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A New Approach to Divergent Series

Me61820 Sun 09 Feb 19:40:04 2020 No.11372338 View Reply Original Report

Quoted By: >>11372340 >>11372346 >>11372358

There is a new way of approaching divergent series using hyperreal numbers.

Me61820

Me61820 Sun 09 Feb 2020 19:40:39 No.11372340 Report

Quoted By: >>11372358 >>11372590

>>11372338
>There is a new way of approaching divergent series using hyperreal numbers.

See article on it here:
https://mindmatters.ai/2020/02/are-divergent-series-really-an-invention-of-the-devil/

Anonymous

Anonymous Sun 09 Feb 2020 19:42:53 No.11372346 Report

Quoted By:

>>11372338
>muh -1/12

Anonymous

Anonymous Sun 09 Feb 2020 19:46:13 No.11372358 Report

Quoted By: >>11372418

>>11372340
>>11372338
Forgot to change your name?

Anonymous

Anonymous Sun 09 Feb 2020 20:13:21 No.11372418 Report

Quoted By:

>>11372358
no, he forgot to post link

Me61820

Me61820 Sun 09 Feb 2020 20:32:19 No.11372466 Report

Quoted By:

For those who aren't familiar with the hyperreal number system, there's a video on it here:

https://www.youtube.com/watch?v=FTXRnEKEn4k&list=PL2Ky2_VcYEd1-UmZxmNXaQh2AjKY8tpvI

Anonymous

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Anonymous Sun 09 Feb 2020 21:26:22 No.11372590 Report

Quoted By: >>11372827

>>11372340
>Doesn't mention Euler
>Therefore, using this system, we can legitimately talk about dividing these two series, and find out that (1 + 2 + 3 + …) / (1 + 3 + 5 + …) is 1/2
S/(S-2S) = -1, not 1/2.
The author might have downs.

Me61820

Me61820 Sun 09 Feb 2020 22:36:59 No.11372827 Report

Quoted By: >>11372835

>>11372590

You should read the paper referenced by the article. (1 + 3 + 5 + ...) is not S - 2S. I can see that you are getting that by multiplying (1 + 2 + 3 + ...) by 2, and then subtracting (i.e., removing the even values). However, as mentioned, this is based on the hyperreal infinities, in which the set that we would need to be subtracting in your case is actually (0 + 2 + 0 + 4 + 0 + 6 + ...) which is a different series (with a different sum) than (2 + 4 + 6 + ...).

The paper goes through how these are determined.

Me61820

Me61820 Sun 09 Feb 2020 22:39:34 No.11372835 Report

Quoted By:

>>11372827
Additionally, even when you subtract (0 + 2 + 0 + 4 + ...) from (1 + 2 + 3 + ...), the resulting series is (1 + 0 + 3 + 0 + 5 + 0 + ...), which is a different series than (1 + 3 + 5 + ...). They look the same, but they are actually distinct based on the hyperreal evaluation method.

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