Isn't infinity a shitty concept in math?

Isn't infinity a shitty concept in math?
I mean, you get retarded shit like
1+2+3+4+... = -1/12
or
1-1+1-1+1-1+... = 1/2
which is OBVIOUSLY wrong, but then they get pretty accepted within the community because it is valid to the current rules of the game and the rules don't get changed no matter how retarded the results that it produces are

Other urls found in this thread:

youtube.com/watch?v=sD0NjbwqlYw
en.wikipedia.org/wiki/Riemann_series_theorem
twitter.com/SFWRedditGifs

to justify that you need more sophisticated theory than basic freshman analysis, brainlet-kun

it's not an equality of numbers. it's an equality over very specific contexts.

>I mean, you get retarded shit like
>1+2+3+4+... = -1/12
>or
>1-1+1-1+1-1+... = 1/2
You obviously do not understand these "equations".
1+2+3+4+... = -1/12 is just plain wrong as anyone with a modest understanding in analysis can tell you, if you consider the standard limit definition of an infinite series. (d(-1/12,x)>1 for all x)

If you ACTUALLY understood what is meant by these things maybe you could actually criticize it, but you are seriously far too uneducated for that.

youtube.com/watch?v=sD0NjbwqlYw

math is a plateaued study where every major conclusion from here on out is going to be purely theoretical btw

what a retarded thing to say, off yourself

GET
USED
TO
IT

1/2=1+1+1+...=1+(1+1)+(1+1+1)+...=1+2+3+...=-1/12
Therefore 1=-6

Boy are you retarded?

Firstly, the sum of 1+2+3... Is divergent not convergent. You simply cannot handle a divergent sum as a convergent sum. You can reach - 1/12 in physics when you use the process of renormalization. With the riemann zeta function it implies that it might be equal to -1/12 as well, but that is NOT the equal as we know it, because riemann zeta function works only when n>1.
Secondly, the sum 1+1-1+1... DOES NOT EXIST. It does not have a limit! I could easily prove it is equal to 13/11....
Again in string theory there is used the process of renormalization!!

No, and you?

Associativity only applies to finite sums.

I AM!

Look up analytic continuation

*your

For you

1+2+3… is a divergent series retard.
It doesn't converge to anything.

Actually it works just fine for convergent sums, finite or not.

Never seen anything from this channel before, thanks for posting, watched a ton of stuff

only when the sum is absolutely convergent

\[Zeta] (s) := \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 1\), \(\[Infinity]\)]\(1/
\*SuperscriptBox[\(n\), \(s\)]\)\)
But this is true only when: s \[Element] \[DoubleStruckCapitalR] and s > 1!!
If you do analitycal continuation that is no longer true!
which means you would have:
\[Zeta] (s) =
2^s \[Pi]^(s - 1)
sin (\[Pi]s/2) \[CapitalGamma] (1 - s) \[Zeta] (1 - s)
and yes if you plug in -1 then you get:
\[Zeta] (-1) =
2^-1 \[Pi]^-2 sin (-\[Pi]/2) \[CapitalGamma] (2) \[Zeta] (2) =
1/2 * 1/\[Pi]^2 *(-1)*\[Pi]^2/6 = -1/12
and hence:
\[Zeta] (-1) = -1/12
but that is NOT equality as we know it as I have said before since \[Zeta] (s) = \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 1\), \(\[Infinity]\)]\(1/
\*SuperscriptBox[\(n\), \(s\)]\)\)
is true ONLY if s \[Element] \[DoubleStruckCapitalR] and s>1

sigh sry.. formating fucked me over

>they get pretty accepted within the community because it is valid to the current rules of the game
[citation needed]

1+2+4+8+...=-1 is more meaningful since ...11111111_(2) =-1 in 2-adic number system.

You're probably the only person ITT who actually understands that zeta function having analytic continuation has nothing to do with values of the sum.

Conditionally converging series can be made to equal any rational number by first substituting each positive number as a sum of positive and negative number, and then rearranging the series. That explains the 1+2+3+...=-1/12. If you are not allowed to do either, 1+2+3... will, of course, diverge to positive infinity.

It does all make sense; you are just brainlet. Study more. Work harder. Communicate better. Keep an open mind and your negative emotions in check.

You're even bigger brainlet than OP if you even assign rearranging of the series to the result of zeta's analytic continuation.

I was awaiting for some brainlet to counterargue me. Delicious.

Suppose you have series S = 1-1+1-1+...

You can make this equal 4 with the following rearrangement of numbers:
S = 1 + 1 + 1 + 1 (-1 + 1) (-1 +1) (-1 + 1)... = 4.
Because the series is infinite, you will never run out of (-1 + 1) pairs, and you will never have to add (-1 -1 -1 -1) at the "end" of the series.

Another example

Say you have S = 1 + 1 + 1 + 1 ...
To make it equal -1/12, simply replace each +1 in the series with:
(-1/12 + 0 + 13/12)
S = (-1/12 + 1 - 1 + 13/12) + (-1/12 + 0 + 13/12) ...
Now rearrange the series
S = -1/12 + 1 - 1 + 1 -1 ... = -1/12
You will have infinite number of +1 and -1 so the series will approach value of -1/12.

>pairs
>rearranging
>what's a fucking bijeciton

jesus fuck you're a retard.

en.wikipedia.org/wiki/Riemann_series_theorem

BTW: 1 + 2 + 3 + ... is not conditionally convergent. Idiot.

Haha losers

he explained exactly HOW it's related, genius

everyone understands the series diverges. while informal, the sum on the left side means zeta(-1) and not the series

That's just me.

No, he didn't. He just gave a functional relationship that the continuation must obey for Re(s) < 1. It has nothing to do with the sum.

it has everything to do with the sum 1/x^s, the analytic continuation is fundamental and unique, as you probably know.

I'm pretty sure you notice and understand the jargon involved (when writing those numbers one doesn't mean the series but the analytic continuation) but are just trying to be contrarian for some reason.

If you're so smart what's your alternative?

Apparently we agree, I thought you were arguing a different point.

He has nothing. He is just angry, brainlet loser. Virtually nobody ever beats the old masters.