/sci/ - Science & Math

[Anonymous No.6301167]

hello science. I learned that 1+2+3+4... = -1/12

How can we conceptualize this? or can humans not??

[Anonymous No.6301175]

>>>/out/

[Anonymous No.6301482]

Its not true

[Anonymous No.6301488]

>>6301167
It's pretty simple. 1 + 2 + 3 + 4 ... is divergent, meaning it goes to infinity.

Divergent series are WEIRD. For instance, think of the series

1 - 1 + 1 -1 + 1 - 1 + 1 ....

What does it equal?

Well, you might say it equals zero-

(1 - 1) + (1 -1) + (1 - 1).....

But, you could use the same argument to say it equals one,

1 + (-1 + 1) + (-1 + 1) ....

And with more fiddling, you could make it equal 1/2 , or you could make it any damn thing you want just by changing the order of the terms.

Essentially, what this means is "equals" doesn't mean a damn thing in the context of divergent series, so certain special peculiar mathematical definitions of "=" have been devised for divergent series. One of these, based on the Reimann-Zeta function and dicking about with imaginary numbers, spits out -1/12 for the sum of all real numbers.

This is, however, extremely misleading for ordinary intuitive defininitions of "1+2+3+4..." and "=" , so for the way you're probably thinking about the problem that answer doesn't really apply.

[Anonymous No.6301491]

>>6301488

Here's a more precise explanation of how the hell you get "-1/12" as the sum of all natural numbers.

http://mathgarage.wordpress.com/2012/11/16/the-sum-of-all-natural-numbers-is-negative-one-twelfth/

But basically, the answer is "1+2+3+4... = -1/12 because we had to make up our own special definition of '='"

[Anonymous No.6301588]

>tfw you have seen the explanations for this and still don't get it

[Anonymous No.6301793]

>>6301167
In the plane of complex numbers, {zeta}(s) =
{Sum from i=1 to n} of 1/(i)^s
Let s = -1 and let n approach infinite, then
{Zeta}(-1) = 1+2+3+4+...
We also know, in the complex plane, {zeta}(-1) = -1/12

[Anonymous No.6301822]

It is kinda like how you can't really define <span class="math">0^0[/spoiler].
<div class="math">\lim_{x \to 0}0^x=0 </div>
but
<div class="math">\lim_{x \to 0}x^0=1</div>
both values would make sense in their way.

-1/12 is the same way, you just take the limit in a different way than just summing up to a bigger and bigger N

[Anonymous No.6301839]

>>6301491
more like a made-up definition of infinite sum
>>6301793
sum doesn't make sense for s=-1. Need analytic continuation.

[Anonymous No.6301843]

>>6301839
The zeta function is analytic continuation...
I'm not using a ramanujan summation

[Anonymous No.6301845]

You know how <span class="math">\frac{1}{1-x}=1+x+x^2+x^3+\dots[/spoiler] for
<span class="math">|x|<1[/spoiler]? Can we put in 2 for <span class="math">x[/spoiler] and get <span class="math">1/(1-2)=-1=1+2+2^2+\dots[/spoiler]? It's ok if we define the sum notation to mean the analytic continuation of the function beyond the domain of convergence. But it isn't literally a sum.

[Anonymous No.6301848]

>>6301845
>|x| < 1
>let x = 2
U wot m8?

[Anonymous No.6301850]

>>6301839
Ok, I agree <span class="math">\zeta(-1)=-1/2[/spoiler], but writing this as a sum is either nonsense or notation that redefines infinite sum.

[Anonymous No.6301852]

>>6301848
Yes, that's the point. You can't really just stick 2 in there, because the sum doesn't converge. You can put it in the left hand side though. That's basically what this -1/2=1+2+3+... thing is doing.

[Anonymous No.6301854]

>>6301850
It does redefine infinite sum, its analytic continuation.
It only works in the complex plane, which involves imaginary numbers

[Anonymous No.6301855]

>>6301822
>both values would make sense in their way.
Well, let's not overlook the fact that
<span class="math">\lim_{x \to {0^+}}x^x=1[/spoiler]

[Anonymous No.6301864]

>>6301852
Except 1+2+3+4+... = -1/12, not -1/2.
The geometric series is not applicable.

[Anonymous No.6301883]

>>6301864
yes, sorry, -1/12. It's an analogy. Not applicable, but illustrates analytic continuation.

[Anonymous No.6301887]

Is this the new .999 = 1?

[Anonymous No.6301889]

Here's basically how it works.

First, look at this series:

S = 1 - 1 + 1 - 1 + 1 - 1 + 1 ...

What happens if we subtract it from one?

1 - S = 1 - ( 1 - 1 + 1 - 1 + 1 - 1 ....

Which is the same as:

1 - S = 1 - 1 + 1 - 1 + 1 - 1 + 1 ...

In other words,

1 - S = S

So,

1 = 2S

So,

S = 1/2.

Makes sense? (This is basically a less-rigorous explanation of Cesaro summation.)

Now, let s be the sum of all natural numbers.

s = 1 + 2 + 3 + 4 + 5 ....

Let's try the same trick.

s - 4s = 1 + 2 + 3 + 4 + 5 .... - 4 - 8 - 12 - 16 - 20 ...

(s - 4s is, of course, obviously - 3s.)

So, -3s = 1 - 2 + 3 - 4 + 5 - 6 ...

(Because we subtracted 4 from 2 , 8 from 4, 12 from 6 ... )

Or, to put it another way,

-3s = 1 - (2 - 3 + 4 - 5 + 6 ...)

And by subtracting the series we just looked at,

-3s = 1 - ( 1 - 2 + 3 - 4 + 5 ...) - ( 1 - 1 + 1 - 1 + 1 - ...)

Since the sum of that alternating series is 1/2 , and we'd already defined s = 1 - 2 + 3 - 4 ...

-3s = 1 + 3s - 1/2

So, s = -1/12.

[Anonymous No.6301892]

>>6301889
One typo:

"and we'd already defined s = 1 - 2 + 3 - 4 ..."

Should be

"and we'd already defined -3s = 1 - 2 + 3 - 4 ..."

[Anonymous No.6301898]

>>6301887
Yes, except that people are not remotely at the point where they can make a good judgement on the boundaries of applicability

[newt !!refi+zqwt0Y No.6301922]

>>6301167
1+2+3+... =/= -1/12

we all know that it's impossible for the sum of positive integers to give a negative number.

The fact is that when people say "1+2+3+4... = -1/12" they don't actually mean "plus" and "equals"

they're just abusing those operations without even letting the reader know that they're being used to mean something totally different from their geneirally defined meaning.

Basically it's physicists abusing maths like normal .
see >>6300815

[Anonymous No.6301929]

>How can we conceptualize this?
Start by understanding that at least one of +, ..., and = means something entirely different from what it's normally used to mean.

Divergent series don't have sums of the normal, familiar sort. They're attaching a value to a divergent series, and arbitrarily calling it a "sum".

[Anonymous No.6301968]

>>6301922
damn newt. how can one person be this buthurt about mathematical notation?

(btw, thanks for using a tripcode sothat i can filter you.)

[Anonymous No.6301973]

>>6301929
there is nothing arbitrary about it, the sum of a convergent series must satisfy 4 conditions, these sums satisfy 3 of them.

[Anonymous No.6301985]

I see all th proofs use the term infinity minus infinity at least at some point in mathematics. Is that where the result stops matching what it was suposed to? Because infinity minus infinity was supposed to be undefined.

[Anonymous No.6301992]

>>6301985
not if you use limits like you should.
the limit of x -> infinity of x-x is 0, not undefined.

[Anonymous No.6301999]

>>6301992
Are you mathematicians just trying to confuse the general population?

[Anonymous No.6302004]

>>6301985
Divergent series are WEIRD. The only reason "1 + 2 + 3 + 4 + 5 .." = -1/12 is because the ordinary definition of 'sum' doesn't have any meaning for a divergent series, in much the same way 1/0 doesn't really mean anything for ordinary division.

So using techniques like zeta-function regularization, analytic continuation, or Ramanujan summing, you can assign meaning to the idea of 'summing a divergent series' , and these give you -1/12 as the sum of the series "1 + 2 + 3 + 4 + 5 + ...". Mathematicians LOVE to do this - if they find a situation where ordinary ideas don't apply, like "sum" or "dimension" , then they try to find a more general system that gives the same answers for all the ordinary cases, but also gives an answer for the cases where the ordinary system wigs out. ("Fractal dimensions" are a similar situation.)

So it's not actually wrong to say that 1 + 2 + 3 + 4 + 5... 'sums' to -1/12.

But it would be ABSOLUTELY wrong to say "if you add up all the natural numbers, you get - 1/12."

[Anonymous No.6302024]

>people who don't actually math trying to math

Jesus, guys, using ambiguous/incorrect notation and then pretending it's somehow deep and meaningful is stupid.

[Anonymous No.6302029]

>>6301992
what's
lim x-> inf x^2-x?
or
lim x-> inf 2x-x?

[Anonymous No.6302032]

>>6301973
>...and the fourth is arbitrarily dismissed.

[Anonymous No.6302046]

>>6301992
you don't limits on divergent functions

[Anonymous No.6302049]

>>6302029

Please write the definition of "limit."

[Anonymous No.6302098]

>>6301588

http://www.youtube.com/watch?v=w-I6XTVZXww

If you don't get this, then you should just stop.

They're still just cheating math, though.

[newt !!kKr6sBQ2r05 No.6302200]

>>6301968
Hurr hurr Oh no what will I do if you filter my tripcode???

yeah, because that works. fucking tard.

[newt !!kKr6sBQ2r05 No.6302205]

>>6302098
that's the worst explanation for it because it actually pretends that what they're doing is valid in normal arithmetic when it isn't at all.

[Anonymous No.6302203]

>>6302200
Are you EK?

[Anonymous No.6302212]

>>6302203
She uses those terrible reaction faces
>>6302205
he specifically says it's not normal limits.
It's not arithmetic no matter what way you look at it.

He basically says it's a Cesàro sum without saying the word Cesàro

Just another tripcode pretending he knows anything

[Anonymous No.6302780]

>>6302212
>She uses those terrible reaction faces

Often she doesn't. She's easily recognizable by her style though.

[Anonymous No.6302813]

>>6302029
do u even algebra?
2x-x = x, this means limit is inf
x^2-x is inf too as x^2 rises faster

[Anonymous No.6302839]

Is diverging series the new 0.99.. = 1?

Also, here's a riddle for the thread: what's wrong with the following "proof"?

S = 1 + 2 + 4 + 8 + ...
S = 1 + 2(1 + 2 + 4 + 8 + ...)
S = 1 + 2S

Which has the solution -1 (or infinity if you consider the fixed points on the associated Riemann sphere)

Thus 1 + 2 + 4 + 8 + ... = -1

????

[Anonymous No.6302851]

It's bullshit.

[Anonymous No.6303531]

>>6302839
The series isn't divergent.