1+2+3+4... = -1/12

[stargroup100]

Posting a thread on this simply because it kinda blew my mind, something math hasn't done for me in a while.

The limit obviously converges to infinity, but the actual total sum of all of the positive integers comes out to be -1/12. Not only can this be derived in multiple ways, but is a statement that is applied and used in string theory.

All I can say is: ...what

[emerald000]

The actual sum does not give -1/12 using normal mathematics. There a lot of videos going on around showing how to get that result using simple algebra on infinite divergent series. That's something you can't do. It's like those proofs of 1 = 2 that divide by 0 to get to their result.

Anyway, there is a branch of mathematics called analytic continuation. Without going into detail, it deals with how analytic functions behave outside of their domain. And there is this well-known function called the Riemann zeta function defined as the sum of 1/n^s for s from 1 to infinity. This function is only defined for s>1. But if we were to take s = -1, we would get 1 + 2 + 3 + .... And using those rules of analytic continuation, you can get a value outside of the range of functions and in this case, the value is -1/12.

So what is the value of the infinite sum? We made a mistake by just asking this question. Divergent sums don't have value until we give them one. The question is not what is the value but what we should define it as? There a couple of ways to go about it and each have their strengths and weaknesses. The usual way involving limits would not give any answer. The analytic continuation (and a couple others) gives -1/12. Since something is more useful that nothing, we decided to give the infinite sum the value -1/12. But neither is more correct than the other.

PS: If you want to read more about it: http://terrytao.wordpress.com/2010/0...-continuation/

[Frederic54]

Originally Posted by stargroup100

but the actual total sum of all of the positive integers

There is no actual value to the sum of all integers as they are infinite, we can just say that it converge to a "thing" we call infinite that represent something that doesn't end

[stargroup100]

Originally Posted by emerald000

...Since something is more useful that nothing, we decided to give the infinite sum the value -1/12. But neither is more correct than the other.

Right, but I was just trying to state it simply in layman's terms. The point is, there are constructs/situations/whatever in which that value is valid, and not only is it valid, but actually used in real world calculations in some form. But in retrospect I should at least be more technical in a critical thinking board lol.

For me, many concepts regarding infinity are not that difficult to accept or comprehend, but this is just something that is totally absurd and ridiculous no matter how I visualize it in my head.

[Dynam0]

Non-Euclidean Geometry is pretty neat too you know

[Zapmeister]

Originally Posted by emerald000

Anyway, there is a branch of mathematics called analytic continuation. Without going into detail, it deals with how analytic functions behave outside of their domain. And there is this well-known function called the Riemann zeta function defined as the sum of 1/n^s for s from 1 to infinity. This function is only defined for s>1. But if we were to take s = -1, we would get 1 + 2 + 3 + .... And using those rules of analytic continuation, you can get a value outside of the range of functions and in this case, the value is -1/12.

this.

it's like saying that just because you can get a maclaurin expansion of, for example:

(1-x)^-2 = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + ...

and the expansion is valid for |x|<1

so you could put x=0.1 and get 1 + 0.2 + 0.03 + 0.004 + ... = 100/81, which is valid

then you put x=10 and get 1 + 20 + 300 + 4000 + ... = 1/81

it's the same idea here, just replace "riemann zeta function" in whatever he said with "(1-x)^-2" and you'll see how ridiculous and nonsensical it is to talk of a divergent series converging to some value just because it has a non series representation outside where it converges

so there'd never be a situation where you'd have to add 1+2+3+4... directly and get -1/12, if you wanted to work out zeta(-1) you'd work out zeta(-1), not try to add the divergent series, which won't make any sense

edit: also non euclidean geometry sucks donkey balls, i did a course on that last year, and it was horrible and made me want to pull out all my hairs and superglue them to my chin. i think it was just badly taught, but maybe i just hate the subject

[stargroup100]

Non-euclidean geometry is pretty cool. I remember telling some guy that with spherical geometry I could make a triangle with 3 right angles and he just couldn't accept it XD

[xThai]

http://skullsinthestars.com/2010/05/...e-weird-redux/

[smartdude1212]

Originally Posted by Dynam0

Non-Euclidean Geometry is pretty neat too you know

Bitch why does my triangle have three right angles WHAT IS THIS MAGIC?

I didn't get a chance to take the Non-Euclidean Geometry course at my university (because I was in Russia oops), but my best math friend explained it all to me in this way.

[jaz_pup]

I haven't studied non-Euclidean Geometry like spherical Geometry, but I did study finite projective Geometry and I find that pretty....interesting lol Parallel lines don't meet? Well make them. Done. haha

[choof]

the concept of an infinite series is sick

if I ever make enough to get to college, I might take some advanced math classes

[SKG_Scintill]

They could have used any symbol to describe the answer to this; the principles are the same. Insert the symbol and continue your mathy ways, or insert "-1/12" and continue your mathy ways. The solution is still based on the assumption that there is a solution to the convergent series. Whether it's -1/12 or a symbol, it doesn't matter.

It's a weird choice to use an actual number, though. As emerald said: it's not normal mathematics and as frederic said: it's not the actual solution of the sum. But it confuses the regular people when it's written as a number instead of a symbol.

Give people pi, people know pi.
Give people the infinite series that results to pi, people don't know pi.
Same here. Accept it for what it is and continue calculating.

[j-rodd123]

math for the sake of math, how cool fun and useful

[nickcool9999]

I'm not going to harp on this topic for that long because math is confusing and people like you only make it more difficult to comprehend. More or less people will give me flack for writing this but I brought to you a logical conclusion as to why this is IMPOSSIBLE, no matter what combination of integers you input (Even if it isn't in this range here):

"1+2+3+4..."
What is a sum? A combination of positive or negative numbers added together to produce a larger or smaller number. For this instance, all of the numbers in this range are both integer and positive, which means:
• there's no possible way to produce a fraction as a result of this calculation.
• there are no numbers in the calculation that are negative.

So in logic, having simply "1" in the combination of numbers has already brought you above -1/12, and there is no such number that brings you below the number 1. The same can be said for the rest of the numbers we have here.

There's probably some stupid math algorithm to justify your answer but the fact of the matter is that infinite numbers summed does not produce a negative rational fraction, unless that is you're trying to be completely incorrect...

Wherever it is that you learned that OP, just ignore it. You're making math harder to justify than it needs to be. -.-

"Non-euclidean geometry is pretty cool. I remember telling some guy that with spherical geometry I could make a triangle with 3 right angles and he just couldn't accept it XD"

Well that isn't really hard to explain at all; just bend the lines. Some people really are dumb when it comes to geometry and I don't really blame them.








PS: If your justification for the sum thing isn't smart please don't use it against me. I'm getting tired of people trying to say that they are correct just because others told them that they are. I get it way too often where I live and that's just an excuse to justify your incorrectness. -.-'...

Mathematical Justification or Critical Thinking (Why this board exists):

ƒ(n,s)=1/n^s where (s=-1),
sub as ƒ(n,-1)=1/n^-1
ƒ(n,-1)=1/(1/n)
ƒ(n,-1)=n

Therefore, if (ƒ(n,-1)=-1/12) and ƒ(n,-1)=n, then n=-1/12. So your math is wrong anyways.

ƒ(-1/12,-1)=-1/12
ƒ(ΣI,-1)≠-1/12
∴ΣI≠-1/12

:P

[ReikonKeiri]

I still don't know why it works, but Numberphile says the value -1/12 for that sum is used for practical quantum physics applications so I have no choice but to believe it.

My theory is black holes.

[nickcool9999]

Originally Posted by ReikonKeiri

I still don't know why it works, but Numberphile says the value -1/12 for that sum is used for practical quantum physics applications so I have no choice but to believe it.

My theory is black holes.

Well that's a cool idea to be true but unfortunately...

It got myth busted.

YEEEEAAAAA lol

[stargroup100]

Originally Posted by nickcool9999

There's probably some stupid math algorithm to justify your answer but the fact of the matter is that infinite numbers summed does not produce a negative rational fraction, unless that is you're trying to be completely incorrect...

Wherever it is that you learned that OP, just ignore it. You're making math harder to justify than it needs to be. -.-

I may have fudged the terminology in the opening post for whatever reason but I assure you, I am no stranger to math.

Obviously, for any finite number of terms, the sum of the integers starting from 1 can't possibly be fractional or negative. However, we're talking about an infinite series here, which characteristically makes the problem different.

The reason why this result can even make sense is because we're talking about different contexts. Mathematical structures in different domains have different properties. When you have problems in which multiple answers can be justified, the conditions for the problem must be explained to obtain a particular solution. While in the sense of a divergent series this goes to infinity, through analytic continuation this statement can be justified. The result is actually used in quantum mechanics, and without it we wouldn't have the level of digital communication we have today.

Just because you don't like the idea doesn't make it false. And whether or not pure mathematics is difficult to justify is irrelevant. Fermat's Last Theorem was notoriously hard to justify, but people spent hundreds of years to do it, and it's an important result.

Originally Posted by nickcool9999

Mathematical Justification or Critical Thinking (Why this board exists):

ƒ(n,s)=1/n^s where (s=-1),
sub as ƒ(n,-1)=1/n^-1
ƒ(n,-1)=1/(1/n)
ƒ(n,-1)=n

Therefore, if (ƒ(n,-1)=-1/12) and ƒ(n,-1)=n, then n=-1/12. So your math is wrong anyways.

ƒ(-1/12,-1)=-1/12
ƒ(ΣI,-1)≠-1/12
∴ΣI≠-1/12

:P

wtf is this

making an assumption with no basis: "Therefore, if (ƒ(n,-1)=-1/12)"
jumping to conclusions: "n=-1/12. So your math is wrong anyways."
another assumption: "ƒ(ΣI,-1)≠-1/12"

clearly math is not your strong suit

[TheSaxRunner05]

Wrecked

[Zapmeister]

i still maintain that it doesn't make sense to talk about a divergent series having a finite value as a sum.

let's go back to my earlier example of

Originally Posted by Zapmeister

(1-x)^-2 = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + ...

and the expansion is valid for |x|<1

of course everyone agrees that if you plug x=0.1 in here you get 1+0.2+0.03+0.004= ... = 100/81, because x=0.1 is inside the radius of convergence.

you're saying: if i plug in x=10, then i get a value of the sum 1+20+300+4000+ ... = 1/81 which is justified by analytic continuation, since there is a unique analytic continuation of the sum 1+2x+3x^2+4x^3... outside its radius of convergence, and this analytic continuation just happens to be equal to (1-x)^-2.

here's the thing. if you wanted to do that, you'd need to tell me what the function you're trying to take out of its radius of convergence is. and because the series is divergent, you're not going to get the same answers whichever way you do it. so you're trying to say that 1+2+3+4... represents the analytic continuation of

1^-s + 2^-s + 3^-s + 4^-s + ... | s=-1

at s=-1, which is the zeta function at -1, which is -1/12.

there is no way to determine that when you wrote 1+2+3+4... you're starting from the zeta function and working things out from that. you could be trying to get it from the analytic continuation of another function.

here's another way to write it. how about this?

1+2+3+4+... = 1 + 2x + 3x^2 + 4x^3 + ... | x=1
= (1-x)^-2 | x=1
= 0^-2 = "infinity"

you're still not convinced?

1+2+3+4... = 1 + 2x + 3x^2 + 4x^3 + ... | x=1
= (1-x)^-2 | x=1
= ( (1-x)^-1)^2 | x=1
= (1 + x + x^2 + x^3 + ...)^2 | x=1
= (1 + 1 + 1 + 1 + ...)^2
= (zeta(0))^2
= 1/4

yeah. none of this would be happening if you used a valid x inside the radius of convergence, since everything would agree on a certain value.

this is the point. there may be occasions when you're trying to work out what the analytic continuation of 1^-s + 2^-s + 3^-s + ... is at s=-1. in these cases you'd write it as zeta(-1), because the zeta function is defined to be this analytic continuation. you'd never write it out as 1+2+3+4+... because nobody knows what you mean when you're writing down the sum of a divergent series. it's complete trash. just like your face (oohh burn)

[Dynam0]

straight up gangst thread