/sci/ - Science & Math » Thread #12252518

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Anonymous Tue 20 Oct 13:10:33 2020 No.12252518 View Reply Original Report

Quoted By: >>12252574 >>12252581 >>12254227 >>12254372 >>12254597 >>12255232

The sum of all positive numbers is -1/12

What the fuck is this, /sci/? I just don't get it, is this even real or am I being trolled by MATH Chads?

Anonymous

Anonymous Tue 20 Oct 2020 13:18:05 No.12252536 Report

Quoted By:

It's easy to confuse yourself with this shit but it's quite simple.

All that the ramanujan summation stuff, cutoff and zeta regularization does, is look at the smoothed curve at x = 0.
What sums usually do is look at the value as x->inf.

It's just a unique value you can assign to a sum, really they have many such values.

https://en.wikipedia.org/wiki/File:Sum1234Summary.svg

$\displaystyle<br />\zeta \neq \Sigma<br />$

https://youtu.be/sD0NjbwqlYw?t=10m

Anonymous

Anonymous Tue 20 Oct 2020 13:36:34 No.12252567 Report

Quoted By: >>12253138 >>12253473 >>12255232

A = 1–1+1–1+1–1?
1-A=1-(1–1+1–1+1–1?)
1-A=1–1+1–1+1–1+1?
1-A =A
1-A+A=A+A
1 = 2A
1/2 = A

B =1–2+3–4+5–6?
A-B = (1–1+1–1+1–1?) — (1–2+3–4+5–6?)
A-B = (1–1+1–1+1–1?) — 1+2–3+4–5+6?
A-B = (1–1) + (–1+2) +(1–3) + (–1+4) + (1–5) + (–1+6)?
A-B = 0+1–2+3–4+5?
A-B = B
A = 2B
1/2 = 2B
1/4 = B

C = 1+2+3+4+5+6?
B-C = (1–2+3–4+5–6?)-(1+2+3+4+5+6?)
B-C = (1-2+3-4+5-6?)-1-2-3-4-5-6?
B-C = (1-1) + (-2-2) + (3-3) + (-4-4) + (5-5) + (-6-6) ?
B-C = 0-4+0-8+0-12?
B-C = -4-8-12?
B-C = -4(1+2+3)?
B-C = -4C
B = -3C
1/4 = -3C
1/-12 = C

Anonymous

Anonymous Tue 20 Oct 2020 13:40:25 No.12252574 Report

Quoted By: >>12252839 >>12253099

>>12252518
Trolled by math Chad's.
By splitting the numbers into arbitrary groups, you're able to apply some tricks to certain number populations, one of which is completely fucking retarded and is never used in real life-
1+1-1+1-1...= 1/2
Because it's either 1 or 0 at any given point, they say it's associated with (=) 1/2.
Nowhere in the universe can you use this concept and not be ridiculed.

Considering sound is something that oscillates between min and max amplitude, with this concept one could argue there is no sound because the average amplitude is zero. Or that wight light is gray because it's either on or off.

Physicists may use this concept, but that's probably why they can't solve gravity or dark matter.

Anonymous

Anonymous Tue 20 Oct 2020 13:43:05 No.12252581 Report

Quoted By:

>>12252518
In short, just learn it for the test and feel free to add little notes about how each step is retarded and why.
One day we'll get the professors to quit wasting our time

Anonymous

Anonymous Tue 20 Oct 2020 14:35:32 No.12252721 Report

Quoted By:

What is the sum of the first n integers?

n(n+1)/2

suppose this function is continuous, where is it equal to 0?

at n=-1 and n=0.

Compute the definite integral of n(n+1)/2 between n=0 to n = -1. (In that order).

Anonymous

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Anonymous Tue 20 Oct 2020 15:16:57 No.12252839 Report

Quoted By: >>12252876

>>12252574
So was Ramanujan just a smart pajeet trolling anglo mathematicians all along?

Anonymous

Anonymous Tue 20 Oct 2020 15:29:13 No.12252876 Report

Quoted By:

>>12252839
No, he just wasted his live ''proving'' meaningless statements about series and got completely btfo by western mathematicians.

Anonymous

Anonymous Tue 20 Oct 2020 16:45:10 No.12253099 Report

Quoted By: >>12253129 >>12253133 >>12253137

>>12252574
So why does anyone care? Is it because it justifies quantum mechanics or something like that? Why does everyone like Ramanujan and his work so much considering? I am a mathlet and don't know how to answer this.

Anonymous

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Anonymous Tue 20 Oct 2020 16:59:57 No.12253129 Report

Quoted By:

>>12253099
Don't get caught up in the opinions of some angry undergrad.

E.g. you have for |x| < 1, you have

$\sum_{k=0}^\infty x^k \cdot = \dfrac{1}{1-x}$

and so with $x = -1+\epsilon$ , you find

$\sum_{k=0}^\infty (-1)^k \cdot (1 - \epsilon)^k = \dfrac{1}{2-\epsilon}$

So from this we e.g. know that $\sum_{k=0}^\infty (-0.99999999)^k \cdot$ is something extremely close to $1/2$ .
(But note that your calulator will very quickly give you numerical precision errors with a number like that)

While we can't plug in $\epsilon = 0$ in clasical analysis to obtain

$\sum_{k=0}^\infty (-1)^k = \dfrac{1}{2}$ ,

it's not like the value unmotivated either.

>So why does anyone care?
You can see math as inferring from axioms. Some theories are useful (e.g. in geometry), some are not. Summation methods for sums that are divergent in classical analysis aren't extremely useful, empirically speaking. But neither is computing cohomology groups of 17 dimensional spheres. Some people care because it's also math.

Anonymous

Anonymous Tue 20 Oct 2020 17:01:04 No.12253133 Report

Quoted By:

>>12253099
Combinatorics, your money would not be secure without it. A lot of the little problems Ramanujan wrote down for fun looking for how many ways you can add up to a number or something like that brought the field forward.

Anonymous

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Anonymous Tue 20 Oct 2020 17:01:55 No.12253137 Report

Quoted By:

>>12253099
Don't get caught up in the opinions of some angry undergrad.

E.g. you have for |x| < 1, you have

$\sum_{k=0}^\infty x^k = \dfrac{1}{1-x}$

and so with $x = -1+\epsilon$ , for positive $\epsilon$ , you find

$\sum_{k=0}^\infty (-1)^k \cdot (1 - \epsilon)^k = \dfrac{1}{2-\epsilon}$

So from this we e.g. know that
$\sum_{k=0}^\infty (-0.99999999)^k$
is something extremely close to $1/2$ .
But note that your calulator will very quickly give you numerical precision errors with a number like that.

While we can't plug in $\epsilon = 0$ in clasical analysis to obtain

$\sum_{k=0}^\infty (-1)^k = \dfrac{1}{2}$ ,

it's not like the value unmotivated either.

>So why does anyone care?
You can see math as inferring from axioms. Some theories are useful (e.g. in geometry), some are not. Summation methods for sums that are divergent in classical analysis aren't extremely useful, empirically speaking. But neither is computing cohomology groups of 17 dimensional spheres. Some people care because it's also math.

Anonymous

Anonymous Tue 20 Oct 2020 17:02:00 No.12253138 Report

Quoted By:

>>12252567
cringe retard

Anonymous

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Anonymous Tue 20 Oct 2020 18:19:50 No.12253473 Report

Quoted By: >>12254109

>>12252567
this makes sense to me, is it wrong?
t. non mathematician

Anonymous

Anonymous Tue 20 Oct 2020 20:53:10 No.12254109 Report

Quoted By:

>>12253473
Yes it's wrong. It's a neat trick, and it's interesting that it turns out that way, but it's not a legitimate way to get to that result. I don't know if this bad logic works in this case for some deep reason or if it's just a coincidence.

One example of something obviously wrong that leaps out at me is that you can't just say A=1/2 :

A = 1 - 1 + 1 - 1 + 1 -1 ...
A = (1 - 1) + (1 - 1) + (1 - 1) ...
A = 0 + 0 + 0 + ...
A = 0

or

A = 1 - 1 + 1 - 1 + 1 -1 ...
A = 1 (-1 + 1) + (-1 + 1) ...
A = 1 + 0 + 0 + ...
A = 1

This sum doesn't actually converge, so you can't just fuck around with algebra and come up with a number that it sums to.

Anonymous

Anonymous Tue 20 Oct 2020 21:17:15 No.12254227 Report

Quoted By:

>>12252518
You got trolled lol, the sum diverges (see any analysis I skript). You can't reorder sums without them being absolut convergent.

Anonymous

Anonymous Tue 20 Oct 2020 21:53:34 No.12254372 Report

Quoted By:

>>12252518
$\zeta (z) := \sum_{n=1}^{\infty} \frac{1}{n^z}$ converges for all complex numbers z with absolute value greater than one.
This function is analytic and it has a unique analytic continuation. This analytic continuation maps 1 to -1/12, so in a sense 1+2+3+... is related to -1/12

Anonymous

Anonymous Tue 20 Oct 2020 22:56:02 No.12254597 Report

Quoted By:

>>12252518
The sum of positive integers is divergent, ie not continuous if it's the value of a function. The -1/12 is what you get when you force a certain function to be continuous at certain values, which is a mathematically valid thing to do for certain classes of complex functions (functions that are defined for imaginary values ie i) this is analytic continuation. The sum of positive integers is one such value of one such function and in that context it does indeed make sense that it is -1/12.

Anonymous

Anonymous Wed 21 Oct 2020 02:04:45 No.12255232 Report

Quoted By:

>>12252518
>The sum of all positive numbers is -1/12
No, fucking hate that youtube video for telling everyone this.
>>12252567
None of those sums are convergent retard.

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