Numberphile defends this

this is called zeta regularization and it has real, useful applications in quantum field theory.

so that's why your parents called you that

-1 is a square root of unity, though.

youtube.com/watch?v=sD0NjbwqlYw

>implying mathematical abstraction doesn’t make sense

I have a feeling you just don’t have a good grasp on what a limit is.

...

thats a load of shit and you know it
there is plenty of counterintuitive stuff throughout all of math and science
good thing people use actual proofs in math and just intuitive conjectures

Fucking stop already.
No the sequence of partial sums of natural numbers up to some n does not converge to -1/12 under the usual definition of convergence.
Yes there does exist a counterintuitive and interesting correspondence between the number -1/12 and this series in several ways (analytical continuation of the Riemann Zeta function, p-adic numbers or smth., some curve fitting and the shit ramanujan did..)

stay mad

>there is plenty of counterintuitive stuff throughout all of math and science
then there is plenty of mistakes

Yes, as human intuition is infallible and the cornerstone of all rigor.

a square root of x is a number y such that x=y^2, and (-1)^2=1
so youre wrong

a square root of x is a number y such that log_y(x) = 2 and logarithms can't have negative base
so youre wrong

NO U

Counter intuitive results are much rarer than intuitive ones.

>The professor who came up with the proof that the sum of all natural numbers is -1/12
They got Ramanujan's reanimated corpse giving the lecture or some shit?

It's a miracle, he was supposed to be dead for an eternity, turns he was only dead for a month and then rose from the grave on the proper date of his death.

>This guy is a legit professor yet Veeky Forums thinks this is a meme?
appeal to authority argument

but I do believe this to be be true

Yes, *a* square root. Unless otherwise stated, the symbol √ or "root" usually denotes the principal square root only.
Equating different branches of the square root is where you are wrong.

The reason for M = 9.999... and 10M = 99.999... then mathfucking them in attempt to prove 0.999... = 1 is the same reason 1 + 2 + 3 + 4... doesn't actually evaluate to -1/12, and the reason is this:
It takes "shifting" into account as a piece of the solution, and disregards all and any work such shifting actually does. It wants to pretend the solution relies in adding 0 ahead of an infinite set, that adding such a zero would bare no difference on the sum of the set, but then misdirects this by comparing the set with the zero to the set without the zero to then compare each index within the sets. Where SetA is [1/4 + 1/8 + 1/16 + 1/32 + ...], SetA2 becomes [1/2 + 1/4 + 1/8 + 1/16 + ...], then fuckery occurs on SetA by dynamically redefining it mid-equation as [0 + 1/4 + 1/8+ 1/16 + ...]

Whether or not the 0 doesn't change the sum of the set bares nothing on the fact we are comparing index values between the sets, so paradoxically subtracting the redefined SetA from SetA2 based on their indicies (((coincidentally))) results 1/2 - 0 = 1/2, and every other index cancels out.
That an infinite sum of ever smaller halves = 1/2.

All the steps of that faggot indian's designated shitting journal for how 1+2+3+4 = -1/12 are retarded and invalid, even ahead of the already retarded concept of shift-fucking with infinite sets.

1/9=0.111...
2/9=0.222...
multiplying works just fine
10/9=1.111...
shifting works just fine

all you've got is m-muh feelings

read a book

Try adding 0.999... and 0.111... again there, retard.

9/9+1/9=10/9

M = 9.999
10M = 99.990
10M - M : 99.990 - 9.999
9M = 89.991
89.991 / 9 = 9.999
M = 9.999

Decimal shifting on a repeating decimal is unintellectual disingenuous horseshit.

0.1 × 10 = 1.0
0.11 × 10 = 1.10
0.111 × 10 = 1.110
0.1111 × 10 = 1.1110
0.11111 × 10 = 1.11110
0.111111 × 10 = 1.111110
0.1111111 × 10 = 1.1111110
•••
[math] 0.\bar{1}_{\frac{1}{9}} × 10 = 0.\bar{1}_{\frac{10}{9}} = 1.\bar{1}_{\frac{1}{9}}, - 0.\bar{1}_{\frac{1}{9}} = \big( 0.\bar{1}_{\frac{1}{9}} × 9 = 0.\bar{9}_{\frac{9}{9}} = 0.\bar{9}_{1} = 0.\bar{9}_{\stackrel{\leftarrow}{1} = 1[/math]

Its easy to confuse 0.999... with 1 because there is no natural way to create the number. You cant get it by multiplying 1/3 three times with reformed repetition, you can't get it by multiplying 1/9 nine times either. You can only craft it from infinite sums, but there are an infinite amount of ways to craft reformed 0.999... decimals, and they're all unequal. For example, [math]\sum_{n=1}^{\infty} \frac{1}{2^n} = 0.\bar{9}_{\frac{1}{2}}[/math] which is an evidently smaller number than [math]\sum_{n=1}^{\infty} \frac{2}{3^n} = 0.\bar{9}_{\frac{2}{3}}[/math] because it can be infinitely proveable for every test of n passed to both sums, the partial sums of the former will always be less than the partial sums of the latter. It is an ironic issue considering the only real way to craft 0.999... would be only 9/10^n, since that is the only sum who's every successive step adds only a 9 while any other x/(x+1)^n adds garbage numbers where the pattern of repeating 9's only exists in front of random numbers, but none-the-less an infinite sum in calculus of [math]\frac{x}{(x+1)^n}[/math] sums to a number who's predominant pattern appears in shorthand as [math]0.\bar{9}[/math]. In any case, you can't start a problem by invoking an arbitrary and undefined [math]0.\bar{9}[/math] number like in M = 0.999...

Lets use the sum of 9/10^n in lieu:

M = 0.999•••{9/10}
10M = 0.999•••{9/10} × 10 = 0.999•••{90/10} = 9.999•••{0}
10M - M = 9.999•••{0} - 0.999•••{9/10}
10M = 8.999•••{10/10}
9M = 8.999•••{10/10} - 0.999•••{9/10}
9M = 8.999•••{1/10}
9M = 0.999•••{81/10}
0.999•••{81/10} ÷ 9 = 0.999•••{9/10}
M = 0.999•••{9/10}

I'm pretty sure in the 2-adics you get that the natural numbers sum to -1/12.

yeah, because marbles obviously accelerate faster than ping pong balls, its just intuitive.

>M = 9.999
is that so...

[math] \frac{1}{3} = 0.\bar{3}_{\frac{1}{3}}[/math] .
[math] 0.\bar{3}_{\frac{1}{3}} × 3 = 0.\bar{9}_{\frac{3}{3}} = 0.\bar{9}_{1} = 0.\bar{9}_{\stackrel{\leftarrow}{1}} = 1[/math]
•
[math]M = \sum_{n=1}^{\infty} \frac{9}{10^n} = 0.\bar{9}_{\frac{9}{10}}[/math]
•
[math]10M = 0.\bar{9}_{\frac{9}{10}} × 10 = 0.\bar{9}_{\frac{90}{10}} = 0.\bar{9}_{9} = 1.\bar{9}_{8} = 2.\bar{9}_{7} = 3.\bar{9}_{6} = 4.\bar{9}_{5} = 5.\bar{9}_{4} = 6.\bar{9}_{3} = 7.\bar{9}_{2} = 8.\bar{9}_{1} = 8.\bar{9}_{\stackrel{\leftarrow}{1}} = 9 [/math]
10M = 9
10M = 8.999...{10/10}
[math]10M - M = 8.\bar{9}_{\frac{10}{10}} - 0.\bar{9}_{\frac{9}{10}} = 8.\bar{9}_{\frac{1}{10}}[/math]
•
[math]9M = 8.\bar{9}_{\frac{1}{10}}[/math]
•
[math]9M = 7.\bar{9}_{\frac{11}{10}} = 6.\bar{9}_{\frac{21}{10}} = 5.\bar{9}_{\frac{31}{10}} = 4.\bar{9}_{\frac{41}{10}} = 3.\bar{9}_{\frac{51}{10}} = 2.\bar{9}_{\frac{61}{10}} = 1.\bar{9}_{\frac{71}{10}} = 0.\bar{9}_{\frac{81}{10}}[/math]
•
[math]\frac{0.\bar{9}_{\frac{81}{10}}}{9} = 0.\bar{9}_{\frac{9}{10}}[/math]
•
[math]M = 0.\bar{9}_{\frac{9}{10}}[/math]

Can y9ou guys fycking calm down with your numbers and shit, im tripping balls and didnt even know this shit was possible?

also why do i think ion numbers?

christ, how much cocaine are you using?

>there is no natural way to create the number
9 * 1/9 is totally natural

>Abstract: Did you know that if you add up all the numbers from one to infinity you get something negative?
STOP

[math]
\sum_{n=1}^{ \infty} \frac{1}{2^n} = 0. \bar{9}_{ \frac{1}{2}}
[/math]

bullshit notation

[math] 0.\bar{1}_{\frac{1}{9}} × 9 = 0.\bar{9}_{\frac{9}{9}} = 0.\bar{9}_{1} = 0.\bar{9}_{\stackrel{\leftarrow}{1}} = 1[/math].

Nope.

1/9 × 9 = 9/9 = 1

2 * 1/9 = 0.222... is ok
3 * 1/9 = 0.333... is ok
10 * 1/9 = 1.111... is ok

you claim something magic happens at 9 * 1/9
based on nothing but m-muh feelings

Its not my feelings that 9/9 = 1

you are literally retarded

and also 0.999...

Nope.

...

t. Deutsche Physik

fucking based

>m-muh feelings

you'll get nowhere with shitty notation

>9/10-1/2=0
fucking kek

I think you don't understand math.

Like, pretty damn sure you don't understand math. Math is a language, my dude. Its not "only what I learned in school", contrary to what you seem to believe.

Yeah, I’ll write that a O(n^2) algorithm is actually a constant -1/12

Fucking brainlets.

ooooooh, so that's why
9/10-1/2=0

grandstand some more faggot

I don't understand why you are arguing that default calculus is obviously incorrect. Consider overdosing on heroin senpai.

>faggot
Why the homophobia?

...

...

>memenujan
Euler arrived at it first - Ramanujan should not be credited for it, but for the fact that he was able to derive it independently and with no education.

bumpkins

I swear to god this forum is filled with obese nerdz

a square root of x is a number y such that x/y = y
so you are wrong

threads like this really weed out who has and hasn't taken analysis. Very telling.

>This guy is a legit professor yet Veeky Forums thinks this is a meme?
>He thinks professors can't be memes.

>1/-1=-1
>1/1=1

Right anyway, default calculus is broken retard shit for brainlets.

I don't understand the problem.

[eqn]\sum_{n=1}^\infty\frac9{10^n}=\ell\\
\lim_{q\to\infty}\sum_{n=1}^q\frac9{10^n}=\ell\\
\lim_{q\to\infty}9\frac{\frac1{10^q}-1}{\frac1{10}-1}=\ell\\
\lim_{q\to\infty}1-\frac1{10^q}=\ell\\
\forall\epsilon>0\;\;\exists N\;\;\forall n>N\;\;\left|1-\frac1{10^n}-\ell\right|0\;\;\exists N\;\;\forall n>N\;\;\frac1{10^n}0\;\;\exists N\;\;\forall n>N\;\;n>-\log_{10}\epsilon[/eqn]
This is true for all [math]N\ge\lceil-\log_{10}\epsilon\rceil[/math].

If [math]\ell>1[/math], then [math]1-\ell0\;\;\exists N\;\;\forall n>N\;\;\frac1{10^n}+\ell-10\;\;\exists N\;\;\forall n>N\;\;\frac1{10^n}0\;\;\exists N\;\;\forall n>N\;\;\frac1{10^n}0\;\;\exists N\;\;\forall n>N\;\;(1-\ell)-\frac1{10^n}0\;\;\exists N\;\;\forall n>N\;\;(1-\ell)-\epsilon0\;\;\exists N\;\;\forall n>N\;\;\frac{1-\ell}20\;\;\exists N\;\;\forall n>N\;\;n

Numberphile is right and you guys are wrong. Deal with it. They've got the best math people in the world with them.

[math]0.\bar{9}[/math] is not a natural number, which is why in classical logical arithmetic solutions it becomes understandable to claim [math]0.\bar{9} = 1[/math] such as the case of multiplying the decimal of [math]\frac{1}{3}[/math] three times. Classically, the answer spits out 0.999..., yet we also know [math]\frac{3}{3} = 1[/math] so it is incredibly easy to then assume [math]0.\bar{9} = 1[/math]. This is not really the case though. A third times three doesnt equal 0.999... at all, it only equals 1, and if you extend a repeating decimal to attach a piece of the infinite problem to it in a similiar way arithmetic done with an imaginary number i will result an answer with i, we can get around the classical logical fallacy that 0.999... = 1

It is important to make note that infinity is not a natural number. You can not get to it, count to it, or approach it, and in any case where you start a problem with infinity already invoked, no intelligble arithmetic performed on it returns expected values as similar arithmetic on any normal or concatenated complex number might.
So when we get an infinite repetition in a decimal, we have to extend the identity of that decimal based on what we know of the values in the problem. That instead of [math]\frac{1}{3} = 0.\bar{3}[/math], you should know it as [math]\frac{1}{3} = 0.\bar{3}_{\frac{1}{3}}[/math], where the decimal evaluation also carries the identity of the work done to achieve it.
It can then be logically solved as [math]0.\bar{3}_{\frac{1}{3}} × 3 \rightarrow 0.\bar{9}_{\frac{3}{3}}[/math].
As stated that there is no natural way to come to [math]\bar{9}[/math] as a final solution beyond infinite sums, you then have to treat it differently than other repeating decimals. [math]0.\bar{9}_{\frac{3}{3}} \rightarrow 0.\bar{9}_{1}[/math] gives us the special property of a single one part to the work directive, and in the case of a single one part, it combines with the infinite repetition of 9's to act as an infinitisemal

This leads to
[math]0.\bar{9}_{1} \rightarrow 0.\bar{9}_{\stackrel{\leftarrow}{1}} = 1[/math], and now the equation of a third times 3 is fully solved with a final unmistakable solution of 1.

nah. [math] 0.\overline{9}=1[/math]
It is far better than arbitrarily carrying metadata.

Then explain after unraveling the limit.

Oh, that's elegant. Nice.

It doesnt equal 1 though, and you can know the reason why via where A and B dont produce any useful further maths but C and D can.

It is literally a brainlet trap of laziness to claim baselessly that [math]0.\bar{9} = 1[/math], and can be further proven why it cant be the case in pic related, which shows that there are infinitely many different and non-equal values that would be classically, incorrectly solved as simply "1".

Classical calculus infinite sums are senseless with the notion of convergence. If a sum 1/2^n can be proven lesser than 9/10^n infinitely many times for infinite tests of n proves that the final solutions cannot be equal, as for such a case to occur would require the probability of them being equal being [math]\frac{1}{\infty}[/math], where even if the probablity was struck would still invoke occams razor that it is more likely the real solution to the comparison is that 1/2^n < 9/10^n because it evaluated as such infinitely many times versus an improbable existence a single time they would have evaluated identically.

Infinity has an aspect of time and rate attributed to it that normal numbers, even complex numbers, do not. To use infinity in lieu of a number but not include any other variables for time and rate is asinine and incorrect, yet that is exactly what classical infinite sums do.

So you either need
[math]\rlap{\infty}_{\sum}[/math] to evaluate without the notion of convergence, or you need to extend an infinite sum with variables [math]\sum_{n=1}^{\stackrel{d.x}{l.y}[/math] to evaluate to a finite limit [math]y[/math] to a finite decimal accuracy [math]x[/math] to produce a precise, detectable convergence/rounding point.

[math]\rlap{\infty}{\sum}_{n=1} \frac{1}{2^n} = 0.\bar{9}_{\frac{1}{2}}[/math]

[math]\sum_{n=1}^{\stackrel{d.9}{l.y}} \frac{1}{2^n} = 1[/math] where we can solve for variables like x or y, where y above would be solved as 34.

In the same setup of [math]\frac{9}{10^n}[/math], y would instead be 10, so even thougg they both converge to 1, we can show 9/10^n does it at a faster rate in fewer steps than 1/2^n

Yeah, isn't this basic numerical analysis?

Oops my bad, y would have been 31 not 34.
At y=31, the sum is 0.9999999995343387126922607421875
Which is 9 significant 9's and a 5 which rounds up. At y=34, it was 10 significant 9's and unecessary since we round on the 10th by [math]d.9[/math]

The metadata is a very bad idea and you are wrong.

This isn't true. They both converge to 1 at the same number of steps.

idk

all i know is that infinity has been cheated in early maths and everyone oughta do their part in fixing it.
To do ifinite work as implied by an infinite limit for the sums 1/2^n and 9/10^n should actually dictate that, if graphed, these line paths never meet and essentially become parallel rather than apparently arbitrarily converging at 1. That if we could zoom in as if peering into a mandlebrot set, no matter how much we zoom in at the further and furthest locations along the lines tending towards 1, they would never actually meet. If we were to have a race between a car that travels at 9/10ths the speed of light against a car that travels 1/2 the speed of light in a straight track that extends from earth through space, but to race infinitely without ever reaching a goal as is infinite work or infinite summation, we could probably always tell that the first car is always way ahead of the second car, but it is virtually meaningless to say the second car will catch up at any point. If any arbitrary finite goal was placed, the first car would win the race ahead of the second car, and for any arbitrary comparison of a moment during an infinite race, the first car would still be ahead of the second car, so in the case of no end goal, no reason for either racer to stop and celebrate, it doesnt make too much sense to extrapolate any real solution from an infinite sum much less a final solution of convergence.

At the second test of equal comparison moments, the first car would be almost double the distance ahead of the second car, and so at an infinite'th test of a comparison moment, both cars would have traveled an infinite distance, but the distance between the two cars is also another infinite distance. This is where infinity breaks as a classical value in math because we've been taught to think [math]\infty + \infty = \infty[/math], that somehow the distance between the two cars doesn't matter because they each travelled an infinite distance, which is really just a psuedo-math way of saying they each travelled "really far idk i cant see them anymore but for some reason i think the second car finally caught up" aka effortless laziness. It is complete irony that infinity would be treated under undefined rules such that it is greater than any number, only to then say it is actually a finite limit of which no number can be greater to give the brain damaged hypothesis that any infinite value is equivocal because they reached an end.

Infinity is absolutely, truthfully paradoxically utilized in the maths as we know them, and only a liberal brainlet would argue otherwise because their entire identity hinges on how much professor cock they sucked to be programmed exactly what they were taught to get good grades so they can feel better than everyone else, despite what they've learned being nothing more than solutions to tests rather than actually understanding the concepts that define the maths they use, unabled to manipulate maths as a communicative language.

Then please explain: given the sets
[eqn]S=\left\{2^{-n}\::\:n\in\mathbb N\right\}\\
T=\left\{9\cdot10^{-n}\::\:n\in\mathbb N\right\}[/eqn]
are you saying that, in your opinion, [math]\sup S\ne\sup T[/math]?

SupS = 2^-1 = 0.5
SupT = 90^-1 = [math]0.0\bar{1}[/math]

Is it you or I who is misunderstanding supremum.

aside from pemdas fuckup, yes they aren't equal

Fuck, I'm an idiot. I forgot the sums. That was meant to read:
[eqn]S=\left\{\sum_{k=1}^n2^{-k}\::\:n\in\mathbb N\right\}\\
T=\left\{9\sum_{k=1}^n10^{-k}\::\:n\in\mathbb N\right\}[/eqn]
Now is it true that [math]\sup S\ne\sup T[/math]?

The largest numbers of these sets are still unequal and i'm not seeing how adding the infinite sum changes that, or rather how the infinite sum's values aren't different than 1/2^n and 9/10^n without needing to invoke sets or supremum.

If its just natural numbers and exclusive from infinity, then yes absolutely without a doubt these values are unequal, cause for any cumulative finite test of k in the infinite sums 1/2^k and 9/10^k, then 9/10^k will always be a larger number.
1: 0.5 < 0.9
2: 0.75 < 0.99
3: 0.875 < 0.999
4: 0.9375 < 0.9999
...
At a maximal real number, there would be a supremum real number amount of 9's in T compared to an infinum real number of 9's followed by various random digits in S, so that T > S

That the partial sums, both stopped at [math]n[/math] terms, are not equal was not up for debate.
What I want to know is: what are the suprema of those two sets?

For T and S to be equal, the final part of S would not be [math]\frac{1}{2^k}[/math], but instead need be [math]\bigg[ \frac{1}{2^k} \bigg] + \frac{1}{2^k}[/math] where
2: 0.75 + [math]\frac{1}{2^2}[/math] = 0.99 + [math]\frac{1}{10^2} = 1, where each adding the smallest part of their division at k would equal, even though the smallest part of S to sum 1 is much larger than the smallest part required of T to sum 1

[math]0.\bar{9}_{\frac{1}{2}}, 0.\bar{9}_{\frac{9}{10}}[/math] are the non-converging sums carrying identities of the work done, and the SupS is still 0.5 while SupT is still 0.9 is it not?

>SupS is still 0.5 while SupT is still 0.9 is it not?
Most certainly not.
0.75 is a member of T, as are 0.875, 0.9375 etc., and 0.5 < 0.875 < 0.9375 etc., so [math]\sup S>0.5[/math].
0.99 is a member of T, as are 0.999, 0.9999 etc., and 0.9 < 0.99 < 0.999 etc., so [math]\sup T>0.9[/math].

>0.75 is a member of T
Of S, of course.

Right so you meant that the set is not the elements of [math]\frac{x}{y^-k}[/math] at k per index, but that the the indicies are the sums of k to that index.

Then i already answered that with the non-convergent precise sums 0.999...{1/2} and 0.999...{9/10}, or
>At a maximal real number, there would be a supremum real number amount of 9's in T compared to an infinum real number of 9's followed by various random digits in S, so that T > S

SupT for n is a set of n size whose each digit is 9 and every index is the past index with another 9.
SupT = [math]\mathbb{U} \sum_{k=1}^{n}
\frac{9}{10^k} = \big{ 9, ... \big}_{n} [/math] and
SupS = [math]\mathbb{U} \sum_{k=1}^{n}
\frac{1}{2^k} = \Bigg{ \big{ 9, ... \big}_{n×0.3} + \bigg{ \sum_{k=n×3}^{n} \frac{1}{2^k} \bigg} \Bigg}_{n}[/math]

Without precisely concatenating as [math]0.\bar{9}_{\frac{1}{2}}[/math], its difficult to precisely extrapolate what a maximal number from S would look, and wont be able to provide you an answer that hasn't already been provided beyond SupT|SupS being element n, which do not equal each other.

Oh no what the fuck happened what did i do fucking latex

...

K needs to start at 0.3n but increment normally by whole n so w/e

The point is that if [math]\sup S=0.\bar9_{\frac12}[/math] and [math]\sup T=0.\bar9_{\frac9{10}}[/math], and as mentioned before [math]0.\bar9_{\frac12}