/sci/ - Science & Math

[Anonymous No.9065732]

Did I make a mistake? Or is analytic continuation just a retarded meme?

[Anonymous No.9065739]

>>9065732
1+2+3+... != -1/12 no matter how many times you watch the numberphile video

[Anonymous No.9065743]

>>9065739
That's not even my question

[Anonymous No.9065747]

>>9065743
Your mistake was at Σn = -1/12 and the guy who replied to you was on point.

[Anonymous No.9065750]

>>9065743
>>9065747
Oh and at Σn^3=ζ(-3)
Τhat's not what ζ(-3) is.

[Anonymous No.9065757]

>>9065750
what is it then?
is the riemann zeta not the summation of the reciprocal of all natural numbers raised to the power of s (in this case -3)?

[Anonymous No.9065759]

>>9065750
I worded >>9065757 really poorly, but you know what i mean

[Anonymous No.9065770]

>>9065757
It is not. The zeta function is defined as Σ1/n^s for s with Re(s)>1 and then, for the rest of the complex plane, you take the (unique) entire function which agrees with Σ1/n^s for s with Re(s)>1.

[Anonymous No.9065772]

>>9065770
For Re(s)<=1 that sum diverges so it can't be a value of a function from C to C.

[Anonymous No.9065790]

>>9065739
>!= as "not equal"
Stay pleb /sci/

[Anonymous No.9065814]

>>9065772
is it not true though that
1 + 2 + 3 + 4 + ...
has an "assigned value" of -1/12 because that is where the analytic continuation of the riemann zeta function lands at -1/12 for s = -1, even it doesn't literally equal it?
And if so, does my problem not remain intact in the sense that multiple values are assigned to the same thing?
I don't know much about this form of math (I dropped out of high school and never studied maths at a higher level) so forgive me if I come across as a complete moron.

[Anonymous No.9065836]

>>9065814
>is it not true though that
1 + 2 + 3 + 4 + ...
Yes, it is true. Since that continuation is unique(it's a bit hard to prove that, but it is true), you can say that Σn corresponds to -1/12. But it is not equal to it and you can't do operations with -1/12 and get truths about Σn. Your picture in the OP shows that.

>forgive me if I come across as a complete moron.
You definitely don't sound like a moron. A moron wouldn't ask such questions.

[Anonymous No.9065891]

>>9065836
To be clear you can use -1/12 to get info about the behavor of Σn, but I haven't studied about that. I remember hearing that it was being used in string theory or something.

You can also take a look at math.stackexchange.com

https://math.stackexchange.com/questions/157045/explanation-of-zeta-function-and-why-1234-1-12

https://math.stackexchange.com/questions/39802/why-does-123-cdots-frac112

[Anonymous No.9065939]

Take a look here as well:

https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/

[Anonymous No.9066226]

>>9065836
Let's do the following proof by contradiction:

>Assume that one can assign [math]\sum{n^s}[/math] the value of [math]\zeta(s)]/math] for Re(s) < 1
>(a) 1 + 2 + 3 ... = -1/12
>(b) 1 + 1 + 1 ... = -1/2
>Subtracting (b) from (a) yields 0 + 1 + 2 + 3 ... = -1/12 - (-1/2)
>0 + 1 + 2 + 3 ... is still 1 + 2 + 3 ...
>5/12 = -1/12

See something wrong there?

[Anonymous No.9066232]

>>9065790
> != isn't 'not equal'

[Anonymous No.9066245]

>>9065732
>Did I make a mistake?
Yes.
>Or is analytic continuation just a retarded meme?
Yes.

[Anonymous No.9066351]

>>9065732
Terry Tao explains the values of zeta at negative values as the constant value appearing in the
(not necessarily convergent) asymptotic expansion using a sufficiently decaying cutoff function.

https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/

[Anonymous No.9066354]

Shit, repost of
>>9065939
My bad
>>9066351
:(
We eat the same food

[Anonymous No.9066363]

>>9066245
He should have XOR'd
OP is sloppy OR he believes the Law of excluded middle.

[Anonymous No.9067005]

>>9066232
Cs fag go home