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File: eq1.png (3 KB, 232x114)
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How does this equal -1/12
does the sum glitch or what
if i have infinite apples to count how do i end up with less then 1
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>>13639336
It happens when you claim you can divide by infinite or divide the sum of a divergent series based on its partial sums, retarded meme.
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>>13639336
It isn't. At least not for any normal definition of summation.
>>
The best treatment of this problem appears in Chapter 1 of Zee's QFT book, in the section on the Casimir effect. Everyone can see the argument that the formula is stupid but a lot of people have never seen the argument about we use the formula anyways.
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>>13639351
>Everyone can see the argument that the formula is stupid
Because they're thinking about normal summation. As far as pure math is concerned, you can come up with all kinds of unintuitive ways to assign a value to an otherwise undefined expression by playing with the definitions, so long as everything is internally consistent. What's surprising, though, is that this alternate take on summation is analogous to something in reality.
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>>13639348
so to get this straight it doesn't actually equal to -1/12 we just pretend
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>>13639391
Well, no one is forcing you to pretend that the plus in that expression means the same thing as regular summation.
>>
The result in Zee shows how you can end up with -1/12 without using summation at all, and then compare to that to the case where you might have used the stupid formula about the infinite sum. It's weird that using one way or the other both makes the same answer in the end.
>>
[math]1 + q + q^2 + q^3 + q^4 +\dots = \frac{1}{1-q}[/math]
therefore
[math]1 + 2 + 4 + 8 + 16 + \dots = -\frac{1}2[/math]
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>>13639445
Does not apply to divergent series.
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>>13639445
Meant to type this, you're cringe.
[math]\sum_{n=0}^{\infty} q^{n} = 1 + q + q^2 + q^3 + q^4 +\dots = \frac{1}{1-q}[/math]
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>>13639391
it goes thru -1/12 towards infinity
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>>13639473
Analytically extend to q = -1
Use zeta(s)*(1-2/2^s) = 1/1^s - 1/2^s + 1/3^s ...

For the -1/12 sum, look at q/(1-q)^2.
Let q go to -1 to get - 1 + 2 - 3 + 4 ... = -1/4.
This gives 1 - 2 + 3 - 4 ... = 1/4.
divide by (1-2/2^(-1)) = -3 to get zeta(-1) = -1/12.
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>>13639525
Regular summation of positive numbers does not result in a negative number. Case closed.
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>>13639551
Exactly SUM = ADDITION = POSITIVE NUMBERS(using postive number)
finally someone
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>>13639551
You will be disappointed to find out computers perform subtraction by adding two "positive" numbers.
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>>13639483
it goes thru -1/12 towards infinity
what do u mean by that
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>>13639551
what is a regular summation?
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>>13639525
is s meant to represent a imaginary number
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>>13639567
example
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>>13639564
>SUM = ADDITION
how do you add infinitely many terms?
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>>13639575
you put an equals sign at the end
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Holy shit, this thread reeks of underage
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>>13639396
so then whats it for what does the equation trying to represent
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>>13639583
1 + 1 + 1 + .. =
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>>13639336
Graph a line of all numbers, look at where that line crosses the Y-axis -> -1/12
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>>13639574
to do 30 - 20 in binary,
30 = ...0011110
20 = ...0010100
It would be a pain in the ass to require separate circuitry for addition and subtraction.
Let -1 = ...1111111
(-1) - 20 = ...1101011
this just flips every bit of 20 (simpler than addition circuitry)
(-1) - 20 + 1 = ...1101100
this uses addition circuitry
30 + (-20) =
...0011110 +
...1101100
=...0001010
this uses addition circuitry
(you keep carrying to the left forever)
You can check this gives the right answer. 2 + 8 = 10
Somehow negative numbers are successfully represented using sums of positive powers of 2.
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>>13639570
>what is a regular summation?
The one you were taught in elementary, where adding a positive number to a positive number always produces a positive number.
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>>13639568
They mean that if you STOP THE COUNT! it can = -1/12
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>>13639567
>computers perform subtraction by adding two "positive" numbers.
Computers perform one's complement modular arithmetic, brainlet.
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>>13639629
right, and you're saying that you can apply this summation to an infinite amount of terms?
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>>13639643
You can't apply it to a divergent series.
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>>13639360
>analogous to something in reality.
What?
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>>13639627
Do u just add all the numbers together and then draw a line?
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>>13639646
you're saying that you can apply this summation to an infinite amount of terms?
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>>13639676
Sometimes.
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>>13639568
look at the pic in
https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF
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>>13639679
how? how do you sum an infinite amount of terms?
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>>13639684
The same way you sum a finite series, except you'd have to go on doing it forever, so you have to invoke the concept of limits to assign a value.
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>>13639683
horizontal bar graph
curve does not fit the data, is below every point

some very questionable shit going on.
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>>13639695
>The same way you sum a finite series
>except it's different
yeah right. I don't need any sort of limits to sum a finite number of terms. in what way is it the same?
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>>13639707
>in what way is it the same?
Has the same properties as sum you'd get from regular addition of a finite number of elements.
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>>13639716
yeah, for example it's always defined, it's associative, it's commutative.. the list goes on
>>
is this board all underage and computer science pajeets?
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>>13639724
>it's always defined
True by definition for a convergent series.

>it's associative, it's commutative
"Sum" is not a synonym for "addition".
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>>13639716
then why does it equal to -1/12 should it equal to
an imaginary number or infinity
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>>13639703
floor(x)*(floor(x)+1)/2
= (x-{x})*(x-{x}+1)/2
=x^2/2 + x(1-2{x})/2 - {x}/2 + {x}^2/2
The x^2 term has average coefficient 1/2
The x term has average coefficient of 0
the constant term has average -(1/2)/2 + (1/3)/2 = -1/12
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>>13639748
>"Sum" is not a synonym for "addition".
but you're saying that it is. you've said summing infinitely many terms is the same thing as summing finitely many terms. but it's not. I don't need any limits to sum finitely many terms.
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>>13639758
>why does it equal to -1/12
The sum of a divergent series is undefined. You can assign some value to that expression, but it's not a normal sum anymore, which should be obvious, because it violates the relationships between sum and summands that you get with normal sums.
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>>13639766
The addition is the same addition, so the regular relationships between the sum and the summands holds.
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>>13639761
Let f(n) be the formula for the nth partial sum.
integrate f(x-e^t)*e^t*dt from -infinity to 0.
This should perform the average of the {x}^k coefficients properly.
Let x=0 for the constant term.
Doing this for the sum of cubes gives 1/120
https://www.wolframalpha.com/input/?i=integrate+e%5E%283t%29*%281-e%5Et%29%5E2%2F4+from+-infinity+to+0
which is zeta(-3)
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>>13639818
You could just integrate f(x-t)dt from 0 to 1
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>>13639703
https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF#Cutoff_regularization
>>
>>13639336
No
No
No.
(((complex analysts))) and (((number theorists))) broke the rules to allow their retarded function to work
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>>13639336
[eqn]\lim_{s \to 0}(\sum_{n = 1}^{\infty}{x^{1-s}})=-\frac{1}{12}[/eqn]
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>>13639880
as long as the circle is complete they can jerk themselves around and make up whatever rules they want...unfortunately they want to be taken seriously.
You either laugh at their foolishness or join the circle.
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>>13639880
Then anyone can create whatever rules and formulas they want. Math formulas and equations should follow order to work
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>>13639336
infty is clearly a half rotation, therefore inf^2 is a full rotation around the projective reals back to zero.
also, [math]1^2+2^2+...+(\inf-1)^2 = 0[/math]
These are definitions that mathfags should be taking for granted.
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>>13639336
>rejects hitomi analysis
>why are there these problems? guess I'll just ignore them
Funny
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>>13639336
You get the averages
Which gets weird when you have an infinite sequence
Watch the numberphile video
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>>13642751
>numberphile
numberphile should be raped in front of their families.
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>>13639564
Just because you don’t understand it doesn’t mean it doesn’t exist. You shouldn’t be taking sides. This is literally an equation that is utilized in other disciplined. Mathematicians know it is -1/12. There are some good explanations on this thread, but most people simply refuse to attempt to understand.
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>>13642754
Yeah but a normie can understand it with some effort
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>>13642759
*disciplines
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>>13639336
+1
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>>13639391
>we just pretend
yeah basically
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>>13642536
Infinity =/= negative infinity

Why would squaring half a rotation give you half a rotation?

Also how do you write things in that math format?
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>>13642769
No
You just don’t understand
It is actually a useful equation
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>>13639336
if you are not Ramanujan the sum is divergent
if you are Ramanujan it should be quite obvious why it is -1/12
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>>13642781
>how do you write things in that math format?
sticky >> https://sites.google.com/site/scienceandmathguide/ >> /sci/ LaTeX Tutorial
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>>13642759
but thats simply not possible dividing infinity to show that an infinite sum equals to a negative number thats not how math works
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>>13642769
then we should pretend about all mathematics 2+2=5 why does this get a exception
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>>13642536
>inf-1 is at the end of a list
lol, into the trash it goes
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>>13642950
>an infinite sum equals
it's not that, retard. lrn2read
>people simply refuse to attempt to understand
this, you're not even trying
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>>13639351
>does something completely different
>btw I didn't do the -1/12 thing that's dumb
This is the best treatment?
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>>13639391
>>13642950
>>13642952
we also "pretend" that 1+1/2+1/4+.. "equals" 2. addition is undefined for infinitely terms, the 2 is not a result of an arithmetic operation, it's a number associated to the expression based on some reasoning. -1/12 is a number associated to 1+2+3+... based on a different reasoning.
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>>13639880
new mathematicians continue to make new rules in order to do new math, and forever there will be other mathematicians to say "reeeee you cant just make new rules" lol
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>>13639336
What is this fuckery? i tried the first one in SymPy and it won't converge. No way the answer is -1/12. No need to try the other two because square and cube obviously won't converge either.
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>>13643019
the -1/12 is not supposed to be the limit of the sequence of partial sums
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>>13642997
>addition is undefined for infinitely terms
lol no it isn't
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>>13643159
yes, it is. addition is a binary operation.
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>>13642997
>addition is undefined for infinitely terms
"Sum" is not synonymous with "addition", brainlet.
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>>13643181
exactly. "sum" of an infinite series is a number assigned to the series based on an ad hoc definition. it's not a result of addition.
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>>13643195
>it's not a result of addition
The same relationships between sum and summands hold, as in summation of a finite series using addition. Why is that so hard for you to wrap your mind around?
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>>13643172
ok sweetie
[math] \displaystyle
\boxed{0 < p < 1} \\
p^n-1 = (p-1)(p^{n-1}+p^{n-2}+ \dots +p+1) \\
\dfrac{p^n-1}{p-1} = \sum \limits_{j=0}^{n-1}p^j \\
\displaystyle
\lim_{n \to \infty} \dfrac{p^n-1}{p-1} = \lim_{n \to \infty} \sum \limits_{j=0}^{n-1}p^j \\
\displaystyle
\dfrac{0-1}{p-1} = \sum \limits_{j=0}^{\infty}p^j \implies \dfrac{1}{1-p} = \sum \limits_{j=0}^{\infty}p^j
[/math]

[math] \displaystyle
p=0.1 \\
\dfrac{1}{1-0.1}=\frac{10}{9} = 1 + \frac{1}{9} \\
\sum_{j=0}^\infty 0.1^j= 1 + \sum_{j=1}^\infty 0.1^j \\
9+1=9+9\sum_{j=1}^\infty 0.1^j \\
1=9\sum_{j=1}^\infty 0.1^j = 0.999...\\
\dfrac{1}{3} = 3 \sum_{j=1}^ \infty 0.1^j = 0.333...
[/math]
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>>13643204
>The same relationships between sum and summands hold, as in summation of a finite series using addition
that doesn't make it the result of addition. what you've written is simply a justification for why is the definition reasonable.
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>>13643220
>that doesn't make it the result of addition
Don't care. I don't know what "the result of addition" means, but that sum has exactly the same relationship with the summands, as do finite sums computed through addition.
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>>13642868
[math]/blacksmiley[/math]
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>>13643232
oops
>>
I know it's pointless, but...

-1/12 is not the limit of the series 1+2+3+...; that series diverges to +infinity.

For, say, the series 1+1/2+1/4+1/8+..., the sequence of partial sums gets arbitrarily close to 2, so we say that the infinite sum "equals' 2. (Again, obviously this is not what happens into the -1/12 case.)

Instead, we use something called analytic continuation on the Reimann Zeta function, which for inputs s with real part greater than 1 equals the infinite sum of 1/m^s -- eg zeta(2) = 1 + 1/4 + 1/9 + 1/16..., which happens to converge to pi^2/6.

It is *not* defined that way for negative m. But it happens to still be uniquely defined alternately -- there's exactly one way to extend it to negative numbers that preserves certain properties (continuity/differentiability). And it happens that if you do that, and plug in -1, you get -1/12.
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>>13639336
It's not equal, it's equivalent from analytic continuation
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>>13639647
Not him, but yes. Tooker's suggestion isn't actually terrible this time, check it.
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>>13643298
It's a very interesting and valid result, so long as you don't try to tell us the "sum" you've computed is the same kind of thing people have in mind when they normally think of a sum.
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>>13643232
[math]
\overset{ \frown \frown}{ \smile}
[/math]
>>
>>13643377
And no decent mathematician will tell you that. Clickbait media articles aiming to wow normies and get some outrage comments -- or, say, a bait thread /sci's seen a bunch already -- will make that equivalence.
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>>13643469
Glad we can all agree that your gay "sum" is not a real sum.
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>>13643490
What makes the usual sum more real than the Ramanujan summation?
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>>13643532
Some serious autism ITT... your "sum" violates the relationships that regular sums have with the summands, as opposed to the sum of an infinite convergent series.
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>>13643617
What relationship?
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>>13643653
For instance, that if the summands are all positive, the sum will be positive and greater than any of the summands.
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>>13643664
Greater or equal*
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>>13642969
give a more detailed explanation
>>13643300
>It's not equal, it's equivalent from analytic continuation
then make a new symbol from it which doesn't collide with normal mathematics
>>13643298
>he sequence of partial sums gets arbitrarily close to 2
why is this sum valid as reaching 2 and everyone agrees that it equals close to 2 but
an infinite sum with all positive integers gets assigned an analytical continuation. Is it because the result is not possible
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>>13639336
$\sum_{n=1}^{\inf}n$ does not converge.
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>>13643716
How do LaTeX?
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>>13643707
>then make a new symbol from it which doesn't collide with normal mathematics
We use '+' for obvious series like 1+1/2+1/4+1/8+... because that's literally what's happening. We use '+' when we say 1+2+3+4+... diverges to infinity. Idiots, and/or mathematicians who were trying to make a point about the zeta function and were taken out of context *by* idiots, may write 1+2+3+... = -1/12, but that doesn't make it true in most contexts.
In other words, it's perfectly fine to use the above notation when talking to someone who already knows we're talking about analytic continuation. 'Overloading' symbols like that happens all the time: we use '+' to represent adding integers, adding real or complex numbers, adding sets of numbers, doing the line-and-curve intersection nonsense that's involved in elliptic curve cryptography, appending strings, adding sequences termwise, adding matrices, expressing the direct sum of two sets (sometimes), Cesaro summation, modular arithmetic, etc. These all have some conceptual bits in common -- some notion of 'adding' -- but the actual operations vary wildly. And that's okay, *if* everyone involved knows what context we're talking about. The -1/12 sum is taken out of context by normies, and so they get confused because they think ordinary integer summation is being used.
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>>13643707
(Cont)
>an infinite sum with all positive integers gets assigned an analytical continuation. Is it because the result is not possible
Yes.
The whole point is the analytic continuation matches for the series where it *does* converge. The sum of 1/m^2, ie 1+1/4+1/9+..., converges, because it gets closer and closer to a real number, pi^2/6. Hence zeta(2) = pi^2/6. However, it turns out that if you define zeta as above for all the m where the summation does converge, then there's only one way to define zeta correctly for all the m where the summation doesn't converge. Then the sum of 1/m^-1 = m, 1+2+3+..., obviously does not converge, but it happens the output of zeta(-1) = -1/12. This is fine, the zeta function is only defined by an infinite sum for re(m) > 1 (we figure out it's value elsewhere with certain identities and symmetries); however, it gets reported as 1+2+3+... = -1/12. That's true only if you already know the context, that we're talking about the zeta function and not ordinary series.
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>>13643806
>'Overloading' symbols like that happens all the time
Does that even count as "overloading the symbol"? The whole thing just seems like a misleading notation for "sum of an infinite series following the implied pattern", such that you can't even break it apart and manipulate like a normal chain of + operations, because it's not even associative anymore.
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>>13643806
>>13643837
If anything, it's the term "sum" that's being "overloaded", and the plus is just a piece of syntax in the notation.
>>
>>13643721
>>13642868
>>
>>13639336
I don't think it does, it's just a number to describe it's properties.
>>
>>13643707
>a more detailed explanation
https://youtu.be/sD0NjbwqlYw?t=10m
>>
>>13643837
The usual infinite sum is not associative as well.
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>>13643946
Thanks!
>>13639336
[eqn]\sum_{n=1}^{\inf}n[/eqn] does not converge.
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>>13643989
That's exactly what I'm saying. It's not just the + being "overloaded". The whole thing is purely notational, and the rules to manipulate that expression are different from the rules to manipulate regular arithmetic expressions. However, the sum of a convergent series is directly related to the regular concept of addition, while the "sum" of a divergent series is not.
>>
>>13643989
>>13644012
Note that it's not just that the '+' is not associative, but you have to be very careful and consider the structure of the entire expression never you manipulate it, so you can't even break it down into independent parts the way you can with regular operators.
>>
>>13643806
>>13643977
When working with negative numbers to continue the series within the zeta function it seems to give out positive numbers which don't converge so an analytical continuation is needed to make the function work. Then wouldn't we need a different function/technique to make negative number converge and also seem that using the same function on positive numbers doesn't work with negative so in way we just have to agree with what we contently have
>>
then the riemann hypothesis is half true or completely false
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>>13642781
>Infinity =/= negative infinity
sure it does. both equal [math]\frac{1}{0}[/math]
>Why would squaring half a rotation give you half a rotation?
It wouldn't necessarily. But summing two of them does.
Since [math]\infty + \infty = 0[/math] and [math]\infty[/math] is even, [math]\infty ^ {\infty} = 0[/math]
>>
[math]\infty^2[/math], my bad
>>
>>13647347
infinity is not even its not a number its a concept
so it cant be even
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>>13647383
concepts have no place in math
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>>13647347
>sure it does
lol, congrats for literally making the largest mistake possible
>>
>>13647383
https://www.youtube.com/watch?v=BBp0bEczCNg
https://www.youtube.com/watch?v=FVZqPaH94qU
>>
>>13639336
You've taken an equation and incorrectly assigned the answer to a different question (about that equation) to it. Your answer is for zero terms, not all of them. Indeed, not any of them. Fun pure math but not an actual equality.
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>>13639445
I think that is only valid for 0<=q<=1. Try it for q = 1/2.
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>>13647488
Infinities of infinity between infinities
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>>13639707
Its the same way if you have infinite time to do infinite additions. Retard. But you lack both the brain power and the time to do something like this. So mathematicians came up with limits to help you get there. Yet, I somehow doubt you can even solve limits.
>>
>>13639336
The series normally diverges. The -1/12 is only true in a very particular sense where you study certain families of divergent series by associating finite values in a specific way. Look up zeta function normalization
>>
>>13648384
>The same way you sum a finite series
>except it's different



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