Why does the sum of all natural numbers equal -1/12? From what

>>7752334

'=' denotes any equivalence relation. people don't use weird squiggly symbols generally. it's no approximation, but "analytic continuation" of a function. It's like saying the geometric series is equal to 1/(1-x) and then continuating it so that the geometric series at 2 is equal to -1 (where it is, in reality, infinity)

The 'natural' answer is that it approaches infinity. More specifically infinity to the second power. So lets say you have 1 + 2 + 3 ... + x. When you have really large values in x, the sum is approximately x^2. There's some series that you can expand it into, don't know if I have the picture, but when you do it you'll get x^2 - 1/12 + something/x + something else/x^2 ...
So, when people say that -1/12 is the answer, where they get that from is ignoring not just the infinitely small parts, but also the infinitely large parts, and only looking at that one constant, -1/12.

The most concise way to put it is -1/12 is the acceptable answer in higher levels of mathmatics. You can do more if you can reduce infinite series into constants than you can if you can only say things like 'yep, that's infinity'

>>7752339
Shut the fuck up, nothing you said is correct

It doesn't, in the most obvious sense.

Divergent series, i.e the sum of all natural numbers can be described by certain summation methods which transform the infinite series

For example the fourier transform of infinite length sinusoids is nice and compact

Sines are not really delta functions, they are just represented that way after that transform

Ok, so it's not an approximation but an "analytical continuation", yet at the end of your post you said that the answer to your example is, in reality, infinity. So it actually is an approximation because the real answer is infinity.

>>7752334
It's easy to confuse yourself with this shit but it's quite simple.

All the ramanujan summation stuff, cutoff and zeta regularization does, 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

>>7752369
>a value or quantity that is nearly but not exactly correct.

-1/12 is not close to infinity

-1/12 is completely correct, under a certain set of rules. Personal think mathematicians play way too loose with the symbols

>>7752334
>Why does the sum of all natural numbers equal -1/12?
It does not. It's just extremely poor notation in the absence of context to make retards think "wow how is this possible, math is so interesting".

>>7752372
Thanks for the explanation.
It just seems strange to use the "=" symbol when you make the agreement that it is only a reduced representation of an operation which makes it easier to work with certain type of problems.
Is it akin to the introduction of i as an imaginary value in the realm of complex numbers?

>>7752390
No; it's strange to use = because it's just incorrect.

You would never denote this particular equivalence with = alone, because it'd be confusing and misleading.

But because it's so "mind-blowingly weird" to have the sum of all numbers literally equal -1/12, the = version gets passed around because it's clickbaitier.

>>7752390
>It just seems strange to use the "=" symbol when you make the agreement that it is only a reduced representation of an operation which makes it easier to work with certain type of problems.
eh. overloaded operators are everywhere, and for the most part you can infer the meaning from context. it's only people who incorrectly write the expression to incorrectly imply a simple summation who are fucking things up.

if someone actually knows what they're doing they'll include the proper [math]\zeta{}()[/math] notation.

>>7752339
You cannot say something tends to infinity, you can only say something grows without bound

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>>7752339
>So lets say you have 1 + 2 + 3 ... + x. When you have really large values in x, the sum is approximately x^2. There's some series that you can expand it into, don't know if I have the picture, but when you do it you'll get x^2 - 1/12 + something/x + something else/x^2 ...
I can try to bridge those

For a function [math] f(n) [/math] on [math] {\mathbb N} [/math] that is sensible for more than just the natural, you can define an average between n and n+1 as follows:

[math] \langle f \rangle(n) := \int_n^{n+1} f(k) dk [/math]

The value [math] \langle f \rangle (n) [/math] equals [math] f(n) [/math] plus a terms that measures the growth to up to the next integer value.

For example, if
[math] f(n)=n^2 [/math],
then
[math] f(3)-f(4)=3^2-4^2=-7 [/math],
while
[math] f(3)-\langle f \rangle (3) = 3^2 - \int_3^{3+1} k^2 dk = - \frac {10} {3} = -3.333 [/math]

We have

[math] 1 + 2 + 3 + 4 + \, … = \sum_{k=0}^\infty k = \lim_{z \to 1} \sum_{k=0}^\infty k \, z^k = \frac {z} {(z-1)^2} [/math]

which is divergent at z=1.

However

[math] \sum_{k=0}^\infty \left( k \, z^k - \langle k \, z^k \rangle \right) = \frac {z} {(z-1)^2} - \frac {1} {\log(z)^2} = - \frac{1}{12} + O(z-1) [/math]

so [math]- \frac{1}{12}[/math] measures the immediate growth of f(n):=n in the above sense.

We can bridge the connection to the finite sum:
Look at

[math] \sum_{k=0}^m \left( k \, z^k - (1-q) \langle k \, z^k \rangle \right) [/math]

Mathematica tells us that this is

[math] \frac{(1-q) (z-1)^2 \left(z^{m+1}-1\right)+z \log (z) \left(z^m \left((m+1) (q-1) (z-1)^2+(m (z-1)-1) \log
(z)\right)+\log (z)\right)}{(z-1)^2 \log ^2(z)} [/math]

Taking the limit z to 1 results in

[math] \frac {1} {2} (m+1) (q\,m+(q-1)) [/math]

and for q to 0 we find the classical result.

On the other hand, taking the limit m to infinity before going with z to 1 gives us

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

>>7752793
and for q to 1 we find the classical result, I mean

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Mathematica

It doesn't in the sense of summation as typically known, it's true in a fashion but the summation sign as used in that expression doesn't mean what you think it does. Watch this for more info

https://youtu.be/LMw0NZDM5B4