![dZqnCIX[1].png](https://i.imgur.com/7Dv9s26.jpg "dZqnCIX[1].png")
You gotta be shitting me.
This is the end result of Mathematicians not respecting the scientific method.
>>8110908
but it is scientifically proven .
try it yourself , add an infinite of sequential integers and see what the result is
>>8110917
>scientifically
Please don't use words you don't understand.
>>8110919
I once added an infinite number of integers together and got -1/12. Clearly it's just science.
>>8110928
Replicate it for me. You know how to make webms right?
I don't understand this meme
>maths
>science
pick one
>>8110938
It's easy just get one grain of sand, and two grains of sand and keep doing this until infinity until you have -1/12 grains of sand. Congrats you did it. Or learn some maths whatever's easiest.
>>8110908
https://youtu.be/SrU9YDoXE88?t=765
perfectly explains to you what mathematics is in three sentences .
>>8110908
String theory is based on this shit, no wonder it is falling apart
>>8110908
Are you trying to repost this -1/12 many times?
>>8111093
>String theory is based on this shit
Here's a experiment:
X = 1+2+3+4+5+6+...
2X = 2+4+6+8+10+...
X-2X = -X = 1+3+5+7+9+...
Therefore,
...-9-7-5-3-1 = 1+2+3+4+5+...
>>8111001
>three sentences
>takes 24 minutes
no nigga
>>8110908
Divergent series bruv, no sum
>>8110908
>math
>scientific method
Please stop posting this thread. I'm guessing you don't even know what math and science are.
>>8111479
Replicate the result in the OP for me.
>>8111453
Underrated post
>>8111453
X = 1+2+3+4+5+6+...
-X = 1+3+5+7+9+...
lol wut? pick one
>>8111486
>>8111453
>>8110908
You've written the thing wrong OP, maybe that's why you (and most of the thread) are confused about all of this.
It's supposed to be
[eqn]\sum^\infty_{n=1} \stackrel{\scriptstyle\Re}{\rightarrow} -\frac{1}{12}[/eqn]
See: https://en.wikipedia.org/wiki/Ramanujan_summation
>>8111453
Pic related left side, there are many nice notions of summations,
but you want to check that the one you use allows for basic notions like linearity to make sense.
>>8110938
In analysis, the sum
[math]0+1+2+3+4+...[/math]
read as
[math]\sum_{n=0}^\infty n [/math]
which is notation for
[math]\lim_{m\to \infty} \sum_{n=0}^m n [/math]
doesn't define a limit.
You may rewrite this as
[math]\sum_{n=0}^\infty n\,1^n [/math]
and so, while
[math]f_{sum}(z) := \sum_{n=0}^\infty n\,z^n [/math]
converges for z in the interval (0,1), the limit z to 1 of this function does not.
If you decide to look at
[math]f_{int}(z) := \int_{n=0}^\infty n\,z^n [/math]
you also get a function that stops being well defined going with z to 1.
However, it's worth pointing out that the difference between the above two functions is always finite (= they have an offset, but diverge equally fast), and the value at z=1 is -1/12.
Pic related, right.
-----
Whether you work out this negative fraction as above or alla Ramanujan (>>8111536) (in both of which cases you compute terms of the Euler–Maclaurin formula),
https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula#Asymptotic_expansion_of_sums
or whether you do the analytic continuation of the Riemann zeta function - in all those cases the number emerges as -1/2 of the second Bernoulli number, which is [math] B_2 = 1/6 [/math]
This is why you also find -1/12 as expansion coefficient in fields like Lie-group theory and algebraic topology.
It just depends how you define what the sum of an infinite series actually is.
>>8111559
at one point I wrote down a "school math" derivation of how -1/12 pops up in the regularized sum in analysis, pic related
>>8111372
?
>>8111483
You want me to watch the numberphile video and write down the steps for you?
>>8111453
I wish we lived 100s of years ago so you could be flogged for this blasphemy
>>8111462
>can't decode t=765
retard
Math isn't science.
Axioms are true because we say they are true. If contradictions appear then we can go back and change them, however OP's example can be logically derived from the relevant axioms, the proof isn't even difficult to follow.
If you personally do not accept an axiom to be true then that is fine, just know that you are severely outvoted by contemporary thought.
>>8112745
>If you personally do not accept an axiom to be true then that is fine, just know that you are severely outvoted by contemporary thought.
Or just pick another axiom.
>>8111453
also,
0 = 2 + 4 + 6 + 8 +...
>>8112745
>however OP's example can be logically derived from the relevant axioms
No, it can't. Not logically. The reason we set it to that value is because of something called analytic continuation.
>>8111559
If anyone is wondering why this gives the correct value you can show that
[eqn]\lim_{x \to 1} \left(x \frac{\mathrm{d}}{\mathrm{d}x} \right)^s \left( \sum_{n=0}^\infty x^n - \int_0^\infty x^n \mathrm{d}n \right)[/eqn]
is the same as
[eqn]\lim_{x \to 0} \sum_{n=0}^\infty \left( n^s + x \frac{n^{s+1}}{s+1} \right) e^{n x}[/eqn]
(Apply derivative, substitute x->e^x then apply L'Hopital's rule.)
It's still not [math]\sum_{n=0}^\infty n^s[/math] though. If the sum was from 1 instead of 0 you could take the limit directly for s < -1 and it would give the familiar sum for [math]\zeta(s)[/math] justifying it as the analytic continuation of the zeta function.
Further simplification of the above gives
[eqn] \lim_{x \to 0} \sum_{n=0}^\infty \frac{1}{s+1} \left( \frac{\mathrm{d}}{\mathrm{d}x} \right)^{s+1} \left( x e^{n x} \right) [/eqn]
[eqn] \frac{-1}{s+1} \lim_{x \to 0} \left( \frac{\mathrm{d}}{\mathrm{d}x} \right)^{s+1} \left( \frac{x}{e^x-1} \right) [/eqn]
[eqn] \frac{-\mathrm{B}(s+1)}{s+1} [/eqn]
Setting s to 1 does indeed give -1/12.
>>8113117
Did you just get motivated by my post and tried around or did you get this from somewhere else?
If the former, I can post some more or results, I play around with the zeta function now and then and defined a bunch of summations
>>8113117
you can't show that a derivative of infinity minus infinity is anything genius
>>8113153
Yeah, just from playing around with what was in your post. I'd definitely be interested in you posting them.
>>8113155
Do you somehow not see the limit.
>>8113166
You say
>If anyone is wondering why this gives the correct value
but I'm not sure if tracing it back to the zeta functions is a good "why" :P
One of my more wacky sums is taking
[math] \sum_{k=0}^\infty a_k [/math]
to a regularized
[math] \sum_{k=0}^\infty a_k-\langle a_k \rangle [/math]
where [math] \langle · \rangle [/math] captures some smooth ("analytical") properties of the expression [math]a_k [/math], pic related.
At the bottom I try to find an expression that interpolates the -1/12 and the finite result
[math]\sum_{k=0}^m k=\frac{1}{2}m(m+1)[/math],
but the expression is rather very ugly.
I've also tried to introduce a sum which basically takes the form of a differential operator, similar to the formal derivatives of path integrals that pull out moments/expectations values out of partition functions in statistical physics and quantum field theory.
The actually useful notion popped out when I tried to look at the expansions, for things like
[math] \dfrac{1}{1-e^{-x}} [/math]
leading to all the curios coefficients, in a more general context.
Austria is playing against Malta now, I'll come back to it later.
