/sci/ - Science & Math

No

>>10839504
No, where did this misunderstanding come from?

If [math] \sum_{n=0}^\infty [/math] is the notation for the sum in analysis, then [math] \sum_{n=0}^\infty n [/math] is divergent.

It can be pointer out that Peano arithmetic doesn't enable you to compute infinite sums, so e.g. to prove things like [math] \sum_{1=0}^\infty \frac{1}{n} = 2 [/math], you need uncountable sets such as [math] {\mathbb R} [/math] and a metric on it - i.e. I huge ballast of mathematical gadgetry.

There are other theories of infinite sums where the result are true, e.g. the Ramanujan sum. In the theory of (different) infinity sums, you can define properties your notion of sum should fulfill, and the sum over n is one that it's particularly hard to give finite meaning to. In comparison, for example, the sum over (-1)^n can be given a value in many theories of infinity sums.
Not all theories giving a value to n are algebraic in that sense, e.g. analytic continuation suggests a value for it too.
But again, the theory with most applications - analysis - doesn't assign a value to the sum over n.

-1/12 pops up in varous related forms also in analysis, e.g. Lie theory and regulatizations.

If this is cause of numberphile don't believe them. They're known for bs info and clickbait. There's a german dude on youtube who shows it's bs.

>>10839504
NO. It's just a bunch of algebraic hocus pocus

If [math] \sum_{n=0}^\infty [/math] is the notation for the sum in analysis, then [math] \sum_{n=0}^\infty n [/math] is divergent.

It can be pointer out that Peano arithmetic doesn't enable you to compute infinite sums, so e.g. to prove things like [math] \sum_{1=0}^\infty \frac{1}{2^n} = 2 [/math] or [math] \sum_{1=0}^\infty \frac{1}{n^2} = \frac{1}{6} \pi^2 [/math], you need uncountable sets such as [math] {\mathbb R} [/math] and a metric on it - i.e. I huge ballast of mathematical gadgetry.

There are other theories of infinite sums where the result are true, e.g. the Ramanujan sum. In the theory of (different) infinity sums, you can define properties your notion of sum should fulfill, and the sum over n is one that it's particularly hard to give finite meaning to. In comparison, for example, the sum over (-1)^n can be given a value in many theories of infinity sums.
Not all theories giving a value to n are algebraic in that sense, e.g. analytic continuation suggests a value for it too.
But again, the theory with most applications - analysis - doesn't assign a value to the sum over n.

-1/12 pops up in various related forms also in analysis, e.g. Lie theory and regularizations. The source is most often the second term in pic related.

>>10839504
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

[math] \displaystyle
\zeta \neq \Sigma
[/math]

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

>>10839555
Is it truth?

>>10840422
You have kept repeating this for years now but you never pointed to that smoothed function that -1/12 is supposedly the limit to 0 of.

theres a funky sum for complex numbers ζ and you can find the formula, actually its an infinite series so its a sum to infinity
anyway you put complex numbers and viola you get some thing
anyway if you put some numbers you get like normal numbers for example if you put -1 you basipalli get reciptrotic value of number

funky sum(x) = 1/1^x + 1/2^x ..
so put x = -1 and you reverse up and down numbers to get
1 + 2 + ...
BUT
the problem is the funky sum doesnt defined for negative numbers, infact not for 0 either
but you can extend the graph of the sum on a plane but you can do it in many ways but the only aprooved one is the one that retains some stuff and that way is to flip it
and it turns out that when you map the flipped extended graph of funky sum, on x=-1 you get y = -1/12

>>10839520
probably something like this
https://www.youtube.com/watch?v=kIq5CZlg8Rg

>>10839504
Yes, but you have to shit in the street between each digit.

>>10839504
what?

>>10839504
Yes it does, see https://math.stackexchange.com/questions/1327812/limit-approach-to-finding-1234-ldots for one example
>>10839512
>>10839520
>>10839555
>t. undergrads
Yes it does, retards

>>10839504
>>10842505
sum of all positive integers*

>>10839504
these are really smart guys that came up with this anon. don't try to question them, when you are a simple everyday citizen. the magnets repel each other. don't ask why

>>10842060
The so-called “funky sum” is defined, you’re just extending the definition of summation in a natural way, much like we come up with natural extensions of addition, multiplication, exponentiation, etc. of naturals, integers, rationals, reals, complex numbers, vector spaces, functions, etc.

>>10839504
Fuzzy maths tricks, which I here from this day dub "Methematics"

>>10839504
What is true of the partial sums is not always true of the total sum. See pic related for example, where the partial sums are all rational but the total sum is irrational.

>>10841759
not him but here you go
https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF#Cutoff_regularization

>>10842526
true, but this has no bearing on the question

>>10842535
It has bearing because people frequently use the erroneous argument that such a sum cannot be negative because the partial sums are positive.

>>10842535
>>10842578
The correct statement would be: IF the ordinary limit of partial sums exists AND the partial sums are positive, then the total sum must be positive.

>>10842578
except that statement is true, the set of nonnegative reals is closed in the metric topology, therefore the limit of any such sequence in that set is contained in that set. in particular, if the partial sums of a series are positive, then the infinite sum is nonnegative or is not defined.
your rational sum counterexample works because the rationals are not a closed subset of the reals.

>>10841759
lrn2read

>>10839504
It is infinity

no, the sum is divergent.
there is a sense in which one can say this, in particular, the riemann zeta function is defined for all complex inputs whose real part is greater than 1. analytic continuation lets us "finish off" the rest of the function in other areas of the complex plane in a single, unique way. it happens that the analytic continuation of the riemann zeta function sends -1 to -1/12. if you pop -1 into the typical riemann zeta function as it is defined to the right of Re z = 1, then it looks like you're saying 1 + 2 + 3 + 4 + ... = -1/12. but you're not actually doing that anymore, instead you're popping -1 into a function which works like an infinite sum on some parts of the complex plane, and like any other function on other parts of the plane.
>then how can you rearrange the sum to get this value!!!
by assuming something nonsensical about the sum 1 - 1 + 1 - ... and letting that coincidentally take you to the value of the riemann zeta function at -1.

>>10839553
Mathologer, that guy is awesome

>>10842534
>>10842612
What's the function?

Yes, it must be of the form [math]-1/12 + O(n)[/math] but what is it?

>>10843419
google it

>>10842505
> negative result with only positive addends

>>10843419
did you actually read that link
>"The constant term of the asymptotic expansion does not depend on f: it is necessarily the same value given by analytic continuation, − 1/12.[1]"
>click on the cited article [1]
>"However, these issues can be resolved by replacing the abruptly truncated partial sums [math] \sum_{n=1}^N n^s [/math] with smoothed sums [math] \sum_{n=1}^\infty \eta(n/N) n^s [/math], where [math] \eta: {\bf R}^+ \rightarrow {\bf R}[/math] is a cutoff function, or more precisely a compactly supported bounded function that equals 1 at 0"
>look at equation 12: [math] \sum_{n=1}^\infty n \eta(n/N) = -\frac{1}{12} + C_{\eta,1} N^2 + O(\frac{1}{N})[/math] with eta as above

wow that was hard

>>10842505
>t. pretending to not be an undergrad when he clearly is
Σn is a divergent sum no matter how you look at it. It does not equal anything. The -1/12 result comes from analytic continuation, which is something entirely different from the sum "equalling" -1/12.

>>10845252
What is the big E doing?

>>10845334
just read the blog post, he spells it all out in excruciating detail

>>10845280
this. i mean even as an undergraduate you acquire awareness of this in calculus II