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/sci/ - Science & Math

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>> No.4415016 [View]

>>4414996
If you really attempt the Putnam problems like you say you do, you know that it was the basic way to go. There are problems in which the "simple" approach does not work, and in these, it could have been suspicious. Here, the candidate is supposed to go that way first, and to notice that it works. There's no reason to look anywhere else. It's like, you're asked to prove that a series of positive reals diverges, you notice that the term doesn't go to 0, why bother proving something more complicated?
But ok, independently from the "did he cheat? did he not cheat" discussion which isn't really more interesting than the science vs religion threads, I get your point and I'll try to be less verbose in my proofs / proof attempts.

>> No.4414990 [View]

>>4414979
Okay. I'll try to make shorter posts and to proof-read them more. Do you consider >>4414610 too long though? I mean... I can skip a few steps, but then it's not as easy to read anymore.

>> No.4414967 [View]

>>4414955
>Especially you post stuff that is wrong and correct it a hundred times by subequent posts.
So it's bad when I post something that is wrong? Then why:
>>4414697
>I simply wait for other anons to try, so I can help them and assist their steps towards it.
Or should I let other anons correct it, because if I correct it myself, it's too much of me and it makes people think I'm megalomaniac?

>> No.4414928 [View]

>>4414901
>1. spam huge amount of latex and expect others to read
How is it not easier to read than the mathematical posts NOT done in LaTeX? I mean, seriously? I don't spam huge LaTeX formulas when there's no need for formulas to solve a problem. And I don't see you going on a crusade against anons or other tripfags using big formulas, when they do (and they sometimes do). I really, really don't think there's a problem with that.
>2. post whole solutions to show off
That's judging on mere intent... I'm thrilled when I'm able to solve a problem so maybe I'm showing off to myself. But "TN5" is a name I only have on /sci/ (since it's derived from the tripcode itself). It's not like I can brag about it or anything. If I wanted to brag, show off, or to build a reputation, I would post as TN5 in other threads too, not as anon.

>> No.4414879 [View]

>>4414846
Well hopefully WE won't lead any thread into derailment anymore if we end up with a way of doing things that satisfies everybody. You're saying that being able to identify a poster:
- Doesn't make the thread easier to read. I disagree: as soon as there are 4 or 5 anons posting, while the mathematical truth of a post is what matters, knowing who posted it helps understanding it faster, since you understand better what it refers to, what notations are used etc,
- Actually makes the thread worse. I can agree that identifiable tripfags from other threads could maybe annoy some. I'm not sure why exactly, but okay. But if the IDs are within a thread? I mean, consider that /b/'s ID system comes to /sci/ right now. Is that a bad thing? I don't think so. Why would it be bad? And if you agree it wouldn't be bad, then why is it so different to have some people with a random ID and some without?

>> No.4414834 [View]

>>4414818
I am saddened by the fact that you think that I think otherwise. I'm not claiming "first" or anything. And, again, I just like the fact that, at least within a discussion, people know which posts are from the same poster. Do you really think that having 3 or 4 people, among the anonymous posters, use a different ID in each putnam thread, would actually make the threads worse than pure anonymity? Maybe I'm not seeing it through, but I really don't think that would hurt. And I would gladly switch to this instead of the "named"-tripfagging.

>> No.4414805 [View]

>>4414784
>4chan's most important concept is the idea of anonymity.
Being able to be anonymous is the main idea indeed. I still don't think that choosing occasionally to be named changes that. I mean, again, the ID system on /b/ does exactly what I'm trying to do here with my trip. I could use something more like an ID by just removing the name and switching trip in every new putnam thread. Assuming the other current tripfag also do it, that would make it kind of anonymous. I wouldn't mind that.

>> No.4414755 [View]

>>4414743
I don't really understand why that would happen. Honestly what would be best is the ID system that is now on /b/, for instance. I haven't really played with it much but it does what I'd want: being able to recognize who's who within a thread. I don't really care about reputation from a thread to another. It's just helpful for the discussion to know who's who when you are working collaboratively.

>> No.4414733 [View]

>>4414697
>I simply wait for other anons to try, so I can help them and assist their steps towards it.
That's nice. I hardly ever see anon taking a work in progress from someone else and adding/correcting it, though. Usually people post either:
- crap,
- a whole solution,
- their first step toward a solution,
but it's almost always independent from what the others did. The only cool discussions about how to tackle a problem that I remember were with mathfag, not with anon (though anon has proved able to do well on Putnam problems more than once).

>>4414704
>1. drop the trip
I only use the trip in the putnam threads or similar threads, I even delete and repost without the trip when I contribute somewhere else and forget to have it. But I like to have it here.
>2. don't cheat with google
Well I don't, but it's not like there's any way it can be proven (in either direction).
>3. don't post full solutions, post hints instead
It's actually a good idea. I'll do that from now on (at least when I convinced myself that I have the solution, because last time I wrote some kind of "hint", it was actually false...).

>> No.4414666 [View]

>>4414630
Wait... because this is a relatively simple Putnam problem which HAS to be attacked from this angle (since neither the perimeter of an ellipse nor the length of the curve of the sine can be simplified properly, and the question is to make them equal), it means that whoever finds the solution googled it? I am pretty sure no one will come up with a solution that doesn't involve writing the two lengths as integrals and matching them without actually computing them.

...when I post my reasoning bit by bit while thinking, I'm spamming and boring, and when I post my solution after properly polishing it for a while, I'm a cheater... People on /sci/ aren't allowed to just be able to solve (some) Putnam problems or what?

>> No.4414628 [View]

>>4414610
Also the condition I end up with is <span class="math">b^2=a^2+c^2[/spoiler]. Cooler in this form.

>> No.4414610 [View]

The length of the curve of <span class="math">y=c\sin(x/a)[/spoiler] over a period is the integral over a period of <span class="math"> \sqrt{
\mathrm{d}x^2 + \mathrm{d}y^2} = \left( \sqrt{1^2+(\mathrm{d}y/
\mathrm{d}x)^2} \right) \mathrm{d}x[/spoiler], so it's
<div class="math">L= \int_{0}^{2\pi a} \sqrt{1+ (c/a)^2\cos^2 (x/a)} \mathrm{d}x = \int_{0}^{2\pi a} \sqrt{1+ (c/a)^2\cos^2 (x/a)} \mathrm{d}x = a\int_{0}^{2\pi} \sqrt{1+ (c/a)^2\cos^2 (u)} \mathrm{d}u</div>
(with <span class="math">u=x/a[/spoiler] and <span class="math">\mathrm{d}x = a\mathrm{d}u[/spoiler])

On the other hand, the perimeter of the ellipse, approached with the same method, is
<div class="math">P= \int_{\theta=0 }^{2\pi} \sqrt{\mathrm{d}x^2+ \mathrm{d}y^2}=\int_0^{2\pi} \sqrt{ a^2\sin^2(\theta) + b^2\cos^2(\theta)}\mathrm{d} \theta =\int_0^{2\pi} \sqrt{ a^2 \left(\sin^2(\theta) + \cos^2(\theta) \right) + (b^2-a^2)\cos^2(\theta)}\mathrm{d} \theta </div> <div class="math">=\int_0^{2\pi} \sqrt{ a^2 + (b^2-a^2)\cos^2(\theta)}\mathrm{d} \theta =a \int_0^{2\pi} \sqrt{1 + ((b/a)^2-1)\cos^2(\theta)}\mathrm{d} \theta</div>

So <span class="math">(b/a)^2-1=(c/a)^2[/spoiler] gives <span class="math">L=P[/spoiler]. I don't know if <span class="math">L=P[/spoiler] is enough or if there is another condition so that a part of the ellipse doesn't "cross" the curve of the sine. I don't know if there are other ways for the two integrals to be equal (well, I don't think so, but I don't know how to prove it). Also, I suck at manipulating integrals, so I'm likely to have done a mistake or two on the way: hopefully the method is still correct.

>> No.4412497 [View]

>>4411993
This one is harder than most. Putnam competitions are made of two times 6 problems of increasing difficulties, this one is the sixth of the first part of the 1995 competition. /sci/'s success rate on the last problems is sensibly lower than on the first ones.

>> No.4411172 [View]

If <span class="math">\pi[/spoiler] contains a part <span class="math">\{a,b,c,d,...\}[/spoiler] of cardinality <span class="math">>=4[/spoiler], then consider the size of the parts <span class="math">p_a,p_b,p_c,p_d[/spoiler] of <span class="math">\pi'[/spoiler] that contain respectively <span class="math">a,b,c,d[/spoiler]. If two of these parts have the same size, we're done. Otherwise, their sizes are different, which is impossible because it would mean that they are of sizes at least 1,2,3 and 4, which totals to 10>9.

Thus let's check what happens when <span class="math">\pi[/spoiler] and <span class="math">\pi'[/spoiler] only contains parts of cardinality at most <span class="math">3[/spoiler]. Consider the function <span class="math">x\mapsto f(x)=(\pi(x),\pi'(x))[/spoiler]. This function has 9 possible outputs:
(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3
,2),(3,3)
If there is <span class="math">x\neq y[/spoiler] such that <span class="math">f(x)=f(y)[/spoiler], then we're done. Otherwise, each of the 9 values of the domain is mapped to a different value in the range of <span class="math">f[/spoiler]: the whole range is reached. That would mean that there are exactly <span class="math">3[/spoiler] elements that are in a parts of size 2 in <span class="math">\pi[/spoiler], which is impossible (that number has to be even).

Probably not the best way to do it, but whatever...

>> No.4410895 [View]

>>4410886
Just ignore him. He doesn't like me and should ignore my posts, he chooses not to, that forces the rest of us to be the mature ones and ignore his. Not a big deal. At least every single of my posts is acknowledged, which is kinda cool.

>> No.4410824 [View]

>>4410517
Yes indeed, and it is wrong. My reasoning was intuitively "We have a random process that adds columns. This process is memoryless and has all the "good" properties that we want on a Markov chain, in particular irreductibility and ergodicity, so the distribution probabilities that we are on a given state converge". If that was true, then my claim would have been correct, because the limit distribution would satisfy that <span class="math">Pr(a'+2=b'+1=c') = 6\cdot Pr(a'=b'=c')[/spoiler]. However, this limit distribution is only guaranteed to exist when the number of states is finite, and we have a countably many states here (for instance, a' can be as small as we want).

I tried thinking of how the number of states could be made finite, maybe by setting a maximum distance from the equilibrium after which we don't move anymore, but I can't get a proof out of it.
Also, by trying to manipulate the probabilities directly and assuming that the statement of the problem is false, I end up with a strong constraint on the probability <span class="math">y_n[/spoiler] that we are, after <span class="math">n[/spoiler] columns, on the state <span class="math">a'+2=b'+1=c'[/spoiler], that is, <span class="math">y_{n+1}\in \{7y_n/12,8y_n/12\}[/spoiler]. Too bad, that was almost enough for a contradiction. The fact that I obtain this by plugging "4" makes me think that 4 is a clever choice and that 5 or 6 wouldn't have worked. But I don't know why.

TL;DR: my proof above doesn't work, and I don't know how to make it work, sorry.

>> No.4408337 [View]

Let's consider <span class="math">a'=a-2n[/spoiler], <span class="math">b'=b-2n[/spoiler] and <span class="math">c'=c-2n[/spoiler]. They can be obtained by asking the <span class="math">n[/spoiler] people to write <span class="math">-1[/spoiler], <span class="math">0[/spoiler] and <span class="math">1[/spoiler] instead of <span class="math">1[/spoiler], <span class="math">2[/spoiler] and <span class="math">3[/spoiler], and following the same rules.

Let's check what can happen to these lines when a new column is written.
The only way to end up with <span class="math">a'=b'=c'[/spoiler] is if we had <span class="math">a'+2=b'+1=c'[/spoiler] before writing the new column, and we luckily got <span class="math">+1[/spoiler] on <span class="math">a'[/spoiler], <span class="math">0[/spoiler] on <span class="math">b'[/spoiler] and <span class="math">-1[/spoiler] on <span class="math">c'[/spoiler] (which happens with probability 1/6).
However, ending up with <span class="math">a'+2=b'+1=c'[/spoiler] happens with probability <span class="math">1[/spoiler] if we had <span class="math">a'=b'=c'[/spoiler], and we can also get there from other states of the system.

The probability to stop with state <span class="math">a'=b'=c'[/spoiler] is therefore asymptotically at least <span class="math">6[/spoiler] times lesser than the probability to stop with state <span class="math">a'+2=b'+1=c'[/spoiler], so we can find <span class="math">n[/spoiler] as large as we want such that the ratio is at least <span class="math">4[/spoiler].

Does this make sense?

>> No.4405321 [View]

>>4405223
Ah, thank you! I was pretty sure it was possible to figure out something.

>> No.4405317 [DELETED]  [View]

>>4405223
Ah, thank you! I was pretty sure it was possible to figure out.

>> No.4405176 [View]

>>4405168
Ah my bad, indeed... Let me try to see if there's another trivial counter-example when we don't assume the functions go to 0.

>> No.4405162 [View]

>>4404292
>I don't know what the going to zero thing is all about
It's about preventing you from using this counter-proof:
Take n=2,
Take <span class="math">x_1=t\mapsto 1[/spoiler], and <span class="math">x_2=t\mapsto t[/spoiler]
<span class="math">x_1'(t)=0= 0\cdot x_1(t) + 0\cdot x_2(t)[/spoiler]
<span class="math">x_2'(t)=1= 1\cdot x_1(t) + 0\cdot x_2(t)[/spoiler]
but <span class="math">x_1[/spoiler] and <span class="math">x_2[/spoiler] are not linearly dependent.

This kind of simple trick can't be done if the functions all go to <span class="math">0[/spoiler].

>> No.4403217 [View]

>>4402671
But you assume that there is a <span class="math">k[/spoiler] for which for each possible cut, the sum of the first <span class="math">k[/spoiler] beads is above <span class="math">k-1[/spoiler]. There is no guarantee that it has to be the same <span class="math">k[/spoiler] for each cut that would make the property fail.

>> No.4401189 [View]

>>4401136
Okay, my bad, that's why I didn't get what you mean when you talked about the forbidden change between the first and last of the chain.

>> No.4401113 [View]

>>4401086
But you aren't allowed to switch two beads. The order must be preserved, only the place at which you cut can change.

My proof says that if you consider, among the n possible cuts, the cut that maximizes the number of beads that you can add together until you finally break the <span class="math">\sum_{i=1}^k x_i\leq k-1[/spoiler] property, then you could have cut one bead further left instead unless that new first bead is greater than 1. If you had cut one bead further left, you would have had a cut that has an even higher number of beads that you can add together until you break the property, which is a contradiction. So that bead is greater than 1. Now if you consider what happens when you cut two beads further left, you'll see that the condition becomes that their sum is greater than 2. Etc. And if you add all of these together, the sum of all the beads will be higher than n+1.

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