Mmmm.. yes, of course. Makes sense

yes

this is a troll/meme right? Like the flat earth meme that everyone pretends to be in on?

Well, you know... sometimes in mathematics you can get unexpected results! XD

it's real

1-1+1-1+1-1+1-1+1-1+1...=1/2

now add

1-1+1-1+1-1+1-1+1-1+1...
- 1-1+1-1+1-1+1-1+1-1+1...
= 1+1+1+1+1+1+1+1+1 = -1/12

and 1+1+1+1+1+1+1+1... = 1+2+3+4+5...

therefore 1+2+3+4+5...=-1/12

could you add parentheses to make it less ambiguous

>I cant read
XD KYS OOLOLOL

>[math]\sum_{n=1}^{\infty}1=\sum_{n=1}^{\infty}n[/math]

What

How did this become such a meme?

plus.maths.org/content/infinity-or-just-112

>But how did this curious, wrong result make it into a physics textbook, as shown in the video? Here is where things really get interesting. Suppose you take two conducting metallic plates and arrange them in a vacuum so that they are parallel to each other. According to classical physics, there shouldn't be any net force acting between the two plates.

>But classical physics doesn't reckon with the weird effects you see when you look at the world at very small scales. To do that, you need quantum physics, which tells us many very strange things. One of them is that the vacuum isn't empty, but seething with activity. So-called virtual particles pop in and out of existence all the time. This activity gives a so called zero point energy: the lowest energy something can have is never zero (see here for more detail).

>So-called virtual particles pop in and out of existence all the time.

The explanation is literally meme tier flat earth BS all on its fucking own!

Huuuuugh...

youtube.com/watch?v=7fGoins7q3s

2(1/2)=2(1-1+1-1+1-1+1...) = 2-2+2-2+2... = 1
1-(2-2+2-2+2...) = 0
Cool

I fucking hate this guy.

Why

That's a great read.

> (Making this new function give you finite values for $x \leq -1$ involves cleverly subtracting another divergent sum, so that the infinity from the first divergent sum minus the infinity from the second divergent sum gives you something finite.)

So really (infinity) zetafunction(-1)-(some specific infinity)= -1/12

It's not clear what the specific infinity, is

gr8 b8 m8 i rel8 2 the deb8 so i r8 8/8

gr1 b2 m3 i rel4 5 the deb6 so i r7 8/9...

this

= gr*b*m*2i*rl*the*d*so*r*/ = -1/12

>checkm8 atheists

>Grone btwo mtrhee i rfour five the debsix so i rseven eight/nine

What did he mean by this?

haha excellent posts my friends! Really enjoying these!

zeta of -1 is -1/12 directly, without any infinities.

Anyway, it's like how for z in (-1,1) you have

[math] \dfrac {1} {1-z} = 1 +z+ z^2 + z^3 + ... [/math]

and if you plug in

z = d-1

where
d = 1/2^999999,
i.e. some unimaginably small number

then
[math] \dfrac {1} {2-d} = 1 + (d - 1) + (d-1)^2 + (d-1)^3 + ... [/math]

you also see the
1-1+1-1+1-1+... = 1/2
result.

If the sum pops up in physics as some inherently simple finite objectQ, but e.g. z=-1 is the edge of a wall and the way the theory is written down doesn't want to deal with it, then people would also say Q at the wall is 1/2, and not incoorporate analytic limits in their language

Is it literally true for sums as defined in analysis? No, only the limit of the sum.

But if you come up with a thoery of physics, the heuristic version of your theory might not allow for the fancy overhead and so whenever you talk to people who aren't experts in theoretical physics and math, you'd go on about arguing why those sums end up this and thay number in some sloppy way. The complex analysis language is also not integral part of it, merely it's tools are used when necessary.
A tiny minority of physicists has ab idea about rebormalization in quantum field theory.

Might be a dumb question, but how do you directly compute values of the zeta function without finding the limit of the summation?

>zeta of -1 is -1/12 directly, without any infinities.

WHAT

but he is a qt user
some videos are oversimplified though and
some of his fans are insufferable

Wow I really didn't minus a twelfth that one coming

Can we start banning people for posting these threads?