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tech / sci.math / Re: the limit of round(x)^2-x^2 as x approaches infinity is 1/12(proof)

SubjectAuthor
* the limit of round(x)^2-x^2 as x approaches infinity is 1/12(proof)Jonh Anderson
`* Re: the limit of round(x)^2-x^2 as x approaches infinity is 1/12(proof)markus...@gmail.com
 `* Re: the limit of round(x)^2-x^2 as x approaches infinity is 1/12(proof)Jonh Anderson
  `- Re: the limit of round(x)^2-x^2 as x approaches infinity is 1/12(proof)Jonh Anderson

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the limit of round(x)^2-x^2 as x approaches infinity is 1/12(proof)

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Subject: the limit of round(x)^2-x^2 as x approaches infinity is 1/12(proof)
From: johnande...@gmail.com (Jonh Anderson)
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 by: Jonh Anderson - Sun, 13 Aug 2023 13:53 UTC

https://docs.google.com/document/d/1qEEXxNjILK0iwgIDSAgeyAPffrWlMA1rSOlNgXTO_CE/edit?usp=sharing

Re: the limit of round(x)^2-x^2 as x approaches infinity is 1/12(proof)

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Subject: Re: the limit of round(x)^2-x^2 as x approaches infinity is 1/12(proof)
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 by: markus...@gmail.com - Mon, 14 Aug 2023 07:45 UTC

söndag 13 augusti 2023 kl. 15:53:22 UTC+2 skrev Jonh Anderson:
> https://docs.google.com/document/d/1qEEXxNjILK0iwgIDSAgeyAPffrWlMA1rSOlNgXTO_CE/edit?usp=sharing
What you have is f(x)=(x-round(x))², which clearly lacks a limit at infinity.

Just pick x=n+3/4. Then f(n) is always 1. No convergence.

Re: the limit of round(x)^2-x^2 as x approaches infinity is 1/12(proof)

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Subject: Re: the limit of round(x)^2-x^2 as x approaches infinity is 1/12(proof)
From: johnande...@gmail.com (Jonh Anderson)
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 by: Jonh Anderson - Mon, 14 Aug 2023 17:16 UTC

On Monday, August 14, 2023 at 2:45:26 AM UTC-5, markus...@gmail.com wrote:
> söndag 13 augusti 2023 kl. 15:53:22 UTC+2 skrev Jonh Anderson:
> > https://docs.google.com/document/d/1qEEXxNjILK0iwgIDSAgeyAPffrWlMA1rSOlNgXTO_CE/edit?usp=sharing
> What you have is f(x)=(x-round(x))², which clearly lacks a limit at infinity.
>
> Just pick x=n+3/4. Then f(n) is always 1. No convergence.
Yeah, sorry for the confusion. on the bottom of the paper i said i knew that this technically didnt actually have a limit. I do bend the definition of limit a bit in this paper, but the basic idea still works, if you smooth out the graph of (x-round(x))² you get 1/12 which doesnt make any sense, even if it technically doesn’t actually have a limit of 1/12.

Re: the limit of round(x)^2-x^2 as x approaches infinity is 1/12(proof)

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Subject: Re: the limit of round(x)^2-x^2 as x approaches infinity is 1/12(proof)
From: johnande...@gmail.com (Jonh Anderson)
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 by: Jonh Anderson - Mon, 14 Aug 2023 17:49 UTC

On Monday, August 14, 2023 at 12:16:44 PM UTC-5, Jonh Anderson wrote:
> On Monday, August 14, 2023 at 2:45:26 AM UTC-5, markus...@gmail.com wrote:
> > söndag 13 augusti 2023 kl. 15:53:22 UTC+2 skrev Jonh Anderson:
> > > https://docs.google.com/document/d/1qEEXxNjILK0iwgIDSAgeyAPffrWlMA1rSOlNgXTO_CE/edit?usp=sharing
> > What you have is f(x)=(x-round(x))², which clearly lacks a limit at infinity.
> >
> > Just pick x=n+3/4. Then f(n) is always 1. No convergence.
> Yeah, sorry for the confusion. on the bottom of the paper i said i knew that this technically didnt actually have a limit. I do bend the definition of limit a bit in this paper, but the basic idea still works, if you smooth out the graph of (x-round(x))² you get 1/12 which doesnt make any sense, even if it technically doesn’t actually have a limit of 1/12.

Although i just wanted to say thank you for reading my paper and giving me feedback, i really do appreciate it. I will rewrite it to use a different term instead of limit. Anyways, here is the new paper, is there any other changes i could make to make it easier to understand or anything like that?
https://docs.google.com/document/d/17G8dvgbkCub0dBUB-lgcBfBpyYzz6L75mdvYFL7OdFY/edit?usp=sharing

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