The Math Thread

[Oleander]

Technically you could say that all real numbers are complex numbers too, they just have imaginary part 0. That's a nice concept to think of when visualizing the complex plane.

I started school when I was 4 (curiously, it happened to be a Religious School). It was mostly coloring and stuff then, but I think I learned a few good things, like the days of the week and how to talk about the weather, and I'm pretty sure I was taught some basic addition with dominoes before I started grade school.

[The Kakama]

School for me was a bit odd.We start school at the age of 7,but I got to go to kindergarten when I was 5.So I was learning until thousands and doing division and times tables and stuff,until I was 7.Turns out,not a lot of people were as lucky as me,so I had to start back to counting until 10.Of course,the next 3 years of maths was a breeze.

[dVanDaHorns]

I was one of those "late-starters" since I was born in February (aka I was the oldest in my class), but if I recall, I started Kindergarten around 5 or 6. After that, I was homeschooled up until grade 9 (The beginning of high school in Canada...not too sure what that equates to in Europe or the rest of the world), where my mother was my teacher for most subjects (Science, history, geography, etc.) and I self-taught myself the few that remained. (Mostly math. I pretty much self-taught myself everything from multiplication up to simple algebra. And then I got ambitious and began looking at a University Calculus textbook...thanks to that, I learned an overly-complicated method of approach to trigonometry when I was in grade 8...for reference, basic trig is taught in grade 10.)
Then I got to high school, where I suddenly realized I was good at math, but I had never noticed previously because I always forced myself on a fast-paced learning curve with no calculator.
The rest of my pre-university math experience was a breeze...and things didn't pick up until I got to university, where my whole class was like me or worse, and so the course was aimed more for our level.
(And, jerkish final aside, I still find it quite fun.)

...And how did we get this far off-topic? O.o

[Oleander]

There's not much on topic I have to say right now; I'm mostly studying infinite sums and products now. I do have a question I've been thinking about lately though. Does the infinite series s(k)/k^2, where s(k) is equal to the sum of the divisors of k except for k itself, converge? The terms are relatively large for small values of k, but for large values they're pretty small. You seem like you're pretty knowledgeable about math, Dvan. Can you solve this? If it doesn't converge, what's the smallest a for which s(k)/k^a converges?

[dVanDaHorns]

There's not much on topic I have to say right now; I'm mostly studying infinite sums and products now. I do have a question I've been thinking about lately though. Does the infinite series s(k)/k^2, where s(k) is equal to the sum of the divisors of k except for k itself, converge? The terms are relatively large for small values of k, but for large values they're pretty small. You seem like you're pretty knowledgeable about math, Dvan. Can you solve this? If it doesn't converge, what's the smallest a for which s(k)/k^a converges?

FINAL FINAL EDIT:

After over 3 hours of work, many interpolations, a bunch of computer number crunching, and a whole lot of nonsense later, (Including posting a thorough, in-depth analysis of why it converges), I realized that it actually doesn't converge.
This can be proven by analyzing a few of the subseries contained within this inconsistent series.

Due to the nature of the numerator, which fluctuates in value all over the place, it is impossible to analyze the series as a whole. However, every term within this series will fall into a different series or another, (primes fall into 1/n^2, all the others fall into at least one other subseries where the numerator follows a series of its own related to the overall series), and thus, if all these subseries converge, then the overall series must converge as well.

However, it is fairly easy to show that not all of the subseries will converge. Take, for example, the subseries s(10^n)/(10^2n), where s(10^n), similar to the notation you used, represents the sum of all the factors.
Just analyzing the first few terms, you get (7/100+117/10000+1340/1000000+24211/100000000+O(5)). Now, compare the numerator to the square root of the denominator. You'll notice that, with the exception of the first term, in all subsequent terms, the numerator is larger. Therefore, comparing it to the power series, (1/n^x, where, if x>1, it converges), you get that it diverges, as it doesn't shrink fast enough. (It is equivalent to 1/n^x, where x<1.)

Thus, the overall series diverges.

[Vortex]

DVan is right. You can also use a Dirichlet series to see that when it exists it converges to ζ(a)*ζ(a-1) - ζ(a-1) = ζ(a-1)*(ζ(a) - 1), and when a=2 you get the multiplying term ζ(1), which is the harmonic series and diverges (I tend to use the more visual word "explodes" for that ).

[Oleander]

Does it really equal that? Where did you find that formula? And I guess given that formula it must converge for a>2.

And the harmonic series doesn't exactly explode, it kinda just leaks out everywhere like an overturned bottle of syrup. It takes over 4.5 billion terms to get past 20!

[dVanDaHorns]

DVan is right. You can also use a Dirichlet series to see that when it exists it converges to ζ(a)*ζ(a-1) - ζ(a-1) = ζ(a-1)*(ζ(a) - 1), and when a=2 you get the multiplying term ζ(1), which is the harmonic series and diverges (I tend to use the more visual word "explodes" for that ).

Dang, forgot about the Dirichlet series...not to mention that I would have been unable to come up with that form...
Oh well, there you have more rigorous proof!

Does it really equal that? Where did you find that formula? And I guess given that formula it must converge for a>2.

And the harmonic series doesn't exactly explode, it kinda just leaks out everywhere like an overturned bottle of syrup. It takes over 4.5 billion terms to get past 20!

It's the Riemann Zeta function he is using, which is a Dirichlet series. The Riemann Zeta function takes the form of sum of 1/n^a, where a is the independent variable.
Anyhow, it's just logic that if the numerator grows too fast, even though the denominator's growing faster, that the series will diverge. Several of the subseries diverge according to this method, and so the sum will grow infinitely, in spite the number of subseries which would converge.

Also, yeah, it's not really explodes, but referencing an explosion provides more action, and makes solving series more epic. That way, once you prove a series diverges, you can get up, turn around, and walk away like a boss, so that you can tell people truthfully that you walked away from an explosion all awesome-like.

[Oleander]

Yes, but how did he get that specific formula for s(k)/k^a in terms of the Riemann Zeta function? That's a pretty fascinating relation.

[Vortex]

Well, it's maybe a bit hard to understand. But if you look at the definition of the Dirichlet convolution (here: http://en.wikipedia.org/wiki/Dirichlet_convolution) and take f(n) as n (the identity) and g(n) as 1 (the constant function 1), then you can see that the convolution just gives the sum of all divisors of n, INCLUDING itself (you can check it if you want, it's easy to see just looking at the constraint below the summation).

Now, if you take the Dirichlet series of a convolution of functions, the result is simply the product of the Dirichlet series of the two functions alone (a known result):

-The Dirichlet series of f(n) is the series f(n)/n^s = n/n^s = 1/n^(s-1), what gives ζ(s-1).

-On the other hand, the Dirichlet series of g(n) is g(n)/n^s = 1/n^s, what as you know gives just ζ(s).

So the product of the two, ζ(s-1)*ζ(s), is the result of summing the convolution, which is the sum of all divisors of n including itself.

Finally, for the sum of all divisors EXCLUDING itself, you just have to substract "n" from the definition of the convolution, and calculate its Dirichlet series (the Dirichlet series of a sum/difference is the sum/difference of the Dirichlet series, like with all series), and you get the series -n/n^s, what gives the extra term -ζ(s-1).

Joining all together, we obtain the same I said above, ζ(s)*ζ(s-1) - ζ(s-1) = ζ(s-1)*(ζ(s) - 1)

And yeah, "exploding" is not precisely adequate in this case, but dVan explained it better than me


EDIT: just found a generalized formula in Wikipedia for the sum of the a-th powers of all divisors including itself here: http://en.wikipedia.org/wiki/Divisor_fu ... _relations (my result appears when a=1)
It's good to know I was right, but I wasted my mind processing power XD