/sci/ - Science & Math

Notice that we skipped yesterday's problem for some reason.

This one is easy (the easiest I've seen, I think), so several people should be able to solve it fast. Maybe we can discuss extensions once we have a few proofs?

Let a_n = 0 for all n. Then (a_n) has the properties mentioned above and sum(a_n) converges.

Problem?

>>4346634
"Positive" here means <span class="math">>0[/spoiler], not <span class="math">\geq 0[/spoiler].

By the way, my proof is simple: just define <span class="math">b_n=\sum_{k=1}^{2^n}a_n[/spoiler], check that <span class="math">b_{n+1}\geq 2b_n[/spoiler] and <span class="math">b_1>0[/spoiler], and notice that <span class="math">b_n[/spoiler] diverges. Done.

I tried thinking of nice extensions but the property is actually quite tight, since you can go to divergences as slow as the harmonic series. I'm not sure if we can find a nice exercise as an extension for this...

Let me know if this works. Let's divide each term by a_1. So, a_1 = 1. Now a_2 + a_3 >= 1. And so a_4 + a_5 + a_6 + a_7 >= 1. And so on. 1 + 1 + 1 + ... diverges.

>>4346651
Yeah, that works for me. Different wording but same idea as in >>4346652 (and I don't think there's a simpler idea).

Aren't these all supposed to be difficult? Some of them are god tier, but this is maybe a first year problem at best.

>>4346680
Yeah this one is easy. Basically the exams have problems of different (increasing?) difficulties, and this one is the first of an exam.

Using the integral test it's quite easy to see that it's covering an infinite area, thus it diverges.

>>4346620

These are really fancy terms to describe something that's really obvious.

>>4347158
I agree that the divergence is rather obvious. I think the statement of the problem is good though. It's clear and concise.

>>4346651
>>4346652
neat proof
i was trying to go for something in the lines that this series is at the very best equivalent to the harmonic series in terms of convergence. something like a_n --> 0 so approximately a_2n + a_2n+1 < 2*a_2n and then a_n < 2*a_2n, which does have a striking resemblance to something like a_n >= 1/n

Suppose the sum converges to <span class="math">L[/spoiler].
<div class="math">L =\sum_{n=1}^\infty a_n\leq \sum_{n=1}^\infty ( a_{2n}+a_{2n+1} )= \sum_{\mathrm{even}\; k\geq 2}a_k +\sum_{\mathrm{odd}\; k\geq 3}a_k = \sum_{k=2}^\infty a_k = L - a_1.</div> This can hold only if <span class="math">a_1 = 0[/spoiler], which contradicts the condition that <span class="math">(a_n)[/spoiler] be a positive sequence.

This is just about the easiest Putnam problem I've ever seen. The first thing I would think to try works, and does so quickly.

I think >>4346651 is a prettier proof since it's constructive yet still brief.

>>4346680
The exam consists of 12 questions and is taken over two three-hour sessions, separated by an hour-long lunch break. The first half consists of problems A1-A6; the second, B1-B6. Each half is designed to present problems in increasing order of difficulty: A1 and B1 are supposed to be the easiest problems, followed by A2 and B2, etc. Sometimes the problem writers don't get it right: in a given year, more participants might solve B4 than B2, say. But generally, the rule is pretty sound. This problem was an A1.

>>4347210
i like your proof more, actually

>>4347210
Yeah maybe I should have posted the B6 from yesterday instead of this one. It was something a bit related to the cutting sticks problems. Didn't really think about it but it was more interesting for sure.

>>4347242
No, you shouldn't have posted anything. You should refrain from posting and stop shitting up threads. I am glad that today's problem was solved before your faggotry entered the thread.

>>4347335
I'm the one who posted the problem, and who gave >>4346652
Strange how you liked my post better when I posted as anonymous.

First noticed this >>4346651. Proved it using induction, but the approach in this thread is much cleaner.

>>4347394
Yes, I liked it more when it isn't obvious that it comes from an arrogant aspie.
Also this time you managed not to spam dozens of posts full of latex.

This one was pretty simple, but it made a pretty interesting point. It's not always immediately easy to tell if a decreasing series diverges, but if you can combine terms together to make a series where every term is greater than the term after it, then series obviously diverges. I think this is a powerful lesson that can be used for proving the divergence of much trickier sums.
Also I think this generalization can be proved in a very similar way:
Let <span class="math"> a_n [/spoiler] be a sequence, such that for some positive integer, <span class="math"> k \ge 2 [/spoiler]: <div class="math"> a_n \le \sum_{i = 0}^{k - 1} a_{k n + i} </div> This series diverges.
I wonder what other generalizations exist. I wonder if there are any cases, where you can make a general statement that a certain term will be less than or equal to the sum of a group of terms later.

BAN MY SMELLY GRANDMA FUTA BALLS ON YOUR MOMS POOPY CUNT

This is now a dubs thred.
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Litrollu

>>4346620
cant this one be proved by just using the definition of series divergence?

wikipedia definition:
>a divergent series is an infinite series that is not convergent, meaning that the infinite sequence of the partial sums of the series does not have a limit.

since we are adding numbers together which are increasing, this falls under the definition of series divergence right away, no?

and since the question explicitly states that the sequence (an) is increasing, we don't even have to prove that.

>> NIGGERDs SUCK! Anonymous 02/09/12(Thu)22:35 No.4348740
BAN MY SMELLY GRANDMA FUTA BALLS ON YOUR MOMS POOPY CUNT

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Litrollu

>>4348754
That proof is inadequate.

>>4348773
Anonymous 02/09/12(Thu)22:41 No.4348754
>> NIGGERDs SUCK! Anonymous 02/09/12(Thu)22:35 No.4348740
BAN MY SMELLY GRANDMA FUTA BALLS ON YOUR MOMS POOPY CUNT

This is now a dubs thred.
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Litrollu

>>4348745
The sequence of a_n is not necessarily increasing.

>> Anonymous 02/09/12(Thu)22:45 No.4348770
I JUST WANNA SMELL UR GRANDMAS SMELLY FILTHY CHEESE LADEN CUNT ON MY 20 INCH MORNING SMELLY FUCKING WOOD
I SMELL LIKE MATHS BUTTHOLE POOPING OUT DICKS ON MY BIG FAT HAIRY CUNT

>>4348792
>>I SMELL LIKE MATHS
gentlemen, we must stop everything. Math has a smell.

Math. Has. a. Smell.

This could clearly revolutionize all fields of mathematics.

SO WHAT MUSTY BUTT??????

CHECKUM

I have thought of a simpler proof, tell me if it works.

Think of the proof that the sum of the inverses of integers diverges: you can group them in infinitely many groups that are >= 1/2 and you therefore show that that the series is >= the sum of infinitely many 1/2.

Here you can pair every element a_2n with the a_(2n+1) and replace it with a_n,

a_1 + (a_2 + a_3) + (a_4 + a_5) + (a_6 + a_7) ...
>=
a_1 + a_1 + a_2 + a_3 ...

So you get a_1 plus the original series, so you can repeat the same grouping and replacing with this one.

a_1 + a_1 + (a_2 + a_3)
>=
a_1 + a_1 + a_1

So we showed that the series is >= (a_1 times infinity).

Since a_1 is positive, we showed that the series diverges.

>>4347722
>Yes, I liked it more when it isn't obvious that it comes from an arrogant aspie.
Arrogant? Am I looking down on anyone? Am I unable to admit when I'm wrong? (Those aren't rhetorical questions)
Maybe I'm annoying you, maybe I write too much, but please, any time you consider something really arrogant in one of my posts, let me know, because I don't think it happens but if it does, I want it fixed.

>that feel when you'll never be able to pin TN5 !/sci/TN5 against a blackboard while he solves the putnam problem of the day and you make sweet love to his ass

>>4349303
3 inch dick + 15 inch deep dorito-fed ass = no buttsex for you

bump

>>4349475
>bumping a sticky

>>4349294
Please stop posting.

Nightmare mode: switch greater than epual to equal and prove it exists, and define its closed form.

>>4349578
>sageing a sticky

\[ \sum{n}{i=0}{2i/n} \]

>>4347210
This was the first thing that came to mind.

>>4346620
>>>/lit/2395708

Ratio test.
Troll harder.

>>4351746
HOLY SHIT, HE'S RIGHT! THE ANSWER WAS "RATIO TEST" ALL ALOING

wtf were you other retards doing anyway?

>>4351811
>>4351746

ok

This was almost shockingly easy for Putnam, I thought of the same thing as >>4346651 almost directly.

>>4352244
1) Why does every problem in these threads seem to be ridiculed as "shockingly easy" by someone who doesn't actually try to answer it?

2) The actual test is TIMED, and even if it only takes you 15 minutes to stumble on to the correct idea (for the easiest problem on the test) and post some vague gesture at a method on 4chan, that's not the same as writing up a full proof by hand (maybe twice if you want to do a rough draft first so you don't risk screwing up your answer form).

>>4352263
This one actually WAS shockingly easy for a putnam problem. Get your panties out of a bunch.

Yo putnam I'm really happy for you but prove me wrong

5 day ban, please, at the very least. I need to study instead of browsing /sci/ and /b/.

I know we have a mod somewhere...

>>4352367
We do, but he's usually too busy deleting all the interesting threads and banning everyone for ridiculous crap.

this is my first time on this board and shit you guys are smart

>>4352614
some of them are

but for every poster that knows anything about math or science there are a dozen kids whose only knowledge on these subjects comes from wikipeida articles far outside of their level of comprehension

Could someone explain to me the Sigma notation being used here?

Correct, I am not very far along in my math career.

>>4353834
sigma just means the sum. For example
<span class="math">\sum_{n=1}^{\infty}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...[/spoiler]

Now if a sequence diverges, like this one and the one in the OP. There will be no number for which, if you added enough terms of the sequence, you could not exceed.

hey mathfags, i'm just a babby but how much math do you need to know to solve these putnam problems? do you need to have taken a class in real analysis or something advanced like that? thanks.

>>4355316
Technically, no. But taking them exposes you to "basic" relations and introduces you to the ever elusive "mathematical maturity."

Real Analysis is basically Math 101.