/sci/ - Science & Math » Thread #13230754

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Now that the dust has settled...

Anonymous Thu 03 Jun 15:38:37 2021 No.13230754 View Reply Original Report

Quoted By: >>13230808 >>13230833 >>13230900 >>13230997 >>13231191 >>13232076 >>13232084 >>13232091 >>13234332 >>13234457 >>13234847

Is 1-1+1-1+1...=1/2?

Anonymous

Anonymous Thu 03 Jun 2021 15:51:00 No.13230796 Report

Quoted By:

if you take the limit of the expectation of choosing a partial sum (s_1,...,s_n) at random as n -> infinity then sure
this is literally the example (Grandi's series) mentioned in the intro to the wiki article on Cesaro summation

Anonymous

Anonymous Thu 03 Jun 2021 15:56:08 No.13230808 Report

Quoted By: >>13230841

>>13230754
You might find out, if you knew what it means. Which you obviously do not.

Anonymous

Anonymous Thu 03 Jun 2021 16:04:16 No.13230833 Report

Quoted By:

>>13230754
it's 1

Anonymous

Anonymous Thu 03 Jun 2021 16:06:28 No.13230841 Report

Quoted By: >>13230969 >>13231010

>>13230808
Ok then what does it mean

Anonymous

Anonymous Thu 03 Jun 2021 16:23:44 No.13230900 Report

Quoted By: >>13232108

>>13230754
$\sum_{n=0}^\infty(-1)^n$ is undefined, you can define it to be literally anything you'd like
troll thread

Anonymous

Anonymous Thu 03 Jun 2021 16:39:20 No.13230969 Report

Quoted By:

>>13230841
The other anon said it: it is not defined.
The good thing is that, since it has not yet been defined, you are free to choose
whatever you like it to be defined as. Then you write a paper and get it published,
and the definition will be your's forever to behold. Unless now that I revealed
this secret, some other anon will beat you to it.

Anonymous

Anonymous Thu 03 Jun 2021 16:47:03 No.13230997 Report

Quoted By: >>13231491

>>13230754
it's a superposition of two states, 1 and 0

Anonymous

Anonymous Thu 03 Jun 2021 16:49:17 No.13231010 Report

Quoted By:

>>13230841
The meaning of the sum is arbitrary on the surface. There's a whole field of math dedicated to sorting this out called summation theory. Essentially there's "stronger" forms of summation than the traditional method, but are still consistent with traditional convergent series, like Borel summation or Cesaro summation.

Grandi's series has no meaning in a traditional sense. However, there are other ways to give it meaning not typically discussed. The way 1/2 or -1/12 are obtained from using traditional algebraic methods is somewhat of a relic from analytic continuation.

Anonymous

Anonymous Thu 03 Jun 2021 17:33:39 No.13231191 Report

Quoted By:

>>13230754
It's an array of numbers.

Anonymous

Anonymous Thu 03 Jun 2021 18:32:12 No.13231491 Report

Quoted By: >>13231512

>>13230997
$\frac{\sqrt{1}+1}{2}$

Anonymous

Anonymous Thu 03 Jun 2021 18:38:07 No.13231512 Report

Quoted By:

>>13231491
take your meds

Anonymous

Anonymous Thu 03 Jun 2021 18:39:53 No.13231518 Report

Quoted By:

The sum does not converge.

The sum fluctuates between 0 and 1.

It is not the average..the sum on the (n)th term is 1 the sum on the (n+1)th term is 0.

The sum will never be 0.5, it will just keep fluctuating between 0 and 1 indefinitely forever.

This means there is no sum for that infinite series, and the sum does not converge.

This is because the limit does not exist. For a sum to be exist, it has to converge to some value. You can check this by testing the limit and seeing if it approaches 0 for the nth term as n approaches infinity.

Hope it made sense.

Anonymous

Anonymous Thu 03 Jun 2021 20:50:41 No.13232076 Report

Quoted By: >>13232080

>>13230754
the sum diverges but might be equal to something in the sense of analytic continuation or something

Anonymous

Anonymous Thu 03 Jun 2021 20:52:05 No.13232080 Report

Quoted By:

>>13232076
kind of like making the the gamma function to calculate fraction factorials which might be silly at first hand but useful

Anonymous

Anonymous Thu 03 Jun 2021 20:52:48 No.13232084 Report

Quoted By:

>>13230754
yes

Anonymous

Anonymous Thu 03 Jun 2021 20:54:44 No.13232088 Report

Quoted By:

1/2. It's either 1 or 0 with fifty percent probability

Anonymous

Anonymous Thu 03 Jun 2021 20:55:24 No.13232091 Report

Quoted By:

>>13230754
It depends. Is infinity even or odd?

Anonymous

Anonymous Thu 03 Jun 2021 21:00:12 No.13232108 Report

Quoted By: >>13232137

>>13230900
It's Cesaro summable, and that has a value of 1/2

Anonymous

Anonymous Thu 03 Jun 2021 21:07:27 No.13232137 Report

Quoted By:

>>13232108
>Cesaro summable
is that the one where you build the sequence of n-th averages?

Anonymous

Anonymous Fri 04 Jun 2021 08:55:00 No.13234332 Report

Quoted By:

>>13230754
infinity is an even number, so 0

Anonymous

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Anonymous Fri 04 Jun 2021 09:42:29 No.13234457 Report

Quoted By: >>13234541 >>13234595

>>13230754
Let A = 1-1+1-...
Let B = 1+1-1+...
B = 1 + A
A + B = 1
A + 1 + A = 1
2A = 0
A = 0

Anonymous

Anonymous Fri 04 Jun 2021 10:03:58 No.13234541 Report

Quoted By:

>>13234457
I feel like I've just been swindled

Anonymous

Anonymous Fri 04 Jun 2021 10:27:56 No.13234595 Report

Quoted By:

>>13234457
Nice try but A + B = 2 in which case A = 1/2

Anonymous

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Anonymous Fri 04 Jun 2021 12:25:32 No.13234787 Report

Quoted By:

$\frac{1}{2+\epsilon} = \frac{1}{1-(-1-\epsilon) } = \sum_n (-1)^n\cdot (1+\epsilon)^n$

Anonymous

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Anonymous Fri 04 Jun 2021 12:27:11 No.13234788 Report

Quoted By:

$\frac{1}{2-\epsilon} = \frac{1}{1-(-1+\epsilon) } = \sum_n (-1)^n\cdot (1-\epsilon)^n$

Anonymous

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Anonymous Fri 04 Jun 2021 13:01:46 No.13234847 Report

Quoted By:

>>13230754
$\sum_i^\infty (-1)^i \in \{0, 1\}$
>status: defined

Anonymous

Anonymous Fri 04 Jun 2021 21:08:28 No.13236679 Report

Quoted By:

$-1/12$ r-right?

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