/sci/ - Science & Math

>>7744283
>meme shit.
No survivors.

>>7744298
>thinking centuries of mathematical rigor is wrong

let me guess you also think banach tarski is a lie?

>>>/r/atheism

>>7744303
Except that sum diverges to infinity my fellow memester. The -1/12 comes from analytically continuing the Zeta function to negative real values.

>>7744307
so basically you're admitting that math is a meme?

>>7744283
The amount of meme shit on /sci/ boggles my mind.
Also, the extent of plebe-shit I see posted

>>7744311
No.

>>7744283

>>7744313
>>7744316
https://www.youtube.com/watch?v=w-I6XTVZXww

https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF

there is empirical evidence that this is the case

mathfags btfo

>>7744319
I didn't say it wasn't true.
I'm saying that, OP posting this old as fuck shit, means he is a complete newfag.
If he is not a newfag, then he is a legitimate faggot. Chokes on dick daily, and so forth.

>>7744325
>homophobia

>>>/pol/

>>7744319
>hur dur memes.
>0+3+9+27+... = -1/2
What you are saying is as wrong as saying the factorial and the Gamma function are the same function. The Gamma function is the analytic continuation of the factorial. They are only equal for certain values. The same type of argument is true for the Zeta function. You have to analytically continue the Zeta function to pop out the -1/12.

>>7744283
The fact that everyone on /sci/ can only refer to things as memes

Doesnt' that also have some relation to p-adic numbers

>>7744338
>_>
You're trying too hard

>>7744341
>>0+3+9+27+... = -1/2

That's wrong
the series is divergent and sums to infinity

Fuck off with your memes.

>>7744353
>>_>
Nm actually you hsould go back to reddit

>>7744316
fuck you, I lost

I haven't heard that since I was in middle school

>>7744355
Exaclty. 0+3+9+27+... = -1/2 is meme math. Just like 1+2+3+4+...= -1/12.

>>7744371
Except you can prove the latter using maths.

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

therefore the integers aren't closed under addition

>>7744391
It's easy to confuse yourself with this shit but it's quite simple.

All of the ramanujan summation stuff, cutoff and zeta regularization do is look at the smoothed curve at x = 0. What sums usually do is to 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/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF#/media/File:Sum1234Summary.svg

>>7744298
999, nice. Can't wait for the new Zero Escape.

>>7744391

>>7744410
>It's just a unique value you can assign to a sum,

so it's like the right hand rule in physics where we assign the angular momentum to be perpendicular by convention because you only need one quantity to discern the direction of the force and not because there's actual momentum going in the perpendicular direction?

>>7744363
Lucky you. A senior said it during her salutatorian speech at graduation at my high school before I graduated.

>>7744423
post pics of her feet

Did you guys know that 9+90+900+9000...=-1? Anyone know why this is?

>>7744440
same reason as this

https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF#/media/File:Sum1234Summary.svg

Another cool divergent series that's also equal to -1/12 is .25 + 3 + 30 + 300 + 3000...

You might nto believe it, but it's true

>>7744440
[math] 9+90+900+9000... = 9 \, \sum_{k=0}^\infty 10^k = 9 \, \frac {1} {1-10} = -1 [/math]

where we used [math] \frac {1} {1-x} = \sum_{k=0}^\infty x^k [/math].

Of course, in the standard metric on the reals that expansion only exists for |x|<1, so you must do a continuation.

The on-and-off sum

[math] \sum_{k=0}^\infty (-1)^k = 1-1+1-1+1-1+... [/math]

is of course also divergent, but if you evaluate

[math] \sum_{k=0}^{10^7} (-0.999999993)^k = 1-0.999999993+(-0.999999993)^2 + ... [/math]

on a computer you'll find a number extremely close to guess what?
[math] \frac {1} {2} [/math].

Note that [math] \frac {1} {1-(-1)} = \frac {1} {2} [/math]

Btw.

[math]1+2+3+4+ .... [/math]

is the limit of z to 1 of

[math] \sum_{k=0}^\infty k \, z^k = z \frac {d} {dz} \sum_{k=0}^\infty z^k = z \frac {d} {dz} \frac {1} {1-z} = \frac{z} { (z-1)^2 } [/math]

while of course

[math] \frac{z} { (z-1)^2 } [/math]

is divergent at z=1. But a regularization of this expression gives you a log in place of [math] (z-1) [/math] and

[math] \frac {z} {(z-1)^2} - \dfrac {1} {\log(z)^2} = - \dfrac {1} {12} + O( (z-1)^1 ) [/math]

because

[math] \frac {r^2} { \log(1+r)^2 } = 1 + r + \frac{1}{12} r^2 + O(r^3) [/math]

>>7744494
I made an image for the limit to z=1 comparion.

I can add: Those numbers are what you generally get when you go from discrete to continuous.

Given a series [math] f(z)=\sum_{k=0}^\infty a_k z^k [/math] , one has

[math] \frac { f(0) } { f(f(0) \, z) } = \frac {1} { 1+\sum_{k=0}^\infty \frac {a_k} {a_0} (a_0\,z)^k } = 1 - a_1 \, z + (a_1\,a_1 - a_0 \, a_2) \, z^2 - (a_1\,a_1\,a_1-2\,a_0\,a_1\,a_2+a_0\,a_0\,a_3)\,z^3 + O(z^3). [/math]

Now consider the finite difference [math] \Delta_h f(x) [/math] of a function f about x, when shifted by h. With the Taylor expansion

[math] \Delta_h f(x) = f(x+h)-f(x)=f'(x)\,h + \sum_{k=2}^\infty \frac {1} {k!} f^{(k)} (x)\,h^k [/math]

we can compute the correction factor, by which the first order approximation fails to capture the finite difference:

[math] \frac {f'(x) \, h} { \Delta_h f(x) } = 1 - \frac {f''(x)} {2!} (\frac {h} {f'(x)})+ (\frac {f''(x) \, f''(x)} {2!\,2!} - \frac {f'(x)\,f'''(x)} {1!\,3!}) (\frac {h} {f'(x)} )^2+O(h^3). [/math]

Spoiler: Note that
[math] \frac {1} {2!\,2!} - \frac {1} {1!\,3!} = \frac{1} {2!\,3!} (3-2) = \frac{1}{12} [/math]

Or, for example, you have the freaky formula

[math] \int_a^b f(n)\,dn = \sum_{n=a}^{b-1} f(n)+ (\lim_{x\to b}-\lim_{x\to a}) \, ( \frac {1} {2} - \frac {1} {12} \frac {d} {dx} +...)f(x) [/math]

E.g. this gives classical identities like

[math] \int_2^4 n^2 dn = 18+ \frac {2} {3} = (2^2+3^2) + \frac {1} {2} (4^2-2^2) - \frac {1} {12} 2 (4^1-2^1) [/math]

>>7744361
:^)
Keep going
You'll tucker yourself out soon kiddo ^_^

>>7744540
go back to gaia online

>>7744283
Guys, does anyone know the answer to this in base 12?
Just curious

>>7744557
-0.1

>>7744423
>salutatorian speech
Good thing high school rewards intellectual sophistication instead of braindead socialite bimbos huh

but what for?

>>7744283
>>7744494
>>7744515
Is there a geometrical presentation for results like these?
Can it be explained with the higher manifold magicks?

>>7746225
https://en.wikipedia.org/wiki/Ramanujan_summation

Infinity

>>7746561
Infinity doesn't boggle my mind. I understand it completely.

>>7746562
No you don't, infinity is inconceivable

Nonlinear pdes are the key to answering almost all the millenium prize problems and it boggles my mind that im so far from studying it

>>7744319

They outright said this occurs in physics but no one has explored the math to figure out why it adds up that way...

...which means they're just blinding accepting it on faith. The same thing will happen that always happens... math will increase its definition and scope of what a number is, create a new system to handle it, and suddenly it'll make sense.

Godfuckingdamn this board sometimes.

>>7744283
Anyone attempting tot sum a divergent infinite series ought to be shot.
>physicists unironically take this seriously

>>7744319
>S - 1/4 = 4S
Yeah, about that.

I'll use S - 1/4 = 4S' for this.

S = 1+2+3+4+... has twice as many terms as 4S' = 0+4+0+8+... = 4(1+2+...) which means that the cardinality of the set consisting of all the terms of S is twice that for S', making S and S' different.

Furthermore, even if we took the steps seriously, and assume S = -1/12,
-1/12 - 1/4 = 4S
4S' = -1/3
4S' - S = -1/3 + 1/12 = -1/4
4S' - S = (0+4+0+8+0+12...) - (1+2+3+4+5+6...) = [(0+4)+(0+8)+(0+12)...] - [(1+2)+(3+4)+(5+6)...] = (4+8+12+...) - (3+7+11+...) = 1+1+1+... = -1/4,
contradicting 1+1+1+... = ζ(0) = -1/2 (which is itself a meme equation).

Too lazy to use LaTeX right now, but it's clear as day that analytic continuation is a meme in general.

>>7746731
Oops, -1/12 - 1/4 = 4S'

>>7744494
this is fucking ridiculous.
You've not shown that the expression
[math]\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k [/math]
is valid for |x| >= 1

>>7744319
>posting a memephile video
K.

>>7744338
>homophobophobia
>>>/tumblr/
>>>/reddit/
>>>/reddit/

>>7744316
I don't get this

mind = boggled

>>7744283
Double negatives, and folding paper. Certain types of spatial transforms just seem to fizzle out into a haze. I can't do it, not even sober.

Double negatives are tied into linguistic processing failures though. In my head I imagine it as dysfunction of the frontmost right chunk of the prefrontal cortex, I'm not sure why. There are periods when I can read certain comparisons over and over, and I just don't get it. Yet a week later, it's fine and quite fluid.

X is to Y as Q is to Z. This shit just tends to fall apart. I can decode the meaning, but I just can't... do anything with it.

It's an awful feeling. I hate waiting. My mother always made me wait as a child, for hours on end with no indication she'd ever show up or get ready. Pretty sure childhood is a good deal of what makes you, after all.

>>7746804
AH. I understand now. I should decode the similarity of each chunk first, then put them together.

I say that like I figured something out, but it's bullshit. It won't work next time either.

>>7744283
This got the be the most retarded notation ever come up with.

>>7744372
No you fucking can't. Your "proof" is like the proof that 1=2 with the hidden assumption that division by 0 is possible.

[eqn]\sum_{n=1}^{\infty}n = -\frac{1}{12}[/eqn] is WRONG.

If anything it's rather something like
[eqn]P\left ( \sum_{n=1}^{\infty}n \right ) = -\frac{1}{12}[/eqn]
where P(x) is some property of x.

>>7746731
>cardinality of the set consisting of all the terms of S is twice that for S'
You can easily make a bijection between the elements. The cardinality is one and the same.

>>7746285
That's geometric in the way of the integrals? How?

>>7744283
>Add a positive integer to a positive integer
>It's now a negative fraction
Mathematicians, everyone.

>>7746810
What the actual fuck lol? Did you just smash your retarded head on a keyboard in order to come up with that? Maybe if you paid more attention in your little local community college class, you'd be able to come up with someone that doesn't make me feel like I'm losing brain cells just looking at it.

>>7746826
>2015
>still disbelieving Cantor

>>7744283
how the fuck is the sum of positive integers equal to -1/12

>>7746901
Great argument! How about you actually tell me what is wrong with it, instead of ad hominem?

>>7744316
mother FUCKER :D

>>7746225
[math] - \frac {1} {12} [/math] is geometric in the sense that it pops up in geometric situations. E.g. it's at the third expansion position here in these formulas

in topology
https://en.wikipedia.org/wiki/Todd_class#Definition

resp. Lie-group theory (boxed Baker–Campbell–Hausdorff formula)
https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula#An_explicit_Baker.E2.80.93Campbell.E2.80.93Hausdorff_formula

But it's rather a combination object and
[math] 12 [/math]
can or should be read as
[math] \prod_{n=0}^3 n! [/math]
It pops up when you go from discrete to smooth.

(And FOR FUCK SAKE, yes (>>7746741 ), for this you can't prove the [math] - \frac {1} {12} [/math] result when you work in the discrete world.)

For example, say you're interested in series coefficients [math]a_n[/math] in some expression
[math]g(z) = \sum_n a_n z^n[/math]
vs. inverse integral transforms, [math]a(t)[/math]?). How do these object compare? What's even the right approach to set up [math]a(t)[/math]?
I guess naively we'd say consider
[math]g(z) = \int a(t) z^t \, dt[/math].
To investigate, we may check the sum and integral of mere monomials
[math]\sum_{n=0}^\infty 1 \, z^n[/math]
and
[math]\int_0^\infty z^t \, dt[/math].
We find they differ by Euler-MacLaurin sum terms. Where the sum converges,

[math] \sum_{n=0}^\infty \, z^n = \frac{1}{1-z} [/math]

while

[math] \int_0^\infty z^t \,dt = \int_0^\infty e^{t \log(z) } \, dt = -\frac{1}{\log(z)} = - \frac{1}{1-z} - \frac{1}{2} + \frac {1} {12} (z-1) + O((z-1)^2). [/math]

To match the integral computation to the sum, i.e. [math] -\log(z)[/math] <=> [math]1-z[/math],
we would first have to perform the conformal mapping [math] z \mapsto e^{z-1}[/math]. Indeed,

[math] \int_0^\infty (e^{z-1})^t \, dt = \int_0^\infty e^{(z-1)t} \, dt = \frac{1}{1-z} [/math].

>>7744316
>2015
>I'm still falling for this shit

(cont.)
So the smooth version the pairing [math] a_n z^n [/math] doesn't so much correspond to the pairing [math] a(t) \, z^t [/math] under the integral, but rather [math] a(t)\,e^{(z-1)t} [/math]. Make that [math] a(t) \, e^{-st} [/math] after a shift [math] z \mapsto 1-s [/math], so that a(t)=1 is connected to 1/s.

If a_0, a_1, a_2, ... is a series, the function
[math]G[a](z) = \sum_{n=0}^\infty a_n z^n [/math]
is called it's generating function. If [math] a(t) [/math] is a function, the corresponding function (shifted by one) is the Laplace transform
[math]L[a](s):=\int_0^\infty a(t)\, e^{-st}\,dt [/math].
For those two only, the constant series/function a_n=a(t)=1 both lead to [math]\frac {1} {1-z}=\frac{1}{s}[/math].

Say you’re again interested in

[math] 1+ (1/2) + (1/4) + (1/8) + ... [/math]

but too drunk to realize what it converges to. You can interpret this as

[math]1+ (1/2) X+ (1/4) X^2+ (1/8) X^3+ ... = \sum_{n=0}^\infty (X/2)^n[/math]

which is valid around X=1. You might remember that this is1/(1-X/2), which can be written as 2/(2-X), and now it’s evident that the series at X=1 is equal to 2. From this we can obtain many new sums converging against 2. Consider a re-parametrization
X(x) := 2 - e^(1-x)
which doesn’t move the point of interest because
X(x=1) = 2 - e^0 = 1
Now

[math] 1+ \frac {1} {2} X(x)+ \frac {1} {4} X(x)^2+ \dots=2 \frac {1} {2-X(x)} =2 \frac {1} {2-(2- \exp(x-1))} =2 \exp(x-1)=2+ \sum_{k=1} ^ \infty \frac {1} {k!} (x-1)^k[/math]

By expressing the sum in a series in (x-1), the result actually converges after the first term (2). Here the sum over the higher terms (x-1)^k also converges at x=1 because every individual term is 0, even. Some -1/12 regularizations do the same, the -1/12 term that’s actually part of the series is pushed to the front.
The rest of those sums don’t converge, though, they have to be cut off - hopefully in a schematically way.

I can say something about Ramanujan summation if you dare me.

>>7746989
>I can say something about Ramanujan summation if you dare me.
ok, go for it

>>7746983
>>7746989
Should I be able to understand this quite well after the introductory analysis courses: "introduction to real functions", and "limits and continuity"?

>>7744283

1 + 2 + 3 + 4 + X = -(1/12)
10 + X = -(1/12)

X = -(1/12) - 10
X = -10.083....

1 + 2 + 3 + 4 + (-10.083) = -(1/12)

There solved it for you OP

>>7747009
I...I...I don't know what to fucking say.

>>7747009
1-1+1-x = 1/2
1-x = 1/2
x = 1/2
1-1+1-(1/2) = 1/2

>>7747025
How about "thank you" you ungrateful twat.

>>7747009
Thank you.

>>7747005
It just some rant, if something seems fuzzy then because it is.

>>7746997
Okay, so only should, I think, again see it under the light that we pass from something discrete to something smooth.

So one starting point is the Bernoulli polynomials [math]B_k(x)[/math]. They are characterized on one side by behaving like monomial building blocks:

[math] B_k’(x)=k\,B_{k-1} [/math]

(several function do this, especially, of course, monomials functions x^k)
and made unique in that picking up smooth information about their coefficients is leading to the most simple algebraic building blocks:

[math] \int_x^{x+1} B_k(y)\, dy=x^n [/math]

The first few are shown in pic related and they can be compactly defined as coefficients of the z-expansion of the function [math] \frac {z} {{e^z} -1} {e} ^{x\,z} [math]

[math] \frac {z} {{e^z} -1} {e} ^{x\,z} = 1 + \frac{1} {1!} (x-\frac{1} {2} ) z + \frac{1} {2!} (x^2-x + \frac{1} {6} ) z^2 + \frac{1} {3!} (x^3-\frac{3 x^2} {2} + \frac{x} {2} + 0) z^3 + O (z^4) = 1 + \sum_{k=1} ^\infty \frac{1} {k!} B_k(x) \,z^k [/math]

I noted as similar generic expansion in (>>7744515). If f is a series with coefficients a_k, and a_0 not zero, then

[math] \frac { f(0) } { f(f(0)\,z) } = \frac { 1 } { \sum_ { k=0 }^\infty \frac { a_k } { a_0 } (a_0\,z)^k } =1-a_1\,z+(a_1\,a_1-a_0\,a_2)\,z^2-(a_1\,a_1\,a_1-2\,a_0\,a_1\,a_2+a_0\,a_0\,a_3)\,z^3+ O (z^3) [/math]

where we compare something of zeroth order f(0) to the full thing f(z).
(the above expression is normalized, f(f(0)·z), so that if f(0) isn’t 1 it factors out)

The particular number pops up when [math]a_n = \frac {1} {(n+1)!}[/math], so that [math] a_1\,a_1-a_0\,a_2 = \frac {1} {2!2!} - \frac {1} {3!} = \frac {1} {12} [/math].

If you take a look at the Todd class pic above (>>7746983), you see that this is exactly about the Bernoullis

(cont.)

>>7747005
It just some rant, if something seems fuzzy then because it is.

>>7746997
Okay, so only should, I think, again see it under the light that we pass from something discrete to something smooth.

So one starting point is the Bernoulli polynomials [math]B_k(x)[/math]. They are characterized on one side by behaving like monomial building blocks:

[math] B_k’(x)=k\,B_{k-1} [/math]

(several function do this, especially, of course, monomials functions x^k)
and made unique in that picking up smooth information about their coefficients is leading to the most simple algebraic building blocks:

[math] \int_x^{x+1} B_k(y)\, dy=x^n [/math]

The first few are shown in pic related and they can be compactly defined as coefficients of the z-expansion of the function [math] \frac {z} {{e^z} -1} {e} ^{x\,z} [/math]

[math] \frac {z} {{e^z} -1} {e} ^{x\,z} = 1 + \frac{1} {1!} (x- \frac {1} {2} ) z + \frac {1} {2!} (x^2-x + \frac{1} {6} ) z^2 [/math]
[math]+ \frac{1} {3!} (x^3 - \frac{3 x^2} {2} + \frac{x} {2} + 0) z^3 + O (z^4) [/math]

[math]= 1 + \sum_{k=1}^\infty \frac{1} {k!} B_k(x) \,z^k [/math]

I noted as similar generic expansion in (>>7744515). If f is a series with coefficients a_k, and a_0 not zero, then

[math] \frac { f(0) } { f(f(0)\,z) } = \frac { 1 } { \sum_ { k=0 }^\infty \frac { a_k } { a_0 } (a_0\,z)^k } =1-a_1\,z+(a_1\,a_1-a_0\,a_2)\,z^2-(a_1\,a_1\,a_1-2\,a_0\,a_1\,a_2+a_0\,a_0\,a_3)\,z^3+ O (z^3) [/math]

where we compare something of zeroth order f(0) to the full thing f(z).
(the above expression is normalized, f(f(0)·z), so that if f(0) isn’t 1 it factors out)

The particular number pops up when [math]a_n = \frac {1} {(n+1)!}[/math], so that [math] a_1\,a_1-a_0\,a_2 = \frac {1} {2!2!} - \frac {1} {3!} = \frac {1} {12} [/math].

If you take a look at the Todd class pic above (>>7746983), you see that this is exactly about the Bernoullis

(cont.)

>>7747043
(const.)

The constant coefficients of the Bernoulli polynomials are called Bernoulli numbers btw., and
[math]B_1(0)[/math] resp. [math]B_2(0)[/math] is plus or minus [math]1/2[/math] resp. [math]1/6[/math], as you see in the pic and the expansion,
and then [math]B_2(0)/2![/math] is [math]1/12[/math].

MacLaurin was interested the approximation of integrals
[math] \int_a^b f(n)\,dn [/math]
by sums
[math] f(a)+f(a+1)+f(a+2)+\dots [/math].
You can do a similar expansion as above (and/or use the fundamental theorem of (h-)calculus) and find the classical Euler–MacLaurin formula
http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula|

[math] \int_a^b f(n) \, dn = \sum_{n=a}^{b-1} f(n)+ (\lim_{x\to b} - \lim_{x\to a} ) ( \frac {1}{2}- \frac {1}{12} \frac {d}{dx}+\dots )f(x) [/math]

It says that a an integral of a function f is a sum of its base points minus Bernoulli-weighted curvature corrections.

The building blocks of analytical functions are monomials [math] f(n)=n^{k-1} [/math] and for those the formula is particularly simple, because most derivatives vanish. The formula immediately tells us that
[math] \int_a^b n^{k-1} \, dn = \frac{b^k}{k} - \frac{a^k}{k} [/math] is also
[math] \sum_{n=a}^{b-1} n^{k-1} [/math]
minus some lower order corrections.

Example 1: With [math] k=3,a=2,b=4 [/math] you get identity

[math] \int_2^4 n^2 dn=(2^2+3^2)+\frac{1}{2}(4^2-2^2)-\frac{1}{12}2(4^1-2^1) [/math]

I mentioned above already. It’s pretty freaky how many numerical identities you know now.

>>7744319
he looks like a rat

Example 1: With [math] k = 3,a = 2,b = 4 [/math] you get identity

[math] \int_2^4 n^2 d n = ( 2^2 + 3^2 ) + \frac {1} {2} ( 4^2 - 2^2 ) - \frac {1} {12} 2 (4^1 - 2^1) [/math]

Example 2: For [math] k = 2,a = 0 [/math] you can easily visualize the result:
The integral [math] \int_0^b n\,dn [/math] is the surface under [math] f(n) = n [/math] , i.e. the triangle area [math] \frac {b^2}{2} [/math] .
The sum of
[math] 0 + 1 + 2 + \dots + (b - 2) + (b - 1) [/math]
corresponds to the surface under a staircase. Hence for ever step, the sum misses [math] \frac {1}{2} [/math] for each step. You get
[math] \int_0^b n = \sum_{n = 0}^{b - 1} n - \frac {b}{2} [/math],
which you might write as
[math] \sum_{n = 0}^{b - 1} n = \frac {b(b - 1)}{2} [/math] .

Since some integrals are easier to solve than sums, you can use the formula also to compute sums. For example, form the above you get the classical formula

[math] \sum_{n = a}^{b - 1} n^{k - 1} = \frac {1}{k}B_k(b) - \frac {1}{k}B_k(a) [/math] ,

where [math] B_k(x) = x^k + \dots [math].
For example, with [math] k = 2,a = 0 [/math] , were [math] B_2(x) = x^2 - x + \frac {1}{6} [/math] , you find

[math] \sum_{n = 0}^{b - 1} n = ( \frac {b^2}{2} - \frac {b}{2} + \frac {1}{12}) - (0 - 0 + \frac {1}{12}) = \frac {b(b-1)}{2} [/math] .

Note that [math] f(n) = n [/math] has no second order curvature, so the [math] \frac {1}{12} [/math] 's cancel.

Now [math] \lim_{b\to\infty} [/math] of [math] \sum_{n = 0}^{b - 1}n [/math] is clearly undefined.

In Ramanujans theory of previously undefined infinite sums, you go back to the Euler–Maclaurin formula and disregard the whole upper bound [math] \lim_{x\to b} [/math] (which contains one of the [math] - \frac {1}{12} [/math] 's) and also the integral if if it's divergent. For [math] \sum_{n = 0}^\infty n [/math] , you're left with the [math] - \lim_{a\to 0} [/math] term, which is [math] - \frac {1}{12} [/math] .

So intuitively, what does the -1/12 or similar results really tell us about the series?
Some fairly important property?

>>7747100

>>7744410

[math] S = \frac{1}{{16\pi {G_{11}}}}\left[ {\int {{d^{11}}x\sqrt { - G} } \left( {R - \frac{1}{2}{{\left| {{F_4}} \right|}^2} + i{{\bar \Psi }_M}{\Gamma ^{MNP}}{\nabla _N}{\Psi _P}} \right) - \frac{1}{6}\int {{A_3} \wedge {F_4} \wedge {F_4}} } \right] [/math]

>>7747100
In view of the Euler-Maclauren formula it's just the statement that the polynomial function
f(n) := n
has no second or higher order terms, [math]n^2[/math], etc.. The fraction [math] \frac {} {1!2!3!} [/math] is just the second order number in the let's "look at it the smooth way" expansion and is what remains as f(n) := n terminates at this point.

it may be really basic but I don't think I've ever come across something as counterintuitive as the fact that (1/x) diverges

>>7744316
FUUUUUUUUU XDDD

>>7747131
Is that something like the 11d sugra action?

>>7747194
Most of it. That is the bosonic part of the action plus a dirac term. However I think there is one more fermionic part, I just don't know it.

>>7746731
>it's clear as day that analytic continuation is a meme in genera
Analytic continuation isn't a meme. Look at the factorial and Gamma function.

>>7747131
Trivial.

>>7744311
>>7744314
[math]lol[/math]

>>7746804
>>7746805
I feel ya, nigga f*m. I have that problem too :'(

>>7746838
together with Riemann integrals

https://en.wikipedia.org/wiki/Riemann_integral

>>7744319
Is string theory still a meme?

>>7744298
goddamn who was the killer in this ending?

>>7744316
FUUUUUUUUUCK

>>7744311
ROASTED

When I see math gimmicks like this it's easier to understand why math majors have trouble finding jobs.

I can't imagine the kind of personality you would need to major in math. it's almost like they sit around telling stories of times they solved a really hard math problem faster than their classmates.

>yfw 4channers complain about memes but post memes like every 2 posts
I'm leaving, going back to reddit...

>>7744283

if t = ∞

line = circle