Mathematician Chat V.floor(π)

[JohnnyMo]

[QUOTE=Joey90;44245072]We know it converges, so it's relatively safe* to mess around with the terms as much as we like.

We want to evaluate:

[; \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + ...;]

Now we can split this up into a sum of sums:

[; \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ...;]

[; + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ...;]

[; + \frac{1}{8} + \frac{1}{16} + ...;]

[;+ \frac{1}{16} + ...;]

[; + ...;]

Then each of these is easily summable to leave

[; 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ...;]

Which you can then sum again to get your final answer.

*never say this[/QUOTE]

Tricky. Thanks. I don't think I've ever had to use rearrangement to solve a sum before, but it's been years since Calc II so how would I even remember?

Shouldn't rearrangement be totally fine since we know it's absolutely convergent?

[Krinkels]

[QUOTE=JohnnyMo1;44245994]Tricky. Thanks. I don't think I've ever had to use rearrangement to solve a sum before, but it's been years since Calc II so how would I even remember?

Shouldn't rearrangement be totally fine since we know it's absolutely convergent?[/QUOTE]

Rearrangement is fine if the function which sends n to the nth index in the rearranged sum is bijective.
In this case it isn't.

[Joey90]

[QUOTE=JohnnyMo1;44245994]Tricky. Thanks. I don't think I've ever had to use rearrangement to solve a sum before, but it's been years since Calc II so how would I even remember?

Shouldn't rearrangement be totally fine since we know it's absolutely convergent?[/QUOTE]
Yeah. Since it's absolutely convergent we can rearrange it, but more importantly we can add or subtract [i]other[/i] convergent series term by term. (This is essentially what I'm doing by breaking up the series)

To add/subtract an infinite number of them requires a bit more care, but since their sum converges as well, it's fine. (By subtracting an arbitrarily large number of series we can get as close as we like to the result)

Generally if everything you're working with is nice and convergent, it'll all 'just work', but I get all paranoid messing around with series, since it's easy to get it wrong without realising :v:

-

If you write it out with summation signs it looks a bit more 'mathematical', but in this case, I think it's clearer to do by showing explicit terms.

[; \sum_{n=1}^{\infty} \frac{n}{2^n} = \sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{1}{2^n} = \sum_{k=1}^{\infty}\sum_{n=k}^{\infty} \frac{1}{2^n} = \sum_{k=1}^{\infty} \frac{1}{2^{k-1}} = 2;]

(The tricky step is swapping the order of summation.)

[mochisushi]

Alright, so it isn't in OEIS, but it is generated from a relatively simple process. I've derived the first 12 numbers in the sequence: 2,2,4,4,8,7,5,4,6,6,8,7.

[Joey90]

[QUOTE=mochisushi;44251776]Alright, so it isn't in OEIS, but it is generated from a relatively simple process. I've derived the first 12 numbers in the sequence: 2,2,4,4,8,7,5,4,6,6,8,7.[/QUOTE]
I can give you an eleventh degree polynomial with those values, but I suspect that's not what you want.

I don't really like trying to 'continue the sequence', since it could basically be anything. At least since you said give an equation it's likely to be mathematical and not completely stupid (like the number of vowels in the nth month or something), but without any context it still ultimately boils down to making a lucky guess (or having seen it before).

Don't let me stop anyone else trying though (and making me look silly if it's exceptionally obvious).

[mochisushi]

[QUOTE=Joey90;44253537]I can give you an eleventh degree polynomial with those values, but I suspect that's not what you want.

I don't really like trying to 'continue the sequence', since it could basically be anything. At least since you said give an equation it's likely to be mathematical and not completely stupid (like the number of vowels in the nth month or something), but without any context it still ultimately boils down to making a lucky guess (or having seen it before).

Don't let me stop anyone else trying though (and making me look silly if it's exceptionally obvious).[/QUOTE]

It's mathematical, and actually rather simple, not anything dumb like you say. Finding another way of deriving it would probably mean a good deal in the world of pure maths. I'll see if there is any way of deriving it other than the original algorithm. It's related to an abstract analysis of the Collatz sequence.

[JohnnyMo]

[QUOTE=Joey90;44253537]I can give you an eleventh degree polynomial with those values, but I suspect that's not what you want.[/QUOTE]

This is why I hate, "Find the next number in the sequence!" questions. ANY OF THE ANSWERS YOU GAVE ME COULD BE CORRECT.

[mochisushi]

[QUOTE=JohnnyMo1;44254309]This is why I hate, "Find the next number in the sequence!" questions. ANY OF THE ANSWERS YOU GAVE ME COULD BE CORRECT.[/QUOTE]

"Find the simplest, most elegant solution to the next number in the sequence" should be the question.

[editline]16th March 2014[/editline]

And even then, it's debatable.

[Krinkels]

[QUOTE=mochisushi;44254361]"Find the simplest, most elegant solution to the next number in the sequence" should be the question.

[editline]16th March 2014[/editline]

And even then, it's debatable.[/QUOTE]

Can you just post whatever you haven't so far? The hints are not at all helpful.

[mochisushi]

[QUOTE=Krinkels;44260776]Can you just post whatever you haven't so far? The hints are not at all helpful.[/QUOTE]

Related to the sequence?

If f(x) is defined to be how long x takes to reach 1 under the collatz procedure, counting the initial x (so f(2) is 2), then the sequence is g(x) for ascending values of x where g(x) is how many steps it takes to repeat a number by repeatedly applying f(x) to itself.

For example, g(3) is 4 because f(3) equals 8, f(8) equals 4 and f(4) equals 3, taking 4 steps to repeat a number.

[Dilbert]

-never mind, had a bit of a brain fart there-

[mochisushi]

[QUOTE=elevate;44271263]-never mind, had a bit of a brain fart there-[/QUOTE]

I welcome brain farts, what was it about?

[Falubii]

They probably already found the answer.

[Hattiwatti]

Had my final math exam today. Highest grade needs 58 points, I would've gotten 61 if I didn't make a very simple mistake of not reading the problem correctly and so I lost 6 points + a few smaller ones. Final score 55 probably. Physics on Friday, we'll see how that goes.
Hopefully well enough so I can get in a university to study physics. They also have a nano-physics program, but they only take like 10 top applicants. I'm interested in it, but I doubt I'll go that good. Will probably get same grade as in mathematics.

Also physics is something I'm interested in, I get OK grades, but I'm unsure if I wanna [B]really[/B] study it for a job.

[ThisIsTheOne]

Is there anyone here with a PhD or doing a PhD in Maths?

[editline]5th April 2014[/editline]

[QUOTE=JohnnyMo1;44233382][url]http://www.playbuzz.com/adamf11/which-fundamental-theorem-are-you[/url]

I'm the fundamental theorem on homomorphisms. Pretty accurate.[/QUOTE]

The Fundamental Theorem of Finitely Generated Abelian Groups

"You're the less general and more down to earth version of the structure theorem of finitely generated modules over a principle ideal domain. You're the life of the party."

[JohnnyMo]

Maybe someday.

Fundamental theorem of finitely generated abelian groups is pretty neat. Finitely generated group are pretty simple to think about once you know it.

[Krinkels]

I'm auditing intro to algebra this semester and it seems like groups are way less complicated than I thought. I wouldn't be surprised if the fundamental theorem of finitely generated abelian groups turned out to be something like 'every group is isomorphic to Z/2Z'.

[ROFLBURGER]

I forgot grade 7 math

sin(2*x)

i want x by itself

or better yet can someone check my logic

I went from
[quote]
Distance = [{(y*39.3701)^2}*sin(2*x)]/600
[/quote]
to
[quote]
sin(2*x) = Distance / [(y*39.3701^2)/600]*600
[/quote]

Anything past sin(2*x) is blurry so I'm not even going to try

[Hattiwatti]

[QUOTE=ROFLBURGER;44466229]I forgot grade 7 math

sin(2*x)

i want x by itself

or better yet can someone check my logic

I went from

to

Anything past sin(2*x) is blurry so I'm not even going to try[/QUOTE]

Distance = [{(y*39.3701)^2}*sin(2*x)]/600 |*600
600*Distance = {(y*39.3701)^2}*sin(2*x) |/{(y*39.3701)^2}
sin(2*x) = 600*Distance/{(y*39.3701)^2}

If you want x out of that it'd be something like
2*x = asin(600*Distance/{(y*39.3701)^2}) + n*2*pi or 2*x = pi - asin(600*Distance/{(y*39.3701)^2}) + n*2*pi (or n*360° if in degrees) | /2
x = (asin(600*Distance/{(y*39.3701)^2}))/2 + n*pi or x = pi /2 - asin(600*Distance/{(y*39.3701)^2})/2 + n*pi
Where n ∊ Z

At least to my knowledge :v:

[ThisIsTheOne]

[QUOTE=JohnnyMo1;44462238]Fundamental theorem of finitely generated abelian groups is pretty neat. Finitely generated group are pretty simple to think about once you know it.[/QUOTE] Yeah, in class the other day we actually proved the structure theorem for finitely generated modules over a PID. We did a proof in a way that actually gives you a sort of algorithm to find what the submodules that it is isomorphic to actually are which was pretty cool.

[JohnnyMo]

constructive proofs are for dirty intuitionists

I kind of which I had taken more algebra before I attempted algebraic topology. There was more group theory than I had expected. Glad I graduated before I took the second semester. That had modules, tensor products, more category theory, and a bunch of other algebra I've never done. I should brush up before I take it if I end up needing it in grad school.

[ThisIsTheOne]

Well we did a non-constructive proof of Hilbert's basis theorem not long after, just to balance it out.

Yeah, some people in that class had taken an Algebraic Geometry course before this one, and it surprised me how much of the content they had already seen before because of it. I'm still not sure area of maths I want to do a PhD in (or even 100% if I will apply), I still feel like I don't know enough to really decide. I still have a while to decide though, and I'm really looking forward Galois Theory next semester.

[JohnnyMo]

Man, I just have zero interest in Galois theory but so many people like it. I don't get it. That and number theory. So many people work on them but they just seem dull.

I guess I'm a topology/geometry person at heart.

[ThisIsTheOne]

Really? I think number theory is amazing, there are so many seemingly "simple" things which have such deep and complex connections.

[JohnnyMo]

I think it's because I've always been a very visual and intuitive learner. Having a picture in my head helps me understand why something is true and that really lends itself to topology and differential geometry. Algebraic geometry is pretty interesting too but it pushes the line sometimes.

[editline]7th April 2014[/editline]

Not that I never like working with equations but factorization, prime ideals and shit... I just don't care. Algebra can be a lot of fun, but I just don't how number theorists do it. Or more to the point, why they do it.

[mochisushi]

The more abstract it gets, the more I'm interested.

[JohnnyMo]

But that's part of it. Number theory doesn't feel very abstract. It feels like dull algebraic manipulation with slight generalization of properties you already know and replacing some shit like "integers" with "ring" instead. Being a bunch of symbols you can't draw a simple picture of doesn't make it more abstract. Algebraic geometry does a bit of a better job with abstraction.

Now category theory... that's abstraction.

[editline]7th April 2014[/editline]

I'm sure my picture of how the field is is inaccurate, but it's never grabbed me enough to generate much interest. I don't care how to factorize a polynomial over a field, but as soon as topology gave me a short and sweet definition of continuity that made no reference to "breaks" in a graph or limits and which works not only in R, but spaces more general even than metric spaces, I was hooked.

[Joey90]

Do you not notice the similarities between Galois Theory and Algebraic Topology though?

[sp]They're both special cases of Category Theory[/sp]

That's why I like both of them - they provide [i]functorial[/i] relationships between two seemingly unrelated branches of maths, which is pretty amazing. The correspondence is so much [i]deeper[/i] than you might originally think: it's not so much that you can follow this weird process to get a group which can sometimes help you prove something. It's that there really is a good correspondence, which commutes with a lot of useful constructions!

I do agree that getting your head round a complicated space is more satisfying than understanding a complicated field though. Sadly, sometimes it's better (and sometimes impossible any other way) to not try and understand it, but just treat it completely abstractly.

[JohnnyMo]

Everything is a special case of category theory basically, but I don't have to like working in any specific category! Like you said, it's just more satisfying.

[ThisIsTheOne]

[QUOTE=JohnnyMo1;44477609]
Not that I never like working with equations but factorization, prime ideals and shit... I just don't care. Algebra can be a lot of fun, but I just don't how number theorists do it. Or more to the point, why they do it.[/QUOTE]

It sounds like you're only talking about algebraic number theory. Of course a lot of it is interconnected, but Number theory is more interesting than that. Things like the Prime Number Theorem and Dirichlet L-series (Riemann Zeta function is one example) all involve lots of complex analysis and are really cool. Then there are really interesting results about the growth rates of arithmetic functions and their averages, and lots of cool facts like Bertrand's postulate and Dirichlet's theorem.

[editline]8th April 2014[/editline]

[QUOTE=mochisushi;44478288]The more abstract it gets, the more I'm interested.[/QUOTE]

I did think like that but sometimes now it can start to feel a little dry if it's too abstract.