \frac{2}{4k^2+8k+3}</div>

>>4332868
I'm attempting. I read the problem, looked at wikipedia to know if the distribution for what I call <span class="math">z[/spoiler] had a name and a simple PDF, and posted it right after. I took maybe a headstart of 3 minutes compared to /sci/, and used a wikipedia page. That's hardly gamebreaking.

For <span class="math">k>1[/spoiler], let we consider
<div class="math">\int_{\frac{2k +1}{2}}^{\frac{2k +3}{2}}\frac{ \mathrm{d}z}{2z^2}</div>
The sum of these for all <span class="math">k>1[/spoiler] will be the contribution of <span class="math">(3/2,5/2)\cup(7/2,9/2)\cup \dots\cup(2k+1,2k+3) \cup\dots[/spoiler] in the probability we're trying to find. The contribution of <span class="math">(0,1/2)[/spoiler] is easier since the PDF is constant equal to <span class="math">1/2[/spoiler] there (so the integral is <span class="math">1/4[/spoiler], which is the probability to obtain <span class="math">0[/spoiler] as the closest integer to <span class="math">x/y[/spoiler]).

So I actually cheated and gave a look at it while posting it. The PDF of the <span class="math">Z=X/Y[/spoiler] is given by what's called a ratio distribution and is (c.f. http://en.wikipedia.org/wiki/Ratio_distribution#Uniform_ratio_distribution):
<span class="math">1/2[/spoiler] for <span class="math">0<z<1[/spoiler]
<span class="math">\frac{1}{2z^2}[/spoiler] for <span class="math">z\geq 1[/spoiler]
<span class="math">0[/spoiler] otherwise.

From there, we write the probabilities for even and odd values of <span class="math">\lfloor z\rfloor[/spoiler] (where I use the <span class="math">\lfloor \rfloor[/spoiler] for the closest integer and not the floor) as sums over <span class="math">(0,1/2)\cup (3/2,5/2)\cup (7/2,9/2)\cup \dots[/spoiler] and its complement in <span class="math">\mathbb{R}^+[/spoiler]. Hopefully the simplifications aren't too complicated.

>>4330705
What pisses me off a bit is that if I had taken 10 minutes to code a search, it would have found the answer pretty quickly. I was too convinced we had to prove no such sequence existed.

>>4330694
Well played.

>>4330474
<span class="math">b_ka_k^2 = b_ma_m^2[/spoiler]
I'm with you on that (and the rest of your work), but it requires the control on the <span class="math">b_k[/spoiler] and <span class="math">b_m[/spoiler] if we want to use it because if not, it's pretty uninformative.

>>4330452
How? The sum of a divisors of a number can easily be strictly above that number. So you could have a number <span class="math">n[/spoiler] so that the sum of a strict subset of its divisors is equal to the <span class="math">n[/spoiler] (still such that the divisors in the subset are by pairs that multiply to <span class="math">n[/spoiler]). That's what two posts already explained.

The question is, is it possible that something like that happens?

>>4330392
Okay, I should avoid the stickies when I've drunk. I've thought about it a little bit and nothing trivial comes to my mind. Assuming that the problems are indeed equivalent (which would look good), proving it is a problem as well (certainly simpler than the open problem of finding an odd perfect number, though). Assuming they aren't equivalent, well we can't even really get a counter example for equivalence until we answer the problem (constructively), so...

>>4330341
>I'm not sure I understand exactly why the first assumption is true though (about the fact that the highest denominator is a perfect number).
Okay never mind that, I understand, I didn't read the post I quoted properly. The equivalence of the two problems is now pretty clear.

Here was last time's thread:
http://archive.installgentoo.net/sci/thread/4289194

Here's the interesting post:
It's highly unlikely such a series of fractions exist because the smallest fraction in that series would be the reciprocal of an odd perfect number, which are believed not to exist.

An odd perfect number's divisors including 1 would sum to itself.

<span class="math">d_0 + d_1 + d_2 + d_3 + ... + d_{n - 1} = d_n ~with \; d_k\cdot d_{n-k} = d_n[/spoiler]
<span class="math">d_0 + d_1 + d_2 + d_3 + ... + d_{n} = 2d_n[/spoiler]
<span class="math">\frac{d_0 + d_1 + d_2 + d_3 + ... + d_{n}}{d_n} = 2[/spoiler]
<span class="math">\frac{1}{d_n} + \frac{1}{d_{n-1}} + \frac{1}{d_{n-2}} + ... + \frac{1}{d_2} + \frac{1}{d_1} + 1 = 2[/spoiler]
<span class="math">\frac{1}{d_n} + \frac{1}{d_{n-1}} + \frac{1}{d_{n-2}} + ... + \frac{1}{d_2} + \frac{1}{d_1} = 1[/spoiler]

I'm not sure I understand exactly why the first assumption is true though (about the fact that the highest denominator is a perfect number).

>>4330251
I think we've already seen this one in a different thread this week. OP forgot to mention "finite" so people got a series, then OP forgot to mention >1 so people answered "1/1". I can't remember if /sci/ eventually got the answer, so I'll think about it again. But I guess that's still googleable easily, though.