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Re:Mathematician-M.D. claims to have solved the 110-year-old... by cbames89 on Wednesday June 27, 2018 @09:43PM (#17413437)
Can someone eli5 the lindelof hypothesis?
Re:Mathematician-M.D. claims to have solved the 110-year-old... by impendia on Wednesday June 27, 2018 @11:30PM (#17413862)
The Riemann zeta function is the function

zeta(s) = 1 + 1/2^s + 1/3^s + 1/4^s + 1/5^s + 1/6^s + ....

For example,

zeta(2) = 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 + ... = pi^2/6.

As a partially tongue-in-cheek example,

zeta(-1) = 1 + 2 + 3 + 4 + 5 + 6 + ... = -1/12.

Obviously it doesn't make sense to add all the positive integers (the series doesn't converge), but if you squint and ignore this, and just do the arithmetic a certain way, you get -1/12.

The original definition I gave is valid when s is a complex number with real part greater than 1. But the Riemann zeta function can be proved to have analytic continuation: zeta(s) makes sense for any complex number s, other than 1. For example, zeta(-1) really equals -1/12.

The zeta function is easy to understand when the real part is greater than 1: the formula I described is enough. Because of the so-called functional equation, it is also easy to understand when the real part is less than 0. But it is in the middle that all of its secrets lie. For example, the notoriously unsolved Riemann Hypothesis stipulates that the "nontrivial" zeroes all have real part 1/2.

The Lindelof Hypothesis stipulates that the zeta function grows very slowly along this line (real part = 1/2). It is very closely related to the Riemann Hypothesis. More technical, and of less direct interest to nonspecialists, but in the same family of problems.

As an example of how much mathematicians care about this, here are the Google search results for "subconvexity bound":

https://www.google.com/search?q=subconvexity+bound

A "subconvexity bound" is any result which approaches the Lindelof Hypothesis, for either the Riemann zeta function or a more general "L-function". A lot of ink has been spilled on proving results weaker than what Fokas is claiming.

Re:Mathematician-M.D. claims to have solved the 110-year-old... by taneq on Thursday June 28, 2018 @02:06AM (#17414464)
> Obviously it doesn't make sense to add all the positive integers (the series doesn't converge), but if you squint and ignore this, and just do the arithmetic a certain way, you get -1/12.

I'm not a pure-maths type person but in my experience, if you get one answer by following simple, well understood maths (like "the sum of two positive integers is a positive integer") and another answer by "squinting and ignoring it", this doesn't mean the simple answer is wrong, it means you did something else wrong (like the hidden divide-by-zero present in your typical "proof that 1 = 2"). Paradoxes point to an error in the formulation of the question.

Re:Mathematician-M.D. claims to have solved the 110-year-old... by rfurmani on Thursday June 28, 2018 @02:34AM (#17414561)
But this isn't just summing up a few positive integers, in which case there's no ambiguity in what the answer is. Once you start summing up infinitely many things you have to bring in some theory and some techniques to justify what the answer is. These techniques generally have a limited scope, but there's a big theory of "divergent series" that shows that if you extend these techniques to more contexts you still get a definition that is compatible with most of what one would want from a limit. For example, taking running averages, or taking it as the coefficients of a series and taking a limit. So classically 1-1+1-1+1... doesn't converge, but if you "squint" and take the running averages of the partial sums (1,0,1,0,...) you would get 1/2. Or if you use the fact that 1+x+x^2+x^3+... = 1/(1-x) and take x=-1 you would get 1/2. Or a myriad other approaches that are perfectly valid with standard convergence, and just all happen to get you that 1-1+1-1+... is 1/2. And yes if you squint very hard you would get that 1+2+4+8+16+... = 1/(1-2)=-1
Re:Mathematician-M.D. claims to have solved the 110-year-old... by taneq on Thursday June 28, 2018 @03:28AM (#17414790)
> So classically 1-1+1-1+1... doesn't converge, but if you "squint" and take the running averages of the partial sums (1,0,1,0,...) you would get 1/2

Why would you take the running average and how is that relevant to the sum?

Re:Mathematician-M.D. claims to have solved the 110-year-old... by Doxin on Thursday June 28, 2018 @03:44AM (#17414858)
He's taking the running average of the partial sums:

    1-1 = 0
    0+1 = 1
    1-1 = 0
    0+1 = etc etc
The partial sum would end up being equal to the final sum by definition when you're done summing all items. Since we're talking about infinite sequences you're never done summing all items so you'll have to do something else to end up with an answer. for example seeing which way the partial sum trends. In this case it trends solidly in the direction of 1/2
Re:Mathematician-M.D. claims to have solved the 110-year-old... by Koshkin on Thursday June 28, 2018 @07:42AM (#17416019)
> In this case it trends solidly in the direction of 1/2

...except that it doesn't; to me, it even looks more like it trends in the opposite direction, i.e. it is trying to stay away from 1/2.

Re:Mathematician-M.D. claims to have solved the 110-year-old... by whatshisface on Thursday June 28, 2018 @08:50AM (#17416402)
The farthest it could get from 1/2 would be if it ran off to + or - infinity. Hovering between +1 and -1 could be compared to an orbit.
Re:Mathematician-M.D. claims to have solved the 110-year-old... by hammock on Thursday June 28, 2018 @07:49AM (#17416064)
>...except that it doesn't; to me, it even looks more like it trends in the opposite direction, i.e. it is trying to stay away from 1/2.

Like two magnets repelling each other: if you were to hold them together and we call that 1/2 - but they are always trying to push away from each other!

Re:Mathematician-M.D. claims to have solved the 110-year-old... by filmor on Thursday June 28, 2018 @02:43AM (#17414590)
The problem with this first approach is that you can get arbitrary results just be reordering the series. Also, the formula you give for the geometric series only works for |x|<1.
Re:Mathematician-M.D. claims to have solved the 110-year-old... by masterjack on Thursday June 28, 2018 @02:57AM (#17414657)
Reordering a divergent series can indeed give you any result you want. This is not reordering though, it’s called Cesàro summation. It won’t always get you an answer, but if it does then the answer is unique and reacts nicely to sums and products.

Talking about it as summation might be misleading since that’s one of those concrete terms that mathematicians like to redefine without anyone’s approval. Picture we have a library that includes many tricks and approaches for taking an infinite series as input and outputs a number. We know it works as expected on every convergent series. But we forgot to put in any preconditions and we’ve let people input things that are not convergent series. But whoa in many cases we are still getting a number out of it, and it’s always the same answer no matter what we do. Maybe that’s something worth studying?

Re:Mathematician-M.D. claims to have solved the 110-year-old... by JadeNB on Thursday June 28, 2018 @10:28AM (#17417124)
> This is not reordering though, it’s called Cesàro summation. It won’t always get you an answer, but if it does then the answer is unique and reacts nicely to sums and products.

It's also probably worth noting that, if the series is genuinely summable, then Cesàro summation gives its sum.

Re:Mathematician-M.D. claims to have solved the 110-year-old... by ur-whale on Thursday June 28, 2018 @05:22AM (#17415308)
I wish impendia had left out of his otherwise excellent explanation the "squinting" part and stuck to explaining that the zeta function is defined by the infinite series where that series converges, and defined where that series diverge using a tool called analytic continuation.

In short:

Where the series obviously converge, use the summing formula.

Where the series is mis-behaving, use analytic continuation instead of resorting to weird infinite series re-ordering tricks (which I've always felt to be borderline offensive from a mathematical rigor pov).

My understanding of analytic continuation is that if a function f of the complex plane is sufficiently well behaved on a certain domain of the plane, it can be "extended" to the rest of the plane in a unique way that preserves the well-behavedness.

In the case of zeta, it can be shown that zeta obeys a functional equation that allows it to be extended everywhere.

A better explanation than mine is here:

https://math.stackexchange.com/questions/437883/what-is-the-...

Re:Mathematician-M.D. claims to have solved the 110-year-old... by OskarS on Thursday June 28, 2018 @02:32AM (#17414554)
The thing impendia is referring to there is analytic continuation, which (as mentioned) is a way to extend the domain of a certain set of functions, like the zeta function. It is perfectly rigorous. His/her language was just a short-hand for "don't worry about the details here, but it does work".

Mathematicians aren't stupid, and more than any other profession, they value rigor. They know what they're doing.

Re:Mathematician-M.D. claims to have solved the 110-year-old... by taneq on Thursday June 28, 2018 @09:37AM (#17416718)
Rigor in logic but, it seems to me, not often rigor in description. There's a lot of willingness to say "if we change rule X to mean something totally other, then you can do this thing" and then just describe it as "you can do this thing". Sure, but "this thing" now means something different.
Re:Mathematician-M.D. claims to have solved the 110-year-old... by hardlianotion on Thursday June 28, 2018 @07:31AM (#17415958)
In this case, it's worth pointing out that the zeta function admits a functional definition in which you can define some points of the function about their "reflection" in the line Re(z) = 1/2. This allows a perfectly rigorous definition of the function for values like -1. One might call the functional form more fundamental and note that the series form only holds true for Re(z) > 1.
Re:Mathematician-M.D. claims to have solved the 110-year-old... by filmor on Thursday June 28, 2018 @02:38AM (#17414573)
In mathematics, the way to the answer is essential. If you are doing /anything/ wrong on your way, your result is void.

It may not be wrong, but nothing is gained.

Re:Mathematician-M.D. claims to have solved the 110-year-old... by vole on Thursday June 28, 2018 @09:46AM (#17416780)
I disagree with this. Often what is gained from a proof is not the fact that a proof is known, but more insights about the original problem. So, a small technical error may "invalidate" a proof but it does not make it meaningless. Just like a bug in software does not make it worthless.
Re:Mathematician-M.D. claims to have solved the 110-year-old... by denzil_correa on Thursday June 28, 2018 @03:17AM (#17414737)
3B1B has a nice visualization of "Riemann zeta function and analytic continuation" which might help in further understanding of this proof.

https://www.youtube.com/watch?v=sD0NjbwqlYw

Re:Mathematician-M.D. claims to have solved the 110-year-old... by alexbeloi on Wednesday June 27, 2018 @10:49PM (#17413692)
Not eli5, but a comparison to the Riemann Hypothesis (RH).

RH says the Riemann-zeta function has no zeros along the line (1/2) + iy in the complex plane.

The Lindelof hypothesis says that the number of zeros between (1/2) + iy and (1/2) + i(y+1) is much smaller (little-o) than log(y) as y grows.

So it can be thought of as a weaker version of RH, but still very very difficult. The fact that Lindelof has been an open problem for over a hundred years (and is an non-trivial weakening of RH) speaks to how difficult RH is as well.

Like RH, Lindelof implies things about primes, and also (like RH) has lots of implications about lots of interesting prime-like (irreducible) objects in different spaces.

Re:Mathematician-M.D. claims to have solved the 110-year-old... by dbaupp on Thursday June 28, 2018 @12:58AM (#17414185)
I think you've flipped the condition: the RH says the Riemann zeta function _only_ has zeros along the line 1/2 + iy. (And, indeed, there are known zeros along this line: 1/2 + 14.135... i.)

The Lindelöf hypothesis is, apparently, equivalent to: the number of zeros with real part greater than 1/2+epsilon and imaginary part between y and y+1 is o(log(y)), for any epsilon > 0. That is, boxes of height 1 starting just off the critical line contain few zeros; the RH implies they contain zero.