Do you agree 1 + 2 + 3 + 4 + 5 + ... = -1/12???

Watch
Bethany221
Badges: 9
Rep:
?
#1
Report Thread starter 6 years ago
#1


:eek:
0
reply
smd4std
Badges: 0
Rep:
?
#2
Report 6 years ago
#2
pretty sure it's bs
0
reply
Noble.
Badges: 17
Rep:
?
#3
Report 6 years ago
#3
No, mathematically this is nonsense. All they've done is redefined what mathematicians have defined as convergence which, before anyone tries arguing, is a very basic definition in real analysis that underpins a significant amount of mathematics - and it just so happens that this re-defining is useful in certain applications.

Clearly, to anyone who is mathematically inclined, the idea of saying the series

S_n = \underbrace{1 -1 + 1 -1 + ... \pm 1}_{\text{n times}}

Converges to \frac{1}{2} is utter nonsense mathematically, there doesn't exist a N \in \mathbb{N} such that for all n > N, |S_n - \frac{1}{2}| < \epsilon for \epsilon \leq \frac{1}{2} so it doesn't converge by the standard definition of convergence.

I won't even get started on how violated mathematicians will feel at the thought of a sum of positive values somehow giving a negative result. Needless to say though, if you start redefining basic definitions in mathematics you're going to get results that don't make mathematical sense, whether they're helpful or not is another matter.

EDIT - TL;DR - They're playing around with sums that diverge, which is a mathematical equivalent of going full-******.
10
reply
UniMastermindBOSS
Badges: 19
Rep:
?
#4
Report 6 years ago
#4
No, I hate those stupid pattern/infinity videos on numberphile.

Ones like this are my favourite

0
reply
genuinelydense
Badges: 11
Rep:
?
#5
Report 6 years ago
#5
1-1=1/2

the reason for this lies deep in quantum mechanix.
0
reply
scrotgrot
Badges: 16
Rep:
?
#6
Report 6 years ago
#6
"Let's just take the average. The answer's 1/2." To translate Noble's post for anyone non-mathsy, that's what's wrong with this. No way is any term of that sequence ever going to be 1/2, not even the infinityth term. That's because it doesn't converge on 1/2, it stays the same distance away ad infinitum.
0
reply
Bethany221
Badges: 9
Rep:
?
#7
Report Thread starter 6 years ago
#7
Bump! Worth consideration defo.


Posted from TSR Mobile
0
reply
Lord Frieza
Badges: 13
Rep:
?
#8
Report 6 years ago
#8
-1/12 x 1 = -1/12
0
reply
QuantumHatter
Badges: 1
Rep:
?
#9
Report 6 years ago
#9
(Original post by Noble.)
No, mathematically this is nonsense. All they've done, and by they I mean mostly physicists, is redefined what mathematicians have defined as convergence which, before anyone tries arguing, is a very basic definition in real analysis that underpins a significant amount of mathematics - and it just so happens that this re-defining is useful in certain applications.

Clearly, to anyone who is mathematically inclined, the idea of saying the series

S_n = \underbrace{1 -1 + 1 -1 + ... \pm 1}_{\text{n times}}

Converges to \frac{1}{2} is utter nonsense mathematically, there doesn't exist a N \in \mathbb{N} such that for all n > N, |S_n - \frac{1}{2}| < \epsilon for \epsilon \leq \frac{1}{2} so it doesn't converge by the standard definition of convergence.

I won't even get started on how violated mathematicians will feel at the thought of a sum of positive values somehow giving a negative result. Needless to say though, if you start redefining basic definitions in mathematics you're going to get results that don't make mathematical sense, whether they're helpful or not is another matter.

EDIT - TL;DR - They're playing around with sums that diverge, which is a mathematical equivalent of going full-******.
I swear this makes perfect sense? If S=1-1+1-1+... then S=1-(1-1+1-1+...) which is the same as S=1-S, then 2S=1 and S=0.5?
0
reply
sbj
Badges: 6
Rep:
?
#10
Report 6 years ago
#10
(Original post by QuantumHatter)
I swear this makes perfect sense? If S=1-1+1-1+... then S=1-(1-1+1-1+...) which is the same as S=1-S, then 2S=1 and S=0.5?
The problem here is your definition of S changes. You say "S=1-1+1-1+... ". So everything from begin (first 1) belongs to S. But then you say "S=1-(1-1+1-1+...)" and define S starting from the second 1. It is no more S.
It is a partial series of S, not S itself because the first 1 is not included.
The problem is the infinity here and that the series doesn't converge.

Because when S=1-S, and S converges to 1, it would mean 1 = 1 - 1 => 1 = 0 and this is wrong.
Or if S converges to 0, it would mean 0 = 1 - 0 = > 0 = 1.

So as you see it does not make any sense because it does not converge and you change the definition of S.

It is always messy when you divide through 0 or play with infinity in any kind of form.
0
reply
L'Evil Fish
Badges: 18
Rep:
?
#11
Report 6 years ago
#11
It's just a load of bull
0
reply
Kvothe the Arcane
Badges: 20
Rep:
?
#12
Report 6 years ago
#12
o.O.
0
reply
QuantumHatter
Badges: 1
Rep:
?
#13
Report 6 years ago
#13
(Original post by sbj)
The problem here is your definition of S changes. You say "S=1-1+1-1+... ". So everything from begin (first 1) belongs to S. But then you say "S=1-(1-1+1-1+...)" and define S starting from the second 1. It is no more S.
It is a partial series of S, not S itself because the first 1 is not included.
The problem is the infinity here and that the series doesn't converge.

Because when S=1-S, and S converges to 1, it would mean 1 = 1 - 1 => 1 = 0 and this is wrong.
Or if S converges to 0, it would mean 0 = 1 - 0 = > 0 = 1.

So as you see it does not make any sense because it does not converge and you change the definition of S.

It is always messy when you divide through 0 or play with infinity in any kind of form.
Infinity is indeed messy, but I do think that my definition does not change. I have lost the first one, but because there are an infinite number of terms, the sum is exactly the same. Infinity minus one is still infinity is it not?
I would agree they wouldn't be the same if they didn't go on to infinity, but they do, so S still equals 0.5 as far as I can see!
1
reply
paradoxicalme
Badges: 16
Rep:
?
#14
Report 6 years ago
#14
*twitch*
0
reply
sbj
Badges: 6
Rep:
?
#15
Report 6 years ago
#15
(Original post by QuantumHatter)
Infinity is indeed messy, but I do think that my definition does not change. I have lost the first one, but because there are an infinite number of terms, the sum is exactly the same. Infinity minus one is still infinity is it not?
I would agree they wouldn't be the same if they didn't go on to infinity, but they do, so S still equals 0.5 as far as I can see!
The bolded part is the problem.
This works 0 = 0 or 5+6 = 11 but this does not work ∞ = ∞.
So is infinity minus one is not the same as inifinity.
Because it is infinite, it has no end. You can't make an equation, it is NOT equal.
My infinity is 10^50 and yours 10^49, is that equal? When do we have an equation, like the same? We never have.
As if "S1=1-1+1-1+... " isn't equal "S2=1-1+1-1+... ", so S1 ≠ S2.
Because you don't know where it ends, there is no result, but you have infinite sums.
I could also make "S=1-1+1-1+... " -> "S=1-1+(1-1+... " -> S = 1 - 1 + S -> 0 = 0 -> No information about S.
But it does not converge, it can't be 1 or 0 or 1/2.
1
reply
sbj
Badges: 6
Rep:
?
#16
Report 6 years ago
#16
(Original post by QuantumHatter)
I Infinity minus one is still infinity is it not?
If this is true, then

∞ = (∞ + 3) | -∞
0 = 3

or

∞ = (∞ - 1) | -∞
0 = -1
0
reply
QuantumHatter
Badges: 1
Rep:
?
#17
Report 6 years ago
#17
(Original post by sbj)
The bolded part is the problem.
This works 0 = 0 or 5+6 = 11 but this does not work ∞ = ∞.
So is infinity minus one is not the same as inifinity.
Because it is infinite, it has no end. You can't make an equation, it is NOT equal.
My infinity is 10^50 and yours 10^49, is that equal? When do we have an equation, like the same? We never have.
As if "S1=1-1+1-1+... " isn't equal "S2=1-1+1-1+... ", so S1 ≠ S2.
Because you don't know where it ends, there is no result, but you have infinite sums.
I could also make "S=1-1+1-1+... " -> "S=1-1+(1-1+... " -> S = 1 - 1 + S -> 0 = 0 -> No information about S.
But it does not converge, it can't be 1 or 0 or 1/2.
Ok, I can see where you are coming from here. But I still think the infinities are the same because it has no end, I thought that was the whole idea of infinity? I'm also not trying to claim infinity as a number, just as a concept of unboundedness, so I didn't mean 0=1
I'm not very good at explaining things but I think this sum is similar to this question "show 1/9 =0.111111..."
Let x= 0.11111111...
10x=1.11111111...
10x-x=9x=1
x=1/9
0
reply
sbj
Badges: 6
Rep:
?
#18
Report 6 years ago
#18
(Original post by QuantumHatter)
Ok, I can see where you are coming from here. But I still think the infinities are the same because it has no end, I thought that was the whole idea of infinity? I'm also not trying to claim infinity as a number, just as a concept of unboundedness, so I didn't mean 0=1
I'm not very good at explaining things but I think this sum is similar to this question "show 1/9 =0.111111..."
Let x= 0.11111111...
10x=1.11111111...
10x-x=9x=1
x=1/9
No, the inifinities are not the same, they can not be. As I showed you 1 message before. This is not the whole idea of infinity. You can not use infinity in normal mathematical equations. The axioms does not apply to infinity. It does not work.
And to the rational number thing.
It is not the same. Do you know why it is even 0.1111111...? Because of the decimal system. If you have another system, maybe dual system or any kind of, you won't have for 1/9 = 0.11111. It is because of the system you use.

And the main difference is, that you do not know I see, that @Noble. told already, that the main problem does not converge.
But your 0.1111... converges. Do you know what geometric series are?
If not check this out or google.
It just works because it converges, like I said. If it converges, then you can make equations because you have a result.
But "S1=1-1+1-1+... " does not converge. There is no result.
0
reply
QuantumHatter
Badges: 1
Rep:
?
#19
Report 6 years ago
#19
(Original post by sbj)
No, the inifinities are not the same, they can not be. As I showed you 1 message before. This is not the whole idea of infinity. You can not use infinity in normal mathematical equations. The axioms does not apply to infinity. It does not work.
And to the rational number thing.
It is not the same. Do you know why it is even 0.1111111...? Because of the decimal system. If you have another system, maybe dual system or any kind of, you won't have for 1/9 = 0.11111. It is because of the system you use.

And the main difference is, that you do not know I see, that @Noble. told already, that the main problem does not converge.
But your 0.1111... converges. Do you know what geometric series are?
If not check this out or google.
It just works because it converges, like I said. If it converges, then you can make equations because you have a result.
But "S1=1-1+1-1+... " does not converge. There is no result.
Ok, you've beaten me. I'm only an A level mathematician. http://math.arizona.edu/~cais/Papers/Expos/div.pdf This paper however continues the fight! What would you say to dispute the seemingly logical deductions here?
0
reply
sbj
Badges: 6
Rep:
?
#20
Report 6 years ago
#20
(Original post by QuantumHatter)
Ok, you've beaten me. I'm only an A level mathematician. http://math.arizona.edu/~cais/Papers/Expos/div.pdf This paper however continues the fight! What would you say to dispute the seemingly logical deductions here?
I don't kniow what an A level mathematician is (I am not from UK) and I just am a mathematics student.
I don't have the time to go this through but in first case I'd say you can't do it because of rearrangement of the series. There he rearranges the sum.
But he proves there something.
So, I searched a bit and found the answer why it "works".
See here.
Although the full series may seem at first sight not to have any meaningful value, it can be manipulated to yield a number of mathematically interesting results, some of which have applications in other fields such as complex analysis, quantum field theory and string theory.
The series can be summed by zeta function regularization. When the real part of s is greater than 1, the Riemann zeta function of s equals the sum Image. This sum diverges when the real part of s is less than or equal to 1, but when s = −1 then the analytic continuation of ζ(s) gives ζ(−1) as −1/12.
So for just a special case it works, not in general.
And here you can understand why:
Ramanujan summation is a technique invented by the mathematician Srinivasa Ramanujan for assigning a sum to infinite divergent series. Although the Ramanujan summation of a divergent series is not a sum in the traditional sense, it has properties which make it mathematically useful in the study of divergent infinite series, for which conventional summation is undefined.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (80)
14.21%
I'm not sure (25)
4.44%
No, I'm going to stick it out for now (179)
31.79%
I have already dropped out (12)
2.13%
I'm not a current university student (267)
47.42%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise